The set values satisfying the inequality
Straight Objective Type
This section contains multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
(A) x∈ 
(B) x∈ 
(C) x∈ 
(D) x∈φ
(C) (–1, 1) (D) (3, 5)
1. (A)
(A) (–1, 1) ∪ (4, 6) (B) (1, 2)
(C) (–∞, –1) ∪ (6, ∞) (D) (2, 4)
⇒ 
x∈(–1, 1) ∪ (4, 6) .
2 If log72 = m, then log4928 is equal to
(A) 2 (1 + 2m) (B) 
(C) [4, ∞) (D) none of these
3. (A)
We have 
and
(C) (–∞, –2] ∪ {2} ∪ (6, ∞) (D) [–2, 1) ∪ (1, 6)
23. (B)
(A) 
(B) 
(C) 
(D) 
31. If S is the set of all real x such that 
> 0 is
(A) 
(B) 
.
23. The solution set of |x2 + x| = x2 + x is given by
⇒ x2 + x ≥ 0
x ≥ 0, x ≤ −1
(C) 
(D) none of these
20. (D)
(A) infinite number of real solution for some 
(B) finite number of real solutions for some 
So no solution for 
(A) 
(B) 
(C) 
(D) None
6. 
Form 1 and 2
3. |x2 – 3x + 2| + |x – 1| = x – 3
⇒ x – 3 ≥ 0 ⇒ x ≥ 3
∴ it has no real solutions.
21. 
, then x belongs to :
20. Given equation is 
(C) 
(D) 
23. (C)
.
21. Greatest negative integral value of x satisfying 
(A) 4 (B) 9
(C) 16 (D) 32
LEVEL II
7. The integral value of x ∈ (–π, π) satisfying the equation
7. 
(x2 –1) cos x ≥ 0
⇒ x2 – 1 ≥ 0 and cos x ≥ 0 or x2 – 1 ≤ 0 and cos x ≤ 0
22. (D)
22. We have, 
or 
but t > 0
22. If P and Q are sum and product respectively of all real values of
x satisfying the equation 
then
(A) |P| + |Q| = 143 (B) |P| + |Q| = 127
P = 8, Q = - 135
(A) (0, ∞) (B) (1, ∞)
(C ) (−∞, 1) (D) (−∞, 2)
⇒ 2x < 2
x < 1.
8. log49 28
= 
(A) 2 (B) 3
(C) 4 (D) 5
20. The value of x satisfying
is
(A) 1 (B) 2
4. (A)
22. The solution set of the equation 
is
21. (C )
21. 
. 
solution set x 
23. If a > 0, then 
simplifies
to
28. log2x + logx2 = 3 + 
⇒ x = 8, x = 21/3
(A) 
(B) 
(C)
(D) 
11. (A)
11. log0.2 
⇒
log0.2 
12. If x ∈ R, the solution set of the inequation 4–x + 0.5 – 7 . 2–x – 4 < 0 is equal to
(A) 
(B) (−2, ∞)
10. Domain of the function 
is
(A) R (B) R+
20. 
on the interval (4, 5) is
(A) increasing (B) decreasing
⇒ 
so
is increasing.
5 How many roots of the following equation 
has
3|x| > 0.
So |x| = 2
(C ) 
(D) None
6. (C)
……..(2)
(C) K1 = K2 = −1 (D) K2 = 1 and K1 = −1
23. (B)
6. If 
then 
(A) 1 (B) 
(A) 
(B) 
(C) 
(D) 
21. (D)
22. The maximum value of the function f(x) = 
is
(A) 
(B) 2
At x = 
maximum value will be 
