Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(b) will help students to clear their doubts quickly.
Intermediate 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(b)
I.
Question 1.
If the quadratic equations ax2 + 2bx + c = 0 and ax2 + 2cx + b = 0, (b ≠ c) have a common root, then show that a + 4b + 4c = 0
Solution:
Let α be the common roots of the equations ax2 + 2bx + c = 0 and ax2 + 2cx + b = 0
aα2 + 2bα + c = 0
aα2 + 2cα + b = 0
on Subtracting,
2α(b – c) + c – b = 0
2α(b – c) = b – c
2α = 1 (b ≠ c)
α = \(\frac{1}{2}\)
Substitute α = \(\frac{1}{2}\) in ax2 + 2bx + c = 0 is
\(a\left(\frac{1}{4}\right)+2 b \frac{1}{2}+c=0\)
⇒ a + 4b + 4c = 0
∴ a + 4b + 4c = 0
Question 2.
If x2 – 6x + 5 = 0 and x2 – 12x + p = 0 have a common root, then find p.
Solution:
Given x2 – 6x + 5 = 0, x2 – 12x + p = 0 have a common root.
If α is the common root then
α2 – 6α + 5 = 0, α2 – 12α + p = 0
α2 – 6α + 5 = 0
⇒ (α – 1) (α – 5) = 0
⇒ α = 1 or 5
If α = 1 then α2 – 12α + p = 0
⇒ 1 – 12 + p = 0
⇒ p = 11
If α = 5 then α2 – 12α + p = 0
⇒ 25 – 60 + p = 0
⇒ p = 35
∴ p = 11 or 35
Question 3.
If x2 – 6x + 5 = 0 and x2 – 3ax + 35 = 0 have a common root, then find a.
Solution:
The roots of the equation x2 – 6x + 5 = 0 are
(x – 1) (x – 5) = 0
⇒ x = 1, x = 5
Case (i): x = 1 is a common root then it is also root for the equation x2 – 3ax + 35 = 0
⇒ 1 – 3a(1) + 35 = 0
⇒ a = 12
Case (ii): x = 5 is a common root then
(5)2 – 3a(5) + 35 = 0
⇒ 60 – 15a = 0
⇒ a = 4
∴ a = 12 (or) a = 4
Question 4.
If the equation x2 + ax + b = 0 and x2 + cx + d = 0 have a common root and the first equation has equal roots, then prove that 2(b + d) = ac.
Solution:
Let α be the common root.
∴ x2 + ax + b = 0 has equal roots.
Its roots are α, α
α + α = -a
⇒ α = \(-\frac{a}{2}\)
α . α = b
⇒ α2 = b
∴ α is a root of x2 + cx + d = 0
⇒ α2 + cα + d = 0
⇒ b + c(\(-\frac{a}{2}\)) + d = 0
⇒ 2(b + d) = ac
Question 5.
Discuss the signs of the following quadratic expressions when x is real.
(i) x2 – 5x + 4
Solution:
x2 – 5x + 4 = (x – 1) (x – 4)
a = 1 > 0
The expression x2 – 5x + 2 is positive if x < 1 or x > 4 and is negative if 1 < x < 4
(ii) x2 – x + 3
Solution:
∆ = b2 – 4ac
= (-1)2 – 4 (1) (3)
= 1 – 12
= -11 < 0
a = 1 > 0, ∆ < 0
⇒ The given expression is positive for all real x.
Question 6.
For what values of x, the following expressions are positive?
(i) x2 – 5x + 6
Solution:
x2 – 5x + 6 = (x – 2) (x – 3)
Roots of x2 – 5x + 6 = 0 are 2, 3 which are real.
The expression x2 – 5x + 6 is positive if x < 2 or x > 3
∴ a = 1 > 0
(ii) 3x2 + 4x + 4
Solution:
Here a = 3, b = 4, c = 4
∆ = b2 – 4ac
= 16 – 48
= -32 < 0
∴ 3x2 + 4x + 4 is positive ∀ x ∈ R
∴ a = 3 > 0 and ∆ < 0
ax2 + bx + c and ‘a’ have same sign ∀ x ∈ R, if ∆ < 0
(iii) 4x – 5x2 + 2
Solution:
Roots of 4x – 5x2 + 2 = 0 are \(\frac{-4 \pm \sqrt{16+40}}{-10}\)
i.e., \(\frac{2 \pm \sqrt{14}}{5}\) which is real
∴ The expression 4x – 5x2 + 2 is positive when
\(\frac{2-\sqrt{14}}{5}<x<\frac{2+\sqrt{14}}{5}\) [∵ a = -5 < 0]
(iv) x2 – 5x + 14
Solution:
Here a = 1, b = -5, c = 14
∆ = b2 – 4ac
= 25 – 56
= -31 < 0
∴ ∆ < 0 ∵ a = 1 > 0 and ∆ < 0
⇒ x2 – 5x + 14 is positive ∀ x ∈ R.
