OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Chapter Test

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S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Chapter Test

Solve the following equations :

Question 1.
5^x+1 + 5^2-x = 5³ + 1
Solution:
Given equation be
5^x+1 + 5^2-x = 5³ + 1 …(1)
putting 5^x = t in equation (1); we have
t x 5 + \(\frac { 25 }{ t }\) = 126
⇒ 5t² – 126t + 25 = 0
⇒ 5t² – 125t – t + 25 = 0
⇒ 5t (t – 25) – 1 (t – 25) = 0
⇒ (5t – 1) (t – 25) = 0
either 5t – 1 = 0 or t – 25 = 0
⇒ t = \(\frac { 1 }{ 5 }\) or t = 25
⇒ 5^x = 5^-1 or 5^x = 5²
⇒ x = – 1 or x = 2
Thus, x = – 1, 2

Question 2.
\(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}\)
Solution:
Given equation be
\(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}\) … (1)
Putting \(\sqrt{\frac{x}{1-x}}\) = t in eqn. (1); we have
t + \(\frac { 1 }{ t }\) = \(\frac { 13 }{ 6 }\)
⇒ 6t² – 13t + 6 = 0
⇒ 6t² – 9t- 4t + 6 = 0
⇒ 3t (2t – 3) – 2 (2t – 3) = 0
⇒ (3t – 3) (2t – 3) = 0
either 3t – 2 = 0 ⇒ 2t – 3 = 0
⇒ t = \(\frac { 2 }{ 3 }\) or t = \(\frac { 3 }{ 2 }\)

Case-I: When t = \(\frac { 2 }{ 3 }\) ⇒ \(\sqrt{\frac{x}{1-x}}=\frac{2}{3}\)
On squaring both sides, we have
\(\frac{x}{1-x}=\frac{4}{9}\) ⇒ 9x = 4 – 4x
⇒ 13x = 4
⇒ x = \(\frac { 4 }{ 13 }\)

Case-II: When t = \(\frac { 3 }{ 2 }\) ⇒ \(\sqrt{\frac{x}{1-x}}=\frac{3}{2}\)
On squaring both sides ; we have
\(\frac { 4 }{ 13 }\) ⇒ 4x = 9 – 9x
⇒ 13x = 9
⇒ x = \(\frac { 9 }{ 13 }\)
Thus, x = \(\frac { 4 }{ 13 }\) or \(\frac { 9 }{ 13 }\)

Question 3.
(x + 1) (x + 2) (x + 3) (x + 4) = 120
Solution:
Given equation be,
(x + 1) (x + 2) (x + 3) (x + 4) = 120
{(x + 1) (x + 4)} {(x + 2) (x + 3)} = 120
⇒ (x² + 5x + 4) (x² + 5x + 6) = 120 … (1)
putting x² + 5x = t in eqn. (1) ; we have
(t + 4)(t + 6) – 120 = 0
⇒ t² + 10t – 96 = 0
⇒ t² + 16t – 6t – 96 = 0
⇒ t(t + 16) – 6(t + 16) = 0
⇒ (t – 6)(t + 16) = 0
either t – 6 = 0 or t + 16 = 0
⇒ t = 6 or t = – 16

Case-I: When t = 6
⇒ x² + 5x = 6
⇒ x² + 5x – 6 = 0
⇒ x² + 6x – x – 6 = 0
⇒ x (x + 6) – 1 (x + 6) = 0
⇒ (x – 1)(x + 6) = 0
⇒ x = 1, – 6

Case-II: When t = – 16
⇒ x² + 5x + 16 = 0
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
⇒ x = \(\frac{-5 \pm \sqrt{25-64}}{2}=\frac{-5 \pm \sqrt{39} i}{2}\)
Hence, x = 1, – 6, \(\frac{-5 \pm \sqrt{39} i}{2}\)

Question 4.
Prove that both the roots of the equation x² – x – 3 = 0 are irrational.
Solution:
Given quadratic eqn. be
x² – x – 3 = 0 …(1)
On comparing eqn. (1) with
ax² + bx + c = 0
we have, a = 1; b = – 1 ; c = – 3
Here discriminant D = b² – 4ac
= (- 1 )² – 4 x 1 x (- 3)
= 1 + 12 = 13 >0
∴ roots are real and distinct
Further D is not a perfect square.
Thus both roots are irrational. Since irrational roots are always occur in conjugate pairs.

