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 ax^{2} + 2bx + c = 0 and ax^{2} + 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 ax^{2} + 2bx + c = 0 and ax^{2} + 2cx + b = 0

aα^{2} + 2bα + c = 0

aα^{2} + 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 ax^{2} + 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 x^{2} – 6x + 5 = 0 and x^{2} – 12x + p = 0 have a common root, then find p.

Solution:

Given x^{2} – 6x + 5 = 0, x^{2} – 12x + p = 0 have a common root.

If α is the common root then

α^{2} – 6α + 5 = 0, α^{2} – 12α + p = 0

α^{2} – 6α + 5 = 0

⇒ (α – 1) (α – 5) = 0

⇒ α = 1 or 5

If α = 1 then α^{2} – 12α + p = 0

⇒ 1 – 12 + p = 0

⇒ p = 11

If α = 5 then α^{2} – 12α + p = 0

⇒ 25 – 60 + p = 0

⇒ p = 35

∴ p = 11 or 35

Question 3.

If x^{2} – 6x + 5 = 0 and x^{2} – 3ax + 35 = 0 have a common root, then find a.

Solution:

The roots of the equation x^{2} – 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 x^{2} – 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 x^{2} + ax + b = 0 and x^{2} + 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.

∴ x^{2} + ax + b = 0 has equal roots.

Its roots are α, α

α + α = -a

⇒ α = \(-\frac{a}{2}\)

α . α = b

⇒ α^{2} = b

∴ α is a root of x^{2} + cx + d = 0

⇒ α^{2} + 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) x^{2} – 5x + 4

Solution:

x^{2} – 5x + 4 = (x – 1) (x – 4)

a = 1 > 0

The expression x^{2} – 5x + 2 is positive if x < 1 or x > 4 and is negative if 1 < x < 4

(ii) x^{2} – x + 3

Solution:

∆ = b^{2} – 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) x^{2} – 5x + 6

Solution:

x^{2} – 5x + 6 = (x – 2) (x – 3)

Roots of x^{2} – 5x + 6 = 0 are 2, 3 which are real.

The expression x^{2} – 5x + 6 is positive if x < 2 or x > 3

∴ a = 1 > 0

(ii) 3x^{2} + 4x + 4

Solution:

Here a = 3, b = 4, c = 4

∆ = b^{2} – 4ac

= 16 – 48

= -32 < 0

∴ 3x^{2} + 4x + 4 is positive ∀ x ∈ R

∴ a = 3 > 0 and ∆ < 0

ax^{2} + bx + c and ‘a’ have same sign ∀ x ∈ R, if ∆ < 0

(iii) 4x – 5x^{2} + 2

Solution:

Roots of 4x – 5x^{2} + 2 = 0 are \(\frac{-4 \pm \sqrt{16+40}}{-10}\)

i.e., \(\frac{2 \pm \sqrt{14}}{5}\) which is real

∴ The expression 4x – 5x^{2} + 2 is positive when

\(\frac{2-\sqrt{14}}{5}<x<\frac{2+\sqrt{14}}{5}\) [∵ a = -5 < 0]

(iv) x^{2} – 5x + 14

Solution:

Here a = 1, b = -5, c = 14

∆ = b^{2} – 4ac

= 25 – 56

= -31 < 0

∴ ∆ < 0 ∵ a = 1 > 0 and ∆ < 0

⇒ x^{2} – 5x + 14 is positive ∀ x ∈ R.

Question 7.

For what values of x, the following expressions are negative?

(i) x^{2} – 7x + 10

Solution:

x^{2} – 7x + 10 = (x – 2)(x – 5)

Roots of x^{2} – 7x + 10 = 0 are 2, 5 which are real.

