If a, b and c are sides of a triangle, then [a + b + c]2 / [ab + bc + ca] always belongs to 1) [1 , 2] 2) [2, 3] 3) [3, 4] 4) [1, 3] Solution: (3) [3, 4] Given that a, b, c are the sides of a triangle. b2 + c2 - a2 = 2bc cos A b2 +... View Article
If log10 (x3 + y3) – log10 (x2 + y2 – xy) ≤ 2, then the maximum value of xy, for all x ≥ 0, y ≥ 0, is 1) 2500 2) 3000 3) 1200 4) 3500 Solution: (1) 2500 log10 (x3 + y3) - log10 (x2 + y2 - xy) ≤ 2 log10 {[x3 + y3] / [x2 + y2 - xy]} ≤ 2 log10... View Article
The solution set of the inequation [x + 11] / [x – 3] > 0 is 1) (- ∞, -11) ∪ (3,∞) 2) (- ∞, - 10 ) ∪ (2,∞) 3) (- 100, - 11) ∪ (1,∞) 4) (- 5, 0) ∪ (3,7) 5) (0, 5) ∪ (- 1, 0) Solution: (1) (- ∞, - 11) ∪... View Article
The minimum of f (x) = |3 – x|+ 7 is 1) 0 2) 6 3) 7 4) 8 Solution: (3) 7 f (x) = |3 - x| = {3 - x + 7, 3 - x ≥ 0 and - (3 - x) + 7, 3 - x < 0 ={10 - x, 3 ≥ x and 4 + x, 3... View Article
If 3 ≤ 3t – 18 ≤ 18, then which one of the following is correct? 1) 15 ≤ 2t + 1≤ 20 2) 8 ≤ t < 12 3) 8 ≤ t + 1≤ 13 4) 21 ≤ 3t ≤ 24 5) t ≤ 7 or t ≥ 12 Solution: (3) 8 ≤ t + 1 ≤ 13 3 ≤ 3t - 18 ≤ 18 ⇒ 3 +... View Article
If a, b > 0 satisfy a3 + b3 = a – b, then 1) a2 + b2 > 1 2) a2 - b < 0 3) a2 + b2 = 1 4) a2 + ab + b2 < 1 Solution: (4) a2 + ab + b2 < 1 a3 + b3 = a - b Let a = 2 / 3, b... View Article
The solution set of the inequality 4-x+(1 / 2) – 7 . (2-x) – 4 < 0, for all x belongs to R, is 1) (- ∞ , 2) 2) (- 2, ∞) 3) (∞ , ∞) 4) (2 , ∞) Solution: (2) (- 2, ∞) 4-x+(1 / 2) - 7 . (2-x) - 4 < 0 2-2x+1 - 7 * (2-x) < 4 2 2-x - 7... View Article
If |2x – 3| < |x + 5|, then x lies in the interval 1) (- 3, 5) 2) (5, 9) 3) (- 2 / 3, 8) 4) ( - 8, 2 / 3) 5) (- 5, 2 / 3) Solution: (3) (- 2 / 3, 8) |2x - 3| < |x + 5| Case 1: If - ∞ <... View Article
The minimum value of the sum of real numbers a-5, a-4, 3a-3, 1, a8 and a10 with a > 0 is 1) 9 2) 8 3) 2 4) 1 Solution: (2) 8 a-5, a-4, 3a-3, 1, a8 and a10 All the numbers are positive as a > 0. ∴ AM > GM (1 / 8) [a-5+... View Article
If x, y and z are three positive real numbers, then minimum values of [y + z] / x + [z + x] / y + [x + y] / z is 1) 1 2) 2 3) 3 4) 6 Solution: (4) 6 [y + z] / x + [z + x] / y + [x + y] / z ⇒ (y / x) + (z / x) + (z / y) + (x / y) + (x / z) + (y / z) ⇒... View Article