Question 5.
For what values of m will the equation x² – 2mx + 1m – 12 = 0 have
(i) equal roots, (it) reciprocal roots ?
Solution:
Given quadratic eqn. be
x² – 2mx + 7m – 12 = 0 …(1)
On comparing eqn. (1) with
ax² + bx + c = 0 ; we have
a = 1 ; b = – 2m ; c = 7m – 12
(i) Since eqn. (1) have equal roots ∴ D = 0
⇒ b² – 4ac = 0
⇒ (- 2m)² – 4 x 1 (7m – 12) = 0
⇒ 7m² – 28m + 48 = 0
⇒ m² – 7m+12 = 0
⇒ (m – 3) (m – 4) = 0
⇒ m = 3, 4

(ii) Since the roots of eqn. (1) are reciprocal to each other.
∴ product of roots = 1
⇒ \(\frac { c }{ a }\) = 1
⇒ \(\frac{7 m-12}{1}\) = 1
⇒ m = \(\frac { 13 }{ 7 }\)

Question 6.
If one root of 2x² – 5x + k = 0 be double the other, find the value of k.
Solution:
Let α, 2α be the roots of eqn.

Question 7.
If α, ß be the roots of the equation x² – x – 1 = 0, determine the value of
(i) α² + ß² and (ii) α³ + ß³.
Solution:
Given a and P are the roots of the eqn. x² – x – 1 =0
∴ α + ß = – (- 1) = 1 ; αß = – 1
(i) ∴ α² + ß² = (α + ß)² – 2αß
= (1)² – 2(- 1) = 3

(ii) α³ + ß³ = (α + ß)³ – 3αß (α + ß)
= 1³ – 3 (- 1) x 1 = 4

Question 8.
If the roots of the equation ax² + bx + c = 0 be in the ratio 3 : 4, show that 12b² = 49ac.
Solution:
Since the roots of the equation ax² + bx + c = 0 be in the ratio 3 : 4.
Let the roots of given equation are 3α, 4α.
∴ 3α + 4α = – \(\frac { b }{ a }\)
⇒ 7α = – \(\frac { b }{ a }\)
⇒ α = – \(\frac { b }{ 7a }\)
Also (3α) (4α) = \(\frac { c }{ a }\)
⇒ 12α² = \(\frac { c }{ a }\)
⇒ 12 \(\left(-\frac{b}{7 a}\right)^2=\frac{c}{a}\)
⇒ \(\frac{12 b^2}{49 a^2}=\frac{c}{a}\)
⇒ 12b² = 49ac

Question 9.
If x is real, prove that the quadratic expression
(i) (x – 2) (x + 3) + 7 is always positive.
(ii) 4x – 3x² – 2 is always negative.
Solution:
(i) The given expression can be written as (x – 2) (x + 3) + 7 = x² + x + 1
On comparing with ax² + bx + c, we have a = 1 ; b = 1 ; c = 1
Here discriminant D = b² – 4ac
= 1² – 4 x 1 x 1 = – 3 < 0 Here a = 1 > 0
∴ x² + x + 1 > 0 for all x.

(ii) Comparing – 3x² + 4x – 2 with ax² + bx + c
We have, a = – 2, b = 4 ; c = – 2
Here D = b² – 4ac = 4² – (- 3) x (- 1)
= 16 – 24 = – 8 < 0
and Here a = – 3 < 0.
Thus – 3x² + 4x – 2 < 0 ∀ x.