∴ The expression x^{2} – 7x + 10 is negative if 2 < x < 5, ∵ a = 1 > 0

(ii) 15 + 4x – 3x^{2}

Solution:

The roots of 15 + 4x – 3x^{2} = 0 are \(\frac{-4 \pm \sqrt{16+180}}{-6}\)

i.e., \(\frac{-5}{3}\), 3

∴ The expression 15 + 4x – 3x^{2} is negative if

-5x < \(\frac{-5}{3}\) or x > 3, ∵ a = -3 < 0

(iii) 2x^{2} + 5x – 3

Solution:

The roots of 2x^{2} + 5x – 3 = 0 are \(\frac{-5 \pm \sqrt{25+24}}{4}\)

i.e., -3, \(\frac{1}{2}\)

∴ The expression 2x^{2} + 5x – 3 is negative if -3 < x < \(\frac{1}{2}\), ∵ a = 2 > 0

(iv) x^{2} – 5x – 6

Solution:

x^{2} – 5x – 6 = (x – 6) (x + 1)

Roots of x^{2} – 5x – 6 = 0 are -1, 6 which are real.

∴ The expression x^{2} – 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 ax^{2} + bx + c = 0 and α < β

(1) If x < α or x > β, ax^{2} + bx + c and ‘a’ have same sign.

(2) If α < x < β, ax^{2} + bx + c and ‘a’ have opposite sign.

(i) x^{2} – 5x + 6

Solution:

x^{2} – 5x + 6 = (x – 2) (x – 3)

(1) If 2 < x < 3, the sign of x^{2} – 5x + 6 is negative, ∵ a = 1 > 0

(2) If x < 2 or x > 3, the sign of x^{2} – 5x + 6 is positive, ∵ a = 1 > 0

Since a > 0, the minimum value of x^{2} – 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 x^{2} – 5x + 6 is \(-\frac{1}{4}\)

(ii) 15 + 4x – 3x^{2}

Solution:

15 + 4x – 3x^{2} = 15 + 9x – 5x – 3x^{2}

= 3(5 + 3x) – x(5 + 3x)

= (3 – x) (5 + 3x)

(1) If \(-\frac{5}{3}\) < x < 3 the sign of 15 + 4x – 3x^{2} is positive, ∵ a = -3 < 0

(2) If x < \(-\frac{5}{3}\) or x > 3, the sign of 15 + 4x – 3x^{2} is negative, ∵ a = -3 < 0

Since a < 0, the maximum value of 15 + 4x – 3x^{2} 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 – 3x^{2} is \(\frac{49}{3}\)

Question 9.

Find the maximum or minimum of the following expressions as x varies over R.

(i) x^{2} – 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 – x^{2} – 32

Solution:

a = -1 < 0,

maximum value = \(\frac{4 a c-b^{2}}{4 a}\)

= \(\frac{128-144}{-4}\)

= 4

(iii) 2x + 5 – 3x^{2}

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) ax^{2} + 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 = \(\frac{x^{2}+x+1}{x^{2}-x+1}\)

⇒ x^{2}y – xy + y = x^{2} + x + 1

⇒ x^{2}y – xy + y – x^{2} – x – 1 = 0

⇒ x^{2} (y – 1) – x(y + 1) + (y – 1) = 0

∴ x is real ⇒ b^{2} – 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 y^{2} = -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 2yx^{2} + 3yx + 6y = x + 2

⇒ 2yx^{2} + (3y – 1)x + (6y – 2) = 0

∴ x is real ⇒ discriminant ≥ 0

⇒ (3y – 1)^{2} – 4(2y)(6y – 2) ≥ 0

⇒ 9y^{2} + 1 – 6y – 48y^{2} + 16y ≥ 0

⇒ -39y^{2} + 10y + 1 ≥ 0

⇒ 39y^{2} – 10y – 1 < 0

⇒ 39y^{2} – 13y + 3y – 1 < 0

⇒ 13y(3y – 1) + 1(3y – 1) ≤ 0

⇒ (3y – 1) (13y + 1) ≤ 0

∴ a = Coeff of y^{2} = 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 = x^{2} + x – 2

⇒ x^{2} + (1 – y)x – 3y – 2 = 0

x ∈ R ⇒ (1 – y^{2}) – 4(-3y – 2) ≥ 0

⇒ 1 + y^{2} – 2y + 12y + 8 ≥ 0

⇒ y^{2} + 10y + 9 ≥ 0

y^{2} + 10y + 9 = 0

⇒ (y + 1) (y + 9) = 0

⇒ y = -1, -9

y^{2} + 10y + 9 ≥ 0

∴ a = Coeff of y^{2} = 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 = \(\frac{2 x^{2}-6 x+5}{x^{2}-3 x+2}\)

