Mahesh

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 4 Ellipse to solve questions creatively.

Intermediate 2nd Year Maths 2B Ellipse Formulas

Definition:
→ A conic with eccentricity less than one is called an ellipse i.e., the locus of a point whose distances from a fixed point and a fixed straight line are in constant ratio ‘e’ which is less than 1, is called an ellipse. The fixed point and fixed straight line are called focus and directrix respectively.

Equation of Ellipse in standard form:
→ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1; a > b, b² = a² (1 – e²) ; e < 1
Foci: (ae, 0), (- ae, 0); directrices x = \(\frac{\mathrm{a}}{\mathrm{e}}\); x = \(\frac{-a}{e}\)
a = Length of the semi-major axis.
b = Length of semi-minor axis.

Various forms of the ellipse:

→ Major axis: along X – axis
Length of major axis: 2a
Minor axis: along Y-axis
Length of minor axis: 2b
Centre: (0, 0)
Foci: S (ae, 0); S’ (- ae, 0)
Directrices: x = \(\frac{\mathrm{a}}{\mathrm{e}}\); x = – \(\frac{\mathrm{a}}{\mathrm{e}}\)
e is given as: b² = a² (1 – e²)
(or) e = \(\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}\), a > b
a < b (or) b > a

→ Major axis: along Y – axis
Length of major axis: 2b
Minor axis: along Y-axis
Length of minor axis: 2a
Centre: (0, 0)
Foci: S (0, be); S’ (0, – be)
Directrices: y = \(\frac{b}{\mathrm{e}}\), y = – \(\frac{b}{\mathrm{e}}\)
e is given as: a² = b² (1 – e²)
(or) e = \(\sqrt{\frac{b^{2}-a^{2}}{b^{2}}} .\)

→ \(\frac{(x-\alpha)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{b^{2}}\) = 1, a > b
Major axis: parallel to X-axis along the line y = β
Length of major axis: 2a
Minor axis: parallel to Y-axis along the line x = α
Length of minor axis: 2b
Center: (α, β)
Foci: S (ae + α, β); S’ ( – ae + α, β)
Directrices: x = α + \(\frac{a}{e}\) ; x = α – \(\frac{a}{e}\)
e is given by b² = a² ( 1 – e²)

→ \(\frac{(x-\alpha)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{b^{2}}\) = 1, a < b
Major axis: parallel to Y-axis along the line x = α
Length of major axis: 2b
Minor axis: parallel to X-axis along the line y = β
Length of minor axis: 2a
Center: (α, β)
Foci: S (α, be + β); S’ (α, -be + β)
Directrices: y – β = \(\frac{b}{\mathrm{e}}\) ; y – β = – \(\frac{b}{\mathrm{e}}\)

→ A line segment joining two points on the ellipse is called a chord of the ellipse. Chord passing through foci is called a focal chord. A focal chord perpendicular to the major axis of the ellipse is cal fed latus rectum. Length of latus rectum = \(\frac{2 b^{2}}{a}\), a > b, Length of the latus rectum = \(\frac{2 a^{2}}{a}\), if b > a

→ If P (x, y) is any point on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 whose foci are S and S’.
Then SP + SP’ = constant = 2a.

→ Parametric equations x = a cos θ; y = b sin θ

→ If P a point lies outside the ellipse, then S₁₁ > 0.

→ If P a point lies on the ellipse, then S₁₁ = 0.

→ If P a point lies inside the ellipse, then S₁₁ < 0.

Ellipse:
A conic is said to be an ellipse if it’s eccentricity e is less than 1.

Equation of an Ellipse in Standard Form:
The equation of an ellipse in the standard form is \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}\) = 1.(a < b)
Proof:

Let S be the focus, e be the eccentricity and L = 0 be the directrix of the ellipse. Let P be a point on the ellipse. Let M, Z be the projections (foot of the perpendiculars) of P, S on the directrix L = 0 respectively. Let N be the projection of P on SZ. Since e < 1, we can divide SZ both internally and externally in the ratio e: 1. Let A, A’ be the points of division of SZ in the ratio e: 1 internally and externally respectively. Let AA’ = 2a. Let C be the midpoint of AA’. The points A, A’ lie on the ellipse and
\(\frac{S A}{A Z}\) = eAZ, \(\frac{\mathrm{SA}^{\prime}}{\mathrm{A}^{\prime} \mathrm{Z}}\) = eA’Z
Now SA + SA’ = eAZ + eA’Z
⇒ AA’ = e(AZ + A’Z)
⇒ 2a = e(CZ – CA + A’C + CZ)
⇒ 2a = e.2CZ (∵ CA = A’C)
⇒ CZ = a/e

Also SA’- SA = eA’Z – eAZ
⇒ A’C + CS – (CA – CS) = e(A’Z – AZ)
⇒ 2CS = eAA’ (∵ CA = A’C)
⇒ 2CS = e2a ⇒ CS = ae
Now PM = NZ = CZ – CN = \(\frac{a}{e}\) – x₁

P lies on the ellipse:
⇒ \(\frac{\mathrm{PS}}{\mathrm{PM}}\) = e ⇒ PS = ePM ⇒ PS² = e²PM²
⇒ (x₁ – ae)² + (y₁ – 0)² = e²\(\left(\frac{\mathrm{a}}{\mathrm{e}}-\mathrm{x}_{1}\right)^{2}\)
⇒ (x₁ – ae)² + y₁² = (a – x₁e)²
⇒ x₁ + ae – 2x₁ae + y₁ = a + x₁e – 2x₁ae
⇒ (1 – e²)x₁² + y₁² = (1 – e²)a²
⇒ \(\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{a^{2}\left(1-e^{2}\right)}\) = 1 ⇒ \(\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}\) = 1
Where b² = a²(1 – e²) > 0
The locus of P is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
∴ The equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Nature of the Curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.

Let be the curve represented by \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1. Then

• The curve is symmetric about the coordinate axes.
• The curve is symmetric about the origin O and hence O is the midpoint of every chord of the ellipse through O. Therefore the origin is the centre of the ellipse.
• Put y = 0 in the equation of the ellipse ⇒ x = a ⇒ x = ±a.
  Thus the curve meets x-axis (Principal axis) at two points A(a, 0), A'(-a, 0). Hence the ellipse has two vertices. The axis AA’ is called major axis. The length of the major axis is AA’ = 2a
• Put x = 0 ⇒ y² = b² ⇒ y = ± b. Thus, the curve meets y-axis (another axis) at two points B(0, b), B'(0, -b). The axis BB’ is called minor axis and the length of the minor axis is BB’ = 2b.
• The focus of the ellipse is S(ae, 0). The image of S with respect to the minor axis is S'(-ae,0). The point S’ is called second focus of the ellipse.
• The directrix of the ellipse is x = a/e. The image of x = a/e with respect to the minor axis is x = -a/e. The line x = -a/e is called second directrix of the ellipse.
• \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
  y² = b²(1 – \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}\)) ⇒ y = \(\frac{b}{a} \sqrt{a^{2}-x^{2}}\)

Thus y has real values only when -a ≤ x ≤ a. Similarly x has real values only when -b ≤ y ≤ b. Thus the curve lies completely with in the rectangle x = ±a, y = ±b. Therefore the ellipse is a closed curve.

Theorem:
The length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 (a > b > 0) is \(\frac{2 b^{2}}{a}\)
The length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 (0 < a < b) is \(\frac{2 b^{2}}{a}\)

Let LL’ be the length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Focus S = (ae, 0)
If SL = 1, then L = (ae, 1)
L is lies on the ellipse ⇒ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
⇒ e + \(\frac{1^{2}}{\mathrm{~b}^{2}}\) = 1 ⇒ \(\frac{1^{2}}{\mathrm{~b}^{2}}\) = 1 – e = \(\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\) ⇒ 1 = \(\frac{b^{4}}{a^{2}}\)
⇒ 1 = \(\frac{b^{2}}{a}\) ⇒ SL = \(\frac{b^{2}}{a}\)
LL’ = 2SL = \(\frac{2 b^{2}}{a}\)

Note: The coordinates of the four ends of the latus recta of the ellipse \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\) = 1 (a < b < 0) are L = (ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)), L’ = (ae, –\(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)), L₁ = (-ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)); L₁‘ = (-ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\))

Note: The coordinates of the four ends of the latus recta of the ellipse \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\) = 1 (0 < a < b) are L = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), be), L’ = (-\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), be), L₁ = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), -be), L₁‘ = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), -be)

Theorem:
If P is a point on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 with foci S and S then PS + PS’ = 2a.
Proof:
Let e be the eccentricity and L = 0, L’ = 0 be the directrices of the ellipse.
Let C be the centre and A, A’ be the vertices of the ellipse.
∴ AA’ = 2a
Foci of the ellipse are S(ae, 0), S'(-ae, 0)
Let P(x₁, y₁) be a point on the ellipse

Let M, M’ be the projections of P on the directrices L = 0, L’ = 0 respectively.
∴ \(\frac{\mathrm{SP}}{\mathrm{PM}}\) = e, \(\frac{\mathrm{S}^{\prime} \mathrm{P}}{\mathrm{PM}^{\prime}}\) = e

Let Z, Z’ be the points of intersection of major axis with directrices.
∴ MM’ = ZZ’ = CZ + CZ’ = 2a/e.
PS + PS’ = ePM + ePM’
= e(PM + PM’) = e(MM’) = e(2a/e) = 2a.

Theorem:
Let P(x₁, y₁) be a point and S ≡ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 = 0 be an ellipse. Then
(i) P lies on the ellipse ⇔ S₁₁ = 0,
(ii) P lies inside the ellipse ⇔ S₁₁ < 0,
(iii) P lies outside the ellipse ⇔ S₁₁ > 0

Theorem:
The equation of the tangent to the ellipse S = 0 at F(x₁, y₁) is S₁ = 0.

Theorem:
The equation of the normal to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 F(x₁, y₁) is \(\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}\) = a² – b².
Proof:
The equation of the tangent to S = 0 at F is S₁ = 0
⇒ \(\frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{~b}^{2}}\) – 1 =
The equation of the normal to S = 0 at F is
\(\frac{\mathrm{y}_{1}}{\mathrm{~b}^{2}}\)(x – x₁) – \(\frac{\mathrm{x}_{1}}{\mathrm{~a}^{2}}\)(y – y₁) = 0
⇒ \(\frac{x_{1}}{b^{2}}-\frac{y_{1}}{a^{2}}=\frac{x_{1} y_{1}}{b^{2}}-\frac{x_{1} y_{1}}{a^{2}}\)
⇒ \(\frac{\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{x}_{1} \mathrm{y}_{1}}\left(\frac{\mathrm{xy}}{\mathrm{b}^{2}}-\frac{\mathrm{y} \mathrm{x}_{1}}{\mathrm{a}^{2}}\right)=\frac{\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{x}_{1} \mathrm{y}_{1}}\left(\frac{\mathrm{x}_{1} \mathrm{y}_{1}}{\mathrm{~b}^{2}}-\frac{\mathrm{x}_{1} \mathrm{y}_{1}}{\mathrm{a}^{2}}\right)\)
⇒ \(\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}\) = a² – b²

Theorem:
The condition that the line y = mx + c may be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 is c² = a²m² + b².
Proof:
Suppose y = mx + c … (1)is a tangent to the e11ipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Let P(x1, y) be the point of contact.
The equation of the tangent at P is
\(\frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{~b}^{2}}\) – 1 = 0 … (2)

Now (1) and (2) represent the same line.
\(\frac{\mathrm{x}_{1}}{\mathrm{a}^{2} \mathrm{~m}}=\frac{\mathrm{y}_{1}}{\mathrm{~b}^{2}(-1)}=\frac{-1}{\mathrm{c}}\) ⇒ x₁ = \(\frac{-a^{2} m}{c}\), y₁ = \(\frac{\mathrm{b}^{2}}{\mathrm{c}}\)
P lies on the line y = mx + c ⇒ y₁ = mx₁ + c
⇒ \(\frac{\mathrm{b}^{2}}{\mathrm{c}}\) = m\(\left(\frac{-a^{2} m}{c}\right)\) + c ⇒ b² = -a²m² + c²
⇒ c² = a²m² + b²

Note:
The equation of a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 may be taken as y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\). The point of contact is \(\left(\frac{-\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\) where c² = a²m² + b²

Theorem:
Two tangents can be drawn to an ellipse from an external point.

Director Circle:
The points of intersection of perpendicular tangents to an ellipse S = 0 lies on a circle, concentric with the ellipse.
Proof:
Equation of the ellipse
S ≡\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 0
Let P(x₁, y₁) be the point of intersection of perpendicular tangents drawn to ellipse.

Let y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) be a tangent to the ellipse S = 0 passing through P.
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
⇒ y₁ – mx₁ = ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
⇒ (y₁ – mx₁)² = a²m² + b²
⇒ y₁² + m²x₁² – 2x₁y₁m = a²m² + b²
⇒ (x₁² – a²)m² -2x₁y₁m + (y₁² -b²) = 0 … (1)

If m₁, m₂ are the slopes of the tangents through P then m1, m2 are the roots of (1).
The tangents through P are perpendicular.
⇒ m₁m₂ = -1 ⇒ \(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\) = -1
⇒ y₁² – b² = -x² + a² ⇒ x²₁ + y²₁ = a² + b²
∴ P lies on x² + y² = a² + b² which is a circle with centre as origin, the centre of the ellipse.

Auxiliary Circle:
Theorem: The feet of the perpendiculars drawn from either of the foci to any tangent to the ellipse S = 0 lies on a circle, concentric with the ellipse.( called auxiliary circle)
Proof:
Equation of the ellipse S ≡ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 = 0
Let P(x₁, y₁) be the foot of the perpendicular drawn from either of the foci to a tangent.
The equation of the tangent to the ellipse S = 0 is y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) … (1)
The equation to the perpendicular from either foci (±ae, 0) on this tangent is
y = –\(\frac{1}{m}\)(x ± ae)

Now P is the point of intersection of (1) and (2)

∴ y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\), y₁ = –\(\frac{1}{m}\)(x₁ ± ae
⇒ y₁ – mx₁ = ±V a2m2 + b2, my₁ + x₁ = ±ae
⇒ (y₁ – mx₁)² + (my₁ + x₁)² = a²m² + b² + a²e²
⇒ y₁² + m²x² – 2x₁y₁m + m²y₁² + x₁² + 2x₁y₁m = a²m² + a²(1 – e²) + a²e²
⇒ x₁²(m² + 1) + y₁²(1 + m²) = a²m² + a²
⇒ (x₁² + y₁²)(m² + 1) = a² (m² + 1)
⇒ x₁² + y₁² = a²
P lies on x² + y² = a² which is a circle with centre as origin, the centre of the ellipse.

Theorem:
The equation to the chord of contact of P(x₁, y₁) with respect to the ellipse S = 0 is S₁ = 0.

Eccentric Angle Definition:
Let P(x, y) be a point on the ellipse with centre C. Let N be the foot of the perpendicular of P on the major axis. Let NP meets the auxiliary circle at P’. Then ∠NCP’ is called eccentric angle of P. The point P’ is called the corresponding point of P.

Parametric Equations: If P(x, y) is a point on the ellipse then x = a cos θ, y = b sin θ where θ is the eccentric angle of P. These equations x = a cos θ, y = b sin θ are called parametric equations of the ellipse. The point P(a cos θ, b sin θ) is simply denoted by θ.

Theorem: The equation of the chord joining the points with eccentric angles α and β on the ellipse S = 0 is \(\frac{x}{a} \cos \frac{\alpha+\beta}{2}+\frac{y}{b} \sin \frac{\alpha+\beta}{2}=\cos \frac{\alpha-\beta}{2}\)
Proof:
Given points on the ellipse are P(a cos α, b sin α), Q(a cos β, b sin β).
Slope of \(\overline{\mathrm{PQ}}\) is \(\frac{\mathrm{b} \sin \alpha-\mathrm{b} \sin \beta}{\mathrm{a} \cos \alpha-\mathrm{a} \cos \beta}=\frac{\mathrm{b}(\sin \alpha-\sin \beta)}{\mathrm{a}(\cos \alpha-\cos \beta)}\)
Equation of \(\overline{\mathrm{PQ}}\) is:

Theorem: The equation of the tangent at P(θ) on the ellipse
S = 0 is \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1.

Theorem: The Equation of The Normal At P(θ) On The Ellipse
S = 0 Is \(\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}\) = a² – b².

Theorem: Four normals can be drawn from any point to the ellipse and the sum of the eccentric angles of their feet is an odd multiple of π.

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 3 Parabola to solve questions creatively.

Intermediate 2nd Year Maths 2B Parabola Formulas

Definition:
→ Conic section: If right circular solid cone is cut by a plane the section of it is called a Conic section.

→ Conic: The locus of a point whose distances from a fixed point and a fixed straight line are in constant ratio ‘e’ is called a conic.

→ Parabola : A conic with eccentricity 7 is called a parabola i.e., the locus of a point, whose distance from fixed point (focus) is equal to the distance from a fixed straight line (directrix) is called a parabola.

→ Axis of Parabola: The line passing through the vertex and the focus and perpendicular to directrix of the parabola is called the axis of the Parabola.

→ Equation of Parabola : General form – Let S(α, β) be the focus and ax + by + c = 0 be directrix then by definition the equation of parabola be,
\(\sqrt{(x-\alpha)^{2}+(y-\beta)^{2}}\) = \(\left|\frac{a x+b y+c}{\sqrt{a^{2}+b^{2}}}\right|\)

Various forms of parabola:
→ y² = 4ax
Axis : X – axis
Focus : (a, 0)
Vertex : (0, 0)
Equation of directrix : x + a = 0

→ y²= – 4ax
Axis: x – axis
Focus: (- a, 0)
Vertex : (0, 0)
Equation of directrix : x – a = 0

→ x² = 4ay (a > 0)
Axis: Y-axis Focus : (0, a)
Vertex : (0, 0)
Equation of directrix: y + a = 0

→ x² = -4ay(a>0)
Axis : Y axis
Focus : (0, – a)
Vertex: (0, 0)
Equation of directrix : y – a = 0

→ (y – k)² = 4a (x – h) a > 0
Axis : y = k’ a line parallel to X- axis
Focus : (a + h, k)
Vertex: (h, k)
Equation of directrix: x + a = h

→ (x – h)² = 4a (y – k) a > 0
Axis: x = h a line parallel to Y-axis
Focus: (h, a + k)
Vertex: (h, k)
Equation of directrix: y + a = k

→ \(\sqrt{(x-h)^{2}+(y-k)^{2}}\) = \(\left|\frac{a x+b y+c}{\sqrt{a^{2}+b^{2}}}\right|\)
Axis: b(x – h) – a(y – k) = 0
Focus: (h, k)
Vertex: Some point A(fig)
Equation of directrix : ax + by + c = 0

Chord: Line segment joining two points of a parabola is called a, chord of the parabola. If this chord passes through focus is called focal chord.
Chord passing through focus and ⊥ to axis is called latus rectum.
Length of latus rectum = 4a

Parametric form of parabola:
For y² = 4ax
Parametric form will be x = at², y = 2at

Definition:
→ The locus of a point which moves in a plane so that its distance from a fixed point bears a constant ratio to its distance from a fixed straight line is called a conic section or conic. The fixed point is called focus, the fixed straight line is called directrix and the constant ratio ‘e’ is called eccentricity of the conic.

If e = 1, then the conic is called a Parabola.
If e < 1, then the conic is called an Ellipse. If e > 1, then the conic is called a hyperbola.
Note: The equation of a conic is of the form ax² + 2hxy + by² + 2gx + 2fy + c = 0.

→ Directrix of the Conic: A line L = 0 passing through the focus of a conic is said to be the principal axis of the conic if it is perpendicular to the directrix of the conic.

→ Vertices: The points of intersection of a conic and its principal axis are called vertices of the conic.

→ Centre: The midpoint o the line segment joining the vertices of a conic is called centre of the conic.

→ Note 1: If a conic has only one vertex then its centre coincides with the vertex.

→ Note 2: If a conic has two vertices then its centre does not coincide either of the vertices. In this case the conic is called a central conic.

→ Standard Form: A conic is said to be in the standard form if the principal axis of the conic is x-axis and the centre of the conic is the origin.

Equation of a Parabola in Standard Form:
The equation of a parabola in the standard form is y² = 4ax.
Proof :
Let S be the focus and L = 0 be the directrix of the parabola.
Let P be a point on the parabola.
Let M, Z be the projections of P, S on the directrix L = 0 respectively.
Let N be the projection of P on SZ.
Let A be the midpoint of SZ.
Therefore, SA = AZ, ⇒ A lies on the parabola. Let AS = a.
Let AS, the principal axis of the parabola as x-axis and Ay perpendicular to SZ as y-axis.
Then S = (a, 0) and the parabola is in the standard form.
Let P = (x₁, y₁).

Now PM = NZ = NA + AZ = x₁ + a
P lies on the parabola ⇒ \(\frac{\mathrm{PS}}{\mathrm{PM}}\) = 1 ⇒ PS = PM
⇒ \(\sqrt{\left(x_{1}-a\right)^{2}+\left(y_{1}-0\right)^{2}}\) = x₁ + a
⇒ (x₁ – a)² + y₁² = (x₁ + a)²
⇒ y₁² = (x₁ + a)² – (x₁ – a)²
⇒ y₁² = 4ax₁
The locus of P is y² = 4ax.
∴ The equation to the parabola is y² = 4ax.

Nature of the Curve y² = 4ax.

(i) The curve is symmetric with respect to the x-axis.
∴ The principal axis (x-axis) is an axis of the parabola.

(ii) y = 0 ⇒ x = 0. Thus the curve meets x-axis at only one point (0, 0).
Hence the parabola has only one vertex.

(iii) If x < 0 then there exists no y ∈ R. Thus the parabola does not lie in the second and third quadrants. (iv) If x > 0 then y² > 0 and hence y has two real values (positive and negative). Thus the parabola lies in the first and fourth quadrants.

(v) x = 0 ⇒ y² = 0 ⇒ y = 0, 0. Thus y-axis meets the parabola in two coincident points and hence y-axis touches the parabola at (0, 0).

(vi) As x → ∞ ⇒ y² → ∞ ⇒ y → ±∞
Thus the curve is not bounded (closed) on the right side of the y-axis.

→ Double Ordinate: A chord passing through a point P on the parabola and perpendicular to the principal axis of the parabola is called the double ordinate of the point P.

→ Focal Chord: A chord of the parabola passing through the focus is called a focal chord.

→ Latus Rectum: A focal chord of a parabola perpendicular to the principal axis of the parabola is called latus rectum. If the latus rectum meets the parabola in L and L’, then LL’ is called length of the latus rectum.

Theorem: The length of the latus rectum of the parabola y² = 4ax is 4a.
Proof:
Let LL’ be the length of the latus rectum of the parabola y² = 4ax.

Let SL = 1, then L = (a, 1)
Since L is a point on the parabola y² = 4ax, therefore 1² = 4a(a)
⇒ 1² = 4a² ⇒ 1 = 2a ⇒ SL = 2a
∴ LL’ = 2SL = 4a.

→ Focal Distance: If P is a point on the parabola with focus S, then SP is called focal distance of P.

Theorem: The focal distance of P(x₁, y₁) on the parabola y² = 4ax is x₁ + a.
Notation: We use the following notation in this chapter
S ≡ y² – 4ax
S₁ ≡ YY₁ – 2a(x + x₁)
S₁₁ = S(x₁, y₁) ≡ y₁² – 4ax₁
S₁₂ ≡ y₁y₂ – 2a(x₁ + x₂)

Note:
Let P(x₁, y₁) be a point and S ≡ y² – 4ax = 0 be a parabola. Then

• P lies on the parabola ⇔ S₁₁ = 0
• P lies inside the parabola ⇔ S₁₁ = 0
• P lies outside the parabola ⇔ S₁₁ = 0

Theorem: The equation of the chord joining the two points A(x₁, y₁), B(x₂, y₂) on the parabola S = 0 is S₁ + S₂ = S₁₂.
Theorem: The equation of the tangent to the parabola S = 0 at P(x₁, y₁) is S₁ = 0.

Normal:
Let S = 0 be a parabola and P be a point on the parabola S = 0. The line passing through P and perpendicular to the tangent of S = 0 at P is called the normal to the parabola S = 0 at P.

Theorem: The equation of the normal to the parabola y² = 4ax at P(x₁, y₁) is y₁(x – x₁) + 2a(y – y₁) = 0.
Proof:
The equation of the tangent to S = 0 at P is S₁ = 0

⇒ yy₁ – 2a(x + x₁) = 0.
⇒ yy₁ – 2ax – 2ax₁ = 0
The equation of the normal to S = 0 at P is:
y₁(x – x₁) + 2a(y – y₁) = 0

Theorem: The condition that the line y = mx + c may be a tangent to the parabola y² = 4ax is c = a/m.
Proof:
Equation of the parabola is y² = 4ax ……….. (1)
Equation of the line is y = mx + c ………. (2)
Solving (1) and (2),
(mx + c)² = 4ax ⇒ m²x² + c² + 2mcx = 4ax
⇒ m²x² + 2(mc – 2a)x + c² = 0
Which is a quadratic equation in x. Therefore it has two roots.
If (2) is a tangent to the parabola, then the roots of the above equation are equal.
⇒ its disc eminent is zero
⇒ 4(mc – 2a)² – 4m²c² = 0
⇒ m²c² + 4a² – 4amc – m²c² = 0
⇒ a² – amc = 0
⇒ a = mc
⇒ C = \(\frac{a}{m}\)

II Method:
Given parabola is y² = 4ax.
Equation of the tangent is y = mx + c ———— (1)
Let P(x₁, y₁) be the point of contact.
The equation of the tangent at P is
yy₁ – 2a(x + x₁) = 0 ⇒ yy₁ = 2ax + 2ax₁ ……. (2)
Now (1) and (2) represent the same line.
∴ \(\frac{\mathrm{y}_{1}}{1}=\frac{2 \mathrm{a}}{\mathrm{m}}=\frac{2 \mathrm{ax}_{1}}{\mathrm{c}}\) ⇒ x₁ = \(\frac{\mathrm{c}}{\mathrm{m}}\), y₁ = \(\frac{\mathrm{2a}}{\mathrm{m}}\)
P lies on the line y = mx + c ⇒ y₁ = mx₁ + c
⇒ \(\frac{2 \mathrm{a}}{\mathrm{m}}=\mathrm{m}\left(\frac{\mathrm{c}}{\mathrm{m}}\right)+\mathrm{c} \Rightarrow \frac{2 \mathrm{a}}{\mathrm{m}}=2 \mathrm{c} \Rightarrow \mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}\)

Note: The equation of a tangent to the parabola y² = 4ax can be taken as y = mx + a/m. And the point of contact is (a/m², 2a/m).

Corollary: The condition that the line lx + my + n = 0 to touch the parabola y² = 4ax is am² = ln.
Proof:
Equation of the parabola is y² = 4ax …………. (1)
Equation of the line is lx + my + n = 0
⇒ y = – \(\frac{1}{\mathrm{~m}}\)x – \(\frac{\mathrm{n}}{\mathrm{m}}\)
But this line is a tangent to the parabola, therefore
C = a/m ⇒ \(-\frac{\mathrm{n}}{\mathrm{m}}=\frac{\mathrm{a}}{-1 / \mathrm{m}} \Rightarrow \frac{\mathrm{n}}{\mathrm{m}}=\frac{\mathrm{am}}{1}\) ⇒ am² = ln
Hence the condition that the line lx + my + n = 0 to touché the parabola y² = 4ax is am² = ln.

Note: The point of contact of lx + my + n = 0 with y² = 4ax is (n/l, – 2am/l).

Theorem: The condition that the line lx + my + n = 0 to touch the parabola x² = 4ax is al² = mn.
Proof:
Given line is lx + my + n = 0 …… (1)
Let P(x₁, y₁) be the point of contact of (1) with the parabola x² = 4ay.
The equation of the tangent at P to the parabola is xx₁ = 2a(y + y₁)
⇒ x₁x – 2ay – 2ay₁ = 0 …… (2)
Now (1) and (2) represent the same line.
∴ \(\frac{\mathrm{x}_{1}}{1}=-\frac{2 \mathrm{a}}{\mathrm{m}}=-\frac{2 \mathrm{ay}_{1}}{\mathrm{n}}\) ⇒ x₁ = – \(-\frac{2 \mathrm{al}}{\mathrm{m}}\), y₁ = \(\frac{n}{m}\)
P lies on the line lx + my + n = 0
⇒ lx₁ + my₁ + n = 0 ⇒ l\(\left(\frac{-2 a l}{m}\right)\) + m\(\left(\frac{\mathrm{n}}{\mathrm{m}}\right)\) + n = 0
⇒ – 2al² + mn + mn = 0 ⇒ al² = mn

Theorem: Two tangents can be drawn to a parabola from an external point.