Question 10.
Draw the graph of the quadratic function x² – 4x + 3 and hence find the roots of the equation x² – 4x + 4 = 0. What is the minimum value of the function ?
Solution:
Let y = x² – 4x + 3
The table of values is given as under :

x	0	1	2	3	4	5
y	3	0	-1	0	3	8

it is clear from graph that, it intersects x- axis at two points (+ 1,0) and (3, 0)
Thus x = 1, 3 are the real solutions of given graph.
Further from graph, Minimum value of y = – 1.

Question 11.
For what real values of a, will be expression x² – ax + 1 – 2a², for the real x, be always positive ?
Solution:
On comparing x² – ax + 1 – 2a² with Ax² + Bx + C,
we have A = 1; B = – a; C = 1 – 2a²
Here discriminant D = B² – 4AC = (- a)² – 4(1 – 2a²)
= a² – 4 + 8 a²
= 9a² – 4
Also A = 1 > 0
∴ given expression be always be positive
if D < 0 if 9a² – 4 < 0 if a² < \(\frac { 4 }{ 9 }\)
if |a| < \(\frac { 2 }{ 3 }\) i.e. if – \(\frac { 2 }{ 3 }\) < a < \(\frac { 2 }{ 3 }\)

Question 12.
If x be real, prove that the value of \(\frac{2 x^2-2 x+4}{x^2-4 x+3}\) cannot lie between – 7 and 1.
Solution:
Let y = \(\frac{2 x^2-2 x+4}{x^2-4 x+3}\)
⇒ y (x² – 4x + 3) = 2x² – 2x + 4
⇒ x² (y – 2) + x (2 – 4y) + 3y – 4 = 0
Since x is real ∴ D > 0
⇒ b² – 4ac ≥ 0
⇒ (2 – 4y)² – 4 (y – 2) (3y – 4) ≥ 0
⇒ 4 [(1 – 2y)² – (y – 2) (3y – 4)] ≥ 0
⇒ [4y² – 4y + 1 – (3t – 10y + 8)] ≥ 0
⇒ y² + 6y – 7 ≥ 0
⇒ (y – 1)(y + 7) ≥ 0
The critical points are given by 1 and – 7 Then by method of intervals, we have y ≤ – 7 or y ≥ 1

Hence y can’t lies between – 7 and 1.
i.e. value of \(\frac{2 x^2-2 x+4}{x^2-4 x+3}\) cannot lie between – 7 and 1.

Question 13.
If the roots of the equation qx² + 2px + 2q = 0 are real and unequal, prove that the roots of the equation (p + q) x² + 2qx +(p – q) = 0 are imaginary.
Solution:
Given quadratic equation be,
qx² + 2px + 2q = 0 …(1)
On comparing eqn. (1) with
ax² + bx + c = 0, we have
a = q; b = 2p ; c = 2q
Since the roots of eqn. (1) are real and unequal.
∴ D > 0
⇒ b² – 4ac > 0
⇒ (2p)² – 4 x q x 2q > 0
⇒ 4p² – 8q² > 0
⇒ p² – 2q² > 0 …(2)
Also given quadratic eqn. be
(p + q)x² + 2qx + (p – q) = 0 …(3)
Here discriminant
D = (2q)² – 4(p + q)(p – q)
= 4 q² – 4(p² – q²)
= 4 [q² – p² + q²]
= 4(2q² – p²)
= – 4 (p² – 2q²) < 0 [using (2)]
Hence the roots of eqn. (3) are imaginary.

Question 14.
If α, ß be the roots of x² – px + q = 0, find the value of α⁵ß⁷ + α⁷ß⁵ in terms of p and q.
Solution:
Since α and ß are the roots of
x² – px + q = 0
∴ α + ß = p; αß = q
Thus, α⁵ß⁷ + α⁷ß⁵ = α⁵ß⁵ (ß² + α²)
= (αß)⁵ [(α + ß)² – 2αß]
= q⁵[P² – 2q]

