Basics & LogarithmseasyFree

Basics & Logarithms: 16x

JEE Maths question with a full step-by-step solution.

Question
If 8x2+16x−51(2x−3)(x+4)>3\dfrac{8x^2+16x-51}{(2x-3)(x+4)} > 3, where xx is such that:
Ax<−4x < -4
B−3<x<32-3 < x < \dfrac{3}{2}
Cx>52x > \dfrac{5}{2}
DAll three correctcorrect
Solution
Step 1: Move 3 to the left and simplify
8x2+16x−51−3(2x−3)(x+4)(2x−3)(x+4)>0=2x2+x−152x2+5x−12>0\frac{8x^2+16x-51 - 3(2x-3)(x+4)}{(2x-3)(x+4)} > 0 = \frac{2x^2+x-15}{2x^2+5x-12} > 0
 Step 2: Factorize
(2x−5)(x+3)(2x−3)(x+4)>0\frac{(2x-5)(x+3)}{(2x-3)(x+4)} > 0
 Step 3: Sign analysis With critical points −4,−3,32,52-4, -3, \dfrac{3}{2}, \dfrac{5}{2}, the expression is positive for x<−4x < -4, −3<x<32-3 < x < \dfrac{3}{2}, and x>52x > \dfrac{5}{2}. All three statements hold. Answer: (4)
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