Inter 1st Year

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(a)

I. Convert the following into the simplest form.

Question 1.
(i) tan(θ – 14π)
Solution:
= tan(14π – θ)
= tan(2 . (7π) – θ)
= tan θ

(ii) \(\cot \left(\frac{21 \pi}{2}-\theta\right)\)
Solution:
\(\cot \left(\frac{21 \pi}{2}-\theta\right)\)
= \(\cot \left(10 \pi+\left(\frac{\pi}{2}-\theta\right)\right)\)
= \(\cot \left(\frac{\pi}{2}-\theta\right)\)
= tan θ

(iii) cosec(5π + θ)
Solution:
cosec(5π + θ) = cosec(2π + (3π + θ))
= cosec(3π + θ)
= cosec(2π + (π + θ))
= cosec(π + θ)
= -cosec θ

(iv) sec(4π – θ?)
Solution:
sec(4π – θ)
= sec(2π + (2π – θ))
= sec(2π – θ)
= sec θ

Question 2.
Find the value of each of the following.
(i) sin(-405°)
Solution:
sin(-405°) = -sin(360° + 45)
= -sin 45°
= \(-\frac{1}{\sqrt{2}}\)

(ii) \(\cos \left(-\frac{7 \pi}{2}\right)\)
Solution:
\(\cos \left(-\frac{7 \pi}{2}\right)\)
= -cos 630°
= -cos(360° + 270°)
= -cos 270°
= -cos(180° + 90°)
= -cos 90°
= 0

(iii) sec(2100°)
Solution:
sec(2100°) = sec (5 × 360° + 300°)
= sec 300°
= sec(360 – 60°)
= sec 60°
= 2

(iv) cot(-315°)
Solution:
cot(-315°) = -cot 315°
= -cot(360° – 45°)
= -cot 45°
= 1

Question 3.
Evaluate.
(i) cos² 45° + cos² 135° + cos² 225° + cos² 315°
Solution:

(ii) \(\sin ^{2} \frac{2 \pi}{3}+\cos ^{2} \frac{5 \pi}{6}-\tan ^{2} \frac{3 \pi}{4}\)
Solution:

(iii) cos 225° – sin 225° + tan 495° – cot 495°
Solution:
cos (180° + 45°) – sin(180° + 45°) + tan(360° + 135°) – cot(360° + 135°)
= -cos 45° + sin 45° – tan 135° + cot 135°
= \(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+1-1\)
= 0

(iv) (cos θ – sin θ) if (a) θ = \(\frac{7 \pi}{4}\) (b) θ = \(\frac{11 \pi}{3}\)
Solution:

Question 4.
(i) If sin θ = \(\frac{-1}{3}\) and θ does not lie in the third quadrant, find the values of (a) cos θ (b) cot θ
Solution:
∵ sin θ = \(\frac{-1}{3}\) and sin θ is negative and θ does not lie in the IIIrd quadrant.
⇒ θ lies in IV quadrant.
∴ In the IV quadrant, cos θ is positive and cot θ is negative.

(ii) If cos θ = t (0 < t < 1) and θ does not lie in the first quadrant, find the values of (a) sin θ (b) tan θ.
Solution:
cos θ = t, (0 < t < 1)
⇒ cos θ is positive and θ does not lie in the first quadrant.
⇒ θ lies in IV quadrant.
(a) sin θ = \(-\sqrt{1-\cos ^{2} \theta}=-\sqrt{1-t^{2}}\)
(b) tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{-\sqrt{1-t^{2}}}{t}\)

(iii) Find the value of sin 330°. cos 120° + cos 210°. sin 300°.
Solution:
sin 330°. cos 120° + cos 210°. sin 300°
= sin(360° – 30°) . cos(180° – 60°) + cos(180° + 30°) . sin(360° – 60°)
= (-sin 30°) (-cos 60°) + (-cos 30°) (-sin 60°)
= sin 30° cos 60° + cos 30° sin 60°
= sin(30° + 60°) [∵ sin A cos B + cos A sin B = sin(A + B)]
= sin (90°)
= 1

(iv) If cosec θ + cot θ = \(\frac{1}{3}\), find cos θ and determine the quadrant in which θ lies.
Solution:
cosec θ + cot θ = \(\frac{1}{3}\)
⇒ cosec θ – cot θ = 3 (∵ cosec²θ – cot²θ = 1)
∴ 2 cosec θ = \(\frac{1}{3}\) + 3 = \(\frac{10}{3}\)
∴ cosec θ = \(\frac{5}{3}\) and sin θ = \(\frac{3}{5}\)
2 cot θ = \(\frac{1}{3}\) – 3 = \(\frac{-8}{3}\)
∴ cot θ = \(\frac{-4}{3}\), tan θ = \(\frac{-3}{4}\)
cos θ = (cot θ) . sin θ = \(\left(\frac{-4}{3}\right)\left(\frac{3}{5}\right)=\frac{-4}{5}\)
∵ sin θ is +ve and cos θ is -ve
⇒ θ lies in II quadrant.

Question 5.
(i) If sin α + cosec α = 2. Find the value of sinⁿα + cosecⁿα, n ∈ z.
Solution:
Given sin α + cosec α = 2
Squaring on both sides
sin²α + cosec²α + 2 = 4
sin²α + cosec²α = 2
sin α + cosec α = 2
Cubing on both sides
sin³α + cosec³α + 3 sin α . cosec α (sin α + cosec α) = 8
sin³α + cosec³α + 3(2) = 8
sin³α + cosec³α = 8 – 6
sin³α + cosec³α = 2
similarly sinⁿα + cosecⁿα = -2

(ii) If sec θ + tan θ = 5. Find the quadrant in which θ lies and find the value of sin θ.
Solution:
sec θ + tan θ = 5
⇒ sec θ – tan θ = \(\frac{1}{5}\) (∵ sec²θ – tan²θ = 1)
2 sec θ = 5 + \(\frac{1}{5}\) = \(\frac{26}{5}\)
sec θ = \(\frac{26}{10}=\frac{13}{5}\)
Aquir 2 tan θ = 5 – \(\frac{1}{5}\) = \(\frac{24}{5}\)
tan θ = \(\frac{24}{10}=\frac{12}{5}\)
Now sin θ = \(\frac{\tan \theta}{\sec \theta}=\frac{\frac{12}{5}}{\frac{13}{5}}=\frac{12}{13}\)
tan θ is +ve, sec θ is +ve
⇒ θ lies in the first quadrant.

II.

Question 1.
Prove that
(i) \(\frac{\cos (\pi-A) \cdot \cot \left(\frac{\pi}{2}+A\right) \cos (-A)}{\tan (\pi+A) \tan \left[\frac{3 \pi}{2}+A\right] \sin (2 \pi-A)}\) = cos A
Solution:
\(\frac{\cos (\pi-A) \cdot \cot \left(\frac{\pi}{2}+A\right) \cos (-A)}{\tan (\pi+A) \tan \left[\frac{3 \pi}{2}+A\right] \sin (2 \pi-A)}\)
= \(\frac{-\cos A(-\tan A) \cos A}{\tan A(-\cot A)(-\sin A)}\)
= cos A

(ii) \(\frac{\sin (3 \pi-A) \cos \left(A-\frac{\pi}{2}\right) \tan \left(\frac{3 \pi}{2}-A\right)}{{cosec}\left(\frac{13 \pi}{2}+A\right) \sec (3 \pi+A) \cot \left(A-\frac{\pi}{2}\right)}\) = cos⁴A
Solution:

(iii) sin 780° sin 480° + cos 240°. cos 300° = \(\frac{1}{2}\)
Solution:
sin(2(360°) + 60°) . sin(360° + 120°) + cos(270° – 30°) . cos(360° – 60°)
= sin 60° . sin 120° – sin 30° . cos 60°
= \(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{1}{2}\)
= \(\frac{3}{4}-\frac{1}{4}\)
= \(\frac{1}{2}\)

(iv) \(\frac{\sin 150^{\circ}-5 \cos 300^{\circ}+7 \tan 225^{\circ}}{\tan 135^{\circ}+3 \sin 210^{\circ}}\) = -2
Solution:

(v) cot(\(\frac{\pi}{20}\)) . cot(\(\frac{3\pi}{20}\)) . cot(\(\frac{5\pi}{20}\)) . cot(\(\frac{7\pi}{20}\)) . cot(\(\frac{9\pi}{20}\)) = 1
Solution:
L.H.S. = cot(\(\frac{\pi}{20}\)) . cot(\(\frac{3\pi}{20}\)) . cot(\(\frac{5\pi}{20}\)) . cot(\(\frac{7\pi}{20}\)) . cot(\(\frac{9\pi}{20}\))
= cot (9°) cot (27°) cot (45°) cot (63°) cot (81°)
= cot (9°) cot (27°) (1) cot (90° – 27°) cot (90° – 9°)
= cot (9°) cot (27°) tan 27° tan 9°
= (tan 9° cot 9°) (tan 27° cot 27°)
= (1) (1)
= 1

Question 2.
Simplify.
(i) \(\frac{\sin \left(-\frac{11 \pi}{3}\right) \tan \left(\frac{35 \pi}{6}\right) \sec \left(-\frac{7 \pi}{3}\right)}{\cot \left(\frac{5 \pi}{4}\right) {cosec}\left(\frac{7 \pi}{4}\right) \cos \left(\frac{17 \pi}{6}\right)}\)
Solution:

(ii) If tan 20° = p, prove that \(\frac{\tan 610^{\circ}+\tan 700^{\circ}}{\tan 560^{\circ}-\tan 470^{\circ}}=\frac{1-p^{2}}{1+p^{2}}\)
Solution:
Given tan 20° = p
L.H.S. = \(\frac{\tan 610^{\circ}+\tan 700^{\circ}}{\tan 560^{\circ}-\tan 470^{\circ}}\)

(iii) If α, β are complementary angles such that b sin α = a, then find the value of (sin α cos β – cos α sin β).
Solution:
∵ α, β are complementary angles
⇒ α + β = 90°
⇒ β = 90° – α
Now sin α cos β – cos α sin β
= sin (α – β)
= sin [(α – (90° – α)]
= sin [2α – 90°]
= -sin (90° – 2α)
= -cos 2α
= -(1 – 2 sin²α) (∵ cos 2α = 1 – 2 sin²α)
= \(-1+2\left(\frac{a}{b}\right)^{2}\) (∵ sin α = \(\frac{a}{b}\))
= \(\frac{-b^{2}+2 a^{2}}{b^{2}}\)
= \(\frac{2 a^{2}-b^{2}}{b^{2}}\)

Question 3.
(i) If cos A = cos B = \(-\frac{1}{2}\), A does not lie in the second quadrant and B does not lie in the third quadrant, then find the value of \(\frac{4 \sin B-3 \tan A}{\tan B+\sin A}\)
Solution:
∵ cos A = \(-\frac{1}{2}\) and A does not lie in second quadrant.
⇒ A lies in the third quadrant, (∵ cos A is -ve)
and cos B = \(-\frac{1}{2}\) and B does not lie in third quadrant.
⇒ B lies in second quadrant.
∵ cos A = \(-\frac{1}{2}\) and A lies in third quadrant.
⇒ A = 240°
∵ cos B = \(-\frac{1}{2}\) and B lies in second quadrant.
⇒ B = 120°

(ii) If 8 tan A = -15 and 25 sin B = -7 and neither A nor B is in the fourth quadrant, then show that sin A cos B + cos A sin B = \(\frac{-304}{425}\)
Solution:
8 tan A = -15 ⇒ tan A = \(\frac{-15}{8}\)
25 sin B = -7 ⇒ sin B = \(-\frac{7}{25}\)
Given neither A nor B is in the fourth quadrant.
Clearly, A is the second quadrant B is the third quadrant
sin A cos B + cos A sin B = \(\left(\frac{15}{17}\right)\left(\frac{-24}{25}\right)+\left(\frac{-8}{17}\right)\left(\frac{-7}{25}\right)\)
= \(\frac{-360}{425}+\frac{56}{425}\)
= \(\frac{-304}{425}\)

(iii) If A, B, C, D are angles of a cyclic quadrilateral, then prove that
(a) sin A – sin C = sin D – sin B
(b) cos A + cos B + cos C + cos D = 0
Solution:
∵ A, B, C, D are angles of a cyclic quadrilateral.
⇒ A + C = 180° and B + D = 180° ……..(1)
C = 180° – A and D = 180° – B
(a) L.H.S. = sin A – sin C
= sin A – sin(180° – A)
= sin A – sin A
= 0
R.H.S. = sin D – sin B
= sin(180°- B) – sin B
= sin B – sin B
= 0
∴ L.H.S. = R.H.S.
i.e., sin A – sin C = sin D – sin B
(b) L.H.S. = cos A + cos B + cos C + cos D
= cos A + cos B + cos(180° – A) + cos(180° – B)
= cos A + cos B – cos A – cos B
= 0
∴ cos A + cos B + cos C + cos D = 0

Question 4.
(i) If a cos θ – b sin θ = c, then show that a sin θ + b sin θ = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\).
Solution:
a cos θ – b sin θ = c
let a sin θ + b cos θ = x
by squaring and adding
(a cos θ – b sin θ)² + (a sin θ + b cos θ)² = c² + x²
a² cos²θ + b² sin²θ – 2ab sin θ cos θ + a² sin²θ + b² cos²θ + 2ab sin θ cos θ = c² + x²
a² + b² = c² + x²
a² + b² – c² = x²
x = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\)
∴ a sin θ + b cos θ = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\)

(ii) If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ±3.
Solution:
3 sin A + 5 cos A = 5
Let 5 sec A – 3 cos A = x
by squaring and adding
(3 sin A + 5 cos A)² + (5 sin A – 3 cos A)² = 5² + x²
9 sin² A + 25 cos² A + 30 sin A cos A + 25 sin² A + 9 cos² A – 30 sin A cos A = 25
9 + 25 = 25 + x²
x² = 9
x = ±3
∴ 5 sin A – 3 cos A = ±3

(iii) If tan²θ = (1 – e²), show that sec θ + tan³θ . cosec θ = (2 – e²)^3/2.
Solution:
tan²θ = 1 – e²
sec²θ = 1 + tan²θ = 2 – e²

III. Prove the following.

Question 1.
(i) \(\frac{(\tan \theta+\sec \theta-1)}{(\tan \theta-\sec \theta+1)}=\frac{1+\sin \theta}{\cos \theta}\)
Solution:

(ii) (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2.
Solution:

(iii) 3(sin θ – cos θ)⁴ + 6(sin θ + cos θ)² + 4(sin⁶θ + cos⁶θ) = 13.
Solution:
(sin θ – cos θ)² = sin²θ + cos²θ – 2 sin θ . cos θ
= 1 – 2 sin θ cos θ
(sin θ – cos θ)⁴ = (1 – 2 sin θ cos θ)²
= 1 + 4 sin²θ cos²θ – 4 sin θ cos θ ……(1)
(sin θ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ
= 1 + 2 sin θ cos θ …….(2)
sin⁶θ + cos⁶θ = (sin²θ +cos²θ)³ – 3 sin²θ cos²θ (sin²θ + cos²θ)
= 1 – 3 sin²θ cos²θ ……..(3)
L.H.S. = 3(1 + 4 sin²θ cos²θ – 4 sin θ cos θ) + 6(1 + 2 sin θ cos θ) + 4(1 – 3 sin²θ cos²θ)
= 3 + 12 sin²θ cos²θ – 12 sin θ cos θ + 6 + 12 sin θ cos θ + 4 – 12 sin²θ cos²θ
= 3 + 6 + 4
= 13
= R.H.S.

