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Basics & Logarithms: Complete Solution Set Inequality

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

Question
The complete solution set of the inequality 3x(2x−5)(x2+x+2)(cos⁡x−2)(x2+x)≤0\dfrac{3^x(2x-5)(x^2+x+2)}{(\cos x-2)(x^2+x)} \leq 0 is:
A(−∞,−1)(-\infty, -1)
B[52,∞)\left[\dfrac{5}{2}, \infty\right)
C[−1,52]\left[-1, \dfrac{5}{2}\right]
D(−1,0)∪[52,∞)(-1, 0) \cup \left[\dfrac{5}{2}, \infty\right)correct
Solution
Step 1: Eliminate the positive factors 3x>03^x > 0, x2+x+2>0x^2+x+2 > 0 (discriminant <0< 0), and cos⁡x−2<0\cos x - 2 < 0 for all xx. Dividing by the negative factor (cos⁡x−2)(\cos x - 2) flips the inequality:
2x−5x(x+1)≥0\frac{2x-5}{x(x+1)} \geq 0
 Step 2: Sign analysis at critical points −1,0,52-1, 0, \dfrac{5}{2} The expression is non-negative on (−1,0)(-1, 0) and on [52,∞)\left[\dfrac{5}{2}, \infty\right):
x∈(−1,0)∪[52,∞)x \in (-1, 0) \cup \left[\frac{5}{2}, \infty\right)
 Answer: (4)
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