OP Malhotra Class 11 Maths Solutions Chapter 11 Inequalities Ex 11(c)

Parents can use OP Malhotra Maths Class 12 Solutions Chapter 11 Inequalities Ex 11(c) to provide additional support to their children.

S Chand Class 11 ICSE Maths Solutions Chapter 11 Inequalities Ex 11(c)

Question 1.
x² – 4x + 3 < 0.
Solution:
x² – 4x + 3 < 0
⇒ (x – 1) (x – 3) < 0
Critical points are given by
x – 1 = 0 and x – 3 = 0
i.e. x = 1, 3
Then by method of intervals, we have

which is the required range.

Question 2.
x² + 5x + 4 > 0.
Solution:
x² + 5x + 4 > 0 ⇒ (x + 1) (x + 4) = 0
Critical points are given by putting
(x – 1) (x – 4) = 0
i.e. x = – 1, – 4
Then by method of intervals, we have

x > – 1 or x < – 4
i.e. x ∈ (- ∞, – 4) u (- 1, ∞)
which is the required range

Question 3.
x² + x – 6 ≥ 0.
Solution:
x² + x – 6 ≥ 0 ⇒ (x – 2) (x + 3) ≥ 0
Then critical points are given by
(x – 2) (x + 3) = 0
i.e., x = 2, – 3
Then by method of intervals, we have

x ≥ 2 or x ≤ – 3
⇒ x ∈ (-∞, – 3] ∪ [2, ∞)
which is the required range.

Question 4.
x² – 16 < 0.
Solution:
Given x² – 16 < 0 ⇒ x² < 16
⇒ | x | < 4 ⇒ – 4 < x < 4
which is the required range.
[∵ | x | < l ⇒ – l < x <l]

Question 5.
x² – 6x + 9 ≥ 0
Solution:
x² – 6x + 9 ≥ 0
⇒ (x – 3)² ≥ 0 which is true for all real x
∴ Range = set of all reals

Question 6.
– x² + 2x + 3 < 0.
Solution:
-x² + 2x + 3 < 0 ⇒ x² – 2x – 3 > 0
⇒ (x + 1) (x- 3) > 0
The critical points are given by
(x + 1) (x – 3) = 0

i.e. x = -1, 3
Then by method of intervals, we have
x < – 1 or x > 3
∴ required range is given by
x ∈ (- ∞, – 1) ∪ (3, ∞).

Question 7.
5x < 2 – 3x².
Solution:
5x < 2 – 3x²
⇒ 3x² + 5x – 2 < 0
⇒ (x + 2) (3x – 1) < 0
⇒ 3 (x + 2) (x – \(\frac { 1 }{ 3 }\)) < 0
⇒ (x + 2) (x – \(\frac { 1 }{ 3 }\)) < 0
Critical points are given by
(x + 2) (x – \(\frac { 1 }{ 3 }\)) = 0
i.e. x = -2, \(\frac { 1 }{ 3 }\)
Then by method of intervals, we have
x < – 2 or x > \(\frac { 1 }{ 3 }\)
which is the required range.

Question 8.
– x² – 4x – 5 < 0.
Solution:
– x² – 4x – 5 < 0 ⇒ x² + 4x + 5 > 0 [if a < b ⇒ – a > – b]
⇒ x² + 4x + 4 + 1 > 0
⇒ (x + 1 )² + 1 > 0
which is true for all x ∈ R
Thus range = set of all real numbers

Question 9.
4x² + 1 > 4x.
Solution:
Given 4x² + 1 > 4x
⇒ 4x² – 4x + 1 > 0 ⇒ (2x – 1)² > 0
which is true for all x ∈ R except x = \(\frac { 1 }{ 2 }\)
[∵ When x = \(\frac { 1 }{ 2 }\) ⇒ 2x – 1 = 0]
∴ required range = R – {\(\frac { 1 }{ 2 }\)}

Question 10.
– x² + x > 0.
Solution:
-x² + x > 0 ⇒ x² – x < 0 [if a > b ⇒ ac < bc if c < 0]
⇒ x (x – 1) < 0
The critical points are given by x (x – 1) = 0
⇒ x = 0, 1
Then by method of intervals, we have
0 < x < 1
∴ Required range is given by x ∈ (0, 1).

Question 11.
6 + x < 2x²
Solution:
Given 6 + x < 2² ⇒ 2x² – x – 6 > 0
⇒ (x – 2) (2x + 3) > 0
⇒ 2(x – 2) [x + \(\frac { 3 }{ 2 }\)] > 0
Critical points are given by
x – 2 = 0 and x + \(\frac { 3 }{ 2 }\) = 0
i.e. x = 2 ,-\(\frac { 3 }{ 2 }\)

Then by method of intervals, we have
x < – \(\frac { 3 }{ 2 }\) or x > 2
which is the required range.