Question 2.
(i) Prove that (sin θ + cosec θ)² + (cos θ + sec θ)² – (tan²θ + cot²θ) = 7.
Solution:
L.H.S. = (sin θ + cosec θ)² + (cos θ + sec θ)² – (tan²θ + cot²θ)
= (sin²θ + cosec²θ + 2 sin θ cosec θ) + (cos²θ + sec²θ + 2 cos θ sec θ) – (tan²θ + cot²θ)
= (sin²θ + cos²θ) + (1 + cot²θ) + (1 + tan²θ) + 4 – tan²θ – cot²θ
= 1 + 1 + 1 + 4
= 7

(ii) cos⁴α + 2 cos²α \(\left(1-\frac{1}{\sec ^{2} \alpha}\right)\) = (1 – sin⁴α)
Solution:
L.H.S. = cos⁴α + 2 cos²α \(\left(1-\frac{1}{\sec ^{2} \alpha}\right)\)
= cos⁴α + 2 cos²α (1 – cos²α)
= cos²α [cos²α + 2 sin²α]
= (1 – sin²α) [cos²α + sin²α + sin²α]
= (1 – sin²α) (1 + sin²α)
= 1 – sin⁴α

(iii) \(\frac{(1+\sin \theta-\cos \theta)^{2}}{(1+\sin \theta+\cos \theta)^{2}}=\frac{1-\cos \theta}{1+\cos \theta}\)
Solution:

(iv) If \(\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}\) = x, then find the value of \(\frac{(1-\cos \theta+\sin \theta)}{(1+\sin \theta)}\)
Solution:

Question 3.
Eliminate θ from the following.
(i) x = a cos³θ; y = b sin³θ
Solution:

(ii) x = a cos⁴θ; y = b sin⁴θ
Solution:

(iii) x = a(sec θ + tan θ); y = b(sec θ – tan θ)
Solution:
\(\frac{x}{a}\) = sec θ + tan θ; \(\frac{y}{b}\) = (sec θ – tan θ)
\(\frac{x}{a} \times \frac{y}{b}\) = (sec θ + tan θ) (sec θ – tan θ)
\(\frac{x y}{a b}\) = sec²θ – tan²θ
\(\frac{x y}{a b}\) = 1
xy = ab

(iv) x = cot θ + tan θ; y = sec θ – cos θ
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Products of Vectors Solutions Exercise 5(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Products of Vectors Solutions Exercise 5(c)

I.

Question 1.
Compute \([\overline{\mathbf{i}}-\overline{\mathbf{j}} \overline{\mathbf{j}}-\overline{\mathbf{k}} \overline{\mathbf{k}}-\overline{\mathbf{i}}]\)
Solution:

Question 2.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-2 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\), \(\bar{c}=\bar{i}+3 \bar{j}-2 \bar{k}\), then compute \(\overline{\mathbf{a}} \cdot(\overline{\mathbf{b}} \times \overline{\mathbf{c}})\).
Solution:

Question 3.
If \(\bar{a}\) = (1, -1, -6), \(\bar{b}\) = (1, -3, 4) and \(\bar{c}\) = (2, -5, 3), then compute the following
(i) \(\overline{\mathbf{a}} \cdot(\bar{b} \times \bar{c})\)
(ii) \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}})\)
(iii) \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}\)
Solution:

Question 4.
Simplify the following.
(i) \((\bar{i}-2 \bar{j}+3 \bar{k}) \times(2 i+j-\bar{k}) \cdot(\bar{j}+\bar{k})\)
(ii) \((2 \bar{i}-3 \bar{j}+\bar{k}) \cdot(\bar{i}-\bar{j}+2 \bar{k}) \cdot(2 \bar{i}+\bar{j}+\bar{k})\)
Solution:

Question 5.
Find the volume of the parallelopiped having coterminous edges.
\(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{i}}-\overline{\mathbf{j}}\) and \(\overline{\mathbf{i}}+\mathbf{2} \overline{\mathbf{j}}-\overline{\mathbf{k}}\)
Solution:
Let \(\overline{\mathrm{a}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{b}}=\overline{\mathrm{i}}-\overline{\mathrm{j}}\) and \(\overline{\mathrm{c}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}-\overline{\mathrm{k}}\)
Volume of the parallelopiped = \([(\bar{a} \bar{b} \bar{c})]\)
= \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 0 \\
1 & 2 & -1
\end{array}\right|\)
= 1(1 – 0) – 1(-1 – 0) + 1(2 + 1)
= 1 + 1 + 3
= 5 cubic units.

Question 6.
Find t for which the vectors \(\mathbf{2} \overline{\mathbf{i}}-\mathbf{3} \overline{\mathbf{j}}+\overline{\mathbf{k}}\), \(\bar{i}+2 \mathbf{j}-3 \bar{k}\) and \(\bar{j}-t \bar{k}\) are coplanar.
Solution:

Question 7.
For non-coplanar vectors, \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) determine p for which the vector \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}, \overline{\mathbf{a}}+\mathbf{p} \overline{\mathbf{b}}+\mathbf{2} \overline{\mathbf{c}}\) and \(-\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\) are coplanar.
Solution:

Question 8.
Determine λ, for which the volume of the parallelopiped having coterminous edges \(\bar{i}+\bar{j}\), \(3 \overline{\mathbf{i}}-\overline{\mathbf{j}}\) and \(3 \bar{j}+\lambda \bar{k}\) is 16 cubic units.
Solution:

Question 9.
Find the volume of the tetrahedron having the edges \(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \quad \mathbf{i}-\overline{\mathbf{j}}\) and \(\bar{i}+2 \bar{j}+\bar{k}\).
Solution:

Question 10.
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be non-coplanar vectors and \(\bar{\alpha}=\bar{a}+2 \bar{b}+3 c, \quad \bar{\beta}=2 \bar{a}+\bar{b}-2 c\) and \(\bar{\gamma}=3 \bar{a}-7 \bar{c}\), then find \(\left[\begin{array}{lll}
\bar{\alpha} & \bar{\beta} & \bar{\gamma}
\end{array}\right]\).
Solution:

Question 11.
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be non-coplanar vectors. If \(|2 \bar{a}-\bar{b}+3 \bar{c}|, \bar{a}+\bar{b}-2 \bar{c},|\bar{a}+\bar{b}-3 \bar{c}|\) = \(\lambda[\overline{\mathbf{a}} \overline{\mathbf{b}} \overline{\mathbf{c}}]\), then find the value of λ.
Solution:

Question 12.
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be non-coplanar vectors, if \(\left[\begin{array}{lll}
\bar{a}+2 \bar{b} & 2 \bar{b}+\bar{c} & 5 \bar{c}+\bar{a}
\end{array}\right]\) = \(\lambda\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\), then find λ.
Solution:

Question 13.
If \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) are non-coplanar vectors, then find the value of \(\frac{(\bar{a}+2 \bar{b}-\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]}{[\bar{a} \bar{b} \bar{c}]}\)
Solution:

Question 14.
If \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) are mutually perpendicular unit vectors, then find the value of \(\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]^{2}\).
Solution:

Question 15.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non-zero vectors and \(\overline{\mathbf{a}}\) is perpendicular to both \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\). If \(|\overline{\mathbf{a}}|\) = 2, \(|\overline{\mathbf{b}}|\) = 3, \(|\overline{\mathbf{c}}|\) = 4 and \((\bar{b}, \bar{c})=\frac{2 \pi}{3}\), then find \(|[\bar{a} \bar{b} \bar{c}]|\).
Solution:

Question 16.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are unit coplanar vectors, then find \(\left[\begin{array}{lll}
2 \bar{a}-\bar{b} & 2 \bar{b}-\bar{c} & 2 \bar{c}-\bar{a}
\end{array}\right]\)
Solution:

II.

Question 1.
If \(\left[\begin{array}{lll}
\bar{b} & \bar{c} & \bar{d}
\end{array}\right]+\left[\begin{array}{lll}
\bar{c} & \bar{a} & \bar{d}
\end{array}\right]+\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{d}
\end{array}\right]\) = \(\left[\begin{array}{lll}
\overline{\mathbf{a}} & \overline{\mathbf{b}} & \overline{\mathbf{c}}
\end{array}\right]\) then show that the points with position vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\bar{d}\) are coplanar.
Solution:

Question 2.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) non-coplanar vectors, then prove that the four points with position vectors \(2 \bar{a}+3 \bar{b}-\bar{c}\), \(\overline{\mathrm{a}}-2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}, 3 \overline{\mathrm{a}}+4 \overline{\mathrm{b}}-2 \overline{\mathrm{c}}\) and \(\bar{a}-6 \bar{b}+6 \bar{c}\) are coplanar.
Solution:
Suppose A, B, C, D are the given points.

The vectors \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}\) are coplanar.
The given points A, B, C, D are coplanar.

Question 3.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) are non-zero and non- collinear vectors and θ ≠ 0, is the angle between \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\). If \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}\) = \(\frac{1}{3}|\bar{b}||\bar{c}|\bar{a}|\), then find sin θ.
Solution:

Question 4.
Find the volume of the tetrahedron whose vertices are (1, 2, 1), (3, 2, 5), (2, -1, 0) and (-1, 0, 1).
Solution:
Let ‘O’ be the given A, B, C, D be the vertices of the ten tetrahedrons. Then

Question 5.
Show that \((\bar{a}+\bar{b}) \cdot(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})\) = \(2\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\)
Solution:

Question 6.
Show that equation of the plane passing through the points with position vectors. \(3 \bar{i}-5 \bar{j}-\overline{\mathbf{k}},-\overline{\mathbf{i}}+5 \bar{j}+7 \overline{\mathbf{k}}\) and parallel to the vector \(3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+7 \overline{\mathbf{k}}\) is 3x + 2y – z = 0.
Solution:
The given plane passes through the points A, B (i.e.,) \(3 \bar{i}-5 \bar{j}-\overline{\mathbf{k}},-\overline{\mathbf{i}}+5 \bar{j}+7 \overline{\mathbf{k}}\) and parallel to the vector \(3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+7 \overline{\mathbf{k}}\)

= x(70 + 8) – y(-28 – 24) + z(4 – 30)
= 78x + 52y – 26z
= 26(3x + 2y – z)
\(\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\) = \(\left|\begin{array}{rrr}
3 & -5 & -1 \\
-4 & 10 & 8 \\
3 & -1 & 7
\end{array}\right|\)
= 3(70 + 8) + 5(-28 – 24) – 1(4 – 30)
= 234 – 260 + 26
= 0
Equation of the required plane is 26(3x + 2y – z) = 0
i.e., 3x + 2y – z = 0

Question 7.
Prove that \(\overline{\mathbf{a}} \times[\overline{\mathbf{a}} \times(\overline{\mathbf{a}} \times \overline{\mathbf{b}})]\) = \((\overline{\mathbf{a}} \cdot \overline{\mathbf{a}})(\overline{\mathbf{b}} \times \overline{\mathbf{a}})\)
Solution:

Question 8.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\bar{d}\) are coplanar vectors, then show that \((\bar{a} \times \bar{b}) \times(\bar{c} \times \bar{d})=0\).
Solution:
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\bar{d}\) are coplanar
⇒ \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}\) is perpendicular to the plane π.
similarly \(\bar{c} \times \bar{d}\) is perpendicular to the plane π.
\(\bar{a} \times \bar{b}\) and \(\bar{c} \times \bar{d}\) are parallel vectors.
⇒ \((\bar{a} \times \bar{b}) \times(\bar{c} \times \bar{d})\) = 0.