Note:
1. If m₁, m₂ are the slopes of the tangents through P, then m₁, m₂ become the roots of
equation (1). Hence m₁ + m₂ = y₁/x₁, m₁m₂ = a/x₁.

2. If P is a point on the parabola S =0 then the roots of equation (1) coincide and hence only one tangent can be drawn to the parabola through P.

3. If P is an internal point to the parabola S = 0 then the roots of (1) are imaginary and hence no tangent can be drawn to the parabola through P.

Theorem: The equation in the chord of contact of P(x₁, y₁) with respect to the parabola S = 0 is S₁ = 0.

Theorem: The equation of the chord of the parabola S = 0 having P(x₁, y₁) as its midpoint is S₁ = S₁₁.

Pair of Tangents:
Theorem: The equation to the pair of tangents to the parabola S = 0 from P(x₁, y₁) is S₁² = s₁₁s.

Parametric Equations of the Parabola:
A point (x, y) on the parabola y² = 4ax can be represented as x = at², y = 2at in a single parameter t. Theses equations are called parametric equations of the parabola y² = 4ax. The point (at², 2at) is simply denoted by t.

Theorem: The equation of the tangent at (at², 2at) to the parabola is y² = 4ax is yt = x + at².
Proof:
Equation of the parabola is y² = 4ax.
Equation of the tangent at (at², 2at) is S₁ = 0.
⇒ (2at)y – 2a(x + at²) = 0
⇒ 2aty = 2a(x + at²) ⇒ yt = x + at².

Theorem: The equation of the normal to the parabola y² = 4ax at the point t is y + xt = 2at + at³.
Proof:
Equation of the parabola is y² = 4ax.
The equation of the tangent at t is:
yt = x + at² = x – yt + at² = 0
The equation of the normal at (at², 2at) is
t(x – at²) + l(y – 2at) = 0
⇒ xt – at³ + y – 2at = 0 ⇒ y + xt = 2at + at³

Theorem: The equation of the chord joining the points t₁ and t₂ on the parabola y² = 4ax is y(t₁ + t₂) = 2x + 2at₁t₂.
Proof:
Equation of the parabola is y² = 4ax.
Given points on the parabola are
P(at₁², 2at₁), Q(at₂², 2at₂) .
Slope of \(\overline{\mathrm{PQ}}\) is
\(\frac{2 \mathrm{at}_{2}-2 \mathrm{at}_{1}}{\mathrm{at}_{2}^{2}-\mathrm{at}_{1}^{2}}=\frac{2 \mathrm{a}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}{\mathrm{a}\left(\mathrm{t}_{2}^{2}-\mathrm{t}_{1}^{2}\right)}=\frac{2}{\mathrm{t}_{1}+\mathrm{t}_{2}}\)
The equation of \(\overline{\mathrm{PQ}}\) is y – 2at₁ = \(\frac{2}{t_{1}+t_{2}}\) (x – at₁²).
⇒ (y – 2at₁) (t₁ + t₂) = 2(x – at₁²)
⇒ y(t₁ + t₂) – 2at₁² – 2at₁t₂ = 2x – 2at₁²
⇒ y(t₁ + t₂) = 2x + 2at₁t₂.

Note: If the chord joining the points t₁ and t₂ on the parabola y² = 4ax is a focal chord then t₁t₂ = – 1.
Proof:
Equation of the parabola is y² = 4ax
Focus S = (a, o)
The equation of the chord is y(t₁ + t₂) = 2x + 2at₁t₂
If this is a focal chord then it passes through the focus (a, 0).
∴ 0 = 2a + 2at₁t₂ ⇒ t₁t₂ = – 1.

Theorem: The point of intersection of the tangents to the parabola y² = 4ax at the
points t₁ and t₂ is (at₁t₂, a[t₂ + t₂]).
Proof:
Equation of the parabola is y² = 4ax
The equation of the tangent at t₁ is yt₁ = x + at₁² ……. (1)
The equation of the tangent at t₂ is
yt₂ = x + at₂² ……….. (2)
(1) – (2) ⇒ y(t₁ – t₂) = a(t₁² – t₂²) ⇒ y = a(t₁ + t₂)
(1) ⇒ a(t₁ + t₂)t₁ = x + at₁²
= at₁² + at₁t₁ = x + at₁² ⇒ x = at₁t₂,
∴ Point of intersection = (at₁t₂, a[t₁ + t₂]).

Theorem: Three normals can be drawn form a point (x₁, y₁) to the parabola y² = 4ax.
Corollary: If the normal at t₁ and t₂, to the parabola y² = 4ax meet on the parabola, then t₁t₂ = 2.
Proof:
Let the normals at t₁ and t₂ meet at t₃ on the parabola.
The equation of the normal at t1 is:
y + xt₁ = 2at₁ + at₁³ ………… (1)
Equation of the chord joining t1 and t3 is:
y(t₁ + t₃) = 2x + 2at₁t₃ ……… (2)

(1) and (2) represent the same line
∴ \(\frac{t_{1}+t_{3}}{1}=\frac{-2}{t_{1}} \Rightarrow t_{3}=-t_{1}-\frac{2}{t_{1}}\)
Similarly t₃ = – t₁ – \(\frac{2}{\mathrm{t}_{2}}\)
∴ \(-\mathrm{t}_{1}-\frac{2}{\mathrm{t}_{1}}=-\mathrm{t}_{2}-\frac{2}{\mathrm{t}_{2}}\) ⇒ t₁ – t₂ \(\frac{2}{\mathrm{t}_{2}}-\frac{2}{\mathrm{t}_{1}}\)
⇒ t₁ – t₂ = \(\frac{2\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}{\mathrm{t}_{1} \mathrm{t}_{2}}\) ⇒ t₁t₂ = 2

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 1 Circle to solve questions creatively.

Intermediate 2nd Year Maths 2B Circle Formulas

→ The locus of a point in a plane such that its distance from a fixed point in the plane is always the same is called a circle.

→ The equation of a circle with centre (h, k) and radius r is (x – h)² + (y – k)² = r²

→ The equation of a circle in standard form is x² + y² = r².

→ The equation of a circle in general form is x² + y² + 2gx + 2fy + c = 0 and its centre is (-g, -f), radius is \(\sqrt{g^{2}+f^{2}-c}\).

→ The intercept made by x² + y² + 2gx + 2fy + c = 0

• on X-axis is 2\(\sqrt{g^{2}-c}\) if g² > c.
• on Y-axis is 2\(\sqrt{f^{2}-c}\) if f² > c.

→ If the extremities of a diameter of a circle are (x₁, y₁) and (x₂, y₂) then its equation is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0

→ The equation of a circle passing through three non-collinear points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
\(\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\) = (x² + y²) + \(\left|\begin{array}{lll}
c_{1} & y_{1} & 1 \\
c_{2} & y_{2} & 1 \\
c_{3} & y_{3} & 1
\end{array}\right|\) x + \(\left|\begin{array}{lll}
x_{1} & C_{1} & 1 \\
x_{2} & C_{2} & 1 \\
x_{3} & C_{3} & 1
\end{array}\right|\) y + \(\left|\begin{array}{lll}
x_{1} & y_{1} & C_{1} \\
x_{2} & y_{2} & C_{2} \\
x_{3} & y_{3} & C_{3}
\end{array}\right|\) = 0.
where c_i = – (x_i² + y_i²)

→ The centre of the circle passing through three non-collinear points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
\(\left[\frac{\left|\begin{array}{lll}
c_{1} & y_{1} & 1 \\
c_{2} & y_{2} & 1 \\
c_{3} & y_{3} & 1
\end{array}\right|}{(-2)\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|}, \frac{\left|\begin{array}{lll}
x_{1} & c_{1} & 1 \\
x_{2} & c_{2} & 1 \\
x_{3} & c_{3} & 1
\end{array}\right|}{(-2)\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|}\right]\)

→ The parametric equations of a circle with centre (h, k) and radius (r ≥ 0) are given by
x = h + r cos θ
y = k + r sin θ 0 ≤ 6 < 2π.

→ A point P(x₁, y₁) is an interior point or on the circumference or an exterior point of a circles S = 0 ⇔ S₁₁ \(\frac{<}{>}\) 0.

→ The power of P(x₁, y₁) with respect to the circle S = 0 is S₁₁.

→ A point P(x₁, y₁) is an interior point or on the circumference or exterior point of the circle S = 0 ⇔ the power of P with respect to S = 0 is negative, zero and positive.

→ If a straight line through a point P(x₁, y₁) meets the circle S = 0 at A and B then the power of P is equal to PA. PB.

→ The length of the tangent from P(x₁, y₁) to S = 0 is \(\sqrt{S_{11}}\).

→ The straight line l = 0 intersects, touches or does not meet the circles = 0 according as l < r, l = r or l > r where l is the perpendicular distance from the centre of the circle to the line l = 0 and r is the radius.

→ For every real value of m the straight line y = mx ± r \(\sqrt{1+m^{2}}\) is a tangent to the circle x² + y² = r².

→ If r is the radius of the circle S = x² + y² + 2gx + 2fy + c = 0 then for every real value of m the straight line y + f = m(n + g) ±r + m2 will be a tangent to the circle.

→ If P(x₁, y₁) and Q(x₂, y₂) are two points on the circle S = 0 then the secant’s \((\stackrel{\leftrightarrow}{P Q})\) equation is S₁ + S₂ = S₁₂

→ The equation of tangent at (x₁, y₁) of the circle S = 0 is S₁ = 0.

→ If θ₁, θ₂ are two points on S = x² + y² + 2gx + 2fy + c = 0 then the equation of the chord joining the points θ₁, θ₂ is
(x + g) cos \(\left(\frac{\theta_{1}+\theta_{2}}{2}\right)\) + (y + f) sin \(\left(\frac{\theta_{1}+\theta_{2}}{2}\right)\) = r cos \(\left(\frac{\theta_{1}-\theta_{2}}{2}\right)\)

→ The equation of the tangent at θ of the circle S = 0 is (x + g) cos θ + (y + f) sin θ = r.

→ The equation of normal at (x₁, y₁) of the circle
S = 0 is (x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0.

→ The chord of contact of P(x₁ y₁) (exterior point) with respect to S = 0 is S₁ = 0.

→ The equation of the polar of a point P(x₁, y₁) with respect to S = 0 is S₁ = 0.

→ The pole of lx + my + n = 0 with respect to S = 0 is
\(\left(-g+\frac{l r^{2}}{l g+m f-n},-f+\frac{m r^{2}}{l g+m f-n}\right)\)

→ Where r is the radius of the circle. The polar of P(x₁, y₁) with respect to S = 0 passes through Q(x₂, y₂) ⇔ the polar of Q with respect to S – 0 passes through P.

→ The points (x₁, y₁) and (x₂, y₂) are conjugate points with respect to S = 0 if S₁₂ = 0

→ Two lines l₁x = m₁y + n₁ = 0, l₂x + m₂y + n₂ = 0 are conjugate with respect to x² + y² = a² ⇔ (l₁l₂ + m₁m₂) = n₁n₂

→ Two points P, Q are said to be inverse points with respect to S = 0 if CP. CQ = r² where C is the centre and r is the radius of the circle S = 0.

→ If (x₁, y₁) is the mid-point of a chord of the circle S = 0 then its chord equation is S₁ = S₁₁.

→ The pair of common tangents to the circles S = 0, S’ = 0 touching at a point on the lines segment \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) (C₁, C₂ are centres of the circles) is called transverse pair of common tangents.

→ The pair of common tangents to the circles S = 0, S’ = 0 intersecting at a point not in \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) is called as direct pair of common tangents.

→ The point of intersection of transvese (direct) common tangents is called internal (external) Centre of similitude.

→ The combined equation of the pair of tangents drawn from an external point P(x₁, y₁) to the circle S = 0 is SS₁₁ = S²₁.

Equation of a Circle:
The equation of the circle with centre C (h, k) and radius r is (x – h)² + (y – k)² = r².
Proof:
Let P(x₁, y₁) be a point on the circle.
P lies in the circle ⇔ PC = r ⇔ \(\sqrt{\left(\mathrm{x}_{1}-\mathrm{h}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{k}\right)^{2}}\) = r
⇔ (x₁ – h)² + (y₁ – k)² = r².

The locus of P is (x – h)² + (y – k)² = r².
∴ The equation of the circle is (x – h)² + (y – k)² = r².

Note: The equation of a circle with centre origin and radius r is (x – 0)² + (y – 0)² = r²
i.e., x² + y² = r² which is the standard equation of the circle.

Note: On expanding equation (1), the equation of a circle is of the form x² + y² + 2gx + 2fy + c = 0.

Theorem: If g² + f² – c ≥ 0, then the equation x² + y² + 2gx + 2fy + c = 0 represents a circle with centre (- g, – f) and radius \(\sqrt{g^{2}+f^{2}-c}\).
Note: If ax² + ay² + 2gx + 2fy + c = 0 represents a circle, then its centre = \(\left(-\frac{g}{a},-\frac{f}{a}\right)\) and its radius \(\frac{\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{ac}}}{|\mathrm{a}|}\).

Theorem: The equation of a circle having the line segment joining A(x₁, y₁) and B(x₂, y₂) as diameter is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0.

Let P(x, y) be any point on the circle. Given points A(x₁, y₁) and B(x₂, y₂).
Now ∠APB = \(\frac{\pi}{2}\). (Angle in a semi circle.)
Slope of AP. Slope of BP = – 1
⇒ \(\frac{y-y_{1}}{x-x_{1}} \frac{y-y_{2}}{x-x_{2}}\) = – 1
⇒ (y – y₂) (y – y₁) = – (x – x₂) (x – x₁) = 0
⇒ (x – x₂) (x – x₁) + (y – y₂) (y – y₁) = 0

Definition: Two circles are said to be concentric if they have same center.

The equation of the circle concentric with the circle x² + y² + 2gx + 2fy + c = 0 is of the form x² + y² + 2gx + 2fy + k = 0.
The equation of the concentric circles differs by constant only.

Parametric Equations of A Circle:

Theorem: If P(x, y) is a point on the circle with centre C(α, β) and radius r, then x = α + r cosθ, y = β + r sin θ where 0 ≤ θ < 2π.

Note: The equations x = α + r cos θ, y = + r sin θ, 0 ≤ θ < 2π are called parametric equations of the circle with centre (α, β) and radius r.

Note: A point on the circle x² + y² = r² is taken in the form (r cos θ, r sin θ). The point (r cos θ, r sin θ) is simply denoted as point θ.

Theorem:
(1) If g² – c > 0 then the intercept made on the x axis by the circle x² + y² + 2gx + 2fy + c = 0 is 2\(\sqrt{g^{2}-a c}\)
(2) If f² – c >0 then the intercept made on the y axis by the circle x² + y² + 2gx + 2fy + c = 0 is 2\(\sqrt{f^{2}-b c}\)

Note: The condition for the x-axis to touch the circle
x² + y² + 2gx + 2fy + c = 0 (c > 0) is g² = c.

Note: The condition of the y-axis to touch the circle
x² + y² + 2gx + 2fy + c = 0 (c > 0) is f² = c.

Position of Point:
Let S = 0 be a circle and P(x₁, y₁) be a point I in the plane of the circle. Then

• P lies inside the circle S = 0 ⇔ S₁₁ < 0
• P lies in the circle S = 0 ⇔ S₁₁ = 0
• Plies outside the circle S = 0 ⇔ S₁₁ = 0

Power of a Point:
Let S = 0 be a circle with centre C and radius r. Let P be a point. Then CP² – r² is called power of P with respect to the circle S = 0.

Theorem: The power of a point P(x₁, y₁) with respect to the circle S = 0 is S₁₁.

Theorem: The length of the tangent drawn from an external point P(x₁, y₁) to the circle s = 0 is \(\sqrt{\mathrm{S}_{11}}\).

Theorem: The equation of the tangent to the circle S = 0 at P(x₁, y₁) is S₁ = 0.

Theorem: The equation of the normal to the circle S = x² + y² + 2gx + 2fy + c = 0 at P(x₁, y₁) is
(y₁ + f) (x – x₁) – (x₁ + g) (y – y₁) = 0.

Corollary: The equation of the normal to the circle x² + y² = a² at P(x₁, y₁) is y₁x – x₁y = 0.

Theorem: The condition that the straight line lx + my + n = 0 may touch the circle x² + y² = a² is n² = a²(l² + m²) and the point of contact is \(\left(\frac{-a^{2} 1}{n}, \frac{-a^{2} m}{n}\right)\).
Proof:
The given line is lx + my + n = 0 …… (1)
The given circle is x² + y² = r² ……. (2)
Centre C = (0, 0), radius = r
Line (1) is a tangent to the circle (2)
⇔ The perpendicular distance from the centre C to the line (1) is equal to the radius r.
⇔ \(\left|\frac{0-n}{\sqrt{1^{2}+m^{2}}}\right|\) = r
⇔ (n)² = r² (l² + m²)

Let P(x₁, y₁) be the point of contact.
Equation of the tangent is S₁ = 0, ⇒ x₁x + y₁y – r² = 0. —- (3)
Equations (1) and (3) are representing the same line, therefore, \(\frac{x_{1}}{l}=\frac{y_{1}}{m}=\frac{-a^{2}}{n}\) ⇒ x₁ = \(\frac{-a^{2} l}{n}\), y₁ = \(\frac{-a^{2} m}{n}\)
Therefore, point of contact is \(\left(\frac{-a^{2} l}{n}, \frac{-a^{2} m}{n}\right)\)

Theorem: The condition for the straight line lx + my + n = 0 may be a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 is (g² + f² – c) (l² + m²) = (lg + mf – n)².
Proof:
The given line is lx + my + n = 0 …….. (1)
The given circle is x² + y² + 2gx + 2fy + c = 0 …….. (2)

Centre C = (- g, – f), radius r = \(\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\)
Line (1) is a tangent to the circle (2)
⇔ The perpendicular distance from the centre C to the line (1) is equal to the radius r.
⇔ \(\left|\frac{-\lg -m f+c}{\sqrt{1^{2}+m^{2}}}\right|\) = \(\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\)
⇔ (lg + mf -n)² = (g² + f² – c) (l² + m²)

Corollary: The condition for the straight line y = mx + c to touch the circle
x² + y² = r² is c² = r²(1 + m²).
The given line is y = mx + c i.e., mx – y + c = 0 … (1)
The given circle is S = x² + y² = r²
Centre C = (0,0), radius = r.
If (1) is a tangent to the circle, then
Radius of the circle = perpendicular distance from centre of the circle to the line.

⇒ r = \(\frac{|c|}{\sqrt{m^{2}+1}}\) ⇒ r² = \(\frac{c^{2}}{m^{2}+1}\) ⇒ r² (m² + 1) = c²

Corollary: If the straight line y = mx + c touches the circle x² + y² = r², then their point of contact is \(\left(-\frac{r^{2} m}{c}, \frac{r^{2}}{c}\right)\).
Proof:
The given line is y = mx + c i.e., mx – y + c = 0 ……. (1)
The given circle is S = x² + y² = r² ……. (2)
Centre C = (0, 0), radius = r
Let P(x₁, y₁) be the point of contact.
Equation of the tangent is S₁ = 0, ⇒ x₁x + y₁y – r² = 0. ………. (3)

Equations (1) and (3) are representing the same line, therefore, \(\frac{x_{1}}{m}=\frac{y_{1}}{-1}=\frac{-r^{2}}{c}\) ⇒ x₁ = \(\frac{-r^{2} m}{c}\), y₁ = \(\frac{r^{2}}{c}\)
Point of contact is (x₁, y₁) = \(\left(-\frac{\mathrm{r}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{r}^{2}}{\mathrm{c}}\right)\)

Theorem: If P(x, y) is a point on the circle with centre C(α, β) and radius r, then x = α + r cos θ, y = β + r sin θ where 0 ≤ θ < 2π.

Note 1: The equations x = α + r cos θ, y = β + r sin θ, 0 ≤ θ < 2π are called parametric equations of the circle with centre (α, β) and radius r.

Note 2: A point on the circle x² + y² = r² is taken in the form (r cosθ, r sin θ). The point (r cosθ, r sin θ) is simply denoted as point θ.

Theorem: The equation of the chord joining two points θ₁ and θ₂ on the circle
x² + y² + 2gx + 2fy + c = 0 is (x + g)cos\(\frac{\theta_{1}+\theta_{2}}{2}\) + (y + f) sin \(\frac{\theta_{1}+\theta_{2}}{2}\) = r cos \(\frac{\theta_{1}+\theta_{2}}{2}\) where r is the radius of the circle.

Note 1: The equation of the chord joining the points θ₁ and θ₂ on the circle x² + y² = r² is x cos\(\frac{\theta_{1}+\theta_{2}}{2}\) + y sin\(\frac{\theta_{1}+\theta_{2}}{2}\) = r cos\(\frac{\theta_{1}-\theta_{2}}{2}\)

Note 2: The equation of the tangent at P(θ) on the circle (x + g) cos θ + (y + f) sin θ = \(\sqrt{g^{2}+f^{2}-c}\).

Note 3: The equation of the tangent at P(θ) on the circle x² + y² = r² is x cos θ + y sin θ = r.

Note 4: The equation of the normal at P(θ) on the circle x² + y² = r² is x sin θ – y cos θ = r.

Theorem:
If a line passing through a point P(x₁, y₁) intersects the circle S = 0 at the points A and B then PA.PB = |S₁₁|.

Corollary:
If the two lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 meet the coordinate axes in four distinct points then those points are concyclic ⇔ a₁a₂ = b₁b₂.

Corollary:
If the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 meet the coordinate axes in four distinct concyclic points then the equation of the circle passing through these concyclic points is (a₁x + b₁y + c₁) (a₂x + b₂y + c₂) – (a₁b₂ + a₂b₁)xy = 0.

Theorem:
Two tangents can be drawn to a circle from an external point.

Note:
If m₁, m₁ are the slopes of tangents drawn to the circle x² + y² = a² from an external point (x₁, y₁) then m₁ + m₂ = \(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}\), m₁m₂ = \(\frac{y_{1}^{2}-a^{2}}{x_{1}^{2}-a^{2}}\).

Theorem:
If θ is the angle between the tangents through a point P to the circle S = 0 then tan = \(\frac{\theta}{2}=\frac{r}{\sqrt{S_{11}}}\) where r is the radius of the circle.
Proof:

Let the two tangents from P to the circle S = 0 touch the circle at Q, R and θ be the angle between
these two tangents. Let C be the centre of the circle. Now QC = r, PQ = \(\sqrt{S_{11}}\) and ∆PQC is a right angled triangle at Q.
∴ tan \(\frac{\theta}{2}=\frac{\mathrm{QC}}{\mathrm{PQ}}=\frac{\mathrm{r}}{\sqrt{\mathrm{S}_{11}}}\)

Theorem: The equation to the chord of contact of P(x₁, y₁) with respect to the circle S = 0 is S₁ = 0.

Theorem: The equation of the polar of the point P(x₁, y₁) with respect to the circle S = 0 is S₁ = 0.

Theorem: The pole of the line lx + my + n = 0 (n ≠ 0) with respect to x² + y² = a² is \(\left(-\frac{1 a^{2}}{n},-\frac{m a^{2}}{n}\right)\)
Proof :
Let P(x₁, y₁) be the pole of lx + my + n = 0 ……. (1)
The polar of P with respect to the circle is:
xx₁ + yy₁ – a² = 0
Now (1) and (2) represent the same line
∴ \(\frac{\mathrm{x}_{1}}{\ell}=\frac{\mathrm{y}_{1}}{\mathrm{~m}}=\frac{-\mathrm{a}^{2}}{\mathrm{n}}\) ⇒ x₁ = \(\frac{-\mathrm{la}^{2}}{\mathrm{n}}\), y₁ = \(\frac{-\mathrm{ma}^{2}}{\mathrm{n}}\)
∴ Pole P = \(\left(-\frac{1 a^{2}}{n},-\frac{m a^{2}}{n}\right)\)

Theorem: If the pole of the line lx + my + n = 0 with respect to the circle x² + y² + 2gx + 21y + c = 0 is (x₁, y₁) then \(\frac{x_{1}+g}{\ell}=\frac{y_{1}+f}{m}=\frac{r^{2}}{\lg +\mathrm{mf}-\mathrm{n}}\) where r is the radius of the circle.
Proof:
Let P(x₁, y₁) be the pole of the line lx + my + n = 0 ……. (1)
The poiar of P with respect to S = 0 is S₁ = 0
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ (x₁ + g)x + (y₁ + f) + gx₁ + fy₁ + c = 0 …….. (2)
Now (1) and (2) represent the same line.
∴ \(\frac{\mathrm{x}_{1}+\mathrm{g}}{\ell}=\frac{\mathrm{y}_{1}+\mathrm{f}}{\mathrm{m}}=\frac{\mathrm{gx} \mathrm{x}_{1}+\mathrm{gy} \mathrm{y}_{1}+\mathrm{c}}{\mathrm{n}}\) = k(say)
\(\frac{\mathrm{x}_{1}+\mathrm{g}}{\ell}\) = k ⇒ x₁ + g = l k ⇒ x₁ = lk – g
\(\frac{\mathrm{y}_{1}+\mathrm{f}}{\mathrm{m}}\) = k ⇒ y₁ + f = m k ⇒ y₁ = mk – f
\(\frac{g x_{1}+g y_{1}+c}{n}\) = k ⇒ gx₁ + gy₁ + c = nk
⇒ g(lk – g) + f(mk – f) + c = nk
⇒ k (lg + mf – n) = g² + f² – c = r² Where r is the radius of the circle ⇒ k = \(\frac{r^{2}}{\lg +\mathrm{mf}-\mathrm{n}}\)
∴ \(\frac{\mathrm{x}_{1}+\mathrm{g}}{\ell}=\frac{\mathrm{y}_{1}+\mathrm{f}}{\mathrm{m}}=\frac{\mathrm{r}^{2}}{\lg +\mathrm{mf}-\mathrm{n}}\)

Theorem: The lines l₁x + m₁y + n₁ = 0 and l₂x + m₂y + n₁y = 0 are conjugate with respect to the circle x² + y₁ + 2gx + 2fy + c = 0 iffr₁ (l₁l₂ + m₁m₂) = (l₁g + m₁f – n₁) (l₂g + m₂f – n₂).

Theorem: The condition for the lines l₁x + m₁y + n₁ = 0 and l₂x + m₂y + n₂ = 0 to be conjugate with respect to the circle x² + y² = a² is a²(l₁l₂ + m₁m₂) = n₁n₂.

Theorem: The equation of the chord of the circle S = 0 having P(x₁, y₁) as its midpoint is S₁ = S₁₁.

Theorem: The length of the chord of the circle S = 0 having P(x₁, y₁) as its midpoint is 2\(\sqrt{\left|S_{11}\right|}\).

Theorem: The equation to the pair of tangents to the circle
S = 0 from P(x₁, y₁) is S²₁ = S₁₁S.
Proof:

Let the tangents from P to the circle S = 0 touch the circle at A and B.
Equation of AB is S₁ = 0.
i.e., x₁x + y₁y + g(x + x₁) + f(y + y₁) + c = 0 ———- (i)
Let Q(x₂, y₂) be any point on these tangents. Now locus of Q will be the equation of the pair of tangents drawn from P.
The line segment PQ is divided by the line AB in the ratio – S₁₁:S₂₂
⇒ \(\frac{P B}{Q B}=\left|\frac{S_{11}}{S_{22}}\right|\) ———— (ii)
But PB = \(\sqrt{S_{11}}\), QB = \(\sqrt{S_{22}}\) ⇒ \(\frac{P B}{Q B}=\frac{\sqrt{S_{11}}}{\sqrt{S_{22}}}\) ———— (iii)
From (ii) and (iii) ⇒ \(\frac{s_{11}^{2}}{s_{22}^{2}}=\frac{S_{11}}{S_{22}}\)
⇒ S₁₁S₂₂ = S²₁₂
Hence locus of Q(x₂, y₂) is S₁₁S = S²₁₂

Touching Circles: Two circles S = 0 and S’ = 0 are said to touch each other if they have a unique point P in common. The common point P is called point of contact of the circles S = 0 and S’ = 0.

Circle – Circle Properties: Let S = 0, S’ = 0 be two circle with centres C₁, C₂ and radii r₁, r₂ respectively.