Question 15.
If the difference between the roots of the equation x² + ax + 1 = 0 is less than \(\sqrt{5}\), then the set of possible values of a is (a) (3, ∞)
(b) (- ∞, – 3)
(c)(- 3, 3)
(d) (- 3, ∞)
Solution:
Let a and P are the roots of eqn. x² + ax + 1 = 0
∴ α + ß = – a ; αß = 1 …(1)
Also, it is given that | α – ß | < \(\sqrt{5}\)
⇒ \(\sqrt{(\alpha+\beta)^2-4 \alpha \beta}<\sqrt{5}\)
⇒ \(\sqrt{(-a)^2-4}<\sqrt{5}\)
⇒ \(\sqrt{a^2-4}<\sqrt{5}\)
On squaring both sides ; we have
a² – 4 < 5
⇒ a² < 9
⇒ | a | < 3
⇒ – 3 < a < 3
⇒ a ∈ (- 3, 3)

Question 16.
Let α, ß be the roots of the equation x² – px + r = 0 and α/2, 2ß be the roots of the equation x² – qx + r = 0, then the value of r is
(a) \(\frac { 2 }{ 9 }\) (p – q)(2q – p)
(b) \(\frac { 2 }{ 9 }\) (q – p)(2p – q)
(c) \(\frac { 2 }{ 9 }\) (q – 2p)(2q – p)
(d) \(\frac { 2 }{ 9 }\) (2p – q) (2q – p)
Solution:
Given a and p are the roots of equation
x² – px + r = 0
∴ α + ß = p ; αß = r …(1)
Also, given \(\frac { α }{ 2 }\), 2ß are the roots of eqn.

Question 17.
α, ß are the roots of ax² + 2bx + c = 0 and α + δ, ß + δ are the roots of Ax² + 2Bx + C = 0, then what is (b² – ac)/(B² – AC) equal to ?
(a) (8/B)²
(b) (a/A)²
(c) (a² b²)/(A²B²)
(d) ab/AB
Solution:
Given α, ß are the roots of ax² + 2bx + c = 0

Question 18.
If α, ß are the roots of the equation x² – 2x – 1 = 0, then what is the value of α²ß^-2 + α^-2ß² ?
(a) – 2
(8) 0
(c) 30
(d) 34
Solution:
Given α and ß are the roots of eqn. x² – 2x – 1 =0

Question 19.
If the roots of the quadratic equation x² + px + q = 0 are tan 30° and tan 15°, then value of 2 + q – p is
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
Given tan 30° and tan 15° are the roots of quadratic equation x² + px + q = 0
∴ tan 30° +tan 15° = – p …(1)
and tan 30° tan 15° = q …(2)
Now tan 45° = tan (30° + 15°)
tan 45° = \(\frac{\tan 30^{\circ}+\tan 15^{\circ}}{1-\tan 30^{\circ} \tan 15^{\circ}}\)
⇒ 1 = \(\frac{-p}{1-q}\) [using (1) and (2)]
⇒ 1 – q = – p
⇒ p – q + 1 = 0
⇒ q – p – 1 = 0
⇒ q – p + 2 = 3

Question 20.
If both the roots of the quadratic equation x² – 2kx + k² + k – 5 = 0 are less than 5, then k lies in the interval
(a) (5, 6]
(b) (6, ∞)
(c) (- ∞, 4)
(d) [4, 5]
Solution:
Given quadratic eqn. be
x² – 2kx + k² + k – 5 = 0 …(1)
since each root of eqn. (1) be less than 5
∴ sum of roots of eqn. (1) be less than 10
⇒ \(\frac { – b }{ a }\) < 10 ⇒ 2k < 10 ⇒ k < 5 …(2)
On comparing eqn. (1) with
ax² + bx + c = 0, we have
a = 1 ; b = – 2k and c = k² + k – 5
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{2 k \pm \sqrt{4 k^2-4\left(k^2+k-5\right)}}{2}\)
⇒ x = k ± \(\sqrt{k^2-k^2+5-k}=k \pm \sqrt{5-k}\)
Further, each root be less than 5.
⇒ k + \(\sqrt{5-k}\) < 5
⇒ \(\sqrt{5-k}\) < 5 – k
On squaring both sides ; we have
5 – k < (5 – k)² ⇒ k² – 10k + 25 + k – 5 > 0
⇒ k² – 9k + 20 > 0
⇒ (k – 4) (k – 5) > 0
Critical points are given by k = 4, 5
Then by method of intervals, we have
k < + 4 or k > 5
But k < 5 ∴ k < 4