Question 9.
Show that \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{d}}=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}](\overline{\mathrm{a}} \cdot \overline{\mathrm{d}})\).
Solution:

Question 10.
Show that \(\bar{a} \cdot[(\bar{b}+\bar{c}) \times(\bar{a}+\bar{b}+\bar{c})]=0\)
Solution:

Question 11.
Find λ in order that the four points A(3, 2, 1), B(4, λ, 5), C(4, 2, -2) and D(6, 5, -1) be coplanar.
Solution:

Question 12.
Find the vector equation of the plane passing through the intersection of planes \(\bar{r} \cdot(2 \bar{i}+2 \bar{j}-3 \bar{k})=7, \bar{r} \cdot(2 \bar{i}+5 \bar{j}+3 \bar{k})=9\) and through the point (2, 1, 3)
Solution:

Question 13.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \(\bar{r} \cdot(\bar{i}+\bar{i}+\bar{k})=2\).
Solution:
Given equation plane is \(\bar{r} \cdot(\bar{i}+\bar{i}+\bar{k})=2\)
Let \(\bar{r}=x \bar{i}+y \bar{j}+z \bar{k}\)
∴ \(\bar{r} \cdot(\bar{i}+\bar{i}+\bar{k})=2\)
\((x \bar{i}+y \bar{j}+z \bar{k}) \cdot(i+j+k)=2\)
x + y + z = 2
Required plane equation is x + y + z = k …….(1)
Equation (1) passes through (a, b, c)
∴ a + b + c = k
Substitute ‘k’ in equation (1)
∴ x + y + z = a + b + c

Question 14.
Find the shortest distance between the lines \(\bar{r}=6 \bar{i}+2 \bar{j}+2 \bar{k}+\lambda, \bar{i}-2 \bar{j}+2 \bar{k}\) and \(\bar{r}=-4 \bar{j}-\bar{k}+\mu=3 \bar{j}-2 \bar{j}-2 \bar{k}\).
Solution:
The first line passes through point A(6, 2, 2) and is parallel to the vector b = i – 2j + 2k.
Second line passes through the point C(-4, 0, -1) and is parallel to the vector d = 3i – 2j – 2k

Question 15.
Find the equation of the plane passing through the line of intersection of the planes \(\bar{r} \cdot(\bar{i}+\bar{j}+\bar{k})=1\) and \(\bar{r} \cdot(2 \bar{i}+3 \bar{i}-\bar{k})+4=0\) and parallel to X-axis.
Solution:
Given the equation of planes are

Since it is parallel to X-axis.

Question 16.
Prove that the four points \(4 \bar{i}+5 \bar{j}+\bar{k}\), \(-(\overline{\mathbf{j}}+\overline{\mathbf{k}}), 3 \overline{\mathbf{i}}+9 \overline{\mathbf{j}}+4 \overline{\mathbf{k}}\) and \(-4 \bar{i}+4 \bar{j}+4 \bar{k}\) are coplanar.
Solution:
Let ‘O’ be the origin. A, B, C, D be the given points. Then

Question 17.
If \(\bar{a}, \bar{b}, \bar{c}\) are non – copianar, then show that the vectors \(\overline{\mathbf{a}}-\overline{\mathbf{b}}, \overline{\mathbf{b}}+\overline{\mathbf{c}}\), \(\overline{\mathbf{c}}+\overline{\mathbf{a}}\) are coplanar.
Solution:

Question 18.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are the position vectors of the points A, B and C respectively, then prove that the vector \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}+\overline{\mathbf{b}} \times \overline{\mathbf{c}}+\overline{\mathbf{c}} \times \overline{\mathbf{a}}\) is perpendicular to the plane of ∆ABC.
Solution:

III.

Question 1.
Show that \((\bar{a} \times(\bar{b} \times \bar{c}) \times \bar{c})=(\bar{a} \cdot \bar{c})(\bar{b} \times \bar{c})\) and \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \cdot(\overline{\mathbf{a}} \times \overline{\mathbf{c}})+(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}})(\overline{\mathbf{a}} \cdot \overline{\mathbf{c}})\) = \((\overline{\mathbf{a}} \cdot \overline{\mathbf{a}})(\overline{\mathbf{b}} \cdot \overline{\mathbf{c}})\)
Solution:

Question 2.
If A = (1, -2, -1), B = (4, 0, -3), C = (1, 2, -1) and D = (2, -4, -5), find the distance between AB and CD.
Solution:

Question 3.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-2 \overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}-\overline{\mathbf{k}}\), find \(\bar{a} \times(\bar{b} \times \bar{c})\) and \(|(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}|\).
Solution:

Question 4.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-2 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(\bar{c}=\bar{i}+3 \bar{j}-2 \bar{k}\), verift that \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}}) \neq(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}\)
Solution:

From (1) and (2), we get
\(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}}) \neq(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}\)
i.e., vector multiplication is not associative.

Question 5.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}-\mathbf{2} \mathbf{j}+\overline{\mathbf{k}}\), \(\bar{c}=-\bar{i}+\bar{j}-4 \bar{k}\) and \(\overline{\mathbf{d}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then compute \(|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{c}} \times \overline{\mathrm{d}})|\)
Solution:

Question 6.
If A = (1, a, a²), B = (1, b, b²) and C = (1, c, c²) are non-coplanar vectors and \(\left|\begin{array}{lll}
a & a^{2} & 1+a^{3} \\
b & b^{2} & 1+b^{3} \\
c & c^{2} & 1+c^{3}
\end{array}\right|\) = 0, then show that abc + 1 = 0
Solution:
\(\bar{A}, \bar{B}, \bar{C}\) are non-coplanar vectors.
∆ = \(\left|\begin{array}{lll}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{array}\right|\) ≠ 0 ………..(1)

∆ + (abc) ∆ = 0; ∆(1 + abc) = 0
∆ ≠ 0 ⇒ 1 + abc = 0

Question 7.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non-zero vectors, then \(|(\overline{\mathbf{a}} \times \mathbf{b} \cdot \overline{\mathbf{c}})|=|\overline{\mathbf{a}}||\mathbf{b}||\overline{\mathbf{c}}|\) \(\Leftrightarrow \overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\overline{\mathbf{b}} \cdot \overline{\mathbf{c}}=\overline{\mathbf{c}} \cdot \overline{\mathbf{a}}=\mathbf{0}\)
Solution:

Question 8.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-\mathbf{2} \overline{\mathbf{j}}+3 \overline{\mathbf{k}}, \quad \mathbf{b}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\), \(\bar{c}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) then find \(|(\bar{a} \times \bar{b}) \times \bar{c}|\) and \(|\overline{\mathbf{a}} \times(\mathbf{b} \times \mathbf{c})|\).
Solution:

Question 9.
If \(|\bar{a}|=1,|\bar{b}|=1,|\bar{c}|=2\) and \(\bar{a} \times(a \times \bar{c})+\bar{b}=0\) then find the angle between \(\bar{a}\) and \(\bar{c}\).
Solution:

Question 10.
Let \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-\overline{\mathbf{k}}, \quad \overline{\mathbf{b}}=\mathbf{x} \overline{\mathbf{i}}+\overline{\mathbf{j}}+(\mathbf{1}-\mathbf{x}) \overline{\mathbf{k}}\) and \(\bar{c}=y \bar{i}+x \bar{j}+(1+x-y) \bar{k}\), prove that the scalar triple product \(\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\) is independent of both x and y.
Solution:

Question 11.
Let \(\overline{\mathbf{b}}=\mathbf{2} \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}, \overline{\mathbf{c}}=\overline{\mathbf{i}}+\mathbf{3} \overline{\mathbf{k}}\). If \(\overline{\mathrm{a}}\) is a unit vector then find the maximum value of \(\left[\begin{array}{lll}
\overline{\mathbf{a}} & \overline{\mathbf{b}} & \bar{c}
\end{array}\right]\).
Solution:
Let \(\bar{a}=x \bar{i}+y \bar{j}+z \bar{k}\) and x² + y² + z² = 1
∵ \(\overline{\mathrm{a}}\) unit vector

Question 12.
Let \(\overline{\mathrm{a}}=\overline{\mathrm{i}}-\overline{\mathrm{i}}, \overline{\mathrm{b}}=\overline{\mathrm{i}}-\overline{\mathrm{k}}, \overline{\mathrm{c}}=\overline{\mathrm{k}}-\overline{\mathrm{i}}\) Find unit vector \(\bar{d}\) such that \(\overline{\mathrm{a}} \cdot \overline{\mathrm{d}}=0=[\bar{b} \overline{\mathrm{c}} \overline{\mathrm{d}}]\)
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Products of Vectors Solutions Exercise 5(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Products of Vectors Solutions Exercise 5(b)

I.

Question 1.
If \(|\overline{\mathbf{p}}|=2,|\overline{\mathbf{q}}|=3\) and \((\bar{p}, \bar{q})=\frac{\pi}{6}\), then find \(|\bar{p} \times \bar{q}|^{2}\).
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\), then find \(|\overline{\mathbf{a}} \times \overline{\mathbf{b}}|\).
Solution:

Question 3.
If \(\bar{a}=2 \bar{i}-3 \bar{j}+\overline{\mathbf{k}}\) and \(\bar{b}=\bar{i}+4 \bar{j}-2 \bar{k}\), then find \((\overline{\mathbf{a}}+\overline{\mathbf{b}}) \times(\overline{\mathbf{a}}-\overline{\mathbf{b}})\).
Solution:

Question 4.
If \(4 \bar{i}+\frac{2 p}{3} \bar{j}+p \bar{k}\) is parallel to the vector \(\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\), find p.
Solution:

Question 5.
Compute \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}}+\overline{\mathbf{c}})+\overline{\mathbf{b}} \times(\overline{\mathbf{c}}+\overline{\mathbf{a}})+\overline{\mathbf{c}} \times(\overline{\mathbf{a}}+\overline{\mathbf{b}})\)
Solution:

Question 6.
If \(\overline{\mathbf{p}}=\mathbf{x} \overline{\mathbf{i}}+\mathbf{y} \overline{\mathbf{j}}+\mathbf{z} \overline{\mathbf{k}}\), find the value of \(|\overline{\boldsymbol{p}} \times \overline{\mathbf{k}}|^{2}\)
Solution:

Question 7.
Compute \(2 \bar{j} \times(3 \bar{i}-4 \bar{k})+(\bar{i}+2 \hat{j}) \times \bar{k}\).
Solution:

Question 8.
Find a unit vector perpendicular to both \(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(2 \bar{i}+\bar{j}+3 \bar{k}\).
Solution:

Question 9.
If θ is the angle between the vectors \(\overline{\mathbf{i}}+\overline{\mathbf{j}}\) and \(\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then find sin ?.
Solution:

Question 10.
Find the area of the parallelogram having \(\bar{a}=2 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=-\overline{\mathbf{i}}+\overline{\mathbf{k}}\) as adjacent sides.
Solution:
Vector area of the parallelogram having \(\bar{a}=2 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=-\overline{\mathbf{i}}+\overline{\mathbf{k}}\) as adjacent sides.

Question 11.
Find the area of the parallelogram, whose diagonals are \(3 \overline{\mathbf{i}}+\overline{\mathbf{j}}-2 \overline{\mathbf{k}}\) and \(\bar{i}-3 \bar{j}+4 \bar{k}\).
Solution:

Question 12.
Find the area of the triangle having \(3 \bar{i}+4 \bar{j}\) and \(-5 \bar{i}+7 \bar{j}\) as two of its sides.
Solution:

Question 13.
Find unit vector perpendicular to the plane determined by the vectors \(\bar{a}=4 \bar{i}+3 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=2 \tilde{i}-6 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}\).
Solution:

Question 14.
Find the area of the triangle whose vertices are A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2).
Solution:
Suppose \(\bar{i}, \bar{j}, \bar{k}\) are unit vectors along the co-ordinate axes.
Position vectors of A, B, C are

II.

Question 1.
If \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}=\overline{\mathbf{0}}\), then prove that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{b}} \times \overline{\mathbf{c}}=\overline{\mathbf{c}} \times \overline{\mathbf{a}}\).
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \bar{i}+\bar{j}-\bar{k}, \quad \bar{b}=-\bar{i}+2 \bar{j}-4 \bar{k}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}+\mathbf{j}+\overline{\mathbf{k}}\), then find \((\bar{a} \times \bar{b}) \cdot(\bar{b} \times \bar{c})\).
Solution:

Question 3.
Find the vector area and the area of the parallelogram having \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\mathbf{2} \overline{\mathbf{k}}\) as adjacent sides.
Solution:

Question 4.
If \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{b}} \times \overline{\mathbf{c}} \neq \overline{\mathbf{0}}\), show that, \(\overline{\mathbf{a}}+\overline{\mathbf{c}}=\mathbf{p} \overline{\mathbf{b}}\), where p is some scalar.
Solution:

Question 5.
Let \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) be vectors, satisfying \(|\overline{\mathbf{a}}|=|\overline{\mathbf{b}}|=5\) and \((\bar{a}, \bar{b})=45^{\circ}\). Find the area of the triangle having \(\overline{\mathbf{a}}-\mathbf{2} \overline{\mathbf{b}}\) and \(3 \bar{a}+2 \bar{b}\) as two of its sides.
Solution:

Question 6.
Find the vector having magnitude ?6 units and perpendicular to both \(\mathbf{2} \overline{\mathbf{i}}-\overline{\mathbf{k}}\) and \(\mathbf{3} \overline{\mathbf{i}}-\overline{\mathbf{j}}-\overline{\mathbf{k}}\).
Solution:

Question 7.
Find a unit vector perpendicular to the plane determined by the points P(1, -1, 2), Q(2, 0, -1) and R(0, 2, 1).
Solution:

Question 8.
If \(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}\) and \(\overline{\mathbf{a}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{c}}, \overline{\mathbf{a}} \neq 0\), then show that \(\overline{\mathbf{b}}=\overline{\mathbf{c}}\).
Solution:

Question 9.
Find a vector of magnitude 3 and perpendicular to both the vector \(\overline{\mathbf{b}}=2 \bar{i}-2 \bar{j}+\bar{k}\) and \(\bar{c}=2 \bar{i}+2 \bar{j}+3 \bar{k}\).
Solution:

Question 10.
If \(|\overline{\mathbf{a}}|\) = 13, \(|\overline{\mathbf{b}}|\) = 5 and \(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\mathbf{6 0}\), then find \(|\overline{\mathbf{a}} \times \overline{\mathbf{b}}|\).
Solution:

Question 11.
Find a unit vector perpendicular to the plane passing through the points (1, 2, 3), (2, -1, 1) and (1, 2, -4).
Solution:
Let ‘O’ be the origin and let A, B, C be the given points.

III.