• If C₁C₂ > r₁ + r₂ then each circle lies completely outside the other circle.
• If C₁C₂ = r₁ + r₂ then the two circles touch each other externally. The point of contact divides C₁C₂ in the ratio r₁ : r₂ internally.
• If |r₁ – r₂| < C₁C₂ < r₁ + r₂ then the two circles intersect at two points P and Q. The chord \(\overline{\mathrm{PQ}}\) is called common chord of the circles.
• If C₁C₂ = |r₁ – r₂| then the two circles touch each other internally. The point of contact divides C₁C₂ in the ratio r₁: r₂ externally.
• If C₁C₁ < |r₁ – r₂] then one circle lies completely inside the other circle.

Common Tangents: A line L = 0 is said to be a common tangent to the circle S = 0, S’ = 0 if L = 0 touches both the circles.

Definition: A common tangent L = 0 of the circles S = 0, S’= 0 is said to be a direct common tangent of the circles if the two circles S = 0, S’ = 0 lie on the same side of L = 0.

Centres of Similitude:
Let S = 0, S’ = 0 be two circles.

• The point of intersection of direct common tangents of S = 0, S’ = 0 is called external centre of similitude.
• The point of intersection of transverse common tangents of S = 0, S’ = 0 is called internal centre of similitude.

Theorem:
Let S = 0, S’ = 0 be two circles with centres C₁, C₂ and radii r₁, r₂ respectively. If A₁ and A₂ are respectively the internal and external centres of similitude circles s = 0, S’ = 0 then

• A₁ divides C₁C₂ in the ratio r₁ : r₂ internally.
• A₂ divides C₁C₂ in the ratio r₁ : r₂ internally.

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Question 1.
A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar.14; May ’13,’06).
Solution:

Question 2.
The probability that a person chosen at random is left handed (in hand writing) is 0.1. What is the probability that in a group of ten people there is one, who is left handed? (TS Mar.’16; AP Mar. ’17 ’15)
Solution:
Here n = 10
p = 0.1
q = 1 – p = 1 – 0.1 = 0.9
P(X = 1) = ¹⁰C₁ (0.1)¹ (0.9)^{10 – 1}
= 10 × 0.1 × (0.9)
= 1 × (0.9)
= (0.9)

Question 3.
A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar.14; May ’13, ’06)
Solution:
Given P(X = 1) = P(X = 2)

Question 4.

is the probability distribution of a random variable X. Find the value of K and the variance of X. ( March 2006) (TS Mar. 17)
Solution:
Sum of the probabilities = 1
0.1 + k + 0.2 + 2k + 0.3 + k = 1
4k + 0.6 = 1
4k = 1 – 0.6 = 0.4
k = \(\frac{0.4}{4}\) = 0.1
Mean = (-2) (0.1) + (-1) k + 0(0.2) + 1(2k) + 2(0.3) + 3k
= -0.2 – k + 0 + 2k + 0.6 + 3k
= 4k + 0.4
=4(0.1) + 0.4
= 0.4 + 0.4
= 0.8
μ = 0.8

∴ Variance = 4(0.1) + 1(k) + 0(0.2) + 1(2k) + 4(0.3) + 9k – μ²
= 0.4 + k + 0 + 2k + 4(0.3) + 9k – μ²
= 12k + 0.4 + 1.2 – (0.8)²
= 12(0.1) + 1.6 – 0.64
= 1.2 + 1.6 – 0.64 .
∴ σ² = 2.8 – 0.64 = 2.16

Question 5.
A random variable X has the following probability distribution. (TS & AP Mar. ‘16)

Find
i) k
ii) the mean and
iii) P(0 < X < 5).
Solution:
Sum of the probabilities = 1
⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
⇒ 10k² + 9k = 1
⇒ 10k² + 9k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (10k – 1) (k + 1) = 0

iii) P(0 < x < 5)
P(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 2k + 3k
= 8k
= 8\(\frac{1}{10}\)
= \(\frac{8}{10}\) = \(\frac{4}{5}\)

Question 6.
The range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c³, P(X = 1) = 4c – 10c², P(X = 2) = 5c – 1
i) Find the value of c
ii) P(X < 1),P(1 < X ≤ 2) and P (0 < X ≤ 3) (AP & TS Mar. 15, 13, ‘11, 07, 05; May ‘11’)
Solution:
P(X = 0) + P(X = 1) + P(X = 2) = 1
3c³ + 4c – 10c² + 5c – 1 = 1
3c³ – 10c² + 9c – 2 = 0
c = 1 satisfy this equation
c = 1 ⇒ P(X = 0) = 3 which is not possible
Dividing with c – 1, we get
3c² – 7c + 2 = 0
(c – 2) (3c – 1) = 0

ii) P(1 < X ≤ 2) = P(X = 2) = 5c – 1
= \(\frac{5}{3}\) – 1 = \(\frac{2}{3}\)

iii)P(0 < X ≤ 3) = P(X = 1) + P(X = 2)
= 4c – 10c² + 5c – 1
= 9c – 10c² – 1
= 9.\(\frac{1}{3}\) – 10.\(\frac{1}{9}\) – 1
= 3 – \(\frac{10}{9}\) – 1 = 2 – \(\frac{10}{9}\) = \(\frac{8}{9}\)

Question 7.
One in 9 ships is likely to be wrecked, when they are set on sail, when 6 ships are on sail, find the probability for
i) Atleast one will arrive safely
ii) Exactly, 3 will arrive safely. (Mar. 2008)
Solution:
p = probability of ship to be wrecked = \(\frac{1}{9}\)

Question 8.
If the mean and variance of a binomial variable X are 2.4 and 1.44 respectively, find P(1 < X ≤ 4). (May ’06)
Solution:
Mean = np = 2.4 …… (1)
Variance = npq = 1.44 …… (2)
Dividing (2) by (1),
\(\frac{\mathrm{npq}}{\mathrm{np}}\) = \(\frac{1.44}{2.4}\)
q = 0.6 = \(\frac{3}{5}\)
2
p = 1 – q = 1 – 0.6 = 0.4 = \(\frac{2}{5}\)
Substituting in (1)

Question 9.
The probability distribution of a random variable X is given below. (AP Mar. ‘17’) (Mar. ‘14; May ‘13)

Find the value of K, and the mean and variance of X.
Solution:

Question 10.
The mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X ≥ 1)
(AP Mar. ‘16 TS Mar. 17 ‘15, ’08)
Solution:
Given distribution ¡s Binomial distribution with mean = np = 4
variance = npq = 3
∴ \(\frac{n p q}{n p}\) = \(\frac{3}{4}\)
⇒ q = \(\frac{3}{4}\)
so that p = 1 – q
= 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
∴ np = 4
n\(\frac{1}{4}\) = 4
⇒ n = 16

Question 11.
A cubical die is thrown. Find the mean and variance of X, giving the number on the face that shows up.
Solution:
Let S be the sample space and X be the random variable associated with S, where P(X) is given by the following table

Question 12.
The probability distribution of a random variable X is given below.

Find the value of k, and the mean and variance of X
Solution:

Question 13.
If x is a random variable with probability distribution. P(X = k) = \(\frac{(k+1) c}{2^{k}}\), k = 0, 1, 2 then find c.
Solution:

Question 14.
Let X be a random variable such that P(X = -2) =P(X = -1) = P(X = 2) = P(X = 1) = \(\frac{1}{6}\) and P(X = 0) = \(\frac{1}{3}\). Find the mean and variance of X.
Solution:

Question 15.
Two dice are rolled at random. Find the probability distribution of the sum of the numbers on them. Find the mean of the random variable.
Solution:
When two dice are rolled, the sample space
S contains 6 × 6 = 36 sample points.
S = {(1, 1), (1, 2) (1, 6), (2, 1), (2, 2) (6, 6)}
Let X denote the sum of the numbers on the tw0 dice.
Then the range of X = {2, 3, 4, ……… 12}
The Prob. distribution of X is given by the following table.

Question 16.
8 coins are tossed simultaneously. Find the probability of getting atleast 6 heads.
Solution:
p = Probability of getting head = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) ; n = 8
P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8)

Question 17.
The mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X ≥ 1)
(A.P. Mar. ’16, T.S. Mar. ’15, ’08)
Solution:
Given distribution is Binomial distribution with mean = np = 4
variance = npq = 3
∴ \(\frac{n p q}{n p}\) = \(\frac{3}{4}\)
⇒ q = \(\frac{3}{4}\)
so that p = 1 – q
= 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
∴ np = 4
n\(\frac{1}{4}\) = 4
⇒ n = 16

Question 18.
The probability that a person chosen at random is left handed (in hand writing) is 0.1. What is the probability that in a group of ten people there is one, who is left handed ? (Mar. 16, AP. Mar. ’15)
Solution:
Here n = 10
p = 0.1
q = 1 – p = 1 – 0.1 = 0.9
P(X = 1) = ¹⁰C₁ (0.1)¹ (0.9)^{10 – 1}
= 10 × 0.1 × (0.9)
= 1 × (0.9)
= (0.9)

Question 19.
In a bõok of 450 pages, there are 400 typographical errors. Assumiñg that the number of errörs per page follow the
poisson law, find the probability that a random sample of 5 pages will contain no typographical error.
Solution:
The average number of errors per page in the book is

The required probability that a random sample of 5 pages will contain no error is
[P(X = 0)]⁵ = \(\left(e^{-8 / 9}\right)^{5}\)

Question 20.
Deficiency of red cells in the blood cells is determined by examining a specimen of blood under a microscope. Suppose a small fixed volume contains on an average 20 red cells for normal persons. Using the poisson distribution, find the probability that a specimen of blood taken from a normal person will contain less than 15 red cells.
Solution:

Question 21.
A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar. ‘14; May ‘06, ‘13)
Solution:
Given P(X = 1) = P(X = 2)

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Probability Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Probability Important Questions

Question 1.
If 4 fair coins are tossed simultaneously, then find the probability that 2 heads and 2 tails appear. (Mar. ’08)
Solution:
4 coins are tossed simultaneously.
Total number of ways = 2⁴ = 16
n(S) = 16
From 4 heads we must get 2 heads.
A number of ways of getting 2 heads.
= ⁴C₂ = \(\frac{4.3}{1.2}\) = 6
∴ n(E) = 6
P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{16}\) = \(\frac{3}{8}\)
∴ Probability of getting 2 heads and 2 tails = \(\frac{3}{8}\)

Question 2.
Find the probability that a non – leap year contains
i) 53 Sundays
ii) 52 Sundays only. (Mar. ’07, May ’06)
Solution:
A non – leap year contains 35 days 52 weeks and 1 day more.
i) We get 53 sundays when the remaining day is Sunday.
Number of days in the week = 7
∴ n(S) = 7
Number of ways getting 53 Sundays.
n(E) = 1
∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{7}\)
∴ Probability of getting 53 Sundays = \(\frac{1}{7}\)

ii) Probability of getting 52 Sundays
P(E) = 1 – P(E)
= 1 – \(\frac{1}{7}\) = \(\frac{6}{7}\)

Question 3.
If one ticket is randomly selected from tickets numbered 1 to 30. Then find the probability that the number on the ticket is a multiple of 3 or 5 (Mar. ‘08)
Solution:
Suppose A is the event of getting a multiple of 3 and B is the event of getting a multiple of 5.
A = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
B = {5, 10, 15, 20, 25, 30}
A ∩ B = {15, 30}

Question 4.
If two numbers are selected randomly from 20 consecutive natural numbers, find the probability that the sum of the two numbers is
(i) an even number
(ii) an odd number. (Mar. ‘08)
Solution:
i) Let A be the event that the sum of the numbers is even when two numbers are selected out of 20 consecutive natural numbers.
In 20 consecutive natural numbers, we have 10 odd and 10 even natural numbers.
The sum of two odd natural numbers is an even number and the sum of two even natural numbers is also an even number

ii) Probability that the sum of two numbers is an odd number
P(\(\bar{A}\)) = 1 – P(A) = 1 – \(\frac{9}{19}\) = \(\frac{10}{19}\)

Question 5.
The probability for a contractor to get a road contract is \(\frac{2}{3}\) and to get a building contract is \(\frac{5}{9}\). The probability to get atleast on contract is \(\frac{4}{5}\). Find the probability to get both the contracts. (AP Mar. ‘16)
Solution:
Suppose A is the event of getting a road contract.
B is the event of getting a building contract

Question 6.
If one card is drawn at random from a pack of cards then show that event of getting an ace and getting heart are independent events. (Mar. 13)
Solution:
Suppose A is the event of getting an ace and B is the event of getting a heart.
∴ P(A) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
P(B) = \(\frac{13}{52}\) = \(\frac{1}{14}\)
A ∩ B is the event of getting a Hearts ace
P(A ∩ B) = \(\frac{1}{52}\) = \(\frac{1}{13}\).\(\frac{1}{4}\) = P(A).P(B)
∴ A and B are independent events.

13 If A. B, are two events with P(A ∪ B) = 0.65 and P(A ∩ B) = 0.15, then find the value of p(A^C) + P (B^C). (TS Mar. ’15, ’13, ’05; May ’11)
Solution:
By addition theorem on probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A) + P(B) = P(A ∪ B) + P(A ∩ B)
= 0.65 + 0.15
= 0.8 —— (1)
P(A^C) + P(B^C) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)]
= 2 – [P(A) + P(B)]
= 2 – 0.8 by (1)
= 1.

Question 7.
Suppose A and B are independent events with P(A) = 0.6, P(B) = 0.7 then compute
(i) P(A ∩ B)
(ii) P(A ∪ B)
(iii) \(\mathbf{P}\left(\frac{\mathbf{B}}{\mathrm{A}}\right)\)
(iv) P(A^C ∩ B^C). (AP Mar. ’17; Mar. ‘14)
Solution:
Giver A, B are independent events and
P(A) 0.6, P(B) = 0.7

i) P(A ∩ B) = P(A) P(B) = 0.6 × 0.7 = 0.42

ii) P(A ∪ B) = 1(A) + P(B) – P(A ∩ B)
= 0.6 + 0.7 – 0.42
= 1.3 – 0.42 = 0.88

iii) \(\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\) = P(B) = 0.7

iv) P(A^C ∩ B^C) = P(A^C). P(B^C)
(A^C & B^C are also independent events)
= [1 – P(A) [1- P(B)]
= (1 – 0.6)(1 – 0.7)
= 0.4 × 0.3 = 0.12

Question 8.
Find the probability of drawing an ace or a spade from a well shuffled pack of 52 cards? (TS Mar. ’17, ’15)
Solution:
Hint : A pack of cards means of pack containing 52 cards, 26 of them are red and 26 of them are black coloured. These 52 cards are divided into 4 sets namely hearts, Spades, Diamonds and Clubs. Each set contains of 13 cards names, A, 2, 3, 4, 5, 6, 7, 8, 9, 10, K, Q, J.
Let E₁ be the event of drawing a spade and E₂ be the event of drawing an ace. E₁, E₂ are not mutually exclusive.
n(A) = 13, n(B) = 4, n(A ∩ B) = 1
= \(\frac{13}{52}\) + \(\frac{4}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

Question 9.
If A, B, C are three events. Show that P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) +
P(A ∩ B ∩ C). (AP Mar. ’15)
Solution:
Write B ∪ C = D then P(A ∪ B ∪ C) = P(A ∪ D)
∴ P(A ∪ D) = P(A) + P(D) – P(A ∩ D)
= [P(A) + P(B ∪ C) – P(A ∩ (B ∪ C)]
= P(A) + P(B) + P(C) – P(B ∩ C) – [P(A ∩ B) ∪ (A ∩ C)
= P(A) + P(B) + P(C) – P(B ∪ C) – [P(A ∩ B) + P(A ∩ C) – P(A ∩ B ∩ D ∩ C]
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P (A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C).

Question 10.
A speaks the truth in 75% of the cases; B in 80% çases. What is the probability that their statements about an incident do not match? (TS & AP Mar.’16)
Solution:
Let E₁, E₂ be the events that A and B respectively speak truth about an incident.

Let E be the event that their statements do not match about the incident. Then this happens in two mutually exclusive ways.
i) A speaks truth, B tells lie
ii) A tells lie, B speaks truth. These two events are represented by E₁ ∩ \(E_{2}^{c}\), \(E_{1}^{c}\), ∩ E₂

Question 11.
A problem in Calculus is given to two students A and B whose chances of solving it are 1/3 and 1/4. What is the probability that the problem will be solved if both of them try independently ? (Mar.’15, ’05)
Solution:
Let E₁ and E₂ denote the events that the problem is solved by A and B respectively.
Given that
P(E₁) = \(\frac{1}{3}\) and P(E₂) = \(\frac{1}{4}\)
Note that these two are independent events. Therefore the required probability
P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)
= P(E₁) + P(E₂) – P(E₁) P(E₂)
(∵ E₁, E₂ are independent)
= \(\frac{1}{3}\) + \(\frac{1}{4}\) – \(\frac{1}{12}\) = \(\frac{1}{2}\)

Question 12.
Suppose A and B are events with P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.3. Find the probability that
i) A does not occur.
ii) neither A nor B occurs. (TS Mar.’17)
Solution:
We have A^C = the event “A does not occur”
(A ∪ B)^C = neither A nor B occurs.
∴ P(A^C) = 1 – P(A) = 1 – 0.5 = 0.5
Since P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.5 + 0.4 – 0.3
= 0.6
P[(A ν B^C)] = 1 – P (A ∪ B)
= 1 – 0.6 = 0.4

Question 13.
In a committee of 25 members, each member is proficient either in Mathematics or in Statistics or in both. If 19 of these are proficient in Mathematics, 16 in Statistics, find the probability that a person selected from the committee is proficient in both. (TS Mar. 16)
Solution:
When a person is chosen at random from the academy consistin9 of 25 members, let A be the event that the person is proficient in Mathematics, B be the event that the person is proficient in Statistics and S be the sample space. Since 19 members are proficient in Mathematics and 16 members aré proficient in Statistics.
P(A) = \(\frac{19}{25}\), P(B) = \(\frac{16}{25}\)
Since every one is either proficient in Mathematics or Statistics or in both

Question 14.
A, B, C are three horses in a race. The probability of A to win the race is twice that of B and probability of B is twice that of C. What are the probabilities of A, B and C to win the race? (Mar. ‘14, ’03)
Solution:
Let A, B, C be the events that the horses A, B, C win the race respectively,
Given P(A) = 2P(B), P(B) = 2P(C)
∴ P(A) = 2P(B) = 2[2P(C)] = 4P(C)
Since the horses A, B and C run the race,
A ∪ B ∪ C = S and A, B, C are mutually disjoint.

Question 15.
If A, B, C are three independent events of an experiment such that P(A ∩ B^C ∩ C^C) = \(\frac{1}{4}\), P(A^C ∩ B^C ∩ C^C) = \(\frac{1}{4}\) then find P(A), P(B) and P(C). (TS Mar. ’15)
Solution:
Since A, B, C are independent events

Question 16.
Addition theorem on probability: Statement: If E₁, E₂ an any two events of a random experiment and P is a probability function, then P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)
Solution:
Proof: Case (i) E₁ ∩ E₂ = φ
Then P(E₁ ∩ E₂) = 0
∴ P(E₁ ∪ E₂) = P(E₁) + P(E₂)
= P(E₁) + P(E₂) – 0
= P(E₁) + P(E₂) – P(E₁ ∩ E₂)

Case (ii) : Suppose E₁ ∩ E₂ ≠ φ
Then E₁ ∪ E₂ = E₁ ∪ (E₂ – E₁) and E₁ ∩ (E₂ – E₁) = 4
∴ P(E₁ ∪ E₂) = P[E₁ ∪ (E₂ – E₁)]
= P(E₁) + P(E₂ – E₁)
= P(E₁) + P[E₂ – (E₁ ∩ E₂)
Since E₂ ∩ (E₁ ∩ E₂) = φ.
= P(E₁) + P(E₂) – P(E₁ ∩ E₂)
Hence P(E₁ ∪ E₂) = P(E₁) + (E₂) – P(E₁ ∩ E₂).

Question 17.
Three Urns have the following composition of balls.
Urn I : 1 White, 2 black
Urn II : 2 White, 1 black
Urn III : 2 White, 2 balck
One of the Urn is selected at random and a ball is drawn. Pt turns out to be white. Find the probability that it come from Urn III. (AP. Mar. ’17)
Solution:
Let E_i be the event of Choosing the Urn i = 1, 2, 3 and P(E_i) be the probability of
choosing the Urn i = 1, 2, 3. Then P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\).
Having choose b the Urn i, the probability of drawing a white ball, P(W/E_i), is given by
P(W/E₁) = \(\frac{1}{3}\),
P(W/E₂) = \(\frac{2}{3}\)
P(W/E₃) = \(\frac{2}{4}\)
We have to find the probability P(E₃/W) by Baye’s theorem.
P(E₃/W) =

Question 18.
Find the Probability, of getting the same number on both the dice when two dice are thrown.
Solution:
Let E be the event of getting the same number on both the dice when two dice are thrown and S be the sample space.
∴ n(S) = 6² = 36
E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
n(E) = 6
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{6}{36}\)
= \(\frac{1}{6}\)

Question 19.
An integer in picked from 1 to 20, both inclusive. Find the probability that it is a prime.
Solution:
Let E be the event that the number picked from 1 to 20 is a prime and S be the sample space.
∴ n(S) = ²⁰C₁ = 20
E = {2, 3, 5, 7, 11, 13, 17, 19}
n(E) = 8
P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{20}\) = \(\frac{2}{5}\)

Question 20.
A bag contains 4 red, 5 black and 6 blue balls. Find the probability that two balls drawn at random simultaneously from the bag are a red and a black ball.
Solution:
Let E be the event that getting a red and black ball when two balls are drawn at random from a bag containing 4 red, 5 black, 6 blue balls and S be the sample space. Total no.of Balls = 4 + 5 + 6 = 15
n(S) = ¹⁵C₂
n(E) = ⁴C₁. ⁵C₁
∴ P(E) = \(\frac{{ }^{4} C_{1} \cdot{ }^{5} C_{1}}{{ }^{15} C_{2}}\)
= \(\frac{4.5}{105}\)
= \(\frac{4}{21}\)

Question 21.
Ten dice are thrown. Find the probability that none of the dice shows the number 1.
Solution:
Let A be the event that none of the dice shows the numbers 1 when the dice thrown.
n(S) = 6¹⁰
n(A) = 5¹⁰
P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac{5^{10}}{6^{10}}\)
= \(\left(\frac{5}{6}\right)^{10}\)

Question 22.
A number x is drawn arbitrarily from the set {1, 2, 3, ……. 100}. What is the probability that (x + \(\frac{100}{x}\)) > 29
Solution:
Here the total number of cases is 100.
Let A be the event that x selected from the set {1, 2, 3, ……. 100} has the property
x + \(\frac{100}{x}\) > 29
Now x + \(\frac{100}{x}\) > 29
⇔ x² – 29x + 100 > 0
⇔ (x – 4) (x – 25) > 0
⇔ x > 25 or x < 4
⇔ x ∈ {1, 2, 3, 26. 27, … 100) = A (say),
so that the number of cases favourable to A is 78
∴ The required probability P(A) = \(\frac{78}{100}\)

Question 23.
Two squares are choosen at random on a chess board. Show that the probability that they have a side in common is \(\frac{1}{18}\).
Solution:
The number of ways of choosing the first square is 64 and that of the second is 63
∴ The number of ways of choosing the first and second squares = 64 × 63
∴ n(S) = 64 × 63
Let E be the event that these squares have a side in common.

If the first square happens to be one of the squares in the four corners of the chess board, the second square (with common side) can be choosen in 2 ways.

If the first square happens to be any one of the remaining 24 squares along the four sides of the chess board other than the corner, the second squäre can be choosen, in 3 ways.

If the first square happens to be any one of the remaining 36 inner squares, then the second square can be choosen in 4 ways.

Hence the number of cases favourable to E is (4 × 2) + (24 × 3) + (36 × 4) = 224
∴ The required probability =
\(\frac{n(E)}{n(S)}\) = \(\frac{224}{64 \times 63}\) = \(\frac{1}{18}\)

Question 24.
A fair coin is tossed 200 times. Find the probability of getting a head an odd number of times.
Solution:
The total number of cases is 2²⁰⁰
The number of favourable cases is
= ²⁰⁰C₁ + ²⁰⁰C₃ + ²⁰⁰C₅ + ….. + ²⁰⁰C₁₉₉
= \(\frac{2^{200}}{2}\)
= 2¹⁹⁹
∴ Probability = \(\frac{2^{191}}{2^{200}}\) = \(\frac{1}{2} .\)

Question 25.
A and B are among 20 persons sit at random along a round table. Find the probability that there are any 6 persons
between A and B.
Solution:
Let ‘A’ occupy any seat around the table. Then theré are 19 seats available for B. But if there are to be six persons between A and B, then B has only two ways to sit.
∴ Probability = [/latex] = \(\frac{2}{19}\)

Question 26.
Out of 30 consecutive integers two are drawn at random. Then what is the probability that their sum is odd.
Solution:
The total number of ways of choosing 2 out of 30 numbers = ³⁰C₂
Out of these 30 numbers, 15 are even and 15 are odd.
For the sum of the choosen two numbers to be odd, one should be odd and the other even.
∴ The number of cases favourable

Question 27.
Out of 1,00,000 new born babies 77,181 survived till the age of 20. Find the probability that a new baby survives till 20 years of age.
Solution:
Here m = 77,181
n = 1,00,000
Required probability = \(\frac{\mathrm{m}}{\mathrm{n}}\)
= \(\frac{77,181}{1,00,000}\)
= 0.77181.

Question 28.
Addition theorem on probability: Statement: If E₁, E₂ an any two events of a random experiment and P is a probability function, then P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂) (Mar. ‘14, ‘13)
Solution:

Question 29.
Find is the probability of throwing a total score of 7 with two dice.
Solution:
Let S be the sample space and A be the event of getting a total score of 7 when two dice are thrown
S = {(1, 1) (1, 2)…. (1, 6), (2, 1) …. (2, 6) … (6, 1), (6, 2) …. (6, 6)}
n(S) = 36
Hint: 6² = 36
A = {(1, 6) (6, 1), (2, 5) (5,2) (3,4) (4, 3)}
n(A) = 6
∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Question 30.
Find the probability of obtaining two tails and one head when three coins are tossed.
Solution:
Lèt S be the sample space and A be the event of getting two tails and one head when three coins are tossed.
n(S) = 2³ = 8
A = [H T T, T H T, T T H]
n(A) = 3
P(A) = \(\frac{3}{8}\)

Question 31.
A page is opened at random from a book containing 200 pages. What is the probability that the number of the page is a perfect square?
Solution:
Let S be the sample space. Let A be the event of getting on the page is perfect square.
n(S) = 200
Let A be the event of drawing a page whose number is perfect square.
A = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196}
n(A) = 14
P(A) = \(\frac{14}{200}\) = \(\frac{7}{100}\) = 0.07

Question 32.
Find the probability of drawing an ace or a spade from a well shuffled pack of 52 cards? (T.S. Mar. ‘15)
Solution:
Hint : A pack of cards means of pack containing 52 cards, 26 of them are red and 26 of them are black coloured. These 52 cards are divided into 4 sets namely Hearts, Spades, Diamonds and Clubs. Each set contains of 13 cards names, A, 2, 3, 4, 5, 6, 7, 8, 9, 10, K, Q, J.
Let E₁ be the event of drawing a spade and E₂ be the event of drawing an ace. E₁, E₂ are not mutually exclusive.
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{13}{52}\) + \(\frac{4}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

If A and B are two events, show that
i) P(A ∩ B^C) = P(A) – P(A ∩ B) and
ii) the probability that one of them occurs is given by
P(A) + P(B) – 2P(A ∩ B)
Solution:
i) We have A = (A ∩ B) ∪ (A ∩ B^C)
and (A ∩ B) ∩ (A ∩ B^C) = φ
∴ P(A) = (A ∩ B) + P (A ∩ B^C)
∴ P(A ∩ B^C) = P(A) – P(A ∩ B) ———(1)

ii) Let E be the event that exactly one of them, (i.e.,) either A or B occws. Given
E = (A – B) ∪ (B – A)
= (A ∩ B^C) ∪ (B ∩ A^C)
∵ So P(E) = P(A ∩ B^C) + P(B ∩ A^C)
(A n BC) n (B n AC)
∵ P(E) = P(A) – P(A ∩ B) + P(B) – P(A ∩ B)
by (1)
∴ P(E) = P(A) + P(B) – 2 P(A ∩ B)

Question 33.
Suppose A and B are events with P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.3. Find the probability that
i) A does not occur,
ii) neither A nor B occurs.
Solution:
We have A^C = the event A does not occurs.
(A ∪ B)^C = neither A nor B occurs.
∴ p(A_C) = 1 – P(A) = 1 – 0.5 = 0.5
Since P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.5 + 0.4 – 0.3
= 0.6
P[(A ν B^C)] = 1 – P (A ∪ B)
= 1 – 0.6 = 0.4

Question 34.
If A, B, C are three events. Show that P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C). (AP. Mar. ’15)
Solution:
Write B ∪ C = D then P(A ∪ B ∪ C) = P(A ∪ D)
∴ P(A ∪ D) = P(A) + P(D) – P(A ∩ D)
= [P(A) + P(B ∪ C) – P(A ∩ (B ∪ C)]
= P(A) + P(B) + P(C) – P(B ∩ C) – [P(A ∩ B) ∪ (A ∩ C)]
= P(A) + P(B) + P(C) – P(B ∩ C) – [P(A ∩ B) + P(A ∩ C) – P(A ∩ B ∩ D ∩ C]
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C).