Question 12.
(i) (x – 4)(x + 6) > 0 ;
(ii) 3 – 2x² > 5x
Solution:
(i) (x – 4) (x + 6) > 0
The critical points are given by
(x – 4)(x + 6) = 0
⇒ x = 4, – 6
Then by method of intervals, we have

which is the required range.

(ii) Given 3 – 2x² > 5x
⇒ 2x² + 5x – 3 < 0
⇒ (x + 3) (2x – 1) < 0
The critical points are given by
(x + 3) (2x – 1) = 0

Then by method of intervals, we have
-3 < x < \(\frac { 1 }{ 2 }\)

Question 13.
Find all real values of x which satisfy x² – 3x + 2 > 0 and x² – 3x – 4 ≤ 0.
Solution:
Given x² – 3x + 2 > 0
⇒ (x – 1) (x – 2) > 0
Critical points are given by
(x – 1) (x – 2) = 0 ⇒ x = 1, 2

Then by method of intervals, we have
x < 1 or x > 2 …(1)
Also, x² – 3x – 4 ≤ 0
⇒ (x + 1)(x – 4) = 0
Critical points are given by
(x + 1) (x – 4) = 0
⇒ x = -1, 4
Then by method of intervals, we have -1 < x < 4 …(2)
Thus the values of x which satisfies both given inequalities are given by (x < 1 or x > 2) n (- 1 < x < 4)
⇒ – 1 < x < 1 or 2 < x < 4

Question 14.
The set of values of x for which the inequalities x² – 3x – 10 < 0, 10x – x² – 16 > 0 hold simultaneously is
(a) (-2, 5)
(b) (2, 8)
(c) (-2,8)
(d) (2, 5).
Solution:
Given x² – 3x – 10 < 0
⇒ (x + 2) (x – 5) < 0

i.e. x = -2,5
Then by method of intervals, we have
– 2 < x < 5 …(1) Also, 10x – x² – 16 > 0
⇒ x² – 10x + 16 < 0
⇒(x – 2) (x – 8) = 0
Then critical points are given by
(x – 2) (x – 8) = 0 ⇒ x = 2, 8
Then by method of intervals, we have
2 < x < 8 …(2)

Thus the set of values of x for which the given inequalities hold simultaneously is given by (- 2 < x < 5) ∩ (2 < x < 8)
i.e. 2 < x < 5 ⇒ x ∈ (2, 5) ∴Ans. (d)
Solve the following inequalities :

Question 15.
\(\frac{x+3}{x-1}\) > x
Solution:

⇒ (x + 1) (x – 3) (x – 1) < 0, x ≠ 1
Critical points are given by
(x + 1)(x – 3)(x – 1) = 0
i.e. x = – 1, 3, 1
Then by method of intervals we have
x < -1 or 1 < x < 3

Question 16.
x + 4 < – \(\frac{2}{x+1}\)
Solution:
Given x + 4 < – \(\frac{2}{x+1}\)
⇒ (x + 4) + \(\frac{2}{x+1}\) < 0
⇒ \(\frac{(x+4)(x+1)+2}{x+1}\) < 0
⇒ \(\frac{x^2+5 x+6}{x+1}\) < 0
⇒ \(\frac{(x+2)(x+3)(x+1)}{(x+1)(x+1)}\) < 0
⇒ (x + 2)(x + 1)(x + 3) < 0, x ≠ – 1 [∵ (x + 1)² > 0]
Critical points are given by putting
(x + 2) (x + 1) (x + 3) = 0
⇒ x = – 1, – 2, – 3
Then by method of intervals, we have
x < – 3 or – 2 < x < – 1

Question 17.
\(\frac{x^2-2 x+3}{x^2-4 x+3}\) > – 3
Solution:
Given \(\frac{x^2-2 x+3}{x^2-4 x+3}\) > – 3

Critical points are given by putting
(x – 2) (2x – 3) (x – 1) (x – 3) = 0
⇒ x = 2, \(\frac { 3 }{ 2 }\), 1, 3
Then by method of intervals, we have
x < 1 or \(\frac { 3 }{ 2 }\) < x < 2 or x > 3

Question 18.
\(\frac{x^2+6 x-11}{x+3}\) < -1.
Solution:

Critical points are given by putting
(x – 1)(x – 8)(x + 3) = 0
⇒ x = 1, 8, – 3
Then by method of intervals, we have
x < – 3 or 1< x < 8

Question 19.
\(\frac{x^2-3 x+24}{x^2-3 x+3}\) < 4
Solution:

since x² – 3x + 3 = x² – 3x + \(\frac { 9 }{ 4 }\) + \(\frac { 3 }{ 4 }\)
= [x – \(\frac { 3 }{ 2 }\)]² + \(\frac { 3 }{ 4 }\) > 0 ∀ x ∈ R
⇒ x² – 3x – 4 > 0
⇒ (x + 1)(x – 4) > 0
Critical points are given by putting
(x + 1)(x – 4) = 0 ⇒ x = -1, 4
Then by method of intervals, we have
x < – 1 or x > 4