Question 1.
If \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) represent the vertices A, B and C respectively of ∆ABC, then prove that \(|(\overline{\mathbf{a}} \times \overline{\mathbf{b}})+(\overline{\mathbf{b}} \times \overline{\mathbf{c}})+(\overline{\mathbf{c}} \times \overline{\mathbf{a}})|\) is twice the area of ∆ABC.
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+4 \overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then compute \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}})\) and verify that it is perpendicular to \(\bar{a}\).
Solution:

Question 3.
If \(\overline{\mathbf{a}}=7 \overline{\mathbf{i}}-2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+8 \overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) then compute \(\overline{\mathbf{a}} \times \mathbf{b}, \overline{\mathbf{a}} \times \overline{\mathbf{c}}\) and \(\bar{a} \times(\bar{b}+\bar{c})\). Verify whether cross product is distributive over vector addition.
Solution:

Question 4.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{c}}=\overline{\mathbf{j}}-\overline{\mathbf{k}}\), then find vector \(\overline{\mathbf{b}}\) such that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{c}}\) and \(\bar{a} \cdot \bar{b}=3\).
Solution:

Question 5.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are three vectors of equal magnitudes and each of them is inclined at an angle of 60° to the others. If \(|\bar{a}+\bar{b}+\bar{c}|=\sqrt{6}\), then find \(|\overline{\mathbf{a}}|\).
Solution:

Question 6.
For any two vectors \(\bar{a}\) and \(\bar{b}\), show that \(\left(1+|\bar{a}|^{2}\right)\left(1+|\bar{b}|^{2}\right)\) = \(|\mathbf{1}-\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}|^{2}+|\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{a}} \times \overline{\mathbf{b}}|^{2}\)
Solution:

Question 7.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are unit vectors such that \(\overline{\mathbf{a}}\) is perpendicular to the plane of \(\overline{\mathbf{b}}, \overline{\mathbf{c}}\) and the angle between \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) is \(\frac{\pi}{3}\), then find \(|\bar{a}+\bar{b}+\bar{c}|\).
Solution:
Given that \(|\bar{a}|=|\bar{b}|=|\bar{c}|=1\)

Question 8.
\(\overline{\mathbf{a}}=3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}, \overline{\mathbf{b}}=-\overline{\mathbf{i}}+3 \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\), \(\bar{c}=4 \bar{i}+5 \bar{j}-2 \bar{k}\) and \(\bar{d}=\bar{i}+3 \bar{j}+5 \bar{k}\) then compute the following.
(i) \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times(\bar{c} \times \bar{d})\) and
(ii) \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \cdot \overline{\mathbf{c}}-(\overline{\mathbf{a}} \times \overline{\mathbf{d}}) \cdot \overline{\mathbf{b}}\)
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Products of Vectors Solutions Exercise 5(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Products of Vectors Solutions Exercise 5(a)

I.

Question 1.
Find the angle between the vectors \(\bar{i}+2 \bar{j}+3 \bar{k}\) and \(3 \bar{i}-\bar{j}+2 \bar{k}\).
Solution:
Let \(\overline{\mathrm{a}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+3 \overline{\mathrm{k}}\) and \(\overline{\mathrm{b}}=3 \overline{\mathrm{i}}-\overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) and ‘θ’ be the angle between them (i.e.,) \((\bar{a}, \bar{b})\) = θ

Question 2.
If the vectors \(\mathbf{2} \overline{\mathbf{i}}+\lambda \overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(4 \bar{i}-2 \bar{j}+2 \bar{k}\) are perpendicular to each other, then find λ.
Solution:

Question 3.
For what values of λ, the vectors \(\overline{\mathbf{i}}-\lambda \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) and \(8 \overline{\mathbf{i}}+6 \overline{\mathbf{j}}-\overline{\mathbf{k}}\) are at right angles?
Solution:

Question 4.
\(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\). Find the vector C such that \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) form the sides of a triangle.
Solution:

Question 5.
Find the angle between the planes \(\bar{r} \cdot(2 \bar{i}-\bar{j}+2 \bar{k})=3\) and \(\overline{\mathrm{r}} \cdot(3 \overline{\mathrm{i}}+6 \bar{j}+\bar{k})=4\).
Solution:

Question 6.
Let \(\overline{\mathbf{e}}_{1}\) and \(\overline{\mathbf{e}}_{2}\) be unit vectors makingangle θ. If \(\frac{1}{2}\left|\bar{e}_{1}-\bar{e}_{2}\right|=\sin \lambda \theta\), then find λ.
Solution:

Question 7.
Let \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=\mathbf{2} \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+\overline{\mathbf{k}}\). Find
(i) The projection vector of \(\overline{\mathbf{b}}\) on \(\overline{\mathbf{a}}\) and its magnitude.
(ii) The vector components of \(\overline{\mathbf{b}}\) in the direction of a and perpendicular to \(\overline{\mathbf{a}}\).
Solution:

Question 8.
Find the equation of the plane through the point (3, -2, 1) and perpendicular to the vector (4, 7, -4).
Solution:

Question 9.
If \(\overline{\mathbf{a}}=2 \bar{i}+2 \bar{j}-3 \bar{k}\); \(\overline{\mathbf{b}}=3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}\), then find the angle between \(2 \overline{\mathbf{a}}+\overline{\mathbf{b}}\) and \(\bar{a}+2 \bar{b}\).
Solution:

II.

Question 1.
Find unit vector parallel to the XOY- plane and perpendicular to the vector \(4 \bar{i}-3 \bar{j}+\bar{k}\).
Solution:
Any vector parallel to XOY-plane will be of the form \(p \bar{i}+q \bar{j}\)
∴ The vector parallel to the XOY-plane and perpendicular to the vector \(4 \bar{i}-3 \bar{j}+\bar{k}\) is \(3 \bar{i}+4 \bar{j}\)
Its magnitude = \(|3 \bar{i}+\overline{4 j}|=\sqrt{9+16}=5\)
∴ Unit vector parallel to the XOY-plane and perpendicular to the vector \(4 \bar{i}-3 \bar{j}+\bar{k}\) is \(\pm \frac{(3 \overline{\mathrm{i}}+4 \overline{\mathrm{j}})}{5}\)

Question 2.
If \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathrm{c}}=0,|\overline{\mathbf{a}}|=3,|\overline{\mathbf{b}}|=5\) and \(|\bar{c}|=7\), then find the angle between \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:

Question 3.
If \(|\overline{\mathbf{a}}|\) = 2, \(|\overline{\mathbf{b}}|\) = 3 and \(|\overline{\mathbf{c}}|\) = 4 and each of \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) is perpendicular to the sum of the other two vectors, then find the magnitude of \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\).
Solution:

Question 4.
Find the equation of the plane passing through the point \(\overline{\mathbf{a}}=\mathbf{2} \overline{\mathbf{i}}+3 \bar{j}-\overline{\mathbf{k}}\) and perpendicular to the vector \(3 \bar{i}-2 \bar{j}-2 \bar{k}\) and the distance of this plane from the origin.
Solution:
Equation of the plane passing through the point \(\overline{\mathbf{a}}=\mathbf{2} \overline{\mathbf{i}}+3 \bar{j}-\overline{\mathbf{k}}\) and perpendicular to the vector \(\bar{n}=3 \bar{i}-2 \bar{j}-2 \bar{k}\) is

Question 5.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\) are the position vectors of four coplanar points such that \((\mathbf{a}-\overline{\mathbf{d}}) \cdot(\bar{b}-\bar{c})=(\bar{b}-\bar{d}) \cdot(\bar{c}-\bar{a})=0\). Show that the point \(\bar{d}\) represents the orthocentre of the triangle with \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) as its vertices.
Solution:
Position vectors of A, B, C, D are \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) respectively.


⇒ BD is perpendicular to AC
∴ BD is another altitude of ∆ABC.
Altitudes AD and BD intersect at D.
∴ D is the orthocentre of ∆ABC.

III.

Question 1.
Show that the points (5, -1, 1), (7, -4, 7), (1, -6, 10) and (-1, -3, 4) are the vertices of a rhombus.
Solution:
Let A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4) are the given points.

∵ AB = BC = CD = DA = 7 units
AC ≠ BD
∴ A, B, C, D points are the vertices of a rhombus.

Question 2.
Let \(\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}, \quad \bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}\) and \(\overline{\mathbf{c}}=\mathbf{3} \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\). Find the vector which is perpendicular to both \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) and whose magnitude is twenty one times the magnitude of \(\overline{\mathbf{c}}\).
Solution:

Question 3.
G is the centroid of ΔABC and a, b, c are the lengths of the sides BC, CA and AB respectively prove that a² + b² + c² = 3 (OA² + OB² + OC²) – 9(OG)² where O is any point.
Solution:
Given that \(\overline{\mathrm{BC}}=\overline{\mathrm{a}}, \overline{\mathrm{CA}}=\overline{\mathrm{b}}, \overline{\mathrm{AB}}=\overline{\mathrm{c}}\)
Let ‘O’ be the origin and let \(\overline{\mathrm{OA}}=\overline{\mathrm{p}}, \overline{\mathrm{OB}}=\overline{\mathrm{q}} \text { and } \overline{\mathrm{OC}}=\overline{\mathrm{r}}\)
Then P.V. of the centroid of ΔABC is

From (1) and (2)
∴ a² + b² + c² = 3(OA² + OB² + OC²) – 9(OG)².

Question 4.
A line makes angles θ₁, θ₂, θ₃, and θ₄ with the diagonals of a cube. Show that cos²θ₁ + cos²θ₂ + cos²θ₃ + cos²θ₄ = \(\frac{4}{3}\).
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(b)

I.

Question 1.
Find the vector equation of the line passing through the point \(2 \bar{i}+3 \bar{j}+\bar{k}\) and parallel to the vector \(4 \bar{i}-2 \bar{j}+3 \bar{k}\).
Solution:

Question 2.
OABC is a parallelogram. If \(\overline{O A}=\bar{a}\) and \(\overline{O C}=\bar{c}\), find the vector equation of the side BC.
Solution:

Question 3.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are the position vectors of the vertices A, B and C respectively of ∆ABC, theind the vector equation of the median through the vertex A.
Solution:

Question 4.
Find the vertor equation of the line joining the points \(2 \bar{i}+\bar{j}+3 \bar{k}\) and \(-4 \bar{i}+3 \bar{j}-\bar{k}\).
Solution:

Question 5.
Find the vector equation of the plane passing through the points \(\overline{\mathbf{i}}-2 \overline{\mathbf{j}}+5 \overline{\mathbf{k}},-5 \overline{\mathbf{j}}-\overline{\mathbf{k}} \text { and }-3 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}\).
Solution:

Question 6.
Find the vector equation of the plane through the points (0, 0, 0), (0, 5, 0) and (2, 0, 1).
Solution:

II.

Question 1.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are noncoplanar find the point of intersection of the line passing through the points \(2 \bar{a}+3 \bar{b}-\bar{c}\), \(3 \bar{a}+4 \bar{b}-2 \bar{c}\) with the line joining the points \(\bar{a}-2 \bar{b}+3 \bar{c}, \bar{a}-6 \bar{b}+6 \bar{c}\).
Solution:

Question 2.
ABCD is a trapezium in which AB and CD are parallel. Prove by vector methods, that the mid points of the sides AB, CD and the intersection of the diagonals are collinear.
Solution:




⇒ M, P, N are collinear
Hence the midpoints of parallel sides of a trapezium and the point of intersection of the diagonals are collinear.

Question 3.
In a quadrilateral ABCD, if the midpoints of one pair of opposite sides and the point of intersection of the diagonals are collinear, using vector methods, prove that the quadrilateral ABCD is a trapezium.
Solution:

III.

Question 1.
Find the vector equation of the plane which passes through the points \(2 \bar{i}+4 \bar{j}+2 \bar{k}, 2 \bar{i}+3 \bar{j}+5 \bar{k}\) and parallel to the vector \(3 \overline{\mathbf{i}}-2 \overline{\mathbf{j}}+\overline{\mathbf{k}}\). Also find the point where this plane meets the line joining the points \(2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+3 \overline{\mathbf{k}}\) and \(4 \bar{i}-2 \bar{j}+3 \bar{k}\).
Solution:

Question 2.
Find the vector equation of the plane passing through points \(4 \overline{\mathbf{i}}-3 \overline{\mathbf{j}}-\overline{\mathbf{k}}\), \(3 \overline{\mathbf{i}}+7 \overline{\mathbf{j}}-10 \overline{\mathbf{k}}\) and \(2 \bar{i}+5 \bar{j}-7 \bar{k}\), and show that the point \(\overline{\mathbf{i}}+2 \bar{j}-3 \overline{\mathbf{k}}\) lies in the plane.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(i) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(i)

Solve the following systems of homogeneous equations.

Question 1.
2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0
Solution:
The Coefficient matrix is \(\left[\begin{array}{ccc}
2 & 3 & -1 \\
1 & -1 & -2 \\
3 & 1 & 3
\end{array}\right]\)
det of \(\left[\begin{array}{ccc}
2 & 3 & -1 \\
1 & -1 & -2 \\
3 & 1 & 3
\end{array}\right]=\left[\begin{array}{ccc}
2 & 3 & -1 \\
1 & -1 & -2 \\
3 & 1 & 3
\end{array}\right]\)
= 2(-3 + 2) – 3(3 + 6) – 1(1 + 3)
= -2 – 27 – 4
= -33 ≠ 0, ρ(A) = 3
Hence the system has the trivial solution x = y = z = 0 only.

Question 2.
3x + y – 2z = 0, x + y + z = 0, x – 2y + z = 0
Hint: If the determinant of the coefficient matrix ≠ 0 then the system has a trivial solution (i.e.) ρ(A) = 3.
Solution:
The coefficient matrix is \(\left[\begin{array}{ccc}
3 & 1 & -2 \\
1 & 1 & 1 \\
1 & -2 & 1
\end{array}\right]\)
\(\left|\begin{array}{ccc}
3 & 1 & -2 \\
1 & 1 & 1 \\
1 & -2 & 1
\end{array}\right|\)
= 3(1 + 2) – 1(1 – 1) – 2(-2 – 1)
= 9 + 6
= 15 ≠ 0, ρ(A) = 3
Hence the system has the trivial solutions x = y = z = 0 only.