Question 35.
MULTIPLICATION THEOREM OF PROBABILITY Statement: If A and B are two events of a random experiment and P(A) > 0 and P(B) > 0 then
P(A ∩ B) = P(A)P\(\left(\frac{\mathbf{B}}{\mathbf{A}}\right)\) = P(B)\(\left(\frac{\mathbf{A}}{\mathbf{B}}\right)\).
Solution:
Proof: Let S be the sample space associated with the random. Experiment and A, B be two events of S such that P(A) > 0 and P(B) > 0.
By the definition of conditional probability, we have

Interchanging A, B we have

Question 36.
A Pair of dice of thrown. Find probability that either of the dice shows 2 when their sum is 6.
Solution:
Let A be the event that 2 appears on either of dice and B be the event that the sum of the two number on the event that the sum of the two numbers on the dice is 6 when two dice is thrown.
A = {(2, 1), (2, 2), (2,3), (2, 4), (2,4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
n(A) = 11

Question 37.
A box contains 4 defective and 6 good bulbs. Two bulbs are drawn at random without replacement. Find the probability that both the bulbs drawn are good.
Solution:
Let a denote the event that drawing a good bulb is the first draw and B denote the event that the second draw is also good when two bulbs are drawn at random without replacement and S be the sample space.
∴ P(A) = \(\frac{6}{10}\) and P\(\left(\frac{B}{A}\right)\) = \(\frac{5}{9}\)

Question 38.
Suppose there are 12 boys and 4 girls in a class. If we choose three children one after another in succession, what is the probability that all the three are boys?
Solution:
Let E, be the event of choosing a boy child in i^th trial (i = 1, 2, 3). We have to find

Question 39.
A speaks the truth in 75% of the cases; B in 80% cases. What is the probability that their statements about an incident do not match? (T.S.& AP. Mar. ‘16)
Solution:
Let E₁, E₂ be the events that A and, B respectively speak truth about an incident.
Then P(E₁) = \(\frac{75}{100}\) = \(\frac{3}{4}\), P(E₂) = \(\frac{80}{100}\) = \(\frac{4}{5}\)
s0 that P\(\left(\mathrm{E}_{1}{ }^{c}\right)\) = \(\frac{1}{4}\), P\(\left(\mathrm{E}_{2}{ }^{c}\right)\) = \(\frac{1}{5}\)
Let E be the event that their statements do not match about the incident. Then this happens in two mutually exclusive ways.
i) A speaks truth, B tells lie
ii) A tells lie, B speaks truth. These two events are represented by E₁ ∩ \(\left(\mathrm{E}_{2}{ }^{c}\right)\), \(\left(\mathrm{E}_{1}{ }^{c}\right)\) ∩ E₂

Question 40.
A problem in Calculus is given to two students A and B whose chances of solving it are 1/3 and 1/4. What is the
probability that the problem will be solved if both of them try independently? (A.P. Mar. ‘15, ‘05).
Solution:
Let E₁ and E₂ denote the events that the problem is solved by A and B respectively.
Given that
P(E₁) = \(\frac{1}{3}\) and P(E₂) = \(\frac{1}{4}\)
Note that these two are independent events. Therefore the required probability
P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)
= P(E₁) + P(E₂) – P(E₁)P(E₂)
(∵ E₁, E₂ are independent)
= \(\frac{1}{3}\) + \(\frac{1}{4}\) – \(\frac{1}{12}\) = \(\frac{1}{2}\)

Question 41.
A and B toss a fair coin 50 times each simultaneously. Then find the probability that both of them will not get tails at the same toss
Solution:
In each toss there are four choices
i) A gets H, B gets H
ii) A gets T, B gets H
iii) A gets H, B gets T
iv) A gets T, B gets T .
Therefore the total number of choices 4⁵⁰ Out of the four cases listed above, (i), (ii) and (iii) are favourable.
(iv) is not favourable to the occurrence of the required event, say E.
∴ P(E) = \(\frac{3^{50}}{4^{50}}\) = \(\left(\frac{3}{4}\right)^{50}\)

Question 42.
If A and B are independent events of a random experiment shöw that A^C and B^C are also independent.
Solution:
If A and B are independent then
P(A ∩ B) = P(A) P(B)
Now P(A^C ∩ B^C) = P[(A ∪ B)^C]
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – [P(A) + P(B) – P(A) P(B)]
= [1 – P(A)] [1 – P(B)] = P(A^C)P(B^C).
∴ A^C, B^C are independent.

Question 43.
A bag contains 10 identical balls, of which 4 are blue and 6 are red. Three balls are taken out at random from the bag one after the other. Find the probability that all the three balls drawn are red without replacement.
Solution:
The probability that the first ball drawn to be red is \(\frac{6}{10}\)
The probability that the second ball drawn to be red is \(\frac{5}{9}\)
The probability that the third ball drawn to be red is \(\frac{4}{8}\)
By the multiplication theorem Required probability = \(\frac{6}{10}\).\(\frac{5}{9}\).\(\frac{4}{8}\)
= \(\frac{1}{6}\).

Question 44.
An urn contains 7 red and 3 black balls. Two balls are drawn without replacement. What is the probability that the second ball is red it is known that the first ball drawn is red.
Solution:
Let R₁ be the event of drawing the first ball is red and R₂ be the event of drawing the second ball also red.

Question 45.
Let A and B be independent events with P(A) = 0.2, P(B) = 0.5.
Find
(i) \(P\left(\frac{A}{B}\right)\)
(ii) \(\mathbf{P}\left(\frac{\mathbf{B}}{\mathbf{A}}\right)\)
(iii) P(A ∩ B)
(iv) P(A ∪ B)
Solution:
Given P(A) = 0.2, P(B) = 0.5 and A, B are independent events.

ii) \(P\left(\frac{B}{A}\right)\) = P(B) = 0.5

iii) P(A ∩ B) = P(A). P(B)
= (0.2) . (0.5)
= 0.1

iv) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.2 + 0.5 – 0.1
= 0.6

Question 46.
Bag B₁ contains 4 white and 2 black balls. Bag B₂ contains 3 white and 4 black balls. A bag is drawn at random and a ball is chosen at random from it. What is the probability that the ball drawn is white.
Solution:
Let E₁, E₂ denote the events or choosing bags B₁ and B₂
∴ P(E₁) = P(E₂) = \(\frac{1}{2}\)
Let w be the even that the ball chosen from the selected bag is white.

Question 47.
BAYES THEOREM : (TS, Mar. 16, A.P. Mar. 15)
Solution:
Statement : Let E₁, E₂, …… E_n are mutually exclusive and exhaustive events or a random experiment with P(E₁) ≠ 0 for
i = 1, 2, 3 , n. Then for any event A or the random experiment with P(A) ≠ 0

By multiplication theorem

Question 48.
Three boxes B₁, B₂ and B₃ contain balls detailed below.

A die is thrown. B₁, is chosen if either 1 or 2 turns up; B₂ is chosen if 3 or 4 turns up
and B₃ is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is chosen at random from this box. If the ball drawn is of red colour, what is the probability that it comes from box B₂?
Solution:
Let P(E_i) be the probability of choosing the box B_i (i = 1, 2, 3).
Then P(E_i) = \(\frac{2}{6}\) = \(\frac{1}{3}\) ; for i = 1, 2, 3
Having chosen the box B., the probability of drawing a red ball, say, P(R/E_i) is given by
P(R/E₁) = \(\frac{2}{5}\). P(R/E₂) = \(\frac{4}{9}\) and P(R/E₃) = \(\frac{2}{9}\)
We have to find the probability P(E₂/R). By Bayeras theorem, we get

Question 49.
An urn contain w white balls and b black balls. Two players Q and R alternately draw a with replacement from the urn. The player that draws a white ball first wins the game. If Q begins the game, find the probability that Q wins the game.
Solution:
Let W denote the event of drawing a white ball at any draw and B that of a black ball.
Then
P(W) = \(\frac{w}{w+b}\), P(B) = \(\frac{b}{w+b}\)
Let E be the event that Q wins the game.
= P(W P(E) BBW P(E) BBBBW P(E) …)
= P(W) + P(BBW) + P(BBBBW) + ….
= P(W) + P(B) P(B) P(W) + P(B) P(B) P(B) P(B) P(W) + ……
= P(W) (1 + P(B)² + P(B)⁴ + ….)

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Measures of Dispersion Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Measures of Dispersion Important Questions

Question 1.
Find the mean deviation about the median for the data 4, 6, 9, 3, 10, 13, 2 (AP Mar. ’17) (TS. Mar. ’15 )
Solution:
Given ungrouped data are 4, 6, 9, 3, 10, 13, 2
Expressing the data in the ascending order of magnitude, we have 2, 3, 4, 6, 9, 10, 13
∴ Median = 6 b (say)
The absolute values are
|6 – 2|, |6 – 3|, |6 – 4|, |6 – 6|, |6 – 9|, |6 – 10|, |6 – 13| = 4, 3, 2, 0, 3, 4, 7

Question 2.
Find the mean deviation from the mean of the following discrete data 6, 7, 10, 12, 13, 4, 12, 16. (Mar. ’14)
Solution:
The A.M. of the given data \(\bar{x}\) = \(\frac{6+7+10+12+13+4+12+16}{8}\) = \(\frac{80}{8}\) = 10
The absolute values of thé deviations are
|6 – 10|, |7 – 10|, |10 – 10|, |12 – 10|, |13 – 10|, |4 – 10|, |12 – 10|, |16 – 10|
= 4, 3, 0, 2, 3, 6, 2, 6
∴ The mean deviation from the mean = \(\frac{1}{n} \sum_{i=1}^{n}\left|x_{i}-\bar{x}\right|\)
= \(\frac{1}{8}\) (4 + 3 + 0 + 2 + 3 + 6 + 2 + 6) = \(\frac{26}{8}\) = 3.25

Question 3.
Find the variance and standard deviation of the following data 5, 12, 3, 18, 6, 8, 2, 10 (A.P. Mar. ’15)
Solution:
A. M. = \(\bar{x}\) = \(\frac{\Sigma x_{i}}{n}\) = \(\frac{5+12+3+18+6+8+2+10}{8}\) = \(\frac{64}{8}\) = 8
Construct the table

Question 4.
Find the mean deviation about the mean for the following data 3, 6, 10, 4, 9, 10 (TS Mar. ’17)
Solution:
The arithmetic mean of the given data

Question 5.
Find the mean deviation about the median for the data 13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17. (AP. Mar. ’16)
Solution:
Given ungrouped data are
13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17
Expressing the data in the ascending order of magnitude, we have
10, 11, 11, 12, 13, 13, 16, 16, 17, 17, 18
∴ Median = 13 = b(say)
The absolute values are
|13 – 10|, |13 – 11|, |13 – 11|, |13 – 12|, |13 – 13|, |13 – 13|, |13 – 16|, |13 – 16|, |13 – 17|, |13 – 17|, |13 – 18|
= 3, 2, 2, 1, 0, 0, 3, 3, 4, 4, 5
Mean deviation from the median

Question 6.
Find the mean deviation about the mean for the following data. (AP Mar. ‘15)

Solution:
Construct the table

Question 7.
Find the mean deviation about the mean of the following data (AP Mar. ’17, ’16)

Solution:
Taking the assumed mean a = 25 and h = 10
Construct the table

Question 8.
Find the variance and standard deviation of the following data

Solution:
Construct the table

Question 9.
Calculate the variance and standard deviation of the following continuous frequency distribution.

Solution:
Let assumed mean A = 65
Then y_i = \(\frac{x_{i}-65}{10}\)
Construct the table

Question 10.
Find the mean deviation from the mean of the following data using step deviation method. (AP Mar. ’16)

Solution:
Let assumed mean A = 35
Then d_i = \(\frac{x_{i}-35}{10}\)
Construct the table

Question 11.
Find the mean deviation from the mean of the following discrete data 6, 7, 10, 12, 13, 4, 12, 16. (Mar – 2014)
Solution:
The A.M. of the given data \(\bar{x}\) = \(\frac{6+7+10+12+13+4+12+16}{8}\)
= \(\frac{80}{8}\) = 10
The absolute values of the deviations are
|6 – 10|, |7 – 10|, |10 – 10|, |12 – 10|, |13 – 10|, |4 – 10|, |12 – 10|, |16 – 10| = 4, 3, 0, 2, 3, 6, 2, 6

Question 12.
Find the mean deviation from the median of discrete data 6, 7, 10, 12, 13, 4, 12, 16
Solution:
Given data points are 6, 7, 10, 12, 13, 4, 12, 16
Expressing the data points in the ascending order of magnitude 4, 6, 7, 10, 12, 12, 13, 16
∴ Median = \(\frac{10+12}{2}\) =11
The absolute values of the deviations are
|11 – 4|, |11 – 6|, |11 – 7|, |11 – 10|, |11 – 12|, |11 – 12|, |11 – 13|, |11 – 16|
= 7, 5, 4, 1, 1, 1, 2, 5
∴ Mean deviation from the median = \(\frac{1}{8}\)(7 + 5 + 4 + 1 + 1 + 1 + 2 + 5)
= \(\frac{26}{8}\) = 3.25

Question 13.
Find the mean about the mean for the following data.

Solution:
Construct the table

Question 14.
Find the mean deviation from the median for the following data

Solution:
Given observations in the ascending order, we get the following distribution.

Median of these observations = 13

Question 15.
Find the mean deviation about the mean for the following data (A.P. Mar. ’15)

Solution:
Construct the table

Question 16.
Find the mean deviation about the median for the following data (A.P. Mar. ’16)

Solution:
Taking the assumed mean a = 25 and h = 10
Construct the table

Question 17.
Find the mean deviation about the median for the following data

Solution:
Construct the table

Question 18.
Find the variance and standard deviation of the following data 5, 12, 3, 18, 6, 8, 2, 10 (A.P. Mar. ’15)
Solution:

Question 19.
Find the variance and standard deviation of the following data

Solution:
Construct the table

Question 20.
Calculate the variance and standard deviation of the following continuous frequency distribution. (Mar. 2014)

Solution:
Let assumed mean A = 65
Then y_i = \(\frac{x_{i}-65}{10}\)
Construct the table

Question 21.
Students of two sections A and B of a class show the following performance in a test (conducted for 100 marks)

Which section of students has greater variability in performance?
Solution:
In section — A variance or the distribution \(\sigma_{1}^{2}\) = 64
⇒ S.D. σ₁ = 8
In section — 8 variance of the distribùtion \(\sigma_{2}^{2}\) = 81
⇒ S.D. σ₂ = 9
Since the average marks of both sections of students is the same i.e., 45
∴ The section with greater S.D. will have more variability. Hence section B has greater variability in the performance.

Question 22.
Lives of two models or refrigerators A and B obtained in a survey are given below:

Which refrigerator model would you suggest to purchase?
Solution:
First to find the mean and variance of the lives of model A and Model B of refrigerators.
Construct the table.



Since C.V. of model B < C.V. of model A
∴ Model B is more consistent than the model A with regard to life in years.
Hence we suggest Model B for purch.

Question 23.
Find the mean deviation from the mean of the following data using step deviation method.

(T.S. Mar. ‘16)
Solution:
Let assumed mean A = 35
Then d_i = \(\frac{x_{i}-35}{10}\)
Construct the table

Question 24.
The following table gives the daily wages of workers in a factory. Compute the standard deviation and the co-efficient of variation of the wages of the workers.

Solution:
To solute this problem using the step deviation method.
Here h = 50
Let assumed mean A = 300
Then y_i = \(\frac{x_{i}-300}{50}\)
Construct the table

Question 25.
An analysis or monthly wages paid to the workers of two firms A and B belonging to the same industry gives the following data.

i) Which firm A or B. has greater variability in industrial wages?
ii) Which firm has large wage bill?
Solution:

∴ firm B has greater variability in industrial wages.

ii) Total wages paid to the workers in firm
A = 500 × 186
= 93,000
Total wages paid to the workers in form
B = 600 × 175
= 1,05,000
Hence firm B has larges wage bill.

Question 26.
The variance of 20 observations in 5. If each of the observation is multiplied by 2, find the variance of the resulting observations.
Solution:
Let x₁, x₂ …, x₂₀ be the given observations and \(\bar{x}\) be their mean.
Given n = 20 and variance = 5

Substituting the values of x_i and in (1), we get

∴ The variance of the resulting observations = \(\frac{1}{20}\) × 400 = 20
= 2² × 5

Question 27.
If each of the observations x₁, x₂ ….. x_n is increased by k, where k is a positive or negative number, then show that the variance remains unchanged.
Solution:
Let \(\overline{\bar{x}}\) be the mean of x₁, x₂ …. x_n. Then their variance is given by \(\sigma_{1}^{2}\) = \(\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\)
If to each observation we add a constant k, then the new (changed) observations will be y_i = x_i + k

Thus the variance of the new observations is the same as that of the original observations.

Question 28.
The scores of two cricketers A and B in 10 innings are given below. Find who is a better run getter and who ¡s a more consistent player.




Cricketer A is a better run getter
Since C.V. of A < C.V. of B
Hence cricketer A is a more consistent players.

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Partial Fractions Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Partial Fractions Important Questions

Question 1.
\(\frac{x+4}{\left(x^{2}-4\right)(x+1)}\) (Mar. 14)
Solution:
\(\frac{x+4}{\left(x^{2}-4\right)(x+1)}\) = \(\frac{A}{x+1}\) + \(\frac{B}{x+2}\) + \(\frac{c}{x-2}\)
Multiplying with (x² – 4) (x + 1)
x + 4 = A(x² – 4) + B(x + 1) (x – 2) + C (x + 1) (x + 2)
x = -1 ⇒ 3 = A(1 – 4) = -3A ⇒ A = -1
x = -2 ⇒ 2 = B(-2 + 1) (-2 – 2)
= 4B ⇒ B = + \(\frac{2}{4}\) = \(\frac{1}{2}\)
x = 2 ⇒ 6 = C(2 + 1)(2 + 2)
= 12C ⇒ C = \(\frac{1}{2}\)

Question 2.
\(\frac{x^{2}-x+1}{(x+1)(x-1)^{2}}\) (TS Mar. 15)
Solution:
Let \(\frac{x^{2}-x+1}{(x+1)(x-1)^{2}}\) = \(\frac{A}{x+1}\) + \(\frac{B}{x-1}\) + \(\frac{C}{(x-1)^{2}}\)
Multiplying with (x + 1) (x – 1)²
x² – x + 1 = A(x – 1)² + B(x + 1) (x – 1) + C(x + 1)
Put x = -1, 1 + 1 + 1 = A(4)
⇒ A = \(\frac{3}{4}\)
Put x = 1, 1 – 1 + 1 = C(2)
⇒ C = + \(\frac{1}{2}\)
Equating the coefficients of x²,
A + B = 1
⇒ B = 1 – A = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)

Question 3.
Resolve the \(\frac{2 x^{2}+3 x+4}{(x-1)\left(x^{2}+2\right)}\) into partial fractions. (AP Mar. ’15, ’11; May ’11) (TS Mar. ’17)
Solution:
Let \(\frac{2 x^{2}+3 x+4}{(x-1)\left(x^{2}+2\right)}\) = \(\frac{A}{x-1}\) + \(\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+2}\)
Multiplying with (x – 1) (x² + 2)
2x² + 3x + 4 = A(x² + 2) + (Bx + C)(x – 1)
x = 1 ⇒ 2 + 3 + 4 = A(1 + 2)
9 = 3A ⇒ A = 3
Equating the coefficients of x²
2 = A + B ⇒ B = 2 – A = 2 – 3 = -1
Equating constants
4 = 2A – C ⇒ C = 2A – 4 = 6 – 4 = 2
\(\frac{2 x^{2}+3 x+4}{(x-1)\left(x^{2}+2\right)}\) = \(\frac{3}{x-1}\) + \(\frac{-x+2}{x^{2}+2}\)

Question 4.
Resolve \(\frac{x^{4}}{(x-1)(x-2)}\) into partial fractions.
Solution:

Equating the coefficients of (x – 1) (x – 2),
15x – 14 = A(x – 2) + B(x – 1)
Put x = 1, 15 – 14 ⇒ A(-1) = A = -1
Put x = 2, 30 – 14 = B(1) ⇒ B = 16
∴ \(\frac{x^{4}}{(x-1)(x-2)}\) = x² + 3x + 7 – \(\frac{1}{x-1}\) + \(\frac{16}{x-2}\)

Question 5.
Resolve \(\frac{x^{2}-3}{(x+2)\left(x^{2}+1\right)}\) into partial fractions. (AP Mar. ’17, ’16)
Solution:
Let \(\frac{x^{2}-3}{(x+2)\left(x^{2}+1\right)}\) = \(\frac{A}{x+2}\) + \(\frac{B x+C}{x^{2}+1}\)
Multiplying with (x + 2) (x² + 1)
x² – 3 = A(x² + 1) + (Bx + C)(x + 2)
x = -2 ⇒ 4 – 3 = A(4 + 1)
1 = 5A ⇒ A = \(\frac{1}{5}\)
Equating the coefficients of x²
1 = A + B ⇒ B = 1 – A
= 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Equating the constants – 3 = A + 2C
2C = -3 – A = -3 – \(\frac{1}{5}\) = – \(\frac{16}{5}\)
C = –\(\frac{8}{5}\)
\(\frac{x^{2}-3}{(x+2)\left(x^{2}+1\right)}\) = \(\frac{1}{5(x+2)}\) + \(\frac{4 x-8}{5\left(x^{2}+1\right)}\)

Question 6.
Resolve \(\frac{1}{(x-1)^{2}(x-2)}\) into partial fractions. (May ’13)
Solution:

Put x = 1 in (1)
1 = A(0) + B(1 – 2) + C(0)
⇒ -B = 1 ⇒ B = -1
Put x = 2 in (1)
⇒ 1 = A(0) + B(0) + C (2 – 1)²
⇒ C = 1
Equating the coefficients of x² in (1)
0 = A + C ⇒ A = -C = -1
A = -1
∴ \(\frac{1}{(x-1)^{2}(x-2)}\) = \(\frac{-1}{x-1}\) – \(\frac{1}{(x-1)^{2}}\) + \(\frac{1}{x-2}\)

Question 7.
Resolve \(\frac{5 x+1}{(x+2)(x-1)}\) into Partial fractions.
Solution:

⇒ 5x + 1 = A(x – 1) + B(x + 2) —– (1)
Put x = 1 in(1)
5(1) + 1,= A(0) + B(1 + 2)
⇒ 3B = 6 ⇒ B = 2
Put x = -2 in (1)
5(-2) + 1 = A (-2 – 1) + B(0)
⇒ -9 = -3A ⇒ A = 3
∴ \(\frac{5 x+1}{(x+2)(x-1)}\) = \(\frac{3}{x+2}\) + \(\frac{2}{x-1}\)

Question 8.
Resolve \(\frac{2 x+3}{5(x+2)(2 x+1)}\) into Partial fractions.
Solution:

Question 9.
Resolve \(\frac{13 x+43}{2 x^{2}+17 x+30}\) into partial fractions.
Solution:
2x² + 17x + 30 = 2x² + 12x + 5x + 30
= 2x(x + 6) + 5(x + 6)
= (x + 6) (2x + 5)
Let \(\frac{13 x+43}{(x+6)(2 x+5)}\) = \(\frac{A}{x+6}\) + \(\frac{B}{2x+5}\)
∴ 13x + 43 = A(2x + 5) + B(x + 6)
putting x = -6
-78 + 43 = A(-12 + 5)
-35 = -7A
⇒ A = 5
Putting x = \(\frac{-5}{2}\)
13\(\left(\frac{-5}{2}\right)\) + 43 = \(B\left(\frac{-5}{2}+6\right)\)
⇒ -65 + 86 = B(—5 + 12)
⇒ 21 = 7B
⇒ B = 3
∴ \(\frac{13 x+43}{2 x^{2}+17 x+30}\) = \(\frac{5}{x+6}\) + \(\frac{3}{2 x+5}\)

Question 10.
Resolve \(\frac{x^{2}+5 x+7}{(x-3)^{3}}\) into partial fractions.
Solution:
Let x – 3 = y ⇒ x = y + 3
\(\frac{x^{2}+5 x+7}{(x-3)^{3}}\) = \(\frac{(y+3)^{2}+5(y+3)+7}{y^{3}}\)

Question 11.
Resolve \(\frac{x^{2}+13 x+15}{(2 x+3)(x+3)^{2}}\) into partial fractions.
Solution:

Question 12.
Resolve \(\frac{1}{(x-1)^{2}(x-2)}\) into Partial fractions.
Solution:

Put x = 1 in (1)
1 = A (0) + B(1 – 2) + C(0)
⇒ -B = 1 ⇒ B = -1
Put x = 2 in (1)
1 = A(0) + B(0) + C(2 – 1)²
⇒ C = 1
Equating the coefficients of x² in (1)
0 = A + C ⇒ A = -C = -1
A = -1
∴ \(\frac{1}{(x-1)^{2}(x-2)}\) = \(\frac{-1}{x-1}\) – \(\frac{1}{(x-1)^{2}}\) + \(\frac{1}{x-2}\)

Question 13.
Resolve \(\frac{3 x-18}{x^{3}(x+3)}\) into Partial fractions.
Solution:

Put x = -3 in(1)
3(-3) – 18 = A(0) + B(0) + C(O) + D (-3)³
⇒ -27D = -27 ⇒ D = 1
Put x = 0 in (1)
3(0) – 18 = A (0) + B(0) + C(0 + 3) + D(0)
⇒ 3C = -18 ⇒ C = -6
Equating the coefficients of x³ in (1)
0 = A + D
⇒ A = -D = -1 ⇒ A = -1
Equating the coefficients of x² in (1)
0 = 3A + B
⇒ B = -3A = -3(-1) = 3 ⇒ B = 31
∴ \(\frac{3 x-18}{x^{3}(x+3)}\) = \(\frac{-1}{x}\) + \(\frac{3}{x^{2}}\) – \(\frac{6}{x^{3}}\) + \(\frac{1}{x+3}\)

Question 14.
Resolve \(\frac{x-1}{(x+1)(x-2)^{2}}\) into Partial fractions.
Solution:

Question 15.
Resolve \(\frac{2 x^{2}+1}{x^{3}-1}\) into partial fractions.
Solution:

Put x = 1 in (1)
2(1) + 1 = A(1 + 1 + 1) + (B + C)(0)
⇒ 3A = 3 ⇒ A = 1
Put x = 0 in (1)
0 + 1 = A(1) + (0 + C)(0 – 1)
⇒ 1 = A – C
⇒ 1 = 1 – C ⇒ C = 0
Equating the coefficients of x² in (1)
2 = A + B
⇒ 2 = 1 + B ⇒ B = 1

Question 16.
Resolve \(\frac{x^{3}+x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+3\right)}\) into partial fractions.
Solution:

Comparing the coefficients of x³, x², x and constant terms
A + C = 1, B + D = 1, 3A + 2C = 0, 3B + 2D = 1
Solve

Question 17.
Resolve \(\frac{3 x^{3}-2 x^{2}-1}{x^{4}+x^{2}+1}\) into Partial fractions.
Solution:

A – B + C + D = 0 —— (3)
B + D = -1 ——- (4) D = -1 – B
Substitute C, D in (2)
-A + B + 3 – A – 1 – B = -2
⇒ -2A = -4 ⇒ A = 2
Substitute C, D in (3)
A – B + 3 – A – 1 – B = 0 ⇒ 2 = 2B ⇒ B = 1
∴ C = 3 – 2 = 1, D = -1 – 1 = -2
Ax + B = 2x + 1, Cx + D = x – 2