Question 21.
If α and ß are the roots of ax² + bx + c = 0 and if px² + qx + r = 0 has roots \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\) then r =
(a) a + 2b
(b) a + b + c
(c) ab + bc + ca
(d) abc
Solution:
Given α and ß are the roots of ax² + bx + c = 0

Hence the quadratic equation having roots
\(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\) be given by x² – Sx + P = 0
⇒ x² + (\(\left(\frac{b+2 c}{c}\right)\))x + \(\frac{a+b+c}{c}\) = 0
⇒ cx² + (b + 2c)x + a + b + c = 0 …(4)
On comparing eqn. (3) and eqn. (4) ; we have
p – c ; q = b +2c and r = a + b + c

Question 22.
The quadratic equations x² – 6x + a = 0 and x² – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is
(a) 1
(b) 4
(c) 3
(d) 2
Solution:
Let α and 4ß are the roots of
x² – 6x + a = 0 …(*)
∴ α + 4ß = 6 …(1)
4αß = a …(2)
Also, let α and 3ß are the roots of
x² – cx + 6 = 0 …(**)
∴ α + 3ß = c …(3)
3αß = 6 …(4)
where α be the common root of both given eqns.
From (2) and (4); we have
\(\frac{4 \alpha \beta}{3 \alpha \beta}=\frac{a}{6}\) ⇒ a = \(\frac { 24 }{ 3 }\) = 8
∴ eqn. (*) becomes : x² – 6x + 8 = 0
⇒ (x – 2) (x – 4) = 0 ⇒ x = 2, 4
Taking α = 2, αß = 4 ⇒ ß = 1
∴ c = 2 + 3 = 5
∴ eqn. (**) becomes ;
x² – 5x + 6 = 0 ⇒ x = 2, 3
Clearly 2 be the common root.
When α = 4 ; 4ß = 2 ⇒ ß = \(\frac { 1 }{ 2 }\)
∴ c = 4 + \(\frac { 3 }{ 2 }\) = \(\frac { 11 }{ 2 }\)
∴ eqn. (**) becomes ;
x² – \(\frac { 11 }{ 2 }\) x + 6 = 0
⇒ 2x² – 11x + 12 = 0
⇒ x = \(\frac{11 \pm 5}{4}\) = 4, \(\frac { 3 }{ 2 }\)
But it is given that other roots of both equations are integers.

Question 23.
If α, ß are the roots of the equation λ(x² – x) + x + 5 = 0 and if λ₁ and λ₂ are two values of λ obtained from \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{5}\), then \(\frac{\lambda_1}{\lambda_2^2}+\frac{\lambda_2}{\lambda_1^2}\) equals
(a) 4192
(b) 4144
(c) 4096
(d) 4048
Solution:
Given α and ß are the roots of eqn.

Question 24.
If α, ß be the roots of x² – a (x – 1) + b = 0, then the value of
\(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-\alpha \beta}+\frac{2}{a+b}\) is
(a) \(\frac{4}{a+b}\)
(b) \(\frac{1}{a+b}\)
(c) 0
(d) – 1
Solution:
Given α and ß are the roots of quadratic eqn.
x² – a (x – 1) + b = 0
⇒ x² – ax + a + b = 0 …(1)
∴ α + ß = a ; αß = a + b
Also, a and P both satisfies eqn. (1)
∴ α² – aα + a + b = 0
⇒ α² – aα = – (a + b)
and ß² – αß + a + b = 0
⇒ ß² – aß = – (a + b)
∴ \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}\)
= \(\frac{1}{-(a+b)}+\frac{1}{-(a+b)}+\frac{2}{a+b}\)
= \(\frac{-2}{a+b}+\frac{2}{a+b}\) = 0