Question 3.
x + y – 2z = 0, 2x + y – 3z = 0, 5x + 4y – 9z = 0
Solution:
The coefficient matrix is \(\left[\begin{array}{rrr}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\) = A (say)
|A| = \(\left|\begin{array}{rrr}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right|\)
= 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5)
= 3 + 3 – 6
= 0
∴ Rank of A = 2 as the submatrix \(\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right]\) is non-singular, ρ(A) < 3
Hence the system has a non-trivial solution.
A = \(\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\)
R₂ → R₂ – 2R₁, R₃ → R₃ – 3R₁
A ~ \(\left[\begin{array}{ccc}
1 & 1 & -2 \\
0 & -1 & 1 \\
0 & -1 & -1
\end{array}\right]\)
The system of equation is equivalent to the given system of equations are x + y – 2z = 0, -y + z = 0
Let z = k
⇒ y = k, x = k
∴ x = y = z = k for real number k.

Question 4.
x + y – z = 0, x – 2y + z = 0, 3x + 6y – 5z = 0
Solution:
Coefficient matrix A = \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
1 & -2 & 1 \\
3 & 6 & -5
\end{array}\right]\)
R₂ → R₂ – R₁, R₃ → R₃ – 3R₁
A ~ \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
0 & -3 & 2 \\
0 & 3 & -2
\end{array}\right]\)
⇒ det A = 0 as R₂, R₃ are identical.
and rank (A) = 2 as the submatrix \(\left[\begin{array}{cc}
1 & 1 \\
0 & -3
\end{array}\right]\) is non-singular.
Hence the system has a non-trivial solution, ∵ ρ(A) < 3
The system of equations equivalent to the given system of equations are
x + y – z = 0
3y – 2z = 0
Let z = k
⇒ y = \(\frac{2 k}{3}\)
x = \(\frac{k}{3}\)
∴ x = \(\frac{k}{3}\), y = \(\frac{2 k}{3}\), z = k for any real number of k.

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(h) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(h)

Solve the following systems of equations.
(i) by using Cramer’s rule and matrix inversion method, when the coefficient matrix is non-singular.
(ii) by using the Gauss-Jordan method. Also, determine whether the system has a unique solution or an infinite number of solutions, or no solution, and find the solutions if exist.

Question 1.
5x – 6y + 4z = 15
7x + 4y – 3z = 19
2x + y + 6z = 46
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
5 & -6 & 4 \\
7 & 4 & -3 \\
2 & 1 & 6
\end{array}\right|\)
= 5(24 + 3) + 6(42 + 6) + 4(7 – 8)
= 135 + 288 – 4
= 419
Δ₁ = \(\left|\begin{array}{ccc}
15 & -6 & 4 \\
19 & 4 & -3 \\
46 & 1 & 6
\end{array}\right|\)
= 15(24 + 3) + 6(114 + 138) + 4(19 – 184)
= 405 + 1512 – 660
= 1917 – 660
= 1257
Δ₂ = \(\left|\begin{array}{ccc}
5 & 15 & 4 \\
7 & 19 & -3 \\
2 & 46 & 6
\end{array}\right|\)
= 5(114 + 138) – 15(42 + 6) + 4(322 – 38)
= 1260 – 720 + 1136
= 1676
Δ₃ = \(\left|\begin{array}{ccc}
5 & -6 & 15 \\
7 & 4 & 19 \\
2 & 1 & 46
\end{array}\right|\)
= 5(184 – 19) + 6(322 – 38) + 15(7 – 8)
= 825 + 1704 – 15
= 2529 – 15
= 2514

Solution is x = 3, y = 4, z = 6.

(ii) Matrix inversion method:



Solution is x = 3, y = 4, z = 6

(iii) Gauss-Jordan method:

∴ Unique solution exists.
Solution is x = 3, y = 4, z = 6.

Question 2.
x + y + z = 1
2x + 2y + 3z = 6
x + 4y + 9z = 3
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(18 – 3) + 1(8 – 2)
= 6 – 15 + 6
= -3
Δ₁ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
6 & 2 & 3 \\
3 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(54 – 9) + 1(24 – 6)
= 6 – 45 + 18
= -21
Δ₂ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 6 & 3 \\
1 & 3 & 9
\end{array}\right|\)
= 1(54 – 9) – 1(18 – 3) + 1(6 – 6)
= 45 – 15
= 30
Δ₃ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 6 \\
1 & 4 & 3
\end{array}\right|\)
= 1(6 – 24) – 1(6 – 6) + 1(8 – 2)
= -18 – 0 + 6
= -12

Solution is x = 7, y = -10, z = 4

(ii) Matrix inversion method:



∴ Solution is x = 7, y = -10, z = 4

(iii) Gauss-Jordan method:
Augmented matrix is A = \(\left[\begin{array}{llll}
1 & 1 & 1 & 1 \\
2 & 2 & 3 & 6 \\
1 & 4 & 9 & 3
\end{array}\right]\)
R₂ → R₂ – 2R₁, R₃ → R₃ – R₁

Unique solution exists.
∴ Solution is x = 7, y = -10, z = 4

Question 3.
x – y + 3z = 5
4x + 2y – z = 0
-x + 3y + z = 5
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
1 & -1 & 3 \\
4 & 2 & -1 \\
-1 & 3 & 1
\end{array}\right|\)
= 1(2 + 3) + 1(4 – 1) + 3(12 + 2)
= 5 + 3 + 42
= 50
Δ₁ = \(\left|\begin{array}{ccc}
5 & -1 & 3 \\
0 & 2 & -1 \\
5 & 3 & 1
\end{array}\right|\)
= 5(2 + 3) + 1(0 + 5) + 3(0 – 10)
= 25 + 5 – 30
= 0
Δ₂ = \(\left|\begin{array}{ccc}
1 & 5 & 3 \\
4 & 0 & -1 \\
-1 & 5 & 1
\end{array}\right|\)
= 1(0 + 5) – 5(4 – 1) + 3(20 – 0)
= 5 – 15 + 60
= 50
Δ₃ = \(\left|\begin{array}{ccc}
1 & -1 & 5 \\
4 & 2 & 0 \\
-1 & 3 & 5
\end{array}\right|\)
= 1(10 – 0) + 1(20 – 0) + 5(12 + 2)
= 10 + 20 + 70
= 100

∴ Solution is x = 0, y = 1, z = 2.

(ii) Matrix inversion method:



Solution is x = 0, y = 1, z = 2

(iii) Gauss Jordan method:

Unique solution exists.
∴ Solution is x = 0, y = 1, z = 2

Question 4.
2x + 6y + 11 = 0
6x + 20y – 6z + 3 = 0
6y – 18z + 1 = 0
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
2 & 6 & 0 \\
6 & 20 & -6 \\
0 & 6 & -18
\end{array}\right|\)
= 2(-360 + 36) – 6(-108 – 0)
= -648 + 648
= 0
∴ Cramer’s rule and matrix inversion method cannot be used.
∵ Δ = 0

(ii) Gauss Jordan method:

ρ(A) = 2, ρ(AB) = 3
ρ(A) ≠ ρ(AB)
∴ The given system of equations does not have a solution.

Question 5.
2x – y + 3z = 9
x + y + z = 6
x – y + z = 2
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & 1 & 1 \\
1 & -1 & 1
\end{array}\right|\)
= 2(1 + 1) + 1(1 – 1) + 3(-1 – 1)
= 4 + 0 – 6
= -2
Δ₁ = \(\left|\begin{array}{ccc}
9 & -1 & 3 \\
6 & 1 & 1 \\
2 & -1 & 1
\end{array}\right|\)
= 9(1 + 1) + 1(6 – 2) + 3(-6 – 2)
= 18 + 4 – 24
= -2
Δ₂ = \(\left|\begin{array}{lll}
2 & 9 & 3 \\
1 & 6 & 1 \\
1 & 2 & 1
\end{array}\right|\)
= 2(6 – 2) – 9(1 – 1) + 3(2 – 6)
= 8 – 0 – 12
= -4
Δ₃ = \(\left|\begin{array}{ccc}
2 & -1 & 9 \\
1 & 1 & 6 \\
1 & -1 & 2
\end{array}\right|\)
= 2(2 + 6) + 1(2 – 6) + 9(-1 – 1)
= 16 – 4 – 18
= -6

Solution is x = 1, y = 2, z = 3.

(ii) Matrix inversion method:


Solution is x = 1, y = 2, z = 3.

(iii) Gauss-Jordan method:

∴ The given equations have a unique solution.
Solution is x = 1, y = 2, z = 3

Question 6.
2x – y + 8z = 13
3x + 4y + 5z = 18
5x – 2y + 7z = 20
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
2 & -1 & 8 \\
3 & 4 & 5 \\
5 & -2 & 7
\end{array}\right|\)
= 2(28 + 10) + 1(21 – 25) + 8(-6 – 20)
= 76 – 4 – 208
= -136
Δ₁ = \(\left|\begin{array}{ccc}
13 & -1 & 8 \\
18 & 4 & 5 \\
20 & -2 & 7
\end{array}\right|\)
= 13(28 + 10) + 1(126 – 100) + 8(-36 – 80)
= 494 + 26 – 928
= -408
Δ₂ = \(\left|\begin{array}{lll}
2 & 13 & 8 \\
3 & 18 & 5 \\
5 & 20 & 7
\end{array}\right|\)
= 2(126 – 100) – 13(21 – 25) + 8(60 – 90)
= 52 + 52 – 240
= -136
Δ₃ = \(\left|\begin{array}{ccc}
2 & -1 & 13 \\
3 & 4 & 18 \\
5 & -2 & 20
\end{array}\right|\)
= 2(80 + 36) + 1(60 – 90) + 13(-6 – 20)
= 232 – 30 – 338
= -136

∴ Solution is x = 3, y = 1, z = 1

(ii) Matrix inversion method:



∴ Solution is x = 3, y = 1, z = 1

(iii) Gauss Jordan method:

∴ The given equations have a unique solution and Solution is x = 3, y = 1, z = 1.

Question 7.
2x – y + 3z = 8
-x + 2y + z = 4
3x + y – 4z = 0
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
-1 & 2 & 1 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-8 – 1) + 1(4 – 3) + 3(-1 – 6)
= -18 + 1 – 21
= -38
Δ₁ = \(\left|\begin{array}{ccc}
8 & -1 & 3 \\
4 & 2 & 1 \\
0 & 1 & -4
\end{array}\right|\)
= 8(-8 – 1) + 1(-16 – 0) + 3(4 – 0)
= -72 – 16 + 12
= -76
Δ₂ = \(\left|\begin{array}{ccc}
2 & 8 & 3 \\
-1 & 4 & 1 \\
3 & 0 & -4
\end{array}\right|\)
= 2(-16 – 0) – 8(4 – 3) + 3(-0 – 12)
= -32 – 8 – 36
= -76
Δ₃ = \(\left|\begin{array}{ccc}
2 & -1 & 8 \\
-1 & 2 & 4 \\
3 & 1 & 0
\end{array}\right|\)
= 2(0 – 4) + 1(0 – 12) + 8(-1 – 6)
= -8 – 12 – 56
= -76

∴ Solution is x = 2, y = 2, z = 2.

(ii) Matrix inversion method:


Solution is x = 2, y = 2, z = 2

(iii) Gauss Jordan method:


∴ The given equations have a unique solution and solution is x = 2, y = 2, z = 2.

Question 8.
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right|\)
= 1(-5 – 7) – 1(-2 – 14) + 1(2 – 10)
= -12 + 16 – 8
= -4
Δ₁ = \(\left|\begin{array}{ccc}
9 & 1 & 1 \\
52 & 5 & 7 \\
0 & 1 & -1
\end{array}\right|\)
= 9(-5 – 7) – 1(-52 – 0) + 1(52 – 0)
= -108 + 52 + 52
= -4
Δ₂ = \(\left|\begin{array}{ccc}
1 & 9 & 1 \\
2 & 52 & 7 \\
2 & 0 & -1
\end{array}\right|\)
= 1(-52 – 0) – 9(-2 – 14) + 1(0 – 104)
= -52 + 144 – 104
= -12
Δ₃ = \(\left|\begin{array}{ccc}
1 & 1 & 9 \\
2 & 5 & 52 \\
2 & 1 & 0
\end{array}\right|\)
= 1(0 – 52) – 1(0 – 104) + 9(2 – 10)
= -52 + 104 – 72
= -20

(ii) Matrix inversion method:


Solution is x = 1, y = 3, z = 5

(iii) Gauss Jordan method:

∴ The given equations have a unique solution and solution is x = 1, y = 3, z = 5.

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(g) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(g)

Examine whether the following systems of equations are consistent or inconsistent and if consistent find the complete solutions.

Question 1.
x + y + z = 4
2x + 5y – 2z = 3
x + 7y – 7z = 5
Solution:

ρ(A) = 2, ρ(AB) = 3
ρ(A) ≠ ρ(AB)
∴ The given system of equations are in consistent.

Question 2.
x + y + z = 6
x – y + z = 2
2x – y + 3z = 9
Solution:

Question 3.
x + y + z = 1
2x + y + z = 2
x + 2y + 2z = 1
Solution:

ρ(A) = 2 = ρ(AB) < 3
The given system of equations are consistent and have infinitely many solutions.
The solutions are given by [(x, y, z) 1x = 1, y + z = 0].

Question 4.
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0
Solution:

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent have a unique solution.
∴ Solution is given by x = 1, y = 3, z = 5.

Question 5.
x + y + z = 6
x + 2y + 3z = 10
x + 2y + 4z = 1
Solution:
Augmented matrix A = \(\left[\begin{array}{cccc}
1 & 1 & 1 & 6 \\
1 & 2 & 3 & 10 \\
1 & 2 & 4 & 1
\end{array}\right]\)
By R₂ → R₂ – R₁, R₃ → R₃ – R₂, we obtain

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent.
They have a unique solution.
∴ Solution is given by x = -7, y = 22, z = -9.

Question 6.
x – 3y – 8z = -10
3x + y – 4z = 0
2x + 5y + 6z = 13
Solution:
The Augmented matrix

ρ(A) = ρ(AB) = 2 < 3
∴ The given system of equations are consistent have infinitely many solutions.
x + y = 2 and y + 2z = 3
Taking z = k, y = 3 – 2z = 3-2k
x = 2 – y
= 2 – (3 – 2k)
= 2 – 3 + 2k
= 2k – 1
∴ The solutions are given by x = -1 + 2k, y = 3 – 2k, z = k where ‘k’ is any scalar.