Question 18.
Resolve \(\frac{x^{4}+24 x^{2}+28}{\left(x^{2}+1\right)^{3}}\) into partial fractions.
Solution:

Question 19.
Resolve \(\frac{x+3}{(1-x)^{2}\left(1+x^{2}\right)}\) into partial fractions.
Solution:
Let \(\frac{x+3}{(1-x)^{2}\left(1+x^{2}\right)}\)
= \(\frac{\mathrm{A}}{1-x}\) + \(\frac{B}{(1-x)^{2}}\) + \(\frac{C x+D}{1+x^{2}}\)
= x + 3 = A(1 – x) (1 – x²) + B(1 + x²) + (Cx + D)(1 – x²)
Comparing the coefficients of like powers of x, we get
A + B + D = 3 → (1)
-A + C – 2D = 1 → (2)
A + B – 2C + D = 0 → (3)
-A + C = 0 → (4)
Solving these equations, we get

Question 20.
Resolve \(\frac{x^{3}}{(2 x-1)(x+2)(x-3)}\) into partial fractions.
Solution:
\(\frac{x^{3}}{(2 x-1)(x+2)(x-3)}\)
= \(\frac{1}{2}\) + \(\frac{A}{2 x-1}\) + \(\frac{B}{x+2}\) + \(\frac{c}{x-3}\)
Multiplying with 2(2x – 1) (x + 2) (x – 3)
2x³ = (2x – 1) (x + 2) (x – 3) + 2A(x + 2)
(x – 3) + 2B (2x – 1) (x – 3) + 2C (2x -1) (x + 2)

Question 21.
Resolve \(\frac{x^{4}}{(x-1)(x-2)}\) into partial fractions. (T.S. Mar. ’16, March – 2013)
Solution:

Equating the coefficients of (x – 1) (x – 2),
15x – 14 = A(x – 2) + B(x – 1)
Put x = 1, 15 – 14 = A(-1) ⇒ A = -1
Put x = 2, 30 – 14 = B(1) ⇒ B = 16
∴ \(\frac{x^{4}}{(x-1)(x-2)}\) = x² + 3x + 7 – \(\frac{1}{x-1}\) + \(\frac{16}{x-2}\)

Question 22.
Find the coefficient of x⁴ in the expansion of \(\frac{3 x}{(x-2)(x+1)}\) in powers of x specifying the interval in which the expansion is valid.
Solution:
\(\frac{3 x}{(x-2)(x+1)}\) = \(\frac{A}{x-2}\) + \(\frac{B}{x+1}\)
Multiplying with (x – 2) (x + 1)
3x = A(x + 1) + B(x – 2)
Put x = -1, -3 = B(-3) ⇒ B = 1
Put x = 2, 6 = A(3) ⇒ A = 2

Question 23.
Find the coefficient of xⁿ in the power series expansion of \(\frac{x}{(x-1)^{2}(x-2)}\) specifying the region in which the expansion is valid.
Solution:
\(\frac{x}{(x-1)^{2}(x-2)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{(x-1)^{2}}\) + \(\frac{C}{x-2}\)
Multiplying with (x – 1)² (x – 2)
x = A(x – 1) (x – 2) + B(x – 2) + C(x – 1)²
Put x = 1, 1 = B(-1) ⇒ B = -1
Put x = 2, 2 = C(1) ⇒ C = 2
Put x = 0, 0 = 2A – 2B + C ⇒ 2A = 2B – C
= -2 – 2 = -4 ⇒ A = -2

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Binomial Theorem Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Binomial Theorem Important Questions

Question 1.
Find the number of terms in the expression of (2x + 3y + z)⁷ [Mar. 14, 13, 07]
Solution:
Number of terms in (a + b + c)ⁿ are \(\frac{(n+1)(n+2)}{2}\), where n is a positive integer. Hence number of terms in (2x + 3y + z)⁷ are \(\frac{(7+1)(7+2)}{2}=\frac{8 \times 9}{2}\) = 36

Question 2.
Prove that C₀ + 2 . C₁ + 4 . C₂ + 8 . C₃ + …………… + 2ⁿ . C_n = 3ⁿ [A.P. Mar. 15; May 07]
Solution:
L.H.S. = C₀ + 2 . C₁ + 4 . C₂ + 8 . C₃ + …………… + 2ⁿ . C_n
= ₀ + C₁(2) + C₂ (2)² + C₃ (2³) + …………… + C_n . 2ⁿ
= (1 + 2)ⁿ = 3ⁿ
Note: (1 + x)ⁿ = C₀ + C₁ . x + C₂ . x² + …………… + C_n . xⁿ

Question 3.
If ²²C_r is the largest binomial coefficient in the expansion of (1 + x)²², find the value of ¹³C_r. [A.P. Mar. 15; May 07]
Solution:
Here n = 22 is an even integer. There is only one largest binomial coefficient and it is
ⁿC_(n/2) = ²²C₁₁ = ²²C_r ⇒ r = 11
∴ ¹³C_r = ¹³C₁₁ = ¹³C₂ = \(\frac{13 \times 12}{1 \times 2}\) = 78

Question 4.
Write down and simplify 6^th term in (\(\frac{2x}{3}\) + \(\frac{3y}{2}\))⁹ [May 13]
Solution:
6^th term in (\(\frac{2x}{3}\) + \(\frac{3y}{2}\))⁹
The general term in (\(\frac{2x}{3}\) + \(\frac{3y}{2}\))⁹ is
T_r+1 = ⁹C_r (\(\frac{2x}{3}\))^9-r (\(\frac{3y}{2}\))^r
Put r = 5
T₆ = ⁹C₅ (\(\frac{2x}{3}\))⁴ (\(\frac{3y}{2}\))⁵
= ⁹C₅ (\(\frac{2}{3}\))⁴ (\(\frac{3}{2}\))⁵ x⁴ y⁵
= \(\frac{9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4} \frac{\left(2^{4}\right)}{3^{4}} \cdot \frac{3^{5}}{2^{5}} \cdot x^{4} y^{5}\)
= 189 x⁴y⁵

Question 5.
If the coefficients of (2r + 4)^th term and (3r + 4)^th term in the expansion of (1 + x)²¹ are equal, find r. [T.S. Mar.15]
Solution:
T_2r+4 in (1 + x)²¹ is
= ²¹C_2r+3 (x)^2r+3 ………………… (1)
T_3r+4 in (1 + x)²¹ is
= ²¹C_3r+3 . (x)^3r+3 ……………….. (2)
⇒ Coefficients are equal
⇒ ²¹C_2r+3 = ²¹C_3r+3
⇒ 21 = (2r + 3) + (3r + 3)
(or)
2r + 3 = 3r + 3
⇒ 5r = 15
⇒ r = 3 (or) r = 0 .
Hence r = 0, 3.

Question 6.
Fin the sum of the infinite series 1 + \(\frac{1}{3}\) + \(\frac{1.3}{3.6}\) + \(\frac{1.3.5}{3.6.9}\) + …………… [T.S. Mar. 15]
Solution:
The series can be written as
S = 1 + \(\frac{1}{1}\). \(\frac{1}{3}\) + \(\frac{1.3}{3.6}\) (\(\frac{1}{3}\))² + \(\frac{1.3.5}{1.2.3}\) (\(\frac{1}{3}\))³ + ……………..
The series of the right is of the form
1 + \(\frac{p}{1}\) (\(\frac{x}{q}\)) + \(\frac{p(p+q)}{1.2}\) (\(\frac{x}{q}\))² + \(\frac{p(p+q)(p+2 q)}{1.2 .3}\) (\(\frac{x}{q}\))³ + ……………
Here p = 1, q = 2, \(\frac{x}{q}\) = \(\frac{1}{3}\) ⇒ x = \(\frac{2}{3}\)
The sum of the given series
S = (1 – x)^-p/q
= (1 – \(\frac{2}{3}\))^-1/2 = (\(\frac{1}{3}\))^-1/2 = \(\sqrt{3}\)

Question 7.
Find the set E of the value of x for which the binomial expansions for the (a + bx)^r are valid. [Mar. 08]
Solution:
(4 + 9x)^-2/3 = 4^-2/3 [1 + \(\frac{9x}{4}\)]^-2/3
The binomial expansion of (4 + 9x)^-2/3 is valid
When |\(\frac{9x}{4}\)| < 1
⇒ |x| < \(\frac{4}{9}\)
⇒ x ∈ (\(\frac{-4}{9}\), \(\frac{4}{9}\))
i.e., E = (\(\frac{-4}{9}\), \(\frac{4}{9}\))

Question 8.
If the 2^nd, 3^rd and 4^th terms in the expansion of (a + x)ⁿ are respectively 240, 720,
1080, find a, x, n. [T.S. Mar. 16]
Solution:
T₂ = 240 ⇒ ⁿC₁ a^n-1 x = 240 …………….. (1)
T₃ = 720 ⇒ ⁿC₂ a^n-2 x² = 720 …………… (2)
T₄ = 1080 ⇒ ⁿC₃ a^n-3 x³ = 1080 …………… (3)
\(\frac{(2)}{(1)} \Rightarrow \frac{{ }^{n} C_{2} a^{n-2} x^{2}}{{ }^{n} C_{1} a^{n-1} x}=\begin{aligned}
&720 \\
&240
\end{aligned}\)
⇒ \(\frac{n-1}{2} \frac{x}{a}\) = 3 ⇒ (n – 1)x = 6a …………………. (4)
\(\frac{(3)}{(2)} \Rightarrow \frac{{ }^{n} C_{3} a^{n-3} x^{3}}{{ }^{n} C_{2} a^{n-2} x^{2}}=\frac{1080}{720}\)
⇒ \(\frac{n-2}{3} \frac{x}{a}=\frac{3}{2}\)
⇒ 2(n – 2)x = 9a …………………… (5)
\(\frac{(4)}{(5)} \Rightarrow \frac{(n-1) x}{2(n-2) x}=\frac{6 a}{9 a} \Rightarrow \frac{n-1}{2 n-4}=\frac{2}{3}\)
⇒ 3n – 3 = 4n – 8
⇒ n = 5
From (4), (5 – 1) x = 6a ⇒ 4x = 6a
⇒ x = \(\frac{3}{2}\) a
Substitute x = \(\frac{3}{2}\) a, n = 5 in (1)
⁵C₁ . a⁴ . \(\frac{3}{2}\) a = 240
5 × \(\frac{3}{2}\) a⁵ = 240
a⁵ = \(\frac{480}{15}\) = 32 = 2⁵
∴ a = 2, x = \(\frac{3}{2}\) a = \(\frac{3}{2}\) (2) = 3
∴ a = 2, x = 3, n = 5.

Question 9.
If the coefficients of r^th, (r + 1)^th, and (r + 2)^nd, terms in the expansion of (1 + x)ⁿ, are in A.P. then show that n² – (4r + 1)n + 4r² – 2 = 0. [T.S. Mar. 15, 08]
Solution:
Coefficient of T_r = ⁿC_r-1
Coefficient of T_r+1 = ⁿC_r
Coefficient of T_r+2 = ⁿC_r+1
Given ⁿC_r-1, ⁿC_r, ⁿC_r+1 are in A.P.
⇒ 2 . ⁿC_r = ⁿC_r-1 + ⁿC_r+1

⇒ (2n – 3r + 2) (r + 1) = (n – r) (n – r + 1)
⇒ 2nr + 2n – 3r² – 3r + 2r + 2 = n² – 2nr + r² + n – r
⇒ n² – 4nr + 4r² – n – 2 = 0
∴ n² – (4r + 1)n + 4r² – 2 = 0

Question 10.
If n is a postive integer, prove that \(\sum_{r=1}^{n} r^{3}\left(\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}\right)^{2}=\frac{(n)(n+1)^{2}(n+2)}{12}\) [Mar. 13]
Solution:

= (n + 1)² Σ r – 2(n + 1) Σ r² + Σ r³
= (n + 1)² \(\frac{(n)(n+1)}{2}\)

Question 11.
Find the set of values of x for which the binomial expansions of the following are valid.
(i) (2 + 3x)^-2/3
(ii) (5 + x)^3/2
(iii) (7 + 3x)^-5
(iv) (4 – \(\frac{x}{3}\))^-1/2 [A.P. Mar. 17; Mar. 16; Mar. 11]
Solution:
(i) (2 + 3x)^-2/3 = [2(1 + \(\frac{3}{2}\)x)]^-2/3
= 2^-2/3 (1 + \(\frac{3}{2}\)x)^-2/3
∴ The binomial expansion of (2 + 3x)^-2/3 is valid when |\(\frac{3}{2}\)x| < 1
(i.e.,) |x| < \(\frac{2}{3}\)
(i.e.,) x ∈ (-\(\frac{2}{3}\), \(\frac{2}{3}\))

ii) (5 + x)^3/2 = [5 (1 + \(\frac{x}{5}\))]^3/2 [T.S. Mar. 17]
= 5^3/2 (1 + \(\frac{x}{5}\))]^3/2
∴ The binomial expansion of (5 + x)^3/2 is valid when \(\frac{x}{5}\) < 1
(i.e.,) |x| < 5
(i.e.,) x ∈ (-5, 5)

iii) (7 + 3x)^-5 = [7 (1 + \(\frac{3}{7}\) x)]^-5
= 7^-5 (1 + \(\frac{3}{7}\) x)]^-5
(7 + 3x)^-5 is valid when \(\frac{3x}{7}\) < 1
⇒ |x| < \(\frac{7}{3}\) ⇒ x ∈ (\(\frac{-7}{3}\), \(\frac{7}{3}\))

iv) (4 – \(\frac{x}{3}\))^-1/2 = [4(1 – \(\frac{x}{3}\))]^-1/2
(4 – \(\frac{x}{3}\))^-1/2 is valid when \(\frac{-x}{12}\) < 1
⇒ |x| < 12
⇒ x ∈ (-12, 12)

Question 12.
Find the sum of the infinite series
\(\frac{3}{4}\) + \(\frac{3.5}{4.8}\) + \(\frac{3.5 .7}{4.8 .12}\) + …… (Mar. 11)
Solution:

Question 13.
If x = \(\frac{1.3}{3.6}\) + \(\frac{1.3 .5}{3.6 .9}\) + \(\frac{1.3 .5 .7}{3.6 .9 .12}\) + …… then prove that 9x² + 24x = 11 (TS Mar. ’16, AP Mar. ’17, ’15)
Solution:

⇒ 3x + 4 = 3\(\sqrt{3}\)
Squaring on both sides
(3x + 4)² = (3\(\sqrt{3}\))²
⇒ 9x² + 24x + 16 = 27
⇒ 9x² + 24x = 11

Question 14.
If x = \(\frac{5}{(2 !) \cdot 3}\) + \(\frac{5.7}{(3 !) \cdot 3^{2}}\) + \(\frac{5.7 .9}{(4 !) \cdot 3^{3}}\) + …… then find the value of x² + 4x. (mar. 13)
Solution:

Question 15.
Find the sum of the infinite series
\(\frac{7}{5}\) (1 + \(\frac{1}{10^{2}}\) + \(\frac{1.3}{1.2}\).\(\frac{1}{10^{4}}\) + \(\frac{1.3 .5}{1.2 .3}\).\(\frac{1}{10^{6}}\) + …….) (AP Mar. ‘16, May 13; Mar. ’05)
Solution:

Question 16.
For n = 0, 1, 2, 3, ….n, prove that

(TS Mar. ’15)
Solution:

Question 17.
If x = \(\frac{1}{5}\) + \(\frac{1.3}{5.10}\) + \(\frac{1.3 .5}{5.10 .15}\) + …… ∞ find 3x² + 6x. (May. ’14, ’07, ’06; May. ’11)
Solution:
Given that

⇒ 3(1 + x)² = 5
⇒ 3x² + 6x + 3= 5
⇒ 3x² + 6x = 2

Question 18.
Write the expansion or (2a + 3b)⁶.
Solution:

Question 19.
Find the 5th term in the expansion of (3x – 4y)⁷.
Solution:
T₅ = T_{4 + 1}
= ⁷C₄ (3x)^{7 – 4} (-4y)⁴
= 35.27x³. 256y⁴
= 241920 x³ y⁴

Question 20.
Find the 4^th term from the end in the expansion (2a + 5b)⁸.
Solution:
(2a + 5b)⁸ expansion contain 9 terms. The fourth term from the end is 6^th term from the beginning.
∴ ⁸C₅ (2a)^{8 – 5} (5b)⁵
= ⁸C₅ (2a)^{8 – 5} (5b)⁵
= ⁸C₅ . 2³. 5⁵ a³ b⁵

Question 21.
Find the middle term of the following expansions
(i) (3a – 5b)⁶
(ii) (2x + 3y)⁷
Solution:
i) Here n = 6 (even)
∴ \(\frac{n}{2}\) + 1 = \(\frac{6}{2}\) + 1 = 4^th term is the middle term
∴ T₄ = T_{3 + 1}
= ⁶C₃ (3a)^{6 – 3} (-5b)³,
= –⁶C₃. 3³. 5³. a³ b³

ii) Here n = 7 (odd)
\(\frac{\mathrm{n}+1}{2}\) = \(\frac{7+1}{2}\) = 4, \(\frac{\mathrm{n}+3}{2}\) = \(\frac{7+3}{2}\) = 5
∴ 4^th, 5^th terms are middle terms.
∴ T₄ = T_{3 + 1} = ⁷C₃ (2x)^{7 – 3} (3y)³ = ⁷C₃ 2⁴ 3³.x⁴.y³
T₅ = T_{4 + 1} = ⁷C₄(2x)^7-4(3y)⁴ = ⁷C₄. 2³.3⁴. x³. y⁴

Question 22.
n is a positive integer then prove that
Solution:
i) C_o + C₁ + C₂ + …….. + C_n = 2ⁿ
ii) a) C_o + C₂ + C₄ + … + C_n = 2^{n – 1} if n is even
(b) C_o + C₂ + C₄ + …. + C_{n – 1} = 2^{n – 1} if n is odd.
iii) (a) C₁ + C₃ + C₅ + …. + C_{n – 1} = 2^{n – 1} if n is even.
(b) C₁ + C₃ + C₅ + …. + C_{n – 1} = 2^{n – 1} if n is odd.
Solution:
We know (1 + x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x²+ …… + ⁿC_n xⁿ
= C₀ + C₁x + C₂x² + …… + C_nxⁿ

Question 23.
Prove that C₀ + 3.C₁ + 5.C₂ + ……… +(2n + 1). C_n = (2n + 2). 2^{n – 1}.
Solution:
Let S = C₀ + 3.C₁ + 5.C₂ + …… + (2n + 1). C_n —— (1)
By writing the terms in (1) in the reverse older, we get

Question 24.
Find the numerically greatest term in the binomial expansion of (1 – 5x)¹² when x = \(\frac{2}{3}\).
Solution:

Question 25.
Compute numerically ireatist term (s) in the expansionly of (3x – 5y)ⁿ when x = \(\frac{3}{4}\), y = \(\frac{2}{7}\) and n = 17
Solution:
Given x = \(\frac{3}{4}\), y = \(\frac{2}{7}\) and n = 17

Question 26.
Find the largest binomial coefficients (s) in the expansion of
(i) (1 + x)¹⁹
(ii) (1 + x)²⁴
Solution:
(i) Here n = 19 is an odd integer. Hence the largest binomial coefficients are

(ii) Here n = 24 is an even integer. Hence the largest binomial coefficient is

Question 27.
If ²²C_r is the largest binomial coefficient in the expansion of (1 + x)²², find the value of ¹³C_r. (A.P. Mar’16, May ‘11)
Solution:
Here n = 22 is an even integer. There is only one largest binomial coefficient and it is

Question 28.
Find the 7^th term in the expansion of \(\left(\frac{4}{x^{3}}+\frac{x^{2}}{2}\right)^{14}\)
Solution:
The general term in the expansion of

Question 29.
Find the 3^rd term from the end in the expansion of \(\left(x^{-2 / 3}-\frac{3}{x^{2}}\right)^{8}\)
Solution:
Comparing with (X + a)ⁿ, we get
X = x^-2/3, a = \(\frac{-3}{x^{2}}\), n = 8
In the given expansion \(\left(x^{-2 / 3}-\frac{3}{x^{2}}\right)^{8}\), we have n + 1 = 8 + 1 = 9 terms
Hence the 3^rd term from the end is 7^th term from the beginning.

Question 30.
Find the coefficient of x⁹ and x¹⁰ in the expansion of \(\left(2 x^{2}-\frac{1}{x}\right)^{2 c}\)
Solution:
If we write X = 2x² and a = –\(\frac{1}{x}\), then the general term in the expansion of

Since r = \(\frac{31}{3}\) which is impossible since r must be a positive integer. Thus ‘there is no term containing x⁹ in the expansion of the given expression. In otherwords the coefficient of x⁹ is ‘0’.
Now, to find the coefficient of x¹⁰.
put 40 – 3r = 10
⇒ r = 10

Question 31.
Find the term independent of x (that is the constant term) in the expansion of

Solution:

Question 32.
If the coefficients of x¹⁰ in the expansion of \(\left(a x^{2}+\frac{1}{b x}\right)^{11}\) is equal to the coefficient of x^-10 in the expansion of \(\left(a x-\frac{1}{b x^{2}}\right)^{11}\) ; find the relation between a and b where a and b are real numbers.
Solution:
The general term in the expansion of \(\left(a x^{2}+\frac{1}{b x}\right)^{11}\) is

To find the coefficient of x¹⁰, put
22 – 3r = 10 ⇒ 3r = 12 ⇒ r = 4
Hence the coefficient of x¹⁰ in

Given that the coefficients are equal.
Hence from (1) and (2), we get

Question 33.
If the k^th term is the middle term in the expansion of \(\left(x^{2}-\frac{1}{2 x}\right)^{20}\), find T_k and T_{k + 3}.
Solution:
The general term in the expansion of

Question 34.
If the coefficients of (2r + 4)^th and (r – 2)^nd terms in the expansion of (1 + x)¹⁸ are equal, find r.
Solution:

Question 35.
Prove that 2.C₀ + 7.C₁ + 12.C₂ + …… + (5n + 2)C_n = (5n + 4)2_{n – 1}.
Solution:
First method:
The coefficients of C₀, C₁, C₂, …., C_n are in A.P. with first term a = 2, C.d (d) = 5

Second method:
General term in LH.S.

Question 36.
Prove that

Solution:

Question 37.
For n = 0, 1, 2, 3 , n, prove that C₀. C_r + C₁. C_{r + 1} + C₂. C_{r + 2} + ……. + C_{n – r}. C_n = ²ⁿC_{n + 1}. (T.S. Mar. ’15)

Solution:
We know that

Question 38.
Prove that

Solution:

Question 39.
Find the numerically greatest term (s) in the expansion of

i) (2 + 3x)¹⁰ when x = \(\frac{11}{8}\)
Solution:

ii) (3x – 4y)¹⁴ when x = 8, y = 3.
Solution:

Question 40.
Prove that 6²ⁿ – 35n – 1 is divisible by 1225 for all natural numbers of n.
Solution:
6²ⁿ – 35n – 1 = (36)ⁿ – 35n – 1
= (35 + 1)ⁿ – 35n – 1

Hence 6²ⁿ – 35n – 1 is divisible by 1225 for all integral values of n.

Question 41.
Suppose that n is a natural number and I, F are respectively the integral part and fractional part of (7 + \(\sqrt{3}\))ⁿ. Then show that
(i) I is an odd integer
(ii) (I + F) (I – F) = 1
Solution:
Given that (7 + 4\(\sqrt{3}\))ⁿ = I + F where I is an integer and 0 < F < 1

= 2k, where k is a positive integer —— (1)
Thus I + F + f is n even integer.
Since I is an integer, we get that F + f is an integer. Also since 0 < F < 1 and 0 < f < 1
⇒ 0 < F + f < 2
∵ F + 1 is an integer
We get F + f = 1
(i.e.,) I – F = f ——— (2)

(i) From (1) I + F + f = 2k
⇒ f = 2k – 1, an odd integer.
(ii) (I + F) (I – F) = (I + F) f
= (7 + 4\(\sqrt{3}\))ⁿ (7 – 4\(\sqrt{3}\))ⁿ
= (49 – 48)ⁿ = 1.

Question 42.
Find the coefficient of x⁶ in (3 + 2x + x²)⁶.
Solution:

Question 43.
If n is a positive integer, then prove that
C_o + \(\frac{C_{1}}{2}\) + \(\frac{C_{2}}{3}\) + ….. + \(\frac{C_{n}}{n+1}\) = \(\frac{2^{n+1}-1}{n+1}\)
Solution:

Question 44.
If n is a positive integer and x is any nonzero real number, then prove that

(May. ’14, May 13, ’05)
Solution:

Question 45.
Prove that

Solution:

Now we can find the term independent of in the L.H.S. of (1).

Suppose n is an even integer, say n = 2k. Then from (2),

When n is odd:
Observe that the expansion in the numerator of (2) contains only even powers of x.
∴ If n is odd, then there is no constant term in (2) (i.e.,) the term indep. of x in

Question 46.
Find the set E of the value of x for which the binomial expansions for the following are valid
(i) (3 – 4x)^3/4
(ii) (2 + 5)x^-1/2
(iii) (7 – 4x)^-5
(iv) (4 + 9x)^-2/3
(iv) (a + bx)^r (Mar. ’08)
Solution:
i) (3 – 4x)^3/4 = 3^3/4\(\left(1-\frac{4 x}{3}\right)^{3 / 4}\)
The binomial expansion of (3 – 4x)^3/4 is valid, when \(\frac{4 x}{3}\) < 1
i.e., |x| < \(\frac{3}{4}\)
i.e., E = \(\left(\frac{-3}{4}, \frac{3}{4}\right)\)

ii) (2 + 5x)^-1/2 = 2^-1/2\(\left(1+\frac{5 x}{2}\right)^{-1 / 2}\)
The binomial expansion of (2 + 5x)^-1/2 is valid when |\(\frac{5 x}{3}\)| < 1 ⇒ |x| < \(\frac{2}{5}\)
i.e., E = (-\(\frac{2}{5}\), \(\frac{2}{5}\))

iii) (7 – 4x)^-5 = 7^-5\(\left(1-\frac{4 x}{7}\right)^{-5}\)
The binomial expansion of (7 – 4x)^-5 is valid when \(\frac{4 x}{7}\) < 1 ⇒ |x| < \(\frac{7}{4}\)
i.e., E = \(\left(\frac{-7}{4}, \frac{7}{4}\right)\)

iv) (4 + 9x)^-2/3 = 4^-2/3 \(\left(1+\frac{9 x}{4}\right)^{-2 / 3}\)
The binomial expansipn of (4 + 9x)^-2/3 is valid
When \(\frac{9 x}{4}\) < 1
⇒ |x| < \(\frac{4}{9}\)
⇒ x ∈ \(\left(\frac{-4}{9}, \frac{4}{9}\right)\)
i.e., E = \(\left(\frac{-4}{9}, \frac{4}{9}\right)\)

v) For any non zero reals a and b, the set of x for which the binomial expansion of (a + bx)^r is valid when r ∉ Z⁺ ∪ {0}, is \(\left(-\frac{|a|}{|b|}, \frac{|a|}{|b|}\right)\)

Question 47.
Find the
(i) 9^th term of \(\left(2+\frac{x}{3}\right)^{-5}\)
(ii) 10^th term of \(\left(1-\frac{3 x}{4}\right)^{4 / 5}\)
(iii) 8^th term of \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\)
(iv) 6^th term of \(\left(3+\frac{2 x}{3}\right)^{3 / 2}\)

(i) 9^th term of \(\left(2+\frac{x}{3}\right)^{-5}\)
Solution:

we get X = \(\frac{x}{6}\), n = 5
The general term in the binomial expansion of (1 + x)^-n is

ii) 10^th term of \(\left(1-\frac{3 x}{4}\right)^{4 / 5}\)
Solution:

iii) 8^th term of \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\)
Solution:

iv) 6^th term of \(\left(3+\frac{2 x}{3}\right)^{3 / 2}\)
Solution:

Question 48.
Write the first 3 terms in the expansion of
(i) \(\left(1+\frac{x}{2}\right)^{-5}\)
(ii) (3 + 4x)^-2/3
(iii) (4 – 5x)^-1/2

i) \(\left(1+\frac{x}{2}\right)^{-5}\)
Solution:
We have

ii) (3 + 4x)^-2/3
Solution:

iii) (4 – 5x)^-1/2
Solution:

iv) (2 – 3x)^-1/3
Solution:

Question 49.
Find the coefficient of x¹² in \(\frac{1+3 x}{(1-4 x)^{4}}\)
Solution:

Question 50.
Find coeff. of x⁶ in the expansion of (1 – 3x)^-2/5
Solution:

Question 51.
Find the sum of the infinite series

Solution:

Question 52.
Find the sum of the series

Solution:

Question 53.
If x = \(\frac{1}{5}\) + \(\frac{1.3}{5.10}\) + \(\frac{1.3 .5}{5.10 .15}\) + ……. ∞ then find 3x² + 6x. (Mar. ’14, ’07, ’06; May. ’11)
Solution:
Given that

Question 54.
Find an approximate value of
i) \(\frac{1}{\sqrt[3]{999}}\)
ii) (627)^1/4
Solution:

Question 55.
If |x| is so small that x3 and higher powers or x can be neglected, find approximate value of \(\frac{(4-7 x)^{1 / 2}}{(3+5 x)^{3}}\)
Solution:

Question 56.
Find an approximate value of \(\sqrt[6]{63}\) correct to 4 decimal places.
Solution:

Question 57.
If |x| is so small thát x² and higher powers of x may be neglected, then find an approximate value of

Solution:

Question 58.
If |x| is so small that x⁴ and higher powers of x may be neglected, then find the approximate value of
\(\sqrt[4]{x^{2}+81}\) – \(\sqrt[4]{x^{2}+16}\)
Solution:

Question 59.
Suppose that x and y are positive and x is very small when compared to y. Then find an approximate value of

Solution:

Question 60.
Expand \(5 \sqrt{5}\) in increasing powers of \(\frac{4}{5}\)
Solution:

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Permutations and Combinations Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Permutations and Combinations Important Questions

Question 1.
If ⁿp₃ = 1320, find n. (Mar. 2005)
Solution:
Hint : ⁿP_r = \(\frac{n !}{(n-r) !}\)
= n(n – 1) (n – 2) …… (n – r + 1)
∴ ⁿP₃ = 1320
= 10 × 132
= 10 × 12 × 11
= 12 × 11 × 10 = ¹²P₃
∴ n = 12

Question 2.
If ⁿP₇ = 42. ⁿP₅, find n. (TS Mar. ’17, ’15, ’11, ’07)
Solution:
ⁿP₇ = 42. ⁿP₅
n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5)(n – 6)
= 42. n(n – 1) (n – 2) (n – 3) (n – 4)
⇒ (n – 5) (n – 6) = 42
⇒ (n – 5)(n – 6) = 7 × 6
⇒ n – 5 = 7 or n – 6 = 6
∴ n = 12

Question 3.
If 10. ⁿC₂ = 3. find n. (AP Mar. ’15)
Solution:

Question 4.
If ₁₅C_{2r – 1} = ¹⁵C_{2r + 4} (Mar. ’14, ’05)
Solution:

Question 5.
If ⁿC₅ = ⁿC₆, then find ¹³C_n  (Mar. ’13)
Solution:

Question 6.
If ⁿP₄ = 1680, find n. (Mar. ’14; May ’06)
Solution:
Given ⁿP₄ = 1680
But ⁿP₄ = n(n – 1) (n – 2) (n -3).
Thus, we are given n(n – 1) (n – 2) (n – 3)
= 1680 = 8 × 7 × 6 × 5.
On comparing the largest integers on both sides, we get n = 8.