Solve the inequality 3x + 2 > – 16, 2x – 3 ≤ 11. 1) (- 6, 7] 2) [ -6, 7) 3) (- 6, 7) 4) [- 6, 7] Solution: (1) (- 6, 7] 3x + 2 > -16 ⇒ 3x > - 16 - 2 ⇒ x > - 18 / 3 ⇒ x > - 6 and... View Article
Solve the inequality 2x – 5 ≤ (4x – 7)/3 1) x ∈ (- ∞ , 4) 2) x ∈ (- ∞ , 4] 3) x ∈ (- ∞ , 8] 4) x ∈ (- ∞ ,- 4] Solution: (2) x ∈ (- ∞ , 4] 2x - 5 ≤ (4x - 7) / 3 ⇒ 6x - 15 ≤ 4x - 7 ⇒... View Article
The number of solutions of the inequation |x – 2| + |x + 2|< 4 is 1) 1 2) 2 3) 4 4) 0 5) 3 Solution: (4) 0 It is given that |x - 2| + |x + 2| < 4 Case 1: when x < - 2 |x + 2| = - x - 2 and |x - 2| =... View Article
If x3 / [(2x – 1) (x + 2) (x – 3)] = A + B / (2x – 1) + C / (x + 2) + D / (x – 3), then A = 1) 1/2 2) -1/50 3) -8/25 4) 27/25 Solution: (1) 1 / 2 x3 / [(2x - 1) (x + 2) (x - 3)] = A + B / (2x - 1) + C / (x + 2) + D / (x - 3) x3 = A... View Article
If [3x + 2] / [(x + 1) (2x2 + 3)] = A / (x + 1) + {Bx + C} / (2x2 + 3)}, then A + C – B = 1) 0 2) 2 3) 3 4) 5 Solution: (2) 2 [3x + 2] / [(x + 1) (2x2 + 3)] = A / (x + 1) + {Bx + C} / (2x2 + 3)} 3x + 2 = A (2x2 + 3) + (Bx + C) (x... View Article
The partial fraction of [3x3 – 8x2 + 10] / [x – 1]4 is 1) 3 / (x - 1) + 1 / (x - 1)2 + 7 / (x - 1)3 + 5 / (x - 1)4 2) 3 / (x - 1) + 1 / (x - 1)2 - 7 / (x - 1)3 + 5 / (x - 1)4 3) 3 / (x - 1) + 1 / (x... View Article
[3x2 + 1] / [x2 – 6x + 8] = 1) 3 - [49] / [2 (x - 4)] - [13] / [2 (x - 2)] 2) [49] / [2 (x - 4)] - [13] / [2 (x - 2)] 3) [- 49] / [2 (x - 4)] + [13] / [2 (x - 2)] 4)... View Article
If [x2 + x + 1] / [x2 + 2x + 1] = A + B / (x + 1) + C / (x + 1)2, then A – B = 1) 4 2) 4C + 1 3) 3C 4) 2 Solution: (4) 2 [x2 + x + 1] / [x + 1]2 = [A (x + 1)2 + B (x + 1) + C] / (x + 1)2 = [A (x2 + 2x + 1) + B (x + 1)... View Article
If a real value of fraction f of a real variable x is such that 1 / [(1 + x2) (1 + x2)] = A / [1 + x] + f (x) / (1 + x2) then f (x) = 1) [1 - x] / 2 2) [x2 + 1] / 2 3) 1 - x 4) None of these Solution: (3) [1 - x] 1 / [(1 + x2) (1 + x2)] = A / [1 + x] + f (x) / (1 + x2) 1 =... View Article
If x is real, then the value of [x + 2] / [2x2 + 3x + 6] is 1) (1 / 13, 1 / 3) 2) (- 1 / 13, 1 / 3) 3) (- 1 / 3, 1 / 13) 4) None of these Solution: (2) (- 1 / 13, 1 / 3) y = [x + 2] / [2x2 + 3x + 6]... View Article