Question 7.
2x + 3y + z = 9
x + 2y + 3z = 6
3x + y + 2z = 8
Solution:

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent have a unique solution.
∴ Solution is given by x = \(\frac{35}{18}\), y = \(\frac{29}{18}\), z = \(\frac{5}{18}\)

Question 8.
x + y + 4z = 6
3x + 2y – 2z = 9
5x + y + 2z = 13
Solution:

∴ ρ(A) = ρ(AB) = 3
∴ The given system of equations are consistent have a unique solution.
∴ Solution is given by x = 2, y = 2, z = \(\frac{1}{2}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(f) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(f)

I. Find the rank of each of the following matrices.

Question 1.
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\) = 0 – 0 = 0
and |1| = 1 ≠ 0
∴ ρ(A) = 1

Question 2.
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1 ≠ 0
∴ ρ(A) = 2

Question 3.
\(\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right|\) = 0 – 0 = 0
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 4.
\(\left[\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right|\) = 0 – 1 = -1 ≠ 0
∴ ρ(A) = 2

Question 5.
\(\left[\begin{array}{ccc}
1 & 0 & -4 \\
2 & -1 & 3
\end{array}\right]\)
Solution:
\(\left|\begin{array}{cc}
1 & -4 \\
2 & 3
\end{array}\right|\) = 3 + 8 = -11 ≠ 0
∴ ρ(A) = 2

Question 6.
\(\left[\begin{array}{lll}
1 & 2 & 6 \\
2 & 4 & 3
\end{array}\right]\)
Solution:
\(\left|\begin{array}{ll}
2 & 6 \\
4 & 3
\end{array}\right|\) = 6 – 24 = -18 ≠ 0
∴ ρ(A) = 2

II.

Question 1.
\(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
= 1(1 – 0) – 0(0 – 0) + 0(0 – 0)
= 1 – 0 + 0
= 1 ≠ 0
∴ ρ(A) = 3

Question 2.
\(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right|\)
= 1(6 – 0) – 2(8 + 1) + 0(0 + 3)
= 6 – 18
= -12 ≠ 0
∴ ρ(A) = 3

Question 3.
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right|\)
= 1(6 – 4) – 2(4 – 3) + 0(8 – 9)
= 2 – 2 + 0
= 0
∴ ρ(A) ≠ 3, ρ(A) < 3
Take \(\left|\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right|\) = 3 – 4 = -1 ≠ 0
∴ ρ(A) = 2

Question 4.
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\), det A = 0, ρ(A) ≠ 3
All 2 × 2 sub-matrix det. is zero
∴ ρ(A) ≠ 2
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 5.
\(\left[\begin{array}{cccc}
1 & 2 & 0 & -1 \\
3 & 4 & 1 & 2 \\
-2 & 3 & 2 & 5
\end{array}\right]\)
Solution:
Take sub-matrix B = \(\left|\begin{array}{ccc}
1 & 2 & 0 \\
3 & 4 & 1 \\
-2 & 3 & 2
\end{array}\right|\)
= 1(8 – 3) – 2(6 + 2)
= 5 – 16
= -11 ≠ 0
Rank of the given matrix is 3.

Question 6.
\(\left[\begin{array}{cccc}
0 & 1 & 1 & -2 \\
4 & 0 & 2 & 5 \\
2 & 1 & 3 & 1
\end{array}\right]\)
Solution:
Take sub matrix A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
4 & 0 & 2 \\
2 & 1 & 3
\end{array}\right]\)
= -1(12 – 4) + 1(4 – 0)
= -8 + 4
= -4 ≠ 0
∴ ρ(A) = 3

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(c)

I.

Question 1.
Find the domains of the following real-valued functions.
(i) f(x) = \(\frac{1}{\left(x^{2}-1\right)(x+3)}\)
Solution:
f(x) = \(\frac{1}{\left(x^{2}-1\right)(x+3)}\) ∈ R
⇔ (x² – 1) (x + 3) ≠ 0
⇔ (x + 1) (x – 1) (x + 3) ≠ 0
⇔ x ≠ -1, 1, -3
∴ Domain of f is R – {-1, 1, -3}

(ii) f(x) = \(\frac{2 x^{2}-5 x+7}{(x-1)(x-2)(x-3)}\)
⇔ (x – 1) (x – 2) (x – 3) ≠ 0
⇔ x ≠ 1, x ≠ 2, x ≠ 3
∴ Domain of f is R – {1, 2, 3}

(iii) f(x) = \(\frac{1}{\log (2-x)}\)
Solution:
f(x) = \(\frac{1}{\log (2-x)}\)
⇔ log (2 – x) ≠ 0 and 2 – x > 0
⇔ (2 – x) ≠ 1 and 2 > x
⇔ x ≠ 1 and x < 2
x ∈ (-∞, 1) ∪ (1, 2) (or) x ∈ (-∞, 2) – {1}
∴ Domain of f is {(-∞, 2) – {1}}

(iv) f(x) = |x – 3|
Solution:
f(x) = |x – 3| ∈ R
⇔ x ∈ R
∴ The domain of f is R

(v) f(x) = \(\sqrt{4 x-x^{2}}\)
Solution:
f(x) = \(\sqrt{4 x-x^{2}}\) ∈ R
⇔ 4x – x² ≥ 0
⇔ x(4 – x) ≥ 0
⇔ x ∈ [0, 4]
∴ Domain of f is [0, 4]

(vi) f(x) = \(\frac{1}{\sqrt{1-x^{2}}}\)
Solution:
f(x) = \(\frac{1}{\sqrt{1-x^{2}}}\) ∈ R
⇔ 1 – x² > 0
⇔ (1 + x) (1 – x) > 0
⇔ x ∈ (-1, 1)
∴ Domain of f is {x/x ∈ (-1, 1)}

(vii) f(x) = \(\frac{3^{x}}{x+1}\)
Solution:
f(x) = \(\frac{3^{x}}{x+1}\) ∈ R
⇔ 3^x ∈ R, ∀ x ∈ R and x + 1 ≠ 0
⇔ x ≠ -1
∴ Domain of f is R – {-1}

(viii) f(x) = \(\sqrt{x^{2}-25}\)
Solution:
f(x) = \(\sqrt{x^{2}-25}\) ∈ R
⇔ x² – 25 ≥ 0
⇔ (x + 5) (x – 5) ≥ 0
⇔ x ∈ (-∞, -5] ∪ [5, ∞)
⇔ x ∈ R – (-5, 5)
∴ Domain of f is R – (- 5, 5)

(ix) f(x) = \(\sqrt{x-[x]}\)
Solution:
f(x) = \(\sqrt{x-[x]}\) ∈ R
⇔ x – [x] ≥ 0
⇔ x ≥ [x]
⇔ x ∈ R
∴ Domain of f is R.

(x) f(x) = \(\sqrt{[x]-x}\)
Solution:
f(x) = \(\sqrt{[x]-x}\) ∈ R
⇔ [x] – x ≥ 0
⇔ [x] ≥ x
⇔ x ≤ [x]
⇔ x ∈ Z
∴ The domain of f is z (Where z denotes a set of integers)

Question 2.
Find the ranges of the following real-valued functions.
(i) log|4 – x²|
Solution:
Let y = f(x) = log|4 – x²|
f(x) ∈ R
⇔ 4 – x² ≠ 0
⇔ x ≠ ±2
∵ y = log|4 – x²|
⇒ |4 – x²| = e^y
∵ e^y > 0 ∀ y ∈ R
∴ The range of f is R.

(ii) \(\sqrt{[x]-x}\)
Solution:
Let y = f(x) = \(\sqrt{[x]-x}\)
f(x) ∈ R
⇔ [x] – x ≥ 0
⇔ x ≤ [x]
⇔ x ∈ z
∴ Domain of f is z. Then range of f is {0}

(iii) \(\frac{\sin \pi[x]}{1+[x]^{2}}\)
Solution:
Let f(x) = \(\frac{\sin \pi[x]}{1+[x]^{2}}\) ∈ R
⇔ x ∈ R
∴ The domain of f is R
For x ∈ R, [x] is an integer,
sin π[x] = 0, ∀ x ∈ R [∵ sin nπ = 0, ∀ n ∈ z]
∴ Range of f is {0}

(iv) \(\frac{x^{2}-4}{x-2}\)
Solution:
Let y = f(x) = \(\frac{x^{2}-4}{x-2}\) ∈ R
⇔ y = \(\frac{(x+2)(x-2)}{x-2}\)
⇔ x ≠ 2
∴ The domain of f is R – {2}
Then y = x + 2, [∵ x ≠ 2 ⇒ y ≠ 4]
Then its range R – {4}

(v) \(\sqrt{9+x^{2}}\)
Solution:
Let y = f(x) = \(\sqrt{9+x^{2}}\) ∈ R
The domain of f is R
When x = 0, f(0) = √9 = 3
For all values of x ∈ R – {0}, f(x) > 3
∴ The range of f is [3, ∞)

Question 3.
If f and g are real-valued functions defined by f(x) = 2x – 1 and g(x) = x² then find
(i) (3f – 2g)(x)
(ii) (fg) (x)
(iii) \(\left(\frac{\sqrt{f}}{g}\right)(x)\)
(iv) (f + g + 2) (x)
Solution:
(i) (3f – 2g)(x)
f(x) = 2x – 1, g(x) = x²
(3f – 2g) (x) = 3f(x) – 2g(x)
= 3(2x – 1) – 2x²
= -2x² + 6x – 3

(ii) (fg) (x)
= f(x) . g(x)
= (2x – 1) (x²)
= 2x³ – x²

(iii) \(\left(\frac{\sqrt{f}}{g}\right)(x)\)
\(\frac{\sqrt{f(x)}}{g(x)}=\frac{\sqrt{2 x-1}}{x^{2}}\)

(iv) (f + g + 2) (x)
= f(x) + g(x) + 2
= (2x – 1) + x² + 2
= x² + 2x + 1
= (x + 1)²

Question 4.
If f = {(1, 2), (2, -3), (3, -1)} then find
(i) 2f
(ii) 2 + f
(iii) f²
(iv) √f
Solution:
Given f = {(1, 2), (2, -3), (3, -1)}
(i) 2f = {(1, 2 × 2), (2, 2(-3), (3, 2(-1))}
= {(1, 4), (2, -6), (3, -2)}

(ii) 2 + f = {(1, 2 + 2), (2, 2 + (-3)), (3, 2 + (-1)}
= {(1, 4), (2, -1), (3, 1)}

(iii) f² = {(1, 22), (2, (-3)2), (3, (-1)2)}
= {(1, 4), (2, 9), (3, 1)}

(iv) √f = {(1, √2)}
∵ √-3 and √-1 are not real

II.

Question 1.
Find the domains of the following real-valued functions.
(i) f(x) = \(\sqrt{x^{2}-3 x+2}\)
Solution:
f(x) = \(\sqrt{x^{2}-3 x+2}\) ∈ R
⇔ x² – 3x + 2 ≥ 0
⇔ (x- 1) (x – 2) ≥ 0
⇔ x ∈ (-∞, 1 ] ∪ [2, ∞]
∴ The domain of f is R – (1, 2)

(ii) f(x) = log(x² – 4x + 3)
Solution:
f(x) = log(x² – 4x + 3) ∈ R
⇔ x² – 4x + 3 > 0
⇔ (x – 1) (x – 3) > 0
⇔ x ∈ (-∞, 1) ∪ (3, ∞)
∴ Domain of f is R – [1, 3]

(iii) f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\)
Solution:
f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\) ∈ R
⇔ 2 + x ≥ 0, 2 – x ≥ 0, x ≠ 0
⇔ x ≥ -2, x ≤ 2, x ≠ 0
⇔ -2 ≤ x ≤ 2, x ≠ 0
⇔ x ∈ [-2, 2] – {0}
Domain of f is [-2, 2] – {0}

(iv) f(x) = \(\frac{1}{\sqrt[3]{(x-2)} \log _{(4-x)} 10}\)
Solution:
f(x) = \(\frac{1}{\sqrt[3]{(x-2)} \log _{(4-x)} 10}\) ∈ R
⇔ 4 – x > 0, 4 – x ≠ 1 and x – 2 ≠ 0
⇔ x < 4, x ≠ 3, x ≠ 2
∴ Domain of f is (-∞, 4) – {2, 3}

(v) f(x) = \(\sqrt{\frac{4-x^{2}}{[x]+2}}\)
Solution:
f(x) = \(\sqrt{\frac{4-x^{2}}{[x]+2}}\) ∈ R
Case (i) 4 – x² ≥ 0 and [x] + 2 > 0 (or) Case (ii) 4 – x² ≤ 0 and [x] + 2 < 0
Case (i): 4 – x² ≥ 0 and [x] + 2 > 0
⇔ (2 – x) (2 + x) ≥ 0 and [x] > -2
⇔ x ∈ [-2, 2] and x ∈ [-1, ∞]
⇔ x ∈ [-1, 2] ……..(1)
Case (ii): 4 – x² ≤ 0 and [x] + 2 < 0
⇔ (2 + x) (2 – x) ≤ 0 and [x] < – 2
⇔ x ∈ (-∞, -2] ∪ [2, ∞] and x ∈ (-∞, -2)
⇔ x ∈ (-∞, -2) ……(2)
from (1) and (2),
Domain of f is (-∞, -2) ∪ [-1, 2]

(vi) f(x) = \(\sqrt{\log _{0.3}\left(x-x^{2}\right)}\)
Solution:
f(x) = \(\sqrt{\log _{0.3}\left(x-x^{2}\right)}\) ∈ R
Then log_0.3(x – x²) ≥ 0
⇒ x – x² ≤ (0.3)⁰
⇒ x – x² ≤ 1
⇒ -x² + x – 1 ≤ 0
⇒ x² – x + 1 ≥ 0
This is true for all x ∈ R ……..(1)
and x – x² ≥ 0
⇒ x² – x ≤ 0
⇒ x(x – 1) ≤ 0
⇒ x ∈ (0, 1) …….(2)
From (1) and (2)
Domain of f is R ^ (0, 1) = (0, 1)
∴ The domain of f is (0, 1)