Question 7.
Find the number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls. (TS Mar.’15)
Solution:
4 boys can be selected from the given 8 boys in ⁸C₄ ways and 3 girls can be selected from the given 5 girls in ⁵C₃ ways. Hence, by the Fundamental principle, the number of required selections is ⁸C₄ x × ⁵C₃ = 70 × 10 = 700.

Question 8.
Find the number of positive divisors of 1080. (AP Mar. ’16)
Solution:
1080 = 2₃ × 3₃ × 5₁.
The number of positive divisors of 1080
= (3 + 1) (3 + 1) (1 + 1)
= 4 × 4 × 2 = 32.

Question 9.
Find the number of different chains that can be prepared using 7 different coloured beads. (AP Mar. 17)
Solution:
We know that the number of circular permutations of hanging type that can be formed using n things is \(\frac{1}{2}\) {(n – 1)!}. Hence the number of chains is \(\frac{1}{2}\) {(7 – 1)’} = \(\frac{1}{2}\) (6!) = 360.

Question 10.
Find the value of ¹⁰C₅ + 2. ¹⁰C₄ + ¹⁰C₃. (TS Mar. ’17)
Solution:

Question 11.
If ^{(n + 1)}P₅ : ⁿP₆ = 2 : 7, find (Mar. ’07)
Solution:

Question 12.
Find the number of ways of preparing a chain with 6 different coloured beads. (T.S Mar. ’16; May ’08)
Solution:
Hint :The number of circular permutations like the garlands of flowers, chains of beads etc., of n things = \(\frac{1}{2}\)(n – 1)!
The number of ways of preparing a chain with 6 different coloured beads.
= \(\frac{1}{2}\) (6 – 1)! = \(\frac{1}{2}\) × 5! = \(\frac{1}{2}\) × 120 = 60

Question 13.
If ⁿP_r = 5040 and ⁿC_r = 210, find n and r. (AP Mar. ‘17, ‘16)
Hint: ⁿP_r = r! ⁿC_r and
ⁿP_r = n (n – 1) (n – 2) ……. [n – (r – 1)]
Solution:

Question 14.
If ¹²C_{r + 1} = ¹²C_{3r – 5}, find r. (TS Mar. ’16, Mar. 2008)
Solution:
¹²C_{r + 1} = ¹²C_{3r – 5}
⇒ r + 1 = 3r – 5 or (r + 1) + (3r – 5) = 12
⇒ 1 + 5 = 2r or 4r – 4 = 12
⇒ 2r = 6 or 4r = 16
⇒ r = 3 or r = 4
∴ r = 3 or 4.

Question 15.
Simplify
(AP Mar. ’17, ’16, ’11)
Solution:

Question 16.
Find the number of ways of selecting 3 vowels and 2 consonants from the letters of the word EQUATION. (May 11; Mar.07)
Solution:
The word EQUATION contains 5 vowels and 3 consonants.
Thè 3 vowels can be selected from 5 vowels in ⁵C₃ = 10 ways.
The 2 consonants can be selected from 3 consonants in ³C₂ = 3 ways.
∴ The required number of ways of selecting 3 vowels and 2 consonants = 1o × 3 = 30

Question 17.
Find the number of ways of selecting 11 member cricket team from 7 bats men, 6 bowlers and 2 wicket keepers so that the team contains 2 wicket keepers and atleast 4 bowlers. (mar. ’14))
Solution:
The required cricket team can have the following compositions

Therefore, the number of ways of selecting the required cricket team
= 315 + 210 + 35 = 560.

Question 18.
i) If ¹²C_{(s + 1)} = ¹²C_{(2s – 5)}, then find s. (Mar.’11)
ii) If ⁿC₂₁ = ⁿC₂₇ find ⁴⁹C_n. (Mar. 06; May 06)
Solution:
i) ¹²C_{s + 1} = ¹C_{2s – 5} ⇒ either s + 1 = 2s – 5 or s + 1 + 2s – 5 = 12
⇒ s = 6 or 3s = 16
∴ s = 6 (since s’ is a non negative integer)

ii) ⁿC₂₁ = ⁿC₂₇ ⇒ n = 21 + 27 = 48
Hence ⁴⁹C_n = ⁴⁹C₄₈ = ⁴⁹C₁ = 49.

Question 19.
Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8 without repetition. (TS Mar. ’15)
Solution:
First Method : The number of 4 digited numbers formed by using the digits 0, 2, 4, 7, 8 without repetition
= ⁵P₄ – ⁴P₃ = 120 – 24 = 96
Out of these 96 numbers,
⁴P₃ – ³P₂ numbers contain 2 in units place
⁴P₃ – ³P₂ numbers contain 2 in tens place
⁴P₃ – ³P₂ numbers contain 2 in hundreds place
⁴P₃ numbers contain 2 in thousands place

∴ The value obtained by adding 2 in all the numbers
= (⁴P₃ – ³P₂) 2 + (⁴P₃ – ³P₂) 20 + (⁴P₃ – ³P₂) 200 + ⁴P₃ × 2000
= ⁴P₃ (2 + 20 + 200 + 2000) – ³P₂ (2 + 20 + 200)
= 24 × (2222) – 6(222)
= 24 × 2 × 1111 – 6 × 2 × 111
Similarly, the value obtained by adding 4 is
24 × 4 × 1111 – 6 × 4 × 111
the value obtained by adding 7 is
24 × 7 × 1111 – 6 × 7 × 111
the value obtained by adding 8 is
24 × 8 × 1111 – 6 × 8 × 111
∴ The sum of all the numbers
= (24 × 2 × 111 – 6 × 2 × 111) + (24 × 4 × 1111 – 6 × 4 × 111) + (24 × 7 × 1111 – 6 × 7 × 111) + (24 × 8 × 1111 – 6 × 8 × 111)
= 24 × 1111 × (2 + 4 + 7 + 8) – 6 × 111 × (2 + 4 + 7 + 8)
= 26664 (21) – 666 (21)
= 21 (26664 – 666)
= 21(25998)
= 5, 45, 958.

(or) Second Method
If Zero is one among the given n digits, then the sum of the r – digited numbers that can be formed using the given ‘n’ distinct digits (r ≤ n ≤ 9) is
^{(n -1)}P_{(r – 1)} × sum of the digits × 111 …. 1 (r times)
– ^{(n – 2)}P_{(r – 2)} × sum of the digits × 111 …. 1 [(r – 1) times]
Hence n = 5, n = 4, digits are {0, 2, 4, 7, 8}
Hence the sum of all 4 digited numbers that can be formed using the digits {0, 2, 4, 7, 8} without repetition is

Question 20.
If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the
dictionary order, then find the ranks of the words (Mar. 11)
i) REMAST
ii) MASTER. (TS Mar. 16; AP Mar. ‘15; May ‘11, ’08, ‘07, ‘06)
Solution:
i) Thé alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words begin with A is 5!
= 120
The number of words begin with E is 5!
= 120
The number of words begin with M is 5!
= 120
The number of words begin with RA is 4!
= 24
The number of words begin with REA is 3!
= 6
The next word is REMAST
Rank of the word REMAST = 3 (120) + 24 + 6 + 1 = 360 + 31 = 391

ii) The alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words begin with A is 5!
= 120
The number of words begin with E is 5!
= 120
The number of words begin with MAE is 3!
= 6
The number of words begin with MAR is 3!
= 6
The number of words begin with MASE is 2!
= 2
The number of words begin with MASR is 2!
= 2
The next word is MASTER.
∴ Rank of the word MASTER = 2(120) + 2(6) + 2(2) + 1
= 240 + 12 + 4 + 1 = 257

Question 21.
Find the number óf ways of arranging the letters of the word. (Mar. ’11, ’06)
i) INDEPENDENCE (May ’13)
ii) MATHEMATICS (Mar. ’13)  (May ’11)
iii) SINGING
iv) PERMUTATION
v) COMBINATION
vi) INTERMEDIATE
Solution:
i) The word INDEPENDENCE contains 12 letters in which there are 3 N’s are alike, 2 D’s are alike, 4 E’s are alike and rest are different.
∴ The number of required arrangements
= \(\frac{(12) !}{4 ! 3 ! 2 !}\)

ii) The word MATHEMATICS contains 11 letters in which there are 2 M’s are alike, 2 A’s are alike, 2 T’s are alike and rest are different.
∴ The number of required arrangements
= \(\frac{7 !}{2 ! 2 ! 2 !}\)

iii) The word SINGING contains 7 letters in which there are 2 l’s are alike, 2 N’s are alike, 2 G’s are alike and rest is different.
∴ The number of required arrangements
= \(\frac{(11) !}{2 !}\)

iv) The word PERMUTATION contains 11 letters in which there are 2 T’s are alike and rest are different.
∴ The number of required arràngements
= \(\frac{(11) !}{2 ! 2 ! 2 !}\)

v) The word COMBINATION contains 11 letters in which there are 2 O’s are alike, 2 l’s are alike, 2 N’s are alike and rest are different.
∴ The number of required arrangements
= \(\frac{(11) !}{2 ! 2 ! 2 !}\)

vi) The word INTERMEDIATE contains 12 letters in which thère are 2 l’s are alike, 2 l’s are alike, 3 E’s are alike and rest are different.
∴ The number of required arrangements
= \(\frac{(12) !}{2 ! 2 ! 3 !}\)

Question 22.
Prove that

(TS Mar. ’17 AP Mar. ’15)
Solution:

Question 23.
If a set A has 12 elements, find the number of subsets of A having
(i) 4 elements
(ii) Atleast 3 elements
(iii) Atmost 3 elements. (May 2007)
Solution:
Number of elements in set A = 12 .
(i) Number of subsets of A with exactly 4 elements = ¹²C₄ = 495

(ii) The required subset contains atleast 3 elements.
Number of subsets of A with exactly 0 elements iš 12C₀,
Number of subsets of A with exactly 1 element is 12C₁.
Number of subsets of A with exactly 2 element is 12C₂.
Total number of subsets of A formed = 212
∴ Number of subsets of A with atleast 3 elements
= (Total number of subsets) – (number of subsets contains 0 or 1 or 2 elements)
= 212 – (12C₀ + 12C₁ + 12C₂)
= 4096 – (1 – 12 + 66) = 4096 – 79 = 4017

(iii) The required sibset contains atmost 3 elements i.e., it may contain 0 or 1 or 2 or 3 elements.
Number of subsets of A with exactly 0 elements is. ¹²C₀
Number of subsets of A with exactly 1 element is 12C₁
Number of subsets of A with exactly 2 elements is 12C₂
Number of subsets of A with exactly 3 elements is 12C₃
∴ Number of subsets of A with atmost 3 elements
= 12C₀ + 12C₁ + 12C₂ + 12C₃
= 1 + 12 + 66 + 220
= 299

Question 24.
Find the number of ways of forming a committee of 5 members out of 6 Indians and 5 Americans so that always the Indians will be in majority in the committee. (TS Mar. ’15, ’13, ’08)
Solution:
Since committee contains majority of Indians, the members of the committee may be of the following types.

The number of selections in type I
= ⁶C₅ × 5C₀ = 6 × 1 = 6
The number of selections in type II
= 6C₄ × 5C₁ = 15 × 5 = 75
The number of selections in type III
= 6C₃ × 5C₂ = 20 10 = 200
= 6C₃ × 5C₂ = 20 10 = 200
∴ The required number ways of selecting a committee = 6 + 75 + 200 = 281.

Question 25.
If the letter of the word PRISON are permuted in all possible ways and the words thus formed are arranged in dictionary order, find the rank of the word. PRISON (AP Mar. ’17 Mar. ‘14, ’05; May ’13)
Solution:
The letters of the given word in dictionary order are
I N O P R S
In the dictionary order, first all the words that begin with I come. If I occupies the first place then the remaining 5 places can be filled with the remaining 5 letters in 5! ways. Thus, there are 5 number of words that begin with I. On proceeding like this we get

Hence the rank of PRISON is
3 × 5! + 3 × 4! + 2 × 2! + 1 × 1
= 360 + 72 + 4 + 1 + 1 = 438.

Question 26.
Find the sum of all 4- digited numbers that can be formed using the digits 1, 3, 5, 7, 9. (Mar. ’13)
Solution:
We know that the number of 4 digited numbers that can be formed using the digits
1, 3, 5, 7, 9 is ⁵P₄ = 120.
We have to find their sum. We first find the sum of the digits in the unitš place of all the 120 numbers. Put 1 in the units place.

The rerpaining 3 places can be filled with the remaining 4 digits in ⁴P₃ ways. Which means that there are ⁴P₃ number of 4 digited numbers with 1 in the units place. Similarly, each of the other digits 3, 5, 7, 9 appears in the units pläce ⁴P₃ times. Hence, by adding all these digits of the units place, we get the sum of the digits in the units place.

Similarly, we get the sum of all digits in 10’s place also as ⁴P₃ × 25. Since it is in 10’s place, its value is
⁴P₃ × 25 × 10.
Like this the values of the sumof the digits in 100’s place and 1000’s place are respectively
⁴P₃ × 25 × 100 and ⁴P₃ × 25 × 1000.
On adding all these sums, we get the sum of all the 4 digited numbers formed by using the digits 1, 3, 5, 7, 9. Hence the required sum is

Question 27.
If the letters of the word EAMCET are permuted in all possible ways and if the words thus formed are arranged in the dictionary order, find the rank of the word EAMCET. (AP Mar. 16) (TS Mar. 17)
Solution:
The dictionary order of the letters of the word EAMCET is
A C E E M T
In the dictionary order first gives the words which begin with the letter A.
If we fill the first place with A, remaining 5 letters can be arranged in \(\frac{5 !}{2 !}\) ways (since there are 2 E’s) on proceeding like this, we get

Question 28.
If \({ }^{\mathrm{n}} \mathrm{P}_{4}\) = 1680, find n. (Mar. ’14, May ’06)
Solution:
Given \({ }^{\mathrm{n}} \mathrm{P}_{4}\) = 1680
But \({ }^{\mathrm{n}} \mathrm{P}_{4}\) = n(n – 1) (n – 2) (n – 3).
Thus, we are given n(n – 1) (n – 2) (n – 3)
= 1680 = 8 × 7 × 6 × 5.
On comparing the largest integers on bothsides, we get n = 8.

Question 29.
If \({ }^{12} \mathrm{P}_{r}\) = 1320, find r.
Solution:
1320 = 12 × 11 × 10 = \({ }^{12} \mathrm{P}_{3}\), Hence r = 3.

Question 30.
If ^{(n + 1)}P₅ : ⁿP₅ = 3 : 2, find n
Solution:

It can be verified that n = 14 satisfies the given equation.

Question 31.
If ⁵⁶P_{(r + 3)} = 30800 : 1, find r.
Solution:


It can be verified that r given equation.

Question 32.
In how many ways 9 mathematics papers can be arranged so that the best and the worst
(i) may come together
(ii) may not come together ?
Solution:
i) If the best and worst papers are treated as one unit, then we have 9 – 2 + 1 = 7 + 1 = 8 papers. Now these can be arranged in (7 + 1) ! ways and the best and worst papers between themselves can be permuted in 2! ways. Therefore the number of arrangements in which best and worst papers come together is 8! 2!.

ii) Total number of ways of arranging 9 mathematics papers is 9!. The best and worst papers come together in 8! 2! ways. Therefore the number of ways they may not come together is 9! – 8! 2! = 8! (9 – 2) = 8! × 7.

Question 33.
Find the number of ways of arranging 6 boys and 6 girls in a row. In how many of these arrangements.
i) all the girls are together
ii) no two girls are together
iii) boys and girls come alternately ?
Solution:
i) 6 boys and 6 girls are altogether 12 persons. They can be arranged in a row in (12)! ways. Treat the 6 girls as one unit. Then we have 6 boys and 1 unit of girls. They can be arranged in 7! ways. Now, the 6 girls can be arranged among themselves in 6! ways. Thus the number of ways in which all 6 girls are together is (7! × 6!).

ii) First arrange the 6 boys in a row in 6! ways. Then we can find 7 gaps between them
(including the beginning gap and the ending gap) as shown below by the letter x :

Thus we have 7 gaps and 6 girls. They can be arranged in ⁷P₆ ways.
Hence, the number of arrangements which no two girls sit together is 6! × ⁷P₆ = 7.6!. 6!.

iii) The row may begin with either a boy or a girl, that is, 2 ways. If it begins with a boy, then odd places will be occupied by boys and even places by girls. The 6 boys can be arranged in 6 odd places in 6! ways and 6 girls in the 6 even places in 6! ways. Thus the number of arrangements in which boys and girls come alternately is 2 × 6! × 6!.

Question 34.
Find the number of 4 – letter words that can be formed using the letters erf the word. MIRACLE. How many of them
i) begin with an vowel
ii) begin and end with vowels
iii) end with a consonant ?
Solution:
The word MIRACLE has 7 letters. The number of 4 letter words that can be formed using these letters = ⁷P₄ = 7 × 6 × 5 × 4 = 840
Now take 4 blanks

i) We can fill the first place with one of the 3 vowels {A, E, 1} in ³P₁ = 3 ways.
Now the remaining 3 places can be filled using the remaining 6 letters in ⁶P₃ = 6 × 5 × 4 = 120 ways.
∴ The number of 4 letter words that begin with an vowel = 3 × 120 = 360 ways.

ii) Fill the first and last places with 2 vowels in ³P₂ = 3 × 2 = 6 ways.
The remaining 2 places can be filled with the remaining 5 letters in ⁵P₂ = 5 × 4 = 20 ways.
∴ The number of 4 letter words that begin and end with vowels = 6 × 20 = 120 ways.

iii) We can fill the last place with one of the 4 consonants {C, L, R, M} in ⁴P₁ = 4 ways. The remaining 3 places can be filled with the remaining 6 letters in ⁶P₃ = 6 × 5 × 4 = 120 ways.
∴ The number of 4 letter words that end with a consonant is = 4 × 120 = 480 ways.

Question 35.
Find the number of ways of permuting the letters of the word PICTURE so that
i) all vowels come together
ii) no two vowels come together
iii) the relative positions of vowels and consonants are not distributed.
Solution:
The word PICTURE has 3 vowels {E, I, U} and 4 consonants {C, P, R, T}
i) Treat the 3 vowels as one unit. Then we can arrange 4 consonants + 1 unit of vowels in 5! ways. Now 3 vowels among themselves can be permuted in 3! ways. Hence the number of permutations in which 3 vowels come together.
= 5! × 3! = 120 × 6 = 720 ways.

ii) No two vowels come together First arrange the 4 consonants in 4! ways. Then in between the vowels, in the beginning and in the ending, there are 5 gaps as shown below by the x letter

In these 5 places we can arrange 3 vowels in ⁵P₃ ways.
∴ The number of words in which no two vowels come together
= 4! × ⁵P₃
= 24 × 5 × 4 × 3 = 1440 ways.

iii) The three vowels can be arranged in their relative positions in 3! ways and the 4 consonants can be arranged in their relative positrons in 4! ways.

The required number of arrangements is 3! 4! = 144

Question 36.
If the letters of the word PRISON are permuted in all possible ways and the words thus formed are arranged in dictionary order, find the rank of the word. PRISON (Mar. ’14, ’05; May. ’13)
Solution:
The letters of the given word in dictionary order are
I N O P R S
In the dictionary order, first all the words that begin with I come. If I occupies the first place then the remaining 5 places can be filled with the remaining 5 letters in 5! ways. Thus, there are 5! number of words that begin with I. On proceeding like this we get

Question 37.
Find the number of 4-digit numbers that can be formed using the digits 2, 3, 5, 6, 8 (without repetition). How many of them are divisible by
(i) 2
(ii) 3
(iii) 4
(iv) 5
(v) 25
Solution:
The number of 4-digit numbers that can be formed using the 5 digits 2, 3, 5, 6, 8 is ⁵P₄ = 120

i) Divisible by 2 : For a number to be divisible by 2, the units place should befilled with an even digit. This can be done in 3 ways (2 or 6 or 8).

Now, the remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ = 24 ways.
Hence, the number of 4 – digit numbers divisible by 2 is 3 × 24 = 72.

ii) Divisible by 3 : A number is divisible by 3 if the sum of the digits n it is a multiple of 3. Since the sum of the given 5 digits is 24, we have to leave either 3 or 6 and use the digits 2, 5, 6, 8 or 2, 3, 5, 8. In each case, we can permute them in 4! ways. Thus the number of 4-digit numbers divisible by 3 is
2 × 4! 48.

iii) Divisible by 4 : A number is divisible by 4 if the number formed by the digits in the last two places (tens and units places) is a multiple of 4;

Thus we fill the last two places (as shown in the figure) with one of
28, 32, 36, 52, 56, 68
That is done in 6 ways. After filling the last two places, we can fill the remaining two places with the remaining 3 digits in
³P₂ = 6 ways.
Thus, the number of 4-digit numbers divisible by 4 is 6 × 6 = 36.

iv) Divisible by 5 : After filling the units place with 5 (one way), the remaining 3 places can be filld with the remaining 4 digits in ⁴P₃ = 24 ways. Hence the number of 4-digit numbers divisible by 5 is 24.

v) Divisible by 25 : Here also we have to fill the last two places (that is, units and tens place) with 25 (one way) as shown below.

Now the remaining 2 places can be filled with the remaining 3 digits in ³P₂ = 6 ways.
Hence the number of 4 – digit numbers divisible by 25 is 6.

Question 38.
Find the sum of all 4- digited numbers that can be formed using the digits 1, 3, 5, 7, 9. (Mar ‘13)
Solution:
We know that the number of 4 digited numbers that can be formed using the digits
1, 3, 5, 7, 9 is ⁵P₄ = 120.
We have to find their sum. We first find the sum of the digits in the units place of all the 120 numbers. Put 1 in the units place.

The remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ ways. Which means that there are ⁴P₃ number of 4 digited numbers with 1 in the units place. Similarly, each of the other digits 3, 5, 7, 9 appears in the units place ⁴P₃ times. Hence, by adding all these digits of the units place, we get the sum of the digits in the units place.

Similarly, we get the sum of all digits in 10’s place also as ⁴P₃ × 25. Since it is in 10’s place, its value is
⁴P₃ × 25 × 10.
Like this the values of the sum of the digits in 100’s place and 1000‘s place are respectively
⁴P₃ × 25 × 100 and ⁴P₃ × 25 × 1000.
On adding all these sums, we get the sum of all the 4 digited numbers formed by using the digits 1, 3, 5, 7, 9. Hence the required sum is

Question 39.
How many four digited numbers can be formed using the digits 1, 2, 5, 7, 8, 9 ? How many of them begin with 9 and end with 2 ?
Solution:
The number of four digited numbers that can be formed using the given digits 1, 2, 5, 7, 8, 9, is ⁶P₄ = 360. Now, the first place and last place can be filled with 9 and 2 in one way.

The remaining 2 places can be filled by the remaining 4 digits 1, 5, 7, 8. Therefore these two places can be filled in 4P2 ways. Hence, the required number of ways 1. ⁴P₂ = 12.

Question 40.
Find the number of injections of a set A with 5 elements to a set B with 7 elements.
Solution:
If a setA has m elements and the set B has n elements (m ≤ n), then the number of injections from A into B = ⁿP_m
∴ The number of injections from set A with 5 elements into set B with 7 elements.
= ⁷P₅ = 2,520.

Question 41.
Find the number of ways in which 4 letters can be put in 4 addressed envelopes so that no letter goes into the envelope meant for it.
Solution:
Required number of ways is 4! \(\left(\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right)\)
= 12 – 4 + 1 = 9

Question 42.
Find the number of 5 letter words that can be formed using the letters of the word ‘MIXTURE’ which begin with an
vowel when repetitions are allowed.
Solution:
The word MIXTURE has 7 letters 3 vowels {E, I, U} and 4 consonants {E, M, R, X} we have to fill up 5 blanks.

Fill the first place with one of the 3 vowels in 3 ways.
Each of the remain 4 places can be filled in 7 ways (since repetition is allowed)
∴ The number of 5 letter words
= 3 × 7 × 7 × 7 × 7 = 3 × 7⁴

Question 43.
a) Find the number of functions from a set A with m elements to a set B with n elements.
Solution:
Let A = {a₁, a₂, ………. a_m} and
B = {b₁, b₂, …., b_n}.
First, to define the image of a1 we have n choices (any element of B). Then to define the image of a₂ we again have n choices (since a₁, a₂ can have the same image). Thus we have n choices for the image of each of the m elements of the set A. Therefore, the number of different ways of defining the images of elements of A (with images in B) is n × n × …. × n(m times) = n^m.

b) Find the number of surjections from a set A with n elements to a set B with 2 elements when n > 1.
Solution:
Let A {a₁, a₂,….., a_n} and b = {x, y}. The total number of functions from A to B is 2ⁿ For a surjection, both the elements x, y of B must be in the range. Therefore, a function is not a surjection if the range contains only x (or y). There are only two such functions.
Hence, the number of surjections from A to B is 2ⁿ – 2.

Question 44.
Find the number of permutations of 4- digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 when repetition is allowed.
Solution:

The number of permutations of 4 digit numbers that can be formed using the given 6 digits = 6 × 6 × 6 × 6 = 6⁴ = 1,296

Question 45.
Find the number of 4- digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 that are divisible by
(i) 2
(ii) 3 when repetition is allowed.
Solution:
(i) Numbers divisible by 2
Take 4 blank palces. First the units place

can be filled by an even digit in 3 ways (2 or 4 or 6). The remaining three places can be filled with the 6 digits in 6 ways each. Thus they can be filled in 6 × 6 × 6 = 6 ways.
Therefore, the number of 4 digited numbers divisible by 2 is 3 × 6³ = 3 × 216 = 648

ii) Numbers divisible by 3
First we fill up the first 3 places with the given 6 digits in 6 ways.