Question 7.
For what values of x, the following expressions are negative?
(i) x2 – 7x + 10
Solution:
x2 – 7x + 10 = (x – 2)(x – 5)
Roots of x2 – 7x + 10 = 0 are 2, 5 which are real.
∴ The expression x2 – 7x + 10 is negative if 2 < x < 5, ∵ a = 1 > 0
(ii) 15 + 4x – 3x2
Solution:
The roots of 15 + 4x – 3x2 = 0 are \(\frac{-4 \pm \sqrt{16+180}}{-6}\)
i.e., \(\frac{-5}{3}\), 3
∴ The expression 15 + 4x – 3x2 is negative if
-5x < \(\frac{-5}{3}\) or x > 3, ∵ a = -3 < 0
(iii) 2x2 + 5x – 3
Solution:
The roots of 2x2 + 5x – 3 = 0 are \(\frac{-5 \pm \sqrt{25+24}}{4}\)
i.e., -3, \(\frac{1}{2}\)
∴ The expression 2x2 + 5x – 3 is negative if -3 < x < \(\frac{1}{2}\), ∵ a = 2 > 0
(iv) x2 – 5x – 6
Solution:
x2 – 5x – 6 = (x – 6) (x + 1)
Roots of x2 – 5x – 6 = 0 are -1, 6 which are real.
∴ The expression x2 – 5x – 6 is negative if -1 < x < 6, ∵ a = 1 > 0
Question 8.
Find the changes in the sign of the following expressions and find their extreme values.
Hint: Let α, β are the roots of ax2 + bx + c = 0 and α < β
(1) If x < α or x > β, ax2 + bx + c and ‘a’ have same sign.
(2) If α < x < β, ax2 + bx + c and ‘a’ have opposite sign.
(i) x2 – 5x + 6
Solution:
x2 – 5x + 6 = (x – 2) (x – 3)
(1) If 2 < x < 3, the sign of x2 – 5x + 6 is negative, ∵ a = 1 > 0
(2) If x < 2 or x > 3, the sign of x2 – 5x + 6 is positive, ∵ a = 1 > 0
Since a > 0, the minimum value of x2 – 5x + 6 is \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(1)(6)-(-5)^{2}}{4(1)}\)
= \(\frac{24-25}{4}\)
= \(-\frac{1}{4}\)
Hence the extreme value of the expression x2 – 5x + 6 is \(-\frac{1}{4}\)
(ii) 15 + 4x – 3x2
Solution:
15 + 4x – 3x2 = 15 + 9x – 5x – 3x2
= 3(5 + 3x) – x(5 + 3x)
= (3 – x) (5 + 3x)
(1) If \(-\frac{5}{3}\) < x < 3 the sign of 15 + 4x – 3x2 is positive, ∵ a = -3 < 0
(2) If x < \(-\frac{5}{3}\) or x > 3, the sign of 15 + 4x – 3x2 is negative, ∵ a = -3 < 0
Since a < 0, the maximum value of 15 + 4x – 3x2 is \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(-3)(15)-16}{4(-3)}\)
= \(\frac{49}{3}\)
Hence the extreme value of the expression 15 + 4x – 3x2 is \(\frac{49}{3}\)
Question 9.
Find the maximum or minimum of the following expressions as x varies over R.
(i) x2 – x + 7
Solution:
a = 1 > 0,
minimum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{28-1}{4}\)
= \(\frac{27}{4}\)
(ii) 12x – x2 – 32
Solution:
a = -1 < 0,
maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{128-144}{-4}\)
= 4
(iii) 2x + 5 – 3x2
Solution:
a = -3 < 0,
maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{(4)(-3)(5)-(2)^{2}}{4 \times-3}\)
= \(\frac{16}{3}\)
(iv) ax2 + bx + a
Solution:
If a < 0, then maximum value = \(\frac{4 a \cdot a-b^{2}}{4 a}\) = \(\frac{4 a^{2}-b^{2}}{4 a}\)
If a > 0, then minimum value = \(\frac{4 a \cdot a-b^{2}}{4 a}\) = \(\frac{4 a^{2}-b^{2}}{4 a}\)
II.