⇒ yx^{2} – 3yx + 2y = 2x^{2} – 6x + 5

⇒ (y – 2)x^{2} + (6 – 3y)x + (2y – 5) = 0

x ∈ R ⇒ (6-3y)^{2} – 4(y – 2) (2y – 5) ≥ 0

⇒ 36 + 9y^{2} – 36y – 4(2y^{2} – 9y + 10) ≥ 0

⇒ 36 + 9y^{2} – 36y – 8y^{2} + 36y – 40 ≥ 0

⇒ y^{2} – 4 ≥ 0

y^{2} – 4 = 0

⇒ y^{2} = 4

⇒ y = ±2

y^{2} – 4 ≥ 0

⇒ y ≤ -2 or y ≥ 2

⇒ y does not lie between -2, 2,

∵ y^{2} 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 = \(\frac{4 x+1}{3 x^{2}+4 x+1}\)

⇒ 3yx^{2} + 4yx + y = 4x + 1

⇒ 3yx^{2} + (4y – 4)x + (y – 1) = 0

x ∈ R ⇒ (4y – 4)^{2} – 4(3y)(y – 1) ≥ 0

⇒ 16y^{2} + 16 – 32y – 12y^{2} + 12y ≥ 0

⇒ 4y^{2} – 20y + 16 ≥ 0

4y^{2} – 20y + 16 = 0

⇒ y^{2} – 5y + 4 = 0

⇒ (y – 1)(y – 4) = 0

⇒ y = 1, 4

4y^{2} – 20y + 16 ≥ 0

⇒ y ≤ 1 or y ≥ 4

⇒ y does not lie between 1 and 4

Since y^{2} 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 = \(\frac{x}{x^{2}-5 x+9}\)

⇒ yx^{2} + (-5y – 1)x + 9y = 0

x ∈ R ⇒ (5y – 1)^{2} – 4y(9y) ≥ 0

⇒ 25y^{2} + 1 + 10y – 36y^{2} ≥ 0

⇒ -11y^{2} + 10y + 1 ≥ 0 ……….(1)

-11y^{2} + 10y + 1 = 0

⇒ -11y^{2} + 11y – y + 1 = 0

⇒ 11y(-y + 1) + 1(-y + 1) = 0

⇒ (-y + 1) (11y + 1) = 0

⇒ y = 1, \(\frac{-1}{11}\)

-11y^{2} + 10y + 1 ≥ 0

∴ y^{2} 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 yx^{2} – 3yx + 2y = x – p

⇒ yx^{2} + (-3y – 1)x + (2y + p) = 0

∵ x is real ⇒ (-3y – 1)^{2} – 4y(2y + p) ≥ 0

⇒ 9y^{2} + 6y + 1 – 8y^{2} – 4py ≥ 0

⇒ y^{2} + (6 – 4p)y + 1 ≥ 0

∵ y is real ⇒ y^{2} + (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

⇒ 4p^{2} – 12p + 8 ≤ 0

⇒ p^{2} – 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 c^{2} ≠ ab and the roots of (c^{2} – ab)x^{2} – 2(a^{2} – bc)x + (b^{2} – ac) = 0 are equal, then show that a^{3} + b^{3} + c^{3} = 3abc or a = 0.

Solution:

Given equation is (c^{2} – ab)x^{2} – 2(a^{2} – bc)x + (b^{2} – ac) = 0

Discriminant = 4(a^{2} – bc)^{2} – 4(c^{2} – ab) (b^{2} – ac)

= 4[(a^{2} – bc)^{2} – (c^{2} – ab) (b^{2} – ac)]

= 4(a^{4} + b^{2}c^{2} – 2a^{2}bc – b^{2}c^{2} + ac^{3} + ab^{3} – a^{2}bc)

= 4(a^{4} + ab^{3} + ac^{3} – 3a^{2}bc)

= 4a(a^{3} + b^{3} + c^{3} – 3abc)

The roots are equal ⇒ discriminant = 0

4a(a^{3} + b^{3} + c^{3} – 3abc) = 0

a = 0 or a^{3} + b^{3} + c^{3} – 3abc = 0

i.e., a = 0 or a^{3} + b^{3} + c^{3} = 3abc