After filling up the first 3 places, if we fill the units place with the given 6 digits, we get 6 consecutive positive integers. Out of these six consecutive integers exactly 2 will be divisible by ‘3’. Hence the units place can be filled in ‘2’ ways. Therefore, the number of 4 digited numbers divisible by 3.
= 6³ × 2 = 216 × 2 = 432.

Question 46.
Find the number of 5 – letter words that can be formed using the letters of the word EXPLAIN that begin and end with a vowel when repetitions are allowed.
Solution:
We can fill the first and last places with vowels each in 3 ways. (E or A or I)

Now each of the remaining 3 places can be filled in 7 ways (using any letter of given 7 letters). Hence the number of 5 letter words
which begin and end with vowels ¡s
= 3² × 7³ = 9 × 343 = 3087

Question 47.
Find the number of ways of arranging the letters of the word SINGING só that
i) they begin and end with I
ii) the two G’s come together .
iii) relative positions of vowels and consonants are not disturbed.
Solution:
The word SINGING has 2 I’s, 2 G’s and 2 N’s and one S. Total 7 letters.

i) First, we fill the first and last places with I’s in \(\frac{2 !}{2 !}\) = 1 way as shown below.

Now we fill the remaining 5 places with the remaining 5 letters in \(\frac{5 !}{2 ! 2 !}\) = 30 ways.
Hence the number of required permutations = 30

ii) Treat two G’s as one unit. Then we have 5 letters 2 I’s, 2 N’s and one S + one unit of 2
G’s = 6 can be arranged in \(\frac{6 !}{2 ! 2 !}\) = \(\frac{720}{2 \times 2}\)
= 180 ways.
Now the two G’s among themselves can be arranged in one way. Hence the number of received permutations = 180 × 1 = 180.

Question 48.
Find the number of ways of arranging the letters of the word a⁴ b³ c⁵ in its expanded form.
Solution:
The expanded form of a⁴ b³ c⁵ is
aaaa bbb ccccc
There are 4 + 3 + 5 = 12 letters
They can be arranged in \(\frac{(12) !}{4 ! 3 ! 5 !}\) ways.

Question 49.
If the letters of the word EAMCET are permuted in all possible ways and if the words thus formed are arranged in the dictionary order, find the rank of the word EAMCET. (A.P. Mar. ’16)
Solution:
The dictionary order of the letters of the word EAMCET is
A C E E M T
In,the dictionary order first gives the words which begin with the letter A.
If we fill the first place with A, remaining 5 letters can be arranged in \(\frac{5 !}{2 !}\) ways (since there
are 2 E’s) on proceeding like this, we get


Hence the rank of the word EAMCET is
= 2 × \(\frac{5 !}{2 !}\) + 2 × 3! + 1
= 120 + 12 + 1 = 133

Question 50.
Find the number of ways of arranging 8 men and 4 women around a circular table. In how many of them
i) all the women come together
ii) no two women come together
Solution:
’Total number of persons = 12 (8 men + 4 women)
Therefore, the number of circular permutations is (11) !

i) Treat the 4 women as one unit. Then we have 8 men + 1 unit of women = 9 entities
which can be arranged around a circle in 8! ways. Now, the 4 women among themselves can be arranged in 4! ways. Thus, the number of required arrangements is 8! × 4!.

ii) First arrange 8 men around a circle in 7! ways. Then there are 8 places in between them as shown in. fig by the symbol x (one place in between any two consecutive men).
Now, the 4 women can be arranged in these 8 places in ⁸P₄ ways.

Therefore, the number of circular arrangements in which no two women come together is 7! × ⁸P₄.

Question 51.
Find the number of ways of seating 5 Indians, 4 Americans and 3 Russians at a round table so that
i) all Indians sit together
ii) no two Russians sit together
iii) persons of same nationality sit together.
Solution:

i) Treat the 5 Indians as a single unit. Then we have 4 Americans, 3 Russians and 1 unit of Indians. That is, 8 entities in total. Which can be arranged at a round table in (8-1)! = 7! ways.
Now, the 5 Indians among themselves can be arranged in 5! ways. Hence, the number of required arrangements is 7! × 5!

ii) First we arrange the 5 Indians + 4 Americans around the table in (9 – 1) ! = 8! ways.
Now, we can find 9 gaps in between these 9 persons (one gap between any two consecutive persons).
The 3 Russians can be arranged in these 9 gaps in 9P3 ways. Hence, the number of required arrangements is 8! × ⁹P₃.

iii) Treat the 5 Indians as one unit, the 4 Americans as the second unit and the 3 Russians as the third unit. These 3 units can be arranged at round table in (3 – 1)! = 2! ways.

Now, the 5 Indians among themselves can be permuted in 5! ways. Similarly, the 4 Americans in 4! ways and the 3 Russians in 3! ways. Hence the number of required arrangements is 2! × 5! × 4! × 3!

Question 52.
Find the number of different chains that can be prepared using 7 different coloured beads.
Solution:
We know that the number of circular permutations of hanging type that can be formed using n things is \(\frac{1}{2}\) {(n – 1)!}. Hence the number of chains is \(\frac{1}{2}\) {(7 – 1)}{
= \(\frac{1}{2}\) (6!) = 360.

Question 53.
Find the number of different ways of preparing a garland using 7 distinct red roses and 4 distinct yellow roses such that no two yellow roses come together.
Solution:
First arrange 7 red roses in a circular form (garland form) in (7 – 1)! = 6! ways. Now, there are 7 gaps and 4 yellow roses can be arranged in these 7 gaps in 4 ⁷P₄ ways.

Thus, the total number of circular permutations is 6! × ⁷P₄.
But, in the case of garlands, clock-wise and anti-clock-wise arrangements look alike. Hence, the number of required ways is.
\(\frac{1}{2}\) (6! × ⁷P₄)

Question 54.
14 persons are seated at a round table. Find the number of ways of selecting two persons out of them who are not seated adjacent to each other.
Solution:
The seating arrangement of given 14 persons at the round table as shown below.

Number of ways of selecting 2 persons out of 14 persons = ¹⁴C₂ = \(\frac{14 \times 13}{1 \times 2}\) = 91.

In the above arrangement two persons sitting adjacent to each other can be selected in 14 ways. (they are a₁ a₂, a₂ a₃ , a₁₃ a₁₄, a₁₄ a₁)
∴ The required no.of ways = 91 – 14 = 77

Question 55.
Find the number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls. (T.S. Mar. ’15)
Solution:
4 boys can be selected from the given 8 boys in ⁸C₄ ways and 3 girls can be selected from the given 5 girls in ⁵C₃ ways. Hence, by the Fundamental principle, the number of required selections is
⁸C₄ × ⁵C₃ = 70 × 10 = 700.

Question 56.
Find the number of ways of selecting 4 English, 3 Telugu and 2 Hindi books out of 7 English, 6 Telugu and 5 Hindi books.
Solution:
The number of ways of selecting
4 English books out of 7 books = ⁷C₄
3 Telugu books out of 6 books = ⁶C₃
2 Hindi books out of 5 books = ⁵C₂
Hence, the number of required ways
⁷C₄ × ⁶C₃ × ⁵C₂ = 35 × 20 × 10 = 7000.

Question 57.
Find the number of ways of forming a committee of 4 members out of 6 boys and 4 girls such that there is atleast one girl in the committee.
Solution:
The number of ways of forming a committee of 4 members out of 10 members (6 boys + 4 girls) is ¹⁰C₄. Out of these, the number of waýs of forming the committee, having no girl is ⁶C₄ (We select all 4 members from boys).
Therefore, the number of ways of forming the committees having atleast one girl is
¹⁰C₄ – ⁶C₄ = 210 – 15 = 195.

Question 58.
Find the number of ways of selecting 11 member cricket team from 7 bats men, 6 bowlers and 2 wicket keepers so that the team contains 2 wicket keepers and atleast 4 bowlers. (Mar. 14)
Solution:
The required cricket team can have the following compositions

Therefore, the number of ways of selecting the required cricket team
= 315 + 210 + 35 = 560.

Question 59.
If a set of ‘n’ parallel lines intersect another set of ‘n’ parallel lines (not parallel to the lines in the first set), then find the number of parallellograms formed in this lattice structure.
Solution:
To form a parallelogram, we hâve to select 2 lines from the first set which can be done in ^mC₂ ways and 2 lines from the second set which can be done in ⁿC₂ ways. Thus, The number of parallelograms formed is ^mC₂ × ⁿC₂.

Question 60.
There are m points in a plane out of which p points are collinear and no three of the points are collinear unless all the three are from these p points. Find the number of different
i) straight lines passing through pairs of distinct points.
ii) triangles formed by joining these points (by line segments).
Solution:
i) From the given m points, by drawing straight lines passing through 2 distinct points at a time, we are supposed to get ^mC₂ number of lines. But, since p out of these m points are collinear, by forming lines passing through these p points 2 at a time we get only one line instead of getting ^PC₂. Therefore, the number of different lines passing through pairs of distinct points is
^mC₂ – ^PC₂ + 1

ii) From the given m points, by joining 3 points at a time, we are supposed to get ^mC₃ number of triangles. Since p out of these m points are collinear by joining these p points 3 at a time we do not get any triangle when as we are supposed to get ^PC₃ number of triangles. Hence the number of triangles formed by joining the given m points
= ^mC₂ – ^PC₃

Question 61.
A teacher wants to take 10 students to a park. He can take exactly 3 students at a time and will not take the same group of 3 students more than once. Find the number of times
(i) each student can go to the park
(ii) the teacher can go to the park.
Solution:
i) To find the number of times a specific student can go to the park, we have to select 2 more students from the remaining 9 students. This can be done in ⁹C₂ ways.
Hence each student can go to the park ⁹C₂ times = \(\frac{9 \times 8}{1 \times 2}\) = 36 times

ii) The no.of times the teacher can go to park = The no.of different ways of selecting 3 students out of 10 = ¹⁰C₃ = 120.

Question 62.
A double decker minibus has 8 seats in the lower deck and 10 seats in the upper deck. Find the number of ways of
arranging 18 persons in the bus if 3 children want to go to the upper deck and 4 old people can not go to the upper deck.
Solution:
Allowing 3 children to the upper deck and 4 old people to the lower deck, we are left with 11 people and 11 seats (7 seats in the upper deck and 4 in the lower deck). We can select 7 people for the upper deck out 11 people in ¹¹C₇ ways. The remaining 4 persons go to the lower deck.

Now we-can arrange 10 persons (3 children and 7 others) in the upper deck and 8 persons (4 old people and 4 others) in the lower deck in (10)! and (8)! ways respectively. Hence the required number of arrangements = ¹¹C₇ × 10! × 8!

Question 63.
Prove that
i) ¹⁰C₃ + ¹⁰C₆ = ¹¹C₄
ii)

Solution:

Question 64.
i) If ¹²C_{(s + 1)} = ¹²C_{(2s – 5)}, then find s. (Mar. ‘11)
ii) If ⁿC₂₁ = ⁿC₂₇ find ⁴⁹C_n. (Mar. ‘06; May 06)
Solution:

Question 65.
If there are 5 alike pens, 6 alike pencils and 7 alike erasers, find the number of ways of selecting any number of (one or more) things out of them.
Solution:
The required number of ways
= (5 + 1 (6 + 1) (7 + 1) – 1 = 335.

Question 66.
Find the number of positive divisors of 1080. (AP. Mar.’16)
Solution:
1080 = 2³ × 3³ × 5¹.
∴ The number of positive divisors of 1080
= (3 + 1) (3 + 1) (1 + 1)
= 4 × 4 × 2 = 32.

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Theory of Equations Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Theory of Equations Important Questions

Question 1.
Form polynomial equation of the lowest degree, with roots 1, -1, 3 (May ’06)
Hint : Equation having roots, α, β, γ is [(x – α)(x – β)(x – γ) = 0
Solution:
Required equation is
(x – 1) (x + 1) (x – 3) = 0
⇒ (x² – 1) (x – 3) = 0
⇒ -3x² – x + 3 = 0

Question 2.
If 1, 1 α are the roots of
x³ – 6x² + 9x – 4 = 0, then find α. (May 11)
Solution:
1, 1, α are roots of x³ – 6x² + 9x – 4 = 0
Sum = 1 + 1 + α = 6
α = 6 – 2 = 4

Question 3.
If -1, 2 and α are the roots of 2x³ + x² – 7x – 6 = 0, then find α (Mar 14, 13)
Solution:
-1, 2, α are roots of 2x³ + x² – 7x – 6 = 0
sum = -1 + 2 + α = –\(\frac{1}{2}\)
α = –\(\frac{1}{2}\) – 1 = –\(\frac{3}{2}\)

Question 4.
If 1, -2 and 3 are roots of x³ – 2x² + ax + 6 = 0, then find α. (Mar. ’04,)
Solution:
1, -2 and 3 are roots of x³ + x² + ax – 6 = 0
x³ – 2x + ax + 6 = 0
⇒ 1(-2) + (-2) 3 + 3. 1 = a
i.e., a = -2 – 6 + 3 = -5

Question 5.
If the product of the roots of 4x³ + 16x² – 9x – a = 0 is 9, then find a. (TS Mar. ’17, ’16)
Solution:
α, β, γ are the roots of
4x³ + 16x² – 9x – a = 0
αβγ = \(\frac{a}{4}\) = 9 ⇒ a = 36

Question 6.
Find the transformed equation whose roots are the negative of the roots of
x⁴ + 5x³ + 11x + 3 = 0
Solution:
Given f(x) = x⁴ + 5x³ + 11x + 3 = 0
We want an equation whose roots are
-α₁, α₂, α₃, α₄,
Required equation f(-x) = 0
⇒ (-x)⁴ + 5 (-x)³ + 11 (-x) + 3 = 0
⇒ x⁴ – 5x³ – 11x + 3 = 0

Question 7.
Form the polynomial equation of degree 3 whose roots are 2, 3 and, 6.
Solution:
The required polynomial equation is,
(x – 2) (x – 3) (x – 6) = 0
⇒ x³ – 11x² + 36x – 36 = 0

Question 8.
Let α, β, γ be the roots of Σα³
Solution:
Σα³ = α³ + β³ + γ³
= (α + β + γ)
= (α² + β² + γ² – αβ – βγ – γα) + 3αβγ
= (-p)(p² – 2q – q) – 3r
= -p(p² – 3q) – 3r
∴ Σα³ = -p³ + 3pq – 3r = 3pq – p³ – 3r

Question 9.
If α, β and 1 are the roots of x³ – 2x² – 5x + 6 = 0, then find α and β (AP Mår ’16, ’08)
Solution:
α, β and 1 are the roots of
x³ – 2x² – 5x + 6 = 0
Sum = α + β + 1 = 2 ⇒ α + β = 1
product = αβ = -6
(α – β)² = (α + β )² – 4αβ = 1 + 24 = 25
α – β = 5
α + β = 1
Adding 2α = 6 ⇒ α = 3
∴ α = 3 and β = -2

Question 10.
If α, β and γ are the roots of x³ – 2x² + 3x – 4 = 0, then find
i) Σα²β²
ii) Σαβ (α + β)
Solution:
Since α, β, γ are the roots of
x³ – 2x² + 3x – 4 = 0
then α + β + γ = 2
αβ + βγ + γα = 3
αβγ = 4

i) Σα²β² = α²β² + β²γ² + γ²α²
= (αβ + βγ + γα)² – 2αβγ(α + β + γ)
= 9 – 2. 2.4 = 9 – 16 = -7

ii) Σαβ (α + β) = α²β + β²γ + γ²α + αβ² + βγ² + γα²
= (αβ + βγ + γα)(α + β + γ) – 3αβγ
= 2.3 – 3.4 = 6 – 12 = -6.

Question 11.
Solve the x³ – 3x² – 6x + 8 = 0 equation, given that the roots of each are in A.P. (‘Mar. ’07)
Solution:
The roots of x³ – 3x² – 6x + 8 = 0 are in A.P
Suppose a – d, a, a + d be the roots
Sum = a – d + a + a + d = 3
3a = 3
⇒ a = 1
∴ (x – 1) is a factor of x³ – 3x² – 6x + 8 = 0

⇒ x² – 2x – 8 = 0
⇒ x² – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ x = 4, -2
∴ The roots are -2, 1, 4.

Question 12.
Solve x⁴ – 4x² + 8x + 35 = 0, given that 2 + i \(\sqrt{3}\) is a root (AP Mar. ’15)
Solution:
Let 2 + i \(\sqrt{3}\) is one root
⇒ 2 – i \(\sqrt{3}\) is another root.
The equation having roots
2 ± i \(\sqrt{3}\) is x² – 4x + 7 = 0
∴ x² – 4x + 7 is a factor of

∴ The roots of the given equation are 2 ± i\(\sqrt{3}\), -2 ± i

Question 13.
Find the polynomial equation whose roots are the reciprocals of the roots of x⁴ – 3x³ + 7x² + 5x – 2 = 0 (TS Mar. ’15, ’11)
Solution:
Given equation is
f(x) = x⁴ – 3x³ + 7x² + 5x – 2 = 0
Required equation is f(\(\frac{1}{x}\)) = 0
i.e. \(\frac{1}{x^{4}}\) – \(\frac{3}{x^{3}}\) + \(\frac{7}{x^{2}}\) + \(\frac{5}{x}\) – 2 = 0
Multiplying with x⁴
⇒ 1 – 3x + 7x² + 5x³ – 2x⁴ = 0
i.e., 2x⁴ – 5x³ – 7x² + 3x – 1 = 0

Question 14.
Find the polynomial equation whose roots are the translates of those of x⁵ – 4x⁴ + 3x² – 4x + 6 = 0 by -3. (TS Mar. 16)
Solution:
Given equation is
f(x) = x⁵ – 4x⁴ + 3x² – 4x + 6 = 0
Required equation is f(x + 3) = 0
(x + 3)⁵ – 4(x + 3)⁴ + 3(x + 3)²

Required equation is
x⁵ + 11x⁴ + 42x³ + 57x² – 13x – 60 = 0

Question 15.
Show that x⁵ – 5x³ + 5x² – 1 = 0 has three equal roots and find this root. (TS Mar. ’17)
Solution:
Let f(x) = x⁵ – 5x³ + 5x² – 1
f'(x) = 5x⁴ – 15x² + 10x
= 5x (x³ – 3x + 2)
f'(1) = 5(1) (1 – 3 + 2) = 0
f(1) = 1 – 5 + 5 – 1 = 0
x – 1 is a factor of f'(x) and f(x)
∴ 1 is a repeated’ root of f(x).

⇒ 1 is a root of above equation
(∵ sum of the coefficients is zero)
∴ 1 is the required root.

Question 16.
Solve the 8x³ – 36x² – 18x + 81 = 0 equation, given that the roots of each are in AP. (Mar. 04’)
Soluution:
Given the roots of 8x³ – 36x² – 18x + 81 = 0 are in AP.
Let the roots be a – d, a, a + d
Sum of the roots = a – d + a + a + d
= \(\frac{36}{8}\) = \(\frac{9}{2}\)
i.e., 3a = \(\frac{9}{2}\) ⇒ a = \(\frac{3}{2}\)
∴ (x – \(\frac{3}{2}\)) is a factor of

⇒ 8x² – 24x – 54 = 0
⇒ 4x² – 12x – 27 = 0
⇒ 4x² – 18x + 6x – 27 = 0
⇒ 2x(2x – 9) + 3 (2x – 9) = 0
⇒ (2x + 3) (2x – 9) = 0
⇒ x = –\(\frac{3}{2}\), \(\frac{9}{2}\)
The roots are –\(\frac{3}{2}\), \(\frac{3}{2}\), \(\frac{9}{2}\)

Question 17.
Solve the 3x³ – 26x² + 52x – 24 = 0 equations, given that the roots of each are in GP.
(TS Mar. ’15)
Solution:
Given equation is 3x³ – 26x² + 52x – 24 = 0
The roots are in G.P.
Suppose \(\frac{a}{r}\), a, ar are the roots.
Product = \(\frac{a}{r}\).a.ar = –\(\left(-\frac{24}{3}\right)\)
a³ = 8 ⇒ a = 2
∴ (x – 2) is a factor of 3x³ – 26x² + 52x – 24

⇒ 3x³ – 20x + 12 = 0
⇒ 3x² – 18x – 2x + 12 = 0
⇒ 3x (x – 6) -2 (x – 6) = 0
⇒ (3x – 2)(x – 6) = 0
⇒ x = \(\frac{2}{3}\), 6
∴ The roots are \(\frac{2}{3}\), 2, 6.

Question 18.
Solve 18x³ + 81x² + 121x + 60 = 0 given that one root is equal to half the sum of the
remaining roots. (May ’11; Mar. ’05)
Solution:
Suppose α, β, γ are the roots of 18x³ + 81x² + 121x + 60 = 0.

Question 19.
Solve the equation
2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0 (AP Mar.17, 16; Mar. ‘08, 07)
Solution:
Given f(x) = 2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0
This is an odd degree reciprocal equation of first type.
∴ -1 is a root.
Dividing f(x) with x + 1

Dividing f(x) by (x + 1) we get
2x⁴ – x³ – 11x² – x + 2 = 0
Dividing by x²

Substituting in (1), required equation is
2(a² – 2) – a – 11 = 0 .
2a² – 4 – a – 11 = 0
2a² – a – 15 = 0
(a – 3)(2a + 5) = 0
a = 3 or \(-\frac{5}{2}\)
Case (i) a = 3

Question 20.
Find the roots of
x⁴ – 16x³ + 86x² – 176x + 105 = 0
Solution:
Let f(x) = x⁴ – 16x³ + 86x² – 176x + 105
Now, f(1) = 1 – 16 + 86 – 176 + 105 = 0
∴ 1 is a root of f(x) = 0
⇒ x – 1 is a factor of f(x)

∴ f(x) = (x – 1) (x³ – 15x² + 71x – 105)
= (x – 1) g(x) where
g(x) = x³ – 15x² + 71x – 105
g(1) = 1 – 15 + 71 – 105 = -48 ≠ 0
g(2) = -15 ≠ 0
g(3) = 27 – 135 + 213 – 105 = 0
∴ 3 is a root of g(x) =0
⇒ x – 3 is a factor of g(x)

∴ g(x) = (x – 3) (x² – 12x + 35)
= (x – 3) (x – 5) (x – 7)
∴ f(x) = (x – 1) (x – 3) (x – 5) (x – 7)
∴ 1, 3, 5, 7 are the roots of f(x) = 0.

Question 21.
Solve 4x³ – 24x² + 23x + 18 = 0, given that the roots of this equation are in arithmetic progression (Mar. ’14; May ’06)
Solution:
Let a – d, a, a + d are the roots of the given equation
Now, sum of the roots
a – d + a + a + d = \(\frac{24}{4}\)
3a = 6
a = 2
Product of the roots (a – d) a (a + d) = \(-\frac{18}{4}\)
a(a² – d) = \(-\frac{9}{2}\)
2(4 – d²) = \(-\frac{9}{2}\)
4(4 – d²) = -9
16 – 4d² = -9
4d² = 25
d = ±\(\frac{5}{2}\)
∴ roots are –\(\frac{1}{2}\), 2 and \(\frac{9}{2}\)

Question 22.
Find the polynomial equation whose roots are the squares of the roots of x⁵ + 4x³ – x² + 11 = 0 by -3. (Mar. ’06)
Solution:
Let f (x) ≡ x⁵ + 4x³ – x² + 11
The required equation is f(x + 3) = 0

The required equation is
x⁵ + 15x⁴ + 94x³ + 305x² + 507x + 353 = 0

Question 23.
Solve the equation 6x⁴ – 35x³ + 62x² – 35x + 6 = 0. (May. 13)
Solution:
We observe that the given equation is an even degree reciprocal equation of class one. On dividing both sides of the given equation by x², we get

Then the above equation reduces to
6(y² – 2) – 35y + 62 = 0
i.e., 6y² – 35y + 50 = 0
i.e., (2y – 5)(3y – 10) = 0.
Hence the roots of 6y² – 35y + 50 = 0 are \(\frac{5}{2}\) and \(\frac{10}{3}\).

Hence the roots of the given equation are \(\frac{1}{2}\), \(\frac{1}{3}\), 2 and 3.

Question 24.
Solve x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0. (Mar. ’13)
Solution:
Given equation is
x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0 is a reciprocal equation of odd degree and of class two.
∴ 1 is a root of the given equation.
⇒ (x – 1) is a factor of
x⁵ + 4x³ + 5x² – 4x² + 1 = 0

Question 25.
Form the polynomial equation of degree 3 whose roots are 2, 3 and 6. (Mar. ’02)
Solution:
The required polynomial equation is,
(x – 2) (x – 3)(x – 6) = 0
⇒ x³ – 11x³ + 36x – 36 = 0

Question 26.
Find the relation between the roots and the coefficients of the cubic equation
3x³ – 10x² + 7x + 10 = 0.
Solution:
3x³ – 10x² + 7x + 1o = 0 ———- (1)
On.comparing (1) with
ax³ + bx² + cx + d = 0,
we have

Question 27.
Write down the relations between the roots and the coefficients of the bi-quadratic equation.
x⁴ – 2x³ + 4x² + 6x – 21 = 0
Solution:
Given equation is
x⁴ – 2x³ + 4x² + 6x – 21 = 0 —— (1)
On comparing (1) with
ax⁴ + bx³ + cx² + dx + c = 0,
we have

Question 28.
If 1, 2, 3 and 4 are the roots of x⁴ + ax³ + bx² + cx + d = 0, then find the
values of a, b, c and d.
Solution:
Given that the roots of the given equation are 1, 2, 3 and 4. Then
x⁴ + ax³ + bx² + cx + d
≡ (x – 1) (x – 2) (x – 3) (x – 4) = 0
≡ x⁴ – 10x³ + 35x² – 50x + 24 = 0
On equating the coefficients of like powers of x, we obtain
a = -10, b = 35, c = -50, d = 24

Question 29.
if a, b, c are the roots of
x³ – px² + qx- r = 0 and r ≠ 0, then find \(\frac{1}{\mathbf{a}^{2}}+\frac{1}{\mathbf{b}^{2}}+\frac{1}{\mathrm{c}^{2}}\) interms of p, q, r.
Solution:
Given that a, b, c are the roots of
x³ – px² + qx – r = 0, then
a + b + c = p, ab + bc + ca = q, abc = r

Question 30.
Find the sum of the squares and the sum of the cubes of the roots of the equation x³ – px² + qx – r = 0 in terms of p, q, r.
Solution:
Let α, β, γ be the roots of the given equation then α + β + γ = p, αβ + βγ + γα = q, αβγ = r
Sum of the squares of the roots is α² + β² + γ²
= (α + β + γ)² – 2(αβ + βγ + γα) = p² – 2q
Sum of the cubes of the roots is α³ + β³ + γ³
= (α + β + γ) (α² + β² + γ² – αβ – βγ – γα) + 3αβγ
= p(p² – 2q – q) + 3r
= p(p² – 3q) + 3r

Question 31.
Obtain the cubic equation, whose roots are the sqüares of the roots of the equation, x³ + p₁x² + P₂x + p₃ = 0
Solution:
The required equation is, f(\(\sqrt{x}\)) =0

Question 32.
Let α, β, γ be the roots of
x³ + px² + qx + r = 0. Then find the
i) Σα²
ii) Σ\(\frac{1}{\alpha}\)
iii) Σα³
iv) Σβ²γ²
v) Σ(α + β) (β + γ) (γ + α)
Solution:
since α, β, γ are the roots of the equation, we have α + β + γ = – p,
αβ + βγ + γα = q, αβγ = -r.

i) Σα²
Solution:
Σα² = α² + β² + γ²
= (α + β + γ) – 2(αβ + βγ + γα)
= p² – 2q

ii) Σ\(\frac{1}{\alpha}\)
Solution:

iii) Σα³
Solution:

iv) Σβ²γ²
Solution:

v) (α + β) (β + γ) (γ + α)
Solution:
We know, α + β + γ = -p
⇒ α + β = -p – r and β + γ = -p – α
= γ + α = -p – β
∴ (α + β) (β + γ)(γ + α)
= (-p – γ) (-p – α) (-p – β)
= -p³ – p²(α + β + γ) – p(αβ + βγ + γα) – αβγ
= -p³ + p³ – pq + r = r – pq .