Question 1.
Determine the range of the following expressions.
(i) \(\frac{x^{2}+x+1}{x^{2}-x+1}\)
Solution:
Let y = \(\frac{x^{2}+x+1}{x^{2}-x+1}\)
⇒ x2y – xy + y = x2 + x + 1
⇒ x2y – xy + y – x2 – x – 1 = 0
⇒ x2 (y – 1) – x(y + 1) + (y – 1) = 0
∴ x is real ⇒ b2 – 4ac ≥ 0
⇒ (y + 1 )2 – 4(y – 1 )2 ≥ 0
⇒ (y + 1)2 – (2y – 2)2 ≥ 0
⇒ (y + 1 + 2y – 2) (y + 1 – 2y + 2) ≥ 0
⇒ (3y – 1) (-y + 3) ≥ 0
⇒ -(3y – 1) (y – 3) ≥ 0
a = Coeff of y2 = -3 < 0
But The expression ≥ 0
⇒ y lies between \(\frac{1}{3}\) and 3
∴ The range of \(\frac{x^{2}+x+1}{x^{2}-x+1}\) is [\(\frac{1}{3}\), 0]
(ii) \(\frac{x+2}{2 x^{2}+3 x+6}\)
Solution:
Let y = \(\frac{x+2}{2 x^{2}+3 x+6}\)
Then 2yx2 + 3yx + 6y = x + 2
⇒ 2yx2 + (3y – 1)x + (6y – 2) = 0
∴ x is real ⇒ discriminant ≥ 0
⇒ (3y – 1)2 – 4(2y)(6y – 2) ≥ 0
⇒ 9y2 + 1 – 6y – 48y2 + 16y ≥ 0
⇒ -39y2 + 10y + 1 ≥ 0
⇒ 39y2 – 10y – 1 < 0
⇒ 39y2 – 13y + 3y – 1 < 0
⇒ 13y(3y – 1) + 1(3y – 1) ≤ 0
⇒ (3y – 1) (13y + 1) ≤ 0
∴ a = Coeff of y2 = 39 > 0 and the exp ≤ 0
⇒ y lies between \(\frac{-1}{13}\) and \(\frac{1}{3}\)
∴ Range of \(\frac{x+2}{2 x^{2}+3 x+6}\) is \(\left[-\frac{1}{13}, \frac{1}{3}\right]\)
(iii) \(\frac{(x-1)(x+2)}{x+3}\)
Solution:
Let y = \(\frac{(x-1)(x+2)}{x+3}\)
⇒ yx + 3y = x2 + x – 2
⇒ x2 + (1 – y)x – 3y – 2 = 0
x ∈ R ⇒ (1 – y2) – 4(-3y – 2) ≥ 0
⇒ 1 + y2 – 2y + 12y + 8 ≥ 0
⇒ y2 + 10y + 9 ≥ 0
y2 + 10y + 9 = 0
⇒ (y + 1) (y + 9) = 0
⇒ y = -1, -9
y2 + 10y + 9 ≥ 0
∴ a = Coeff of y2 = 1 > 0 and exp ≥ 0
⇒ y ≤ -9 or y ≥ -1
∴ Range = (-∞, -9] ∪ [-1, ∞)
(iv) \(\frac{2 x^{2}-6 x+5}{x^{2}-3 x+2}\)
Solution:
Let y = \(\frac{2 x^{2}-6 x+5}{x^{2}-3 x+2}\)
⇒ yx2 – 3yx + 2y = 2x2 – 6x + 5
⇒ (y – 2)x2 + (6 – 3y)x + (2y – 5) = 0
x ∈ R ⇒ (6-3y)2 – 4(y – 2) (2y – 5) ≥ 0
⇒ 36 + 9y2 – 36y – 4(2y2 – 9y + 10) ≥ 0
⇒ 36 + 9y2 – 36y – 8y2 + 36y – 40 ≥ 0
⇒ y2 – 4 ≥ 0
y2 – 4 = 0
⇒ y2 = 4
⇒ y = ±2
y2 – 4 ≥ 0
⇒ y ≤ -2 or y ≥ 2
⇒ y does not lie between -2, 2,
∵ y2 Coeff is > 0 and exp is also ≥ 0
∴ Range of \(\frac{2 x^{2}-6 x+5}{x^{2}-3 x+2}\) is (-∞, -2] ∪ [2, ∞)
Question 2.