(vii) f(x) = \(\frac{1}{x+|x|}\)
Solution:
f(x) = \(\frac{1}{x+|x|}\) ∈ R
⇔ x + |x| ≠ 0
⇔ x ∈ (0, ∞)
∵ |x| = x, if x ≥ 0
|x| = -x, if x < 0
∴ The domain of f is (0, ∞)

Question 2.
Prove that the real valued function f(x) = \(\frac{x}{e^{x}-1}+\frac{x}{2}+1\) is an even function on R \ {0}.
Solution:
f(x) ∈ R, e^x – 1 ≠ 0
⇒ e^x ≠ 1
⇒ x ≠ 0

⇒ f(x) is an even function on R – {0}

Question 3.
Find the domain and range of the following functions.
(i) f(x) = \(\frac{\tan \pi[x]}{1+\sin \pi[x]+\left[x^{2}\right]}\)
Solution:
f(x) = \(\frac{\tan \pi[x]}{1+\sin \pi[x]+\left[x^{2}\right]}\) ∈ R
⇔ x ∈ R, since [x] is an integer tan π[x] and sin π[x] each is zero for ∀ x ∈ R and f(x) ∈ R
Domain of f is R
Its range = {0}

(ii) f(x) = \(\frac{x}{2-3 x}\)
Solution:

(iii) f(x) = |x| + |1 + x|
Solution:
f(x) = |x| + |1 + x| ∈ R
⇔ x ∈ R
∴ Domain of f is R
∵ |x| = x, if x ≥ 0
= -x, if x < 0
|1 + x| = 1 + x, if x ≥ -1
= -(1 + x) if x < -1
For x = 0, f(0) = |0| + |1 + 0| = 1
x = 1, f(1) = |1| + |1 + 1| = 1 + 2 = 3
x = 2, f(2) = |2| + |1 + 2| = 2 + 3 = 5
x = -2, f(-2) = |-2| + |1 + (-2)| = 2 + 1 = 3
x = -1, f(-1) = |-1| + |1 +(-1)| = 1 + 0 = 1
∴ The range of f is [1, ∞]

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(b)

I.

Question 1.
If f(x) = e^x and g(x) = log_ex, then show that f o g = g o f and find f^-1 and g^-1.
Solution:
Given f(x) = e^x and g(x) = log_ex
Now (f o g) (x) = f(g(x))
= f(log_ex) [∵ g(x) = \(\log _{e} x\)]
= \(e^{\left(\log _{e} x\right)}\)
= x
∴ (fog) (x) = x ………(1)
and (g o f) (x) = g(f(x))
= g(e^x) [∵ f(x) = e^x]
= log_e (e^x) [∵ g(x) = log_ex]
= x log_e (e)
= x(1)
= x
∴ (g o f) (x) = x …….(2)
From (1) and (2)
f o g = g o f
Given f(x) = e^x
Let y = f(x) = e^x ⇒ x = f^-1(y)
and y = e^x ⇒ x = log_e (y)
∴ f^-1(y) = log_e (y) ⇒ f^-1(x) = log_e (x)
Let y = g(x) = log_e (x)
∵ y = g(x) ⇒ x = g^-1(y)
∵ y = log_e (x) ⇒ x = e^y
∴ g^-1(y) = e^y ⇒ g^-1(x) = e^x
∴ f^-1(x) = log_e (x) and g^-1(x) = e^x

Question 2.
If f(y) = \(\frac{y}{\sqrt{1-y^{2}}}\), g(y) = \(\frac{y}{\sqrt{1+y^{2}}}\) then show that (fog) (y) = y
Solution:
f(y) = \(\frac{y}{\sqrt{1-y^{2}}}\) and g(y) = \(\frac{y}{\sqrt{1+y^{2}}}\)
Now, (fog) (y) = f(g(y))

∴ (fog) (y) = y

Question 3.
If f : R → R, g : R → R are defined by f(x) = 2x² + 3 and g(x) = 3x – 2, then find
(i) (fog)(x)
(ii) (gof) (x)
(iii) (fof) (0)
(iv) go(fof) (3)
Solution:
f : R → R, g : R → R and f(x) = 2x² + 3; g(x) = 3x – 2
(i) (f o g) (x) = f(g(x))
= f(3x – 2) [∵ g(x) = 3x – 2]
= 2(3x- 2)² + 3 [∵ f(x) = 2x² + 3]
= 2(9x² – 12x + 4) + 3
= 18x² – 24x + 8 + 3
= 18x² – 24x + 11

(ii) (gof) (x) = g(f(x))
= g(2x² + 3) [∵ f(x) = 2x² + 3]
= 3(2x² + 3) – 2 [∵ g(x) = 3x – 2]
= 6x² + 9 – 2
= 6x² + 7

(iii) (fof) (0) = f(f(0))
= f(2(0) + 3) [∵ f(x) = 2x² + 3]
= f(3)
= 2(3)² + 3
= 18 + 3
= 21

(iv) g o (f o f) (3)
= g o (f (f(3)))
= g o (f (2(3)² + 3)) [∵ f(x) = 2x² + 3]
= g o (f(21))
= g(f(21))
= g(2(21)² + 3)
= g(885)
= 3(885) – 2 [∵ g(x) = 3x – 2]
= 2653

Question 4.
If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x² + 1, then find
(i) (f o f) (x² + 1)
(ii) f o g (2)
(iii) g o f (2a – 3)
Solution:
f : R → R, g : R → R and f(x) = 3x – 1 ; g(x) = x² + 1
(i) (f o f) (x² + 1)
= f(f(x² + 1))
= f[3(x² + 1) – 1] [∵ f(x) = 3x – 1]
= f(3x² + 2)
= 3(3x² + 2) – 1
= 9x² + 5

(ii) (f o g) (2)
= f(g(2))
= f(2² + 1) [∵ g(x) = x² + 1]
= f(5)
= 3(5) – 1
= 14 [∵ f(x) = 3x – 1]

(iii) (g o f) (2a – 3)
= g(f(2a – 3))
= g[3(2a – 3) – 1] [∵ f(x) = 3x – 1]
= g(6a – 10)
= (6a – 10)² + 1 [∵ g(x) = x² + 1]
= 36a² – 120a + 100 + 1
= 36a² – 120a + 101

Question 5.
If f(x) = \(\frac{1}{x}\), g(x) = √x for all x ∈ (0, ∞) then find (g o f) (x).
Solution:
f(x) = \(\frac{1}{x}\), g(x) = √x, ∀ x ∈ (0, ∞)
(g o f) (x) = g(f(x))
= g(\(\frac{1}{x}\)) [∵ f(x) = \(\frac{1}{x}\)]
= \(\sqrt{\frac{1}{x}}\)
= \(\frac{1}{\sqrt{x}}\) [∵ g(x) = √x]
∴ (gof) (x) = \(\frac{1}{\sqrt{x}}\)

Question 6.
f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) for all x ∈ R, find (g o f) (x).
Solution:
f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) ∀ x ∈ R
(g o f) (x) = g(f(x))
= g(2x – 1) [∵ f(x) = 2x – 1]
= \(\frac{(2 x-1)+1}{2}\)
= x [∵ g(x) = \(\frac{x+1}{2}\)]
∴ (g o f) (x) = x

Question 7.
If f(x) = 2, g(x) = x², h(x) = 2x for all x ∈ R, then find (f o (g o h)) (x).
Solution:
f(x) = 2, g(x) = x², h(x) = 2x, ∀ x ∈ R
[f o (g o h) (x)]
= [f o g (h(x))]
= f o g (2x) [∵ h(x) = 2x]
= f[g(2x)]
= f((2x)²) [∵ g(x) = x²]
= f(4x²) = 2 [∵ f(x) = 2]
∴ [f o (g o h) (x)] = 2

Question 8.
Find the inverse of the following functions.
(i) a, b ∈ R, f : R → R defined by f(x) = ax + b, (a ≠ 0).
Solution:
a, b ∈ R, f : R → R and f(x) = ax + b, a ≠ 0
Let y = f(x) = ax + b
⇒ y = f(x)
⇒ x = f^-1(y) ……..(i)
and y = ax + b
⇒ x = \(\frac{y-b}{a}\) ……..(ii)
From (i) and (ii)
f^-1(y) = \(\frac{y-b}{a}\)
⇒ f^-1(x) = \(\frac{x-b}{a}\)

(ii) f : R → (0, ∞) defined by f(x) = 5^x
Solution:
f : R → (0, ∞) and f(x) = 5^x
Let y = f (x) = 5^x
y = f(x) ⇒ x = f^-1(y) ……(i)
and y = 5^x ⇒ log₅ (y) = x ……..(ii)
From (i) and (ii)
f^-1(y) = log₅(y) ⇒ f^-1(x) = log5 (x)

(iii) f : (0, ∞) → R defined by f(x) = log₂ (x).
Solution:
f : (0, ∞) → R and f(x) = log₂ (x)
Let y = f(x) = log₂ (x)
∵ y = f(x) ⇒ x = f^-1(y) ……..(i)
and y = log₂(x) ⇒ x = 2y
From (i) and (ii)
f^-1(y) = 2y ⇒ f^-1(x) = 2x

Question 9.
If f(x) = 1 + x + x² + …… for |x| < 1 then show that f^-1(x) = \(\frac{x-1}{x}\)
Solution:
f(x) = 1 + x + x² + ……..

Question 10.
If f : [1, ∞) ⇒ [1, ∞) defined by f(x) = \(2^{x(x-1)}\) then find f^-1(x).
Solution:

II.

Question 1.
If f(x) = \(\frac{x-1}{x+1}\), x ≠ ±1, then verify (f o f^-1) (x) = x.
Solution:
Given f(x) = \(\frac{x-1}{x+1}\), x ≠ ±1
Let y = f(x) = \(\frac{x-1}{x+1}\)
∵ y = f(x) ⇒ x = f^-1(y) ……(i)
and y = \(\frac{x-1}{x+1}\)

Question 2.
If A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f : A → B, g : B → C are defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}, then show that f and g are bijective functions and (gof)^-1 = f^-1 o g^-1.
Solution:
A = {1, 2, 3}, B = {α, β, γ},
f : A → B and f = {(1, α), (2, γ), (3, β)}
⇒ f(1) = α, f(2) = γ, f(3) = β
∵ Distinct elements of A have distinct f – images in B, f: A → B is an injective function.
Range of f = {α, γ, β} = B(co-domain)
∴ f : A → B is a surjective function.
Hence f : A → B is a bijective function.
B = {α, β, γ}, C = {p, q, r}, g : B → C and g : {(α, q), (β, r), (γ, p)}
⇒ g(α) = q, g(β) = r, g(γ) = p
∴ Distinct elements of B have distinct g – images in C, g : B → C is an injective function.
Range of g = {q, r, p} = C, (co-domain)
∴ g : B → C is a surjective function
Hence g : B → C is a bijective function
Now f = {(1, α), (2, γ), (3, β)}
g = {(α, q), (β, r), (γ, p)}
g o f = {(1, q), (2, p), (3, r)}
∴ (g o f)^-1 = {(q, 1), (r, 3), (p, 2)} ………(1)
g^-1 = {(q, α), (r, β), (p, γ)}
f^-1 = {(α, 1), (γ, 2),(β, 3)}
Now f^-1 o g^-1 = {(q, 1), (r, 3), (p, 2)} …….(2)
From eq’s (1) and (2)
(gof)^-1 = f^-1 o g^-1

Question 3.
If f : R → R, g : R → R defined by f(x) = 3x – 2, g(x) = x² + 1, then find
(i) (g o f^-1) (2)
(ii) (g o f)(x – 1)
Solution:
f : R → R, g : R → R and f(x) = 3x – 2
f is a bijective function ⇒ its inverse exists
Let y = f(x) = 3x – 2
∵ y = f(x) ⇒ x = f^-1(y) …….(i)
and y = 3x – 2
⇒ x = \(\frac{y+2}{3}\) ……..(ii)
From (i) and (ii)
f^-1(y) = \(\frac{y+2}{3}\)
⇒ f^-1(x) = \(\frac{x+2}{3}\)
Now (gof^-1) (2)
= g(f^-1(2))

∴ (g o f^-1) (2) = \(\frac{25}{9}\)

(ii) (g o f) (x -1)
= g(f(x – 1))
= g(3(x – 1) – 2) [∵ f(x) = 3x – 2]
= g(3x – 5)
= (3x – 5)² + 1 [∵ g(x) = x² + 1]
= 9x² – 30x + 26
∴ (g o f) (x – 1) = 9x² – 30x + 26

Question 4.
Let f = {(1, a), (2, c), (4, d), (3, b)} and g^-1 = {(2, a), (4, b), (1, c), (3, d)} then show that (gof)^-1 = f^-1 o g^-1
Solution:
f = {(1, a), (2, c), (4, d), (3, b)}
∴ f^-1 = {(a, 1), (c, 2), (d, 4), (b, 3)}
g^-1 = {(2, a), (4, b), (1, c), (3, d)}
∴ g = {(a, 2), (b, 4), (c, 1), (d, 3)}
(g o f) = {(1, 2), (2, 1), (4, 3), (3, 4)}
∴ (gof)^-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ……….(1)
f^-1 o g^-1 = {(2, 1), (4, 3), (1, 2), (3, 4)} ……..(2)
From eq’s (1) and (2), we observe (gof)^-1 = f^-1 o g^-1