Question 33.
Let α, β, γ be the roots of
x³ + ax² + bx + c = 0 then find Σα² + Σβ²
Solution:
Since α, β, γ are roots of the given equation,

Question 34.
If α, β, γ are the roots of x³ + px² + qx + r = 0, then form the cubic equation whose roots are .
α(β + γ), β(γ + α), γ(α + β)
Solution:
Let α, β, γ be the roots of the given equation.
we have, α + β + γ = – p, αβ + βγ + γα = q, αβγ = -r
Let y = α(β + γ)
= αβ + αγ + γβ – βγ

Question 35.
Solve x³ – 3x² – 16x + 48 = 0
Solution:
Let f(x) = x³ – 3x² – 16x + 48
by inspection, f(3) = 0
Hence 3 is a root of 1(x) = 0
Now we divide f(x) by (x – 3)

Question 36.
Find the roots of
x⁴ – 16x³ + 86x² – 176x + 105 = 0 (Mar. ‘02)
Solution:
Let f(x) = x⁴ – 16x³ + 86x² – 176x + 105
Now if (1) = 1 – 16 + 86 – 176 + 105 = 0
∴ 1 is a root of f(x) = 0
⇒ x – 1 is a factor of f(x)

∴ g(x) = (x – 3)(x² – 12x + 35)
= (x – 3) (x – 5) (x – 7)
∴ f(x) = (x – 1) (x – 3) (x – 5) (x – 7)
∴ 1, 3, 5, 7 are the roots of f(x) = 0.

Question 37.
Solve x³ – 7x² + 36 = 0, given one root being twice the other.
Solution:
Let α, β, γ be the root of the equation
x³ – 7x² + 36 = 0 and
let β = 2α
Now, we have, α + β + γ = 7
⇒ 3α + γ = 7 —— (1)
αβ + βγ + γα = 0
⇒ 2α² + 3αγ = 0 —— (2)
αβγ = -36 ⇒ 2α²γ = -36 —— (3)
From (1) and (2), we have
2α² + 3α(7 – 3α) = 0
i.e., α² – 3α = 0 (or) α(α – 3) = 0
∴ α = 0 or α = 3
Since α = 0 does not satisfy the given equation.
∴ α = 3, so β = 6 and γ = -2
∴ The roots are 3, 6, -2.

Question 38.
Given that 2 is a root of
x³ – 6x² + 3x + 1o = 0, find the other roots.
Solution:
Let f(x) = x³ – 6x² + 3x + 10
Since 2 is a root of f(x) = 0, we divide f(x) by (x – 2)

∴ -1, 2, and 5 are the roots of the given equation.

Question 39.
Given that two roots of
4x³ + 20x² – 23x + 6 = 0 are equal, find all the roots of the given equation.
Solution:
Let α, β, γ are the roots of
4x³ + 20x² – 23x + 6 = 0
Given two roots are equal, let α = β

⇒ 12α² + 4α – 23 = 0
⇒ (2α – 1) (6α + 23) = 0
α = \(\frac{1}{2}\), α = \(\frac{-23}{6}\)
On verfication, we get that is a root of (1)
α = \(\frac{1}{2}\) is roots of (1)
(2) ⇒ roots are \(\frac{1}{2}\), \(\frac{1}{2}\), -6

Question 40.
Given that the sum of two roots of
x⁴ – 2x³ + 4x² + 6x – 21 = 0 is zero find the roots of the equation.
Solution:
Let α, β, γ, δ are the roots of given equation, since sum of two is zero.
α + β = 0
Now α + β + γ + δ = 2 ⇒ γ + δ = 2
Let αβ = p, γδ = q
The equation having the roots α, β is

∴ Roots are –\(\sqrt{3}\), \(\sqrt{3}\), 1 – i \(\sqrt{6}\) and 1 + i \(\sqrt{6}\)

Question 41.
Solve 4x³ – 24x² + 23x + 18 = 0, given that the roots of this equation are in arithmetic progression. (Mar. 14, May ‘06)
Solution:
Let a – d, a, a + d are the roots of the given equation
Now, sum of the roots
a – d + a + a + d = \(\frac{24}{4}\)
3a = 6
a = 2
Product of the roots (a – d) a (a + d) = \(\frac{-18}{4}\)
a(a² – d²) = \(\frac{-9}{2}\)
2(4 – d²) = \(\frac{-9}{2}\)
4(4 – d²) = -9
16 – d² = -9
4d² = 25
d = ± \(\frac{5}{2}\)
∴ roots are –\(\frac{1}{2}\), 2 and \(\frac{9}{2}\)

Question 42.
Solve x³ – 7x² + 14x – 8 = 0, given that the roots are in geometric progression.
Solution:
Let \(\frac{a}{r}\), a, ar be the roots of the given equation. Then .

Hence a = 2. On substituting a = 2 in (1), we obtain
\(\frac{2}{r}\) + 2 + 2r = 7
i.e., 2r² – 5r + 2 = 0
i.e., (r – 2) (2r – 1) = 0
Therefore r = 2 or r = \(\frac{1}{2}\)
Hence the roots of the given equation are 1, 2 and 4.

Question 43.
Solve x⁴ – 5x³ + 5x² + 5x – 6 = 0 given that the product of two of its roots is 3.
Solution:
Let α, β, γ, δ be the roots of the given equation.
Product of the roots αβγδ = -6
Given αβ = 3 (∵ Product of two roots is 3)
∴ α, β, γ, δ = -6
γδ = -2
Let α + β, γ + δ = q
The equation having the roots α,β is
x² – (α + β) x + αβ = 0
x² + px + 3 = 0
The equation having the roots γ, δ is
x² – (γ + δ)x + γδ = 0
x² – qx – 2 = 0
∴ x⁴ – 5x³ + 5x² + 5x – 6
= (x² – px + 3)(x² – qx – 2)
= x⁴ – (p + q)x³ + (1 + pq)x² + (2p – 3q)x – 6
Comparing the like terms,
p + q = 5, 2p – 3q = 5
∴ 2p – 3q = 5
3p + 3q = 15
5p = 20 ⇒ p = 4
∴ q = 1
Now x² – 4x + 3 = 0 ⇒ (x – 3)(x – 1) = 0
⇒ x = 1, 3
x² – x – 2 = 0 ⇒ (x – 2)(x + 1) = 0
⇒ x = -1, 2
∴ The roots are -1, 2, 1, 3

Question 44.
Solve x⁶ + 4x³ – 2x² – 12x + 9 = 0, Given that it has two pairš of equal roots.
Solution:
Given equation is
x⁴ + 4x³ – 2x² – 12x + 9 = 0
Let the roots be α, α, β, β
Sum of the roots, 2(α + β) = -4
⇒ α + β = -2
Let αβ = p
The equation having roots α, β is
x² – (α + β)x + αβ = 0
i.e. x² + 2x + p = 0
∴ x⁴ + 4x³ – 2x² – 12x + 9
= [x² – (α + β)x + αβ]²
= (x² + 2x + p)²
= x⁴ + 4x³ + (2p + 4)x² + 4px + p²
Comparing coefficients of x on both sides
4p = -12 ⇒ p = – 3
x² + 2x + p = 0 ⇒ x² + 2x – 3 = 0
⇒ (x + 3)(x – 1) = 0
⇒ x = -3, 1
∴ The roots of the given equation are -3, -3, 1, 1

Question 45.
Prove that the sum of any two of the roots of the equation x⁴ + px³ + qx² + rx + s = 0 is equal to the sum of the remaining two roots of the equation iff p³ – 4pq + 8r = 0.
Solution:
Suppose that the sum of two of the roots of the given equation is equal to the sum of the remaining two roots.
Let α, β, γ, δ be roots of the given equation such that α + β = γ + δ

From these equations, we have

Then equations (1), (2) and (4) are satisfied. In view of (5), equation (3) is also satisfied. Hence (x² + bx + c)(x² + bx + d) = x⁴ + 2bx³ + (b² + c + d)x² + b(c + d)x + cd = x⁴ + px³ + qx² + rx + s
Hence the roots of the given equation are α₁, β₁, γ₁ and δ₁. where α₁ and β₁ are the roots of the equations x² + bx + c = 0 and γ₁ and δ₁ are those of the equation x² + bx + d = 0.
We have α₁ + β₁ = -b = γ₁ + δ₁.

Question 46.
Form the polynomial equation of degree 4 whose roots are
4 + \(\sqrt{3}\), 4 – \(\sqrt{3}\), 2 + i and 2 – i
Solution:
The equation having roots 4 + \(\sqrt{3}\), 4 – \(\sqrt{3}\) is
x² – 8x + 13 = 0
The equation having roots 2 + i, 2 – i is
x² – 4x + 5 = 0.
The required equation is
(x² – 8x + 13) (x² – 4x + 5) = 0
∴ x⁴ – 12x³ + 50x² – 92x + 65 = 0

Question 47.
Solve 6x⁴ – 13x³ – 35x² – x + 3 = 0 given that one of its root is 2 + \(\sqrt{3}\).
Solution:
2 + \(\sqrt{3}\) is a root 2 – \(\sqrt{3}\) is also a root.
The equation having roots

Question 48.
Find the polynomial equation of degree 4 whose roots are the negatives of the roots of x⁴ – 6x³ + 7x² – 2x + 1 = 0
Solution:
Let f(x) ≡ x⁴ – 6x³ + 7x² – 2x + 1
The required equation is f(-x) = 0
i.e., (-x)⁴ – 6(-x)³ + 7(-x)² + 2(-x) + 1 = 0
∴ x⁴ + 6x³ + 7x² + 2x + 1 = 0

Question 49.
Find the algebraic equation of the degree 4 whose roots are 3 times the roots of the equation
6x⁴ – 7x³ + 8x² – 7x + 2 = 0
Solution:
Let f(x) ≡ 6x⁴ – 7x³ + 8x² – 7x +2
The required equation is f\(\left(\frac{x}{3}\right)\) = o

Question 50.
Form the equation whose roots are m times the roots of the equation x³ + \(\frac{x^{2}}{4}\) – \(\frac{x}{16}\) + \(\frac{1}{72}\) = 0 and deduce the case when m = 12.
Solution:

Question 51.
Find the algebraic equation of degree 5 whose roots are the translates of the roots of x⁵ + 4x³ – x² + 11 = 0 by -3. (Mar. ’06)
Solution:
Let f (x) ≡ x⁵ + 4x³ – x² + 11
The required equation is f(x + 3) = 0

The required equation is
x⁵ + 15x⁴ + 94x³ + 305x² + 507x + 353 = 0

Question 52.
Find the algebraic equation of degree 4 whose roots are the translates of the roots
4x⁴ + 32x³ + 83x² + 76x + 21 = 0 by 2.
Solution:
Let f(x) ≡ 4x⁴ + 32x³ + 83x² + 76x + 21
The required equation is f(x – 2) = 0

The required equation is
4x⁴ – 13x² + 9 = 0

Question 53.
Find the polynomial equation whose roots are the reciprocals of the roots of the equation
x⁴ + 3x³ – 6x² + 2x -4 = 0
Solution:
Let f(x) ≡ x⁴ + 3x³ – 6x² + 2x – 4

Question 54.
Find the polynomial equation whose roots are the squares of the roots of x³ – x² + 8x – 6 = 0
Solution:
Let f(x) ≡ x³ – x² + 8x – 6 .
The required equation is f(\(\sqrt{x}\)) = o

Squaring on both sides
⇒ x(x² + 16x + 64) = x² + 12x + 36
= x³ + 6x² + 64x – x² – 12x – 36 = 0
∴ x³ + 15x² + 52x – 36 = 0

Question 55.
Show that 2x³ + 5x² + 5x + 2 = 0 is a reciprocal equation of class one.
Solution:
Given equation is 2x³ + 5x² + 5x + 2 = 0
P₀ = 2, p₁ = 5, P₂ = 5, p₃ = 2
Here P₀ = p₃, p₁ = p₂
∴ The equation 2x³ + 5x² + 5x + 2 = 0 is a reciprocal equation of class one.

Question 56.
Solve the equation
4x³ – 13x² – 13x + 4 = 0
Solution:
4x³ – 13x² – 13x + 4 = 0 is a reciprocal equation of first class and of odd degree.
Thus -1 is a root of the 9iven equation.

4x² – 17x + 4 = 0 ⇒ 4x² – 16x – x + 4 = 0
⇒ 4x(x – 4) – 1 (x – 4) = 0
⇒ (x – 4)(4x – 1) = 0
⇒ x = 4 or \(\frac{1}{4}\)
The roots are -1, 4, \(\frac{1}{4}\)

Question 57.
Solve the equation 6x⁴ – 35x³ + 62x² – 35x + 6 = 0. (May ‘13)
Solution:
We observe that the given equation is an even degree reciprocal equation of class one.
On dividing both sides 6f the given equation by x², we get

Then the above equation reduces to
6(y² – 2) – 35y + 62 = 0
i.e., 6y² – 35y + 50 = 0
i.e., (2y – 5) (3y – 10) = 0.
Hence the roots of 6y² – 35y + 50 = 0 are \(\frac{5}{2}\) and \(\frac{10}{3}\).
Therefore x + \(\frac{1}{x}\) = \(\frac{5}{2}\) and x + \(\frac{1}{x}\) = \(\frac{10}{3}\)
i.e., 2x² – 5x + 2 = 0 and 3x² – 10x + 3 = 0.
The roots of these equations are respectively

Hence the roots of the given equation are \(\frac{1}{2}\), \(\frac{1}{3}\), 2 and 3.

Question 58.
Solve x⁵ – 5x⁴ – 9x³ – 9x² + 5x – 1 = 0. (Mar ’13)
Solution:
Given equation is
x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0 is a reciprocal equation’of odd degree and of class two.
∴ 1 is a root of the given equation.
⇒ (x – 1) is a factor of
x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0

Question 59.
Solve the equation
6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0
Solution:
Given equation is
6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0 is a reciprocal equation of second class and of even degree.
∴ x² – 1 is a factor of
6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0

∴ (1) becomes 6(y² – 2) – 25(y) + 37 = 0
⇒ 6y² – 12 – 25y + 37 = 0
⇒ 6y² – 25y + 25 = 0
⇒ 6y² – 15y – 10y + 25 = 0
⇒ 3y(2y – 5) – 5(2y – 5) = 0
⇒ (2y – 5)(3y – 5) = 0

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 1.
Find the value of (1 – i)⁸  (Mar. ’07)
Solution:

Question 2.
If x = cis θ, then find the value of [x⁶ + \(\frac{1}{x^{6}}\)]
Solution:
∵ x = cos θ + i sin θ
⇒ x⁶ = (cos θ + i sin θ)⁶
= cos 6θ + i sin 6θ
⇒ \(\frac{1}{x^{6}}\) = cos 6θ – i sin 6θ
∴ x⁶ + \(\frac{1}{x^{6}}\) = 2 cos 6θ

Question 3.
If A, B, C are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find the value of XYZ. (AP Mar. ‘16, ’15)
Solution:
∴ A, B, C are angles of a triangle
⇒ A + B + C = 180° ——- (1)
x = cis A, y = cis B, Z = cis C
⇒ xyz = cis(A + B + C)
= cos(A + B + C) + i sin(A + B + C)
= cos(180°) + i sin (180°) .
= -1 + i(0) = -1
∴ xyz = -1

Question 4.
If 1, ω, ω² are the cube roots of unity, then prove that \(\frac{1}{2+\omega}\) – \(\frac{1}{1+2 \omega}\) = \(\frac{1}{1+\omega}\)
Solution:
ω is a cube root of unity
1 + ω + ω² = 0 and ω³ = 1

Question 5.
(2 – ω) (2 – ω²) (2 – ω¹⁰) (2 – w¹¹) = 49. (TS Mar. ’17)
Solution:
∵ 1, ω, ω² are the cube roots of unity,
ω³ = 1 and 1 + ω + ω² = 0
2 – ω¹⁰ = 2 – ω⁹ . ω
= 2 – (ω³)³ . ω²
= 2 – (1)³ ω² = 2 – ω²
(2 – ω)(2 – ω²) = 4 – 2ω – 2ω² + ω³
= 4 – 2(ω + ω²) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1 = 7
∴ (2 – ω)(2 – ω²)(2 – ω¹⁰)(2 – ω¹¹)
= (2 – ω) (2 – ω²) (2 – ω) (2 – ω²)
= ((2 – ω) (2 – ω²))²
= 7² = 49

Question 6.
If α, β are the roots of the equation x² – 2x + 4 = 0 then for any n ∈ N show that αⁿ + βⁿ = 2^{n + 1} cos \(\left(\frac{n \pi}{3}\right)\) (Mar. ’14)
Solution:

Question 7.
Show that one value of

is -1.
Solution:

Question 8.
If n is a positive integer, show that (1 + i)ⁿ + (1 – i)ⁿ = 2^{\(\frac{n+2}{2}\)} cos \(\left(\frac{\mathrm{n} \pi}{4}\right)\). (A.P) (Mar. ’15)
Solution:

Question 9.
If n is an integer then show that (1 + cos θ + i sin θ)ⁿ + (1 + cos θ – i sin θ)ⁿ = 2^{n + 1} cosⁿ (θ/2) cos \(\left(\frac{n \theta}{2}\right)\) (May. ’11) (TS & AP Mar. ’17)
Solution:
L.H.S.

Question 10.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ. Prove that cos² α + cos² β + cos² γ = \(\frac{3}{2}\) = sin² α + sin² β + ² γ. (AP. Mar. ’16; TS Mar. ’15, ‘13)
Solution:
Let x = cos α + i sin α
y = cos β + i sin β
z = cos γ + i sin γ
∴ x + y + z = (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i. 0 = 0
(x + y + z)² = 0
⇒ x² + y² + z² + 2(xy + yz + zx) = 0
x² + y² + z² = -2(xy + yz + zx)


-i(sin α + sin β + sin γ) = 0 – i. 0 = 0
Substituting in (1)
x² + y² + z² = 0.
(cos α + i sin α)² + (cos β + i sin β)² + (cos γ + i sinγ)² = 0
(cos 2α + i sin 2α) + (cos β + i sin 2β) + (cos 2γ + i sin 2γ) = 0
(cos 2α + cos 2β + cos 2γ) + i (sin 2α + sin 2β + sin 2γ) = 0
Equating real parts
cos 2α + cos 2β + cos 2γ = 0
2cos²α – 1 + 2 cos²β – 1 + 2 cos²γ – 1 = 0
2 (cos²α + cos²β + cos²γ) = 3
cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
∵ ²α + cos²β + cos²γ = \(\frac{3}{2}\)
⇒ (1 – sin²α) + (1 – sin²β) + (1 – sin²γ) = \(\frac{3}{2}\)
⇒ sin²α + sin²β + sin²γ = \(\frac{3}{2}\)
∴ cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
= sin²α + sin²β + sin²β

Question 11.
If 1, ω, ω² are the cube roots of unity prove that
i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶ = 128
= (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
ii) (a + b) (aω + bω²) (aω² + bω) = a³ + b³
iii) x² + 4x + 7 = 0 where x = ω – ω² – 2.
Solution:
∵ 1, ω, ω² are the cube roots of unity
⇒ 1 + ω + ω² = 0 and ω³ = 1

i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶
= (-ω – ω)⁶ + (-ω² – ω²)⁶
= (-2ω)⁶ + (-2ω²)⁶
= 2⁶(ω⁶ + ω¹²)
= 2⁶(1 + 1) = 2⁶ × 2 = 2⁷ = 128

Again (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
= (1 + ω² – ω)⁷ + (1 + ω – ω²)⁷
= ( -ω – ω)⁷ + (-ω² – ω²)⁷
= (-2ω)⁷ + (-2ω²)⁷
= (-2)⁷ (ω⁷ + ω¹⁴)
= (-2)⁷ (ω + ω²)
= -(2)⁷ (-1)
= 2⁷ = 128

ii) (a+b) (a]ω + bω²)(aω² + bω) (AP) (Mar. ’17)
Solution:
= (a + b) [a²ω³ + abω⁴)4 + abω² + b²ω³]
= (a + b) [a² + ab(ω² + ω⁴) + b²]
= (a + b) [a² + ab(ω² + ω) + b²]
= (a + b) (a² – ab + b²) = a³ + b³

iii) x = ω – ω² – 2
⇒ x + 2 = ω – ω²
⇒ (x + 2)² = ω² + ω⁴ – 2ω³
⇒ x² + 4x + 4 = ω² + ω – 2
⇒ x² + 4x + 4 = (-1) – 2 = -3
⇒ x² + 4x + 7 = 0

Question 12.
Simplify \(\frac{(\cos \alpha+i \sin \alpha)^{4}}{(\sin \beta+i \cos \beta)^{8}}\)
Solution:


= (cos α + i sin α)⁴ (cos β – i sin β)^-8 [i⁸ = (i²)⁴ = (-1)⁴ = 1]
= (cos 4α + i sin 4α) (cos 8β + i sin 8β)
= cos (4α + 8β) + i sin (4α + 8β)

Question 13.
If m, n are integers and x = cos α + i sin α, y = cos β + i sin β, then prove that
x^m yⁿ + \(\frac{1}{x^{m} y^{n}}\) = 2 cos (mα + nβ) and
x^m yⁿ – \(\frac{1}{x^{m} y^{n}}\) = 2i sin (mα + nβ).
Solution:
∵ x = cos α + i sin α, y = cos β + i sin β
⇒ x^m = (cos α + i sin α)^m = cos mα + i sin mα
yⁿ = (cos β + i sin β)ⁿ = cos nβ + i sin nβ
∴ x^m yⁿ = (cos mα + i sin mα)(cos nβ + i sin nβ)
= cos (mα + nβ) + i sin (mα + nβ) ——— (1)
\(\frac{1}{x^{m} \cdot y^{n}}\) = cos (mα + nβ) – i sin (mα + nβ) —— (2)
By adding (1) and (2).
x^myⁿ + \(\frac{1}{x^{m} y^{n}}\) = 2 cos (mα + nβ)
By subtracting (2) from (1)
x^myⁿ – \(\frac{1}{x^{m} y^{n}}\) = 2 sin (mα + nβ)

Question 14.
If n is a positive integer, show that (1 + i)ⁿ + (1 – i)ⁿ = 2^{\(\frac{n+2}{2}\)} cos \(\left(\frac{n \pi}{4}\right)\) (A.P.) (Mar. ‘15)
Solution:

Question 15.
If n is an integer then show that (1 + cos θ + i sin θ)ⁿ + (1 + cos θ – i sin θ)ⁿ = 2^{n + 1} + cosⁿ (θ/2) cos \(\left(\frac{\mathrm{n} \theta}{2}\right)\). (May ’11)
Solution:
L.H.S.

Question 16.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ, Prove that cos² α + cos² β + cos² γ = \(\frac{3}{2}\) = sin² α +
sin² β + sin² γ. (A.P. Mar. ‘16, T.S. Mar. ‘15, ’13)
Solution:
Let x = cos α + i sin α
y = cos β + i sin β
z = cos γ + i sin γ
∴ x + y + z = (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i.0 = 0
(x + y + z)² = 0
⇒ x² + y² + z² + 2(xy + yz + zx) = 0
x² + y² + z² = -2(xy + yz + zx)

Similarly \(\frac{1}{y}\) = cos β – i sin β
\(\frac{1}{z}\) = cos γ – i sin γ
∴ \(\frac{1}{x}\) + \(\frac{1}{y}\) + \(\frac{1}{z}\) = (cos α + cos β + cos γ) – i(sin α + sin β + sin γ) = 0 – i. 0 = 0
Substituting in (1)
x² + y² + z² = 0
(cos α + i sin β)² + (cos β + i sin β)² + (cos γ + i sin γ)² = o
(cos 2α + i sin 2α) + (cos 2β + i sin 2β) + (cos 2γ + i sin 2γ) = 0
(cos 2α + cos 2β + cos 2γ) + i (sin 2α + sin 2β + sin 2γ) = 0
Equating real parts
cos 2α + cos 2β + cos 2γ = 0
2 cos²α – 1 + 2 cos²β – 1 + 2 cos²γ – 1 = 0
2 (cos²α + cos²β + cos²γ) = 3
cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
∵ cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
⇒ (1 – sin²α) + (1 – sin²β) + (1 – sin²γ) = \(\frac{3}{2}\)
⇒ sin²α + sin²β + sin²γ = 3 – \(\frac{3}{2}\) = \(\frac{3}{2}\)
∴ cos² α + cos²β + cos²γ = \(\frac{3}{2}\)
= sin² α + sin²β + sin²β

Question 17.
Find all the values of (\(\sqrt{3}\) + i)^1/4.
Solution:

Question 18.
Find all the roots of the equation
x¹¹ – x⁷ + x⁴ – 1 = 0.
Solution:
x¹¹ – x⁷ + x⁴ – 1 = 0
⇒ x⁷(x⁴ – 1) + 1 (x⁴ – 1) = 0
⇒ (x⁴ – 1) (x⁷ + 1) = 0
Case(i) : x⁴ – 1 = 0
x⁴ = 1 = (cos o + i sin 0)
⇒ x⁴ = (cos 2kπ + i sin 2kπ)
∴ x = (cos 2kπ + i sin 2kπ)^1/4
⇒ x = cis \(\left(\frac{2 k \pi}{4}\right)\) = cis \(\frac{\mathrm{k} \pi}{2}\), k = 0, 1, 2, 3.

Case (ii): x⁷ + 1 = 0
⇒ x⁷ = -1 = cos π + i sin π
⇒ x⁷ = cos (2kπ + n) + i sin (2kπ + π)
∴ x = [cos (2k + 1)π + i sin (2k + 1)π]^1/7
⇒ x = cis(2k + 1)\(\frac{\pi}{7}\), k = 0, 1, 2, 3, 4, 5, 6
The values of x are

Question 19.
If 1, ω, ω² are the cube roots of unity prove that (TS. Mar. ‘16)
i) (1 – ω + (ω²)⁶ + (1 – ω² + ω)⁶ = 128
= (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
ii) (a + b)(aω + bω²)(aω² + bω) = a³ + b³
iii) x² + 4x + 7 = 0 where x = ω – ω² – 2.
Solution:
∵ 1, ω, ω² are the cube roots of unity
⇒ 1 + ω + ω² = 0 and ω³ = 1
i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶
= (-ω – ω)⁶ +(-ω² – ω²)⁶
= (-2ω)⁶ + (-2w²)⁶
= 2⁶ (ω⁶ + ω¹²)
= 2⁶(1 + 1) = 2⁶ × 2 = 2⁷ = 128 .
Again (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
= (1 + ω² – ω)⁷ + (1 + ω – ω²)⁷
= (- ω – ω) + (-ω² – ω²)⁷
= (-2ω)⁷ + (-2ω²)⁷
= (-2)⁷ (ω⁷ + ω¹⁴)
= (-2)⁷(ω + ω²)
= -(2)⁷ (-1)
= 2⁷ = 128

ii) (a + b) (aω + bω²) (aω² + bω)
= (a + b) [a²ω³ + abω⁴ + abω² + b²ω³]
= (a + b) [a² + ab(ω² + ω⁴) + b²]
= (a + b) [a² + ab(ω² + ω) + b²]
= (a + b) (a² – ab + b²) = a³ + b³

iii) x = ω – ω² – 2
= x + 2 = ω – ω²
(x + 2)² = ω² – 2ω³
⇒ x² +4x + 4 = ω² + ω – 2
⇒ x² + 4x + 4 = (-1) – 2 = -3.
⇒ x² + 4x + 7 = 0

Question 20.
If α, β are the roots of the equation x² + x + 1 = 0 then prove that α⁴ + β⁴ + α^-1β^-1 = 0
Solution:
Since α, β are the complex cube roots of unity.
We take α = ω, β = ω²
∴ α⁴ + β⁴ + α^-1 + β^-1
= ω⁴ + (ω²)⁴ + \(\frac{1}{\omega} \cdot \frac{1}{\omega^{2}}\)
= ω + ω² + \(\frac{1}{\omega^{3}}\)
= (-1) + \(\frac{1}{1}\)
= -1 + 1 = 0
∴ α⁴ + β⁴ + α^-1 + β^-1 = 0

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Complex Numbers Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Complex Numbers Important Questions

Question 1.
If z₁ = -1, z₂ = i then find Arg \(\left(\frac{z_{1}}{z_{2}}\right)\) (AP Mar. 17) (TS Mar.’ 16; May ‘11)
Solution:
Z₁ = -1 = cos π + i sin π
⇒ Arg z₁ = π
z₂ = i = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
⇒ Arg z₂ = \(\frac{\pi}{2}\)
⇒ Arg \(\left(\frac{z_{1}}{z_{2}}\right)\) = Arg z₁ – Arg z₂ = π – \(\frac{\pi}{2}\).
= \(\frac{\pi}{2}\)