Prove that \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4, if x is real.
Solution:
Let y = \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\)
⇒ y = \(\frac{x+1+3 x+1-1}{(3 x+1)(x+1)}\)
⇒ y = \(\frac{4 x+1}{3 x^{2}+4 x+1}\)
⇒ 3yx2 + 4yx + y = 4x + 1
⇒ 3yx2 + (4y – 4)x + (y – 1) = 0
x ∈ R ⇒ (4y – 4)2 – 4(3y)(y – 1) ≥ 0
⇒ 16y2 + 16 – 32y – 12y2 + 12y ≥ 0
⇒ 4y2 – 20y + 16 ≥ 0
4y2 – 20y + 16 = 0
⇒ y2 – 5y + 4 = 0
⇒ (y – 1)(y – 4) = 0
⇒ y = 1, 4
4y2 – 20y + 16 ≥ 0
⇒ y ≤ 1 or y ≥ 4
⇒ y does not lie between 1 and 4
Since y2 Coeff is the and exp ≥ 0.
Question 3.
If x is real, prove that \(\frac{x}{x^{2}-5 x+9}\) lies between 1 and \(\frac{-1}{11}\).
Solution:
Let y = \(\frac{x}{x^{2}-5 x+9}\)
⇒ yx2 + (-5y – 1)x + 9y = 0
x ∈ R ⇒ (5y – 1)2 – 4y(9y) ≥ 0
⇒ 25y2 + 1 + 10y – 36y2 ≥ 0
⇒ -11y2 + 10y + 1 ≥ 0 ……….(1)
-11y2 + 10y + 1 = 0
⇒ -11y2 + 11y – y + 1 = 0
⇒ 11y(-y + 1) + 1(-y + 1) = 0
⇒ (-y + 1) (11y + 1) = 0
⇒ y = 1, \(\frac{-1}{11}\)
-11y2 + 10y + 1 ≥ 0
∴ y2 Coeff is -ve, but the exp is ≥ 0 from (1)
⇒ \(\frac{-1}{11}\) ≤ y ≤ 1
⇒ y lies between 1 and \(\frac{-1}{11}\)
Question 4.
If the expression \(\frac{x-p}{x^{2}-3 x+2}\) takes all real value for x ∈ R, then find the bounts for p.
Solution:
Let y = \(\frac{x-p}{x^{2}-3 x+2}\), given y is real
Then yx2 – 3yx + 2y = x – p
⇒ yx2 + (-3y – 1)x + (2y + p) = 0
∵ x is real ⇒ (-3y – 1)2 – 4y(2y + p) ≥ 0
⇒ 9y2 + 6y + 1 – 8y2 – 4py ≥ 0
⇒ y2 + (6 – 4p)y + 1 ≥ 0
∵ y is real ⇒ y2 + (6 – 4p)y + 1 ≥ 0
⇒ The roots are imaginary or real and equal
⇒ ∆ ≤ 0
⇒ (6 – 4p)2 – 4 ≤ 0
⇒ 4(3 – 2p)2 – 4 ≤ 0
⇒ (3 – 2p)2 – 1 ≤ 0
⇒ 4p2 – 12p + 8 ≤ 0
⇒ p2 – 3p + 2 ≤ 0
⇒ (p – 1)(p – 2) ≤ 0
If p = 1 or p = 2 then \(\frac{x-p}{x^{2}-3 x+2}\) is not defined.
∴ 1 < p < 2
Question 5.
If c2 ≠ ab and the roots of (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0 are equal, then show that a3 + b3 + c3 = 3abc or a = 0.
Solution:
Given equation is (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0
Discriminant = 4(a2 – bc)2 – 4(c2 – ab) (b2 – ac)
= 4[(a2 – bc)2 – (c2 – ab) (b2 – ac)]
= 4(a4 + b2c2 – 2a2bc – b2c2 + ac3 + ab3 – a2bc)
= 4(a4 + ab3 + ac3 – 3a2bc)
= 4a(a3 + b3 + c3 – 3abc)
The roots are equal ⇒ discriminant = 0
4a(a3 + b3 + c3 – 3abc) = 0
a = 0 or a3 + b3 + c3 – 3abc = 0
i.e., a = 0 or a3 + b3 + c3 = 3abc