Question 5.
Let f : R → R, g : R → R be defined by f(x) = 2x – 3, g(x) = x³ + 5 then find (f o g)^-1 (x).
Solution:
f : R → R, g : R → R and f(x) = 2x – 3 and g(x) = x³ + 5
Now (fog) (x) = f(g(x))
= f(x³ + 5) [∵ g(x) = x² + 5]
= 2(x³ + 5) – 3 [∵ f(x) = 2x – 3]
= 2x³ + 7
∴ (f o g) (x) = 2x³ + 7
Let y = (f o g) (x) = 2x³ + 7
∵ y = (fog)(x)
⇒ x = (fog)^-1 (y) …….(1)
and y = 2x³ + 7
⇒ x³ = \(\frac{y-7}{2}\)
⇒ x = \(\left(\frac{y-7}{2}\right)^{\frac{1}{3}}\) …..(2)
From eq’s (1) and (2),
(f o g)^-1 (y) = \(\left(\frac{y-7}{2}\right)^{\frac{1}{3}}\)
∴ (f o g)^-1 (x) = \(\left(\frac{x-7}{2}\right)^{\frac{1}{3}}\)

Question 6.
Let f(x) = x², g(x) = 2^x. Then solve the equation (f o g) (x) = (g o f) (x)
Solution:
Given f(x) = x² and g(x) = 2^x
Now (f o g) (x) = f(g(x))
= f(2^x) [∵ g(x) = 2^x]
= (2^x)²
= 2^2x [∵ f(x) = x²]
∴ (f o g) (x) = 2^2x ……(1)
and (g o f) (x) = g(f(x))
= g(x²) [∵ f(x) = x²]
= \((2)^{x^{2}}\) [∵ g(x) = 2^x]
∴ (g o f) (x) = \((2)^{x^{2}}\)
∵ (f o g) (x) = (g o f) (x)
⇒ 2^2x = \((2)^{x^{2}}\)
⇒ 2^x = x²
⇒ x² – 2x = 0
⇒ x(x – 2) = 0
⇒ x = 0, x = 2
∴ x = 0, 2

Question 7.
If f(x) = \(\frac{x+1}{x-1}\), (x ≠ ±1) then find (fofof) (x) and (fofofof) (x).
Solution:
f(x) = \(\frac{x+1}{x-1}\), (x ≠ ±1)
(i) (fofof) (x) = (fof) [f(x)]
= (fof) \(\left(\frac{x+1}{x-1}\right)\) [∵ f(x) = \(\left(\frac{x+1}{x-1}\right)\)]

(ii) (fofofof) (x) = f[(f o f o f) (x)]
= f [f(x)] {from (1)}

In the above problem if a number of f is even its answer is x and if a number of f is odd its answer is f(x).

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(a)

I.

Question 1.
If the function f is defined by

then find the values of
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
(v) f(-5)
Solution:
(i) f(3)
For x > 1, f(x) = x + 2
∴ f(3) = 3 + 2 = 5

(ii) f(0)
For -1 ≤ x ≤ 1, f(x) = 2
∴ f(0) = 2

(iii) f(-1.5)
For -3 < x < -1, f(x) = x – 1
∴ f(-1.5) = -1.5 – 1 = -2.5

(iv) f(2) + f(-2) For x > 1, f(x) = x + 2
∴ f(2) = 2 + 2 = 4
For -3 < x < -1, f(x) = x – 1
∴ f(-2)= -2 – 1 = -3
f(2) + f(-2) = 4 + (-3) = 1

(v) f(-5) is not defined, since domain of x is {X/X ∈ (-3, ∞)}

Question 2.
If f: R{0}R is defined by f(x) = \(x^{3}-\frac{1}{x^{3}}\); then show that f(x) + \(f\left(\frac{1}{x}\right)\) = 0.
Solution:
Given f(x) = \(x^{3}-\frac{1}{x^{3}}\) ……(i)
Now \(f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^{3}-\frac{1}{\left(\frac{1}{x}\right)^{3}}=\frac{1}{x^{3}}-x^{3}\) ……(2)
Add (1) and (2)
\(f(x)+f\left(\frac{1}{x}\right)=\left(x^{3}-\frac{1}{x^{3}}\right)+\left(\frac{1}{x^{3}}-x^{3}\right)\) = 0
∴ f(x) + \(f\left(\frac{1}{x}\right)\) = 0

Question 3.
If f : R → R is defined by f(x) = \(\frac{1-x^{2}}{1+x^{2}}\), then show that f(tan θ) = cos 2θ.
Solution:
Given f(x) = \(\frac{1-x^{2}}{1+x^{2}}\)

∴ f(tan θ) = cos 2θ

Question 4.
If f : R\{±1} → R is defined by f(x) = \(\log \left|\frac{1+x}{1-x}\right|\), then show that \(f\left(\frac{2 x}{1+x^{2}}\right)\) = 2f(x)
Solution:
f : R\{±1} → R and f(x) = \(\log \left|\frac{1+x}{1-x}\right|\)

∴ \(f\left(\frac{2 x}{1+x^{2}}\right)\) = 2f(x)

Question 5.
If A = {-2, -1, 0, 1, 2} and f : A → B is a surjection defined by f(x) = x² + x + 1, then find B.
Solution:
A = {-2, -1, 0, 1, 2} and f : A → B, f(x) = x² + x + 1
f : A → B is a surjection
f(-2) = (-2)² + (-2) + 1
= 4 – 2 + 1
= 3
f(-1) = (-1)² + (-1) + 1
= 1 – 1 + 1
= 1
f(0) = 0² + 0 + 1
= 0 + 0 + 1
= 1
f(1) = 1² + 1 + 1
= 1 + 1 + 1
= 3
f(2) = 2² + 2 + 1
= 4 + 2 + 1
= 7
∴ B = f(A) = {3, 1, 7}

Question 6.
If A = {1, 2, 3, 4} and f : A → R is a function defined by f(x) = \(\frac{x^{2}-x+1}{x+1}\), then find the range of f.
Solution:
A= {1, 2, 3, 4}

∴ Range of f = f(A) = \(\left\{\frac{1}{2}, 1, \frac{7}{4}, \frac{13}{5}\right\}\)

Question 7.
If f(x + y) = f(xy) ∀ x, y ∈ R then prove that f is a constant function.
Solution:
Given f(x + y) = f(x y), x, y ∈ R
take x = y = 0
⇒ f(0) = f(0) ………(1)
Let x = 1, y = 0
⇒ f(1) = f(0) ……..(2)
Let x = 1, y = 1
⇒ f(2) = f(1) ………(3)
from (1), (2), (3)
f(0) = f(1) = f(2)
⇒ f(0) = f(2)
Similarly f(3) = f(0)
f(4) = f(0)
and so on
f(n) = f(0)
∴ f is a constant function

II.

Question 1.
If A = {x | -1 ≤ x ≤ 1}, f(x) = x², g(x) = x³, which of the following are surjections?
(i) f : A → A
(ii) g : A → A
Solution:
(i) ∵ A = {x | -1 ≤ x ≤ 1} and f(x) = x²
This implies f(x) is a function from A to A
(i.e.,) f : A → A
Now let y ∈ A
If f(x) = y then x² = y
x = √y
So, if y = -1 then x = √-1 ∉ A
∴ f : A → A is not a surjection.

(ii) ∵ A = {x | -1 ≤ x ≤ 1} and g(x) = x³
⇒ g : A → A
Let y ∈ A. Then g(x) = y
⇒ x³ = y
⇒ x = \((y)^{1 / 3}\) ∈ A
So if y = -1 then x = -1 ∈ A
y = 0, then x = 0 ∈ A
y = 1, then x = 1 ∈ A
∴ g : A → A is a surjections.

Question 2.
Which of the following are injections or surjections or bijections? Justify your answers.
(i) f : R → R defined by f(x) = \(\frac{2 x+1}{3}\)
Solution:
f(x) = \(\frac{2 x+1}{3}\)
Let x₁, x₂ ∈ R
∵ f(x₁) = f(x₂)
⇒ \(\frac{2 x_{1}+1}{3}=\frac{2 x_{2}+1}{3}\)
⇒ 2x₁ + 1 = 2x₂ + 1
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂
∵ f(x₁) = f(x₂) ⇒ x₁ = x₂, ∀ x₁, x₂ ∈ R
So f(x) = \(\frac{2 x+1}{3}\), f : R → R is an injection
If y ∈ R (co-domain) then y = \(\frac{2 x+1}{3}\)
⇒ x = \(\frac{3 y-1}{2}\)
Then f(x) = \(\frac{2 x+1}{3}=\frac{2\left(\frac{3 y-1}{2}\right)+1}{3}=y\)
∴ f is a surjection
∴ f : R → R defined by f(x) = \(\frac{2 x+1}{3}\) is a bijection

(ii) f : R → (0, ∞) defined by f(x) = 2x
Solution:
Let x₁, x₂ ∈ R
∵ f(x₁) = f(x₂)
⇒ \(2^{x_{1}}=2^{x_{2}}\)
⇒ x₁ = x₂
∴ f(x₁) = f(x₂) ⇒ x₁ = x₂ ∀ x₁, x₂ ∈ R
∴ f(x) = 2x, f : R → (0, ∞) is injection
If y ∈ (0, ∞) and y = 2x ⇒ x = log₂ (y)
Then f(x) = 2x
= \(2^{\log _{2}(y)}\)
= y
∴ f is a surjection
Hence f is a bijection.

(iii) f : (0, ∞) → R defined by f(x) = log_ex
Solution:
Let x₁, x₂ e (0, ∞)
f(x₁) = f(x₂)
⇒ \(\log _{e}\left(x_{1}\right)=\log _{e}\left(x_{2}\right)\)
⇒ x₁ = x₂
∵ f(x₁) = f(x₂)
⇒ x₁ = x₂ ∀ x₁, x₂ ∈ (0, ∞)
∴ f(x) is injection.
Let y ∈ R.
y = log_ex ⇒ x = e^y
Then f(x) = log_ex
= log_e(e^y)
= y . log_ee
= y(1)
= y
∴ f is a surjection.
∴ f is a bijection.

(iv) f : [0, ∞) → [0, ∞) defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ [0, ∞) (i.e.,) domain of f.
Now f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = x₂
∵ x₁, x₂ ≥ 0
∴ f(x) = x², f : {0, ∞) → {0, ∞) is injection
Let y ∈ (0, ∞), co-domain of f
Let y = x² ⇒ x = √y, ∵ y ≥ 0
Then f(x) = x²
= \((\sqrt{y})^{2}\)
= y
∴ f is surjection.
Hence f is a bijection.

(v) f : R → [0, ∞) defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ R.
f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = ±x₂, ∵ x₁, x₂ ∈ R
Hence f is not injection
Let y ∈ [0, ∞)
y = x²
⇒ x = ±√y, where y ∈ [0, ∞)
Then f(x) = x²
= \((\sqrt{y})^{2}\)
= y
∴ f is surjection
Hence f is not a bijection

(vi) f : R → R defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ R, (domain of f)
∴ f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = ±x₂, ∵ x₁, x₂ ∈ R
∴ f(x) is not injection
For elements that belong to (-∞, 0) codomain of f has no pre-image in f.
∴ f is not a surjection
Hence f is neither injection nor surjection.

Question 3.
Is g = {(1, 1) (2, 3) (3, 5) (4, 7)} is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}. If this is given by the formula g(x) = ax + b, then find a and b.
Solution:
A = {1, 2, 3, 4}; B = {1, 3, 5, 7}
g : {(1, 1), (2, 3), (3, 5), (4, 7)}
∵ g(1) = 1, g(2) = 3, g(3) = 5, g(4) = 7
So for each element a ∈ A, there exists a unique b ∈ B such (a, b) ∈ g
∴ g : A → B is a function
Given g(x) = ax + b, ∀ x ∈ A
g(1) = (a) + b = 1
⇒ a + b = 1 ……..(1)
g(2) = 2a + b = 3
⇒ 2a + b = 3 …….(2)
Solve (1) and (2)
a = 2, b = -1

Question 4.
If the function f : R → R defined by f(x) = \(\frac{3^{x}+3^{-x}}{2}\), then show that f(x + y) + f(x – y) = 2f(x) f(y).
Solution:
f : R → R and f(x) = \(\frac{3^{x}+3^{-x}}{2}\)

∴ f(x + y) + f(x – y) = 2 f(x).f(y)

Question 5.
If the function f : R → R defined by f(x) = \(\frac{4^{x}}{4^{x}+2}\), then show that f(1 – x) = 1 – f(x) and hence reduce the value of \(f\left(\frac{1}{4}\right)+2 f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)\)
Solution:


∴ f(1 – x) = 1 – f(x)

Question 6.
If the function f : {-1, 1} → {0, 2), defined by f(x) = ax + b is a surjection, then find a and b.
Solution:
f : {-1, 1} → {0, 2} and f(x) = ax + b is a surjection
Given f(-1) = 0 and f(1) = 2 (or) f(-1) = 2, f(1) = 0
Case (i):
f(-1) = 0 and f(1) = 2
a(-1) + b = 0 ⇒ -a + b = 0 ……..(1)
a(1) + b = 2 ⇒ a + b = 2 ……(2)
Solve eq’s (1) and (2), we get a = 1, b = 1
Case (ii):
f(-1) = 2 and f(1) = 0
a(-1) + b = 2 ⇒ -a + b = 2 ……(3)
a(1) + b = 0 ⇒ a + b = 0 ……….(4)
Solve eq’s (3) and (4), we get a = -1, b = 1
Hence a = ±1 and b = 1

Question 7.
If f(x) = cos (log x), then show that \(f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]=0\)
Solution:
Given f(x) = cos(log x)
\(f\left(\frac{1}{x}\right)=\cos \left(\log \left(\frac{1}{x}\right)\right)\)
= cos(log 1 – log x)
= cos(-log x)
= cos (log x) (∵ log 1 = 0)
Similarly
\(f\left(\frac{1}{y}\right)\) = cos (log y)
\(f\left(\frac{x}{y}\right)=\cos \log \left(\frac{x}{y}\right)\)
= cos (log x – log y)
and f(x y) = cos log (x y) = cos (log x + log y)
\(f\left(\frac{x}{y}\right)\) + f(x y) = cos (log x – log y) + cos (log x + log y)
= 2 cos (log x) cos (log y)
[∵ cos (A – B) + cos (A + B) = 2 cos A . cos B]
LHS = \(f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]\)
= cos (log x) cos (log y) – \(\frac{1}{2}\) [2 cos (log x) cos (log y)]
= 0