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Basics & Logarithms: Set Real Negative Positive Contains

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

Question
If SS is the set of all real xx such that x2(5−x)(1−2x)(5x+1)(x+2)\dfrac{x^2(5-x)(1-2x)}{(5x+1)(x+2)} is negative and 3x+16x3+x2−x\dfrac{3x+1}{6x^3+x^2-x} is positive, then SS contains:
A(1,4)(1, 4)correct
B(5,11)(5, 11)
C(−32,12)\left(-\dfrac{3}{2}, \dfrac{1}{2}\right)
D(−10,−4)(-10, -4)
Solution
Step 1: Solve the second condition
3x+1x(3x−1)(2x+1)>0  ⟹  x∈(−∞,−12)∪(−13,0)∪(13,∞)\frac{3x+1}{x(3x-1)(2x+1)} > 0 \implies x \in \left(-\infty, -\frac{1}{2}\right) \cup \left(-\frac{1}{3}, 0\right) \cup \left(\frac{1}{3}, \infty\right)
 Step 2: Solve the first condition
x2(5−x)(1−2x)(5x+1)(x+2)<0  ⟹  x∈(−2,−15)∪(12,5)\frac{x^2(5-x)(1-2x)}{(5x+1)(x+2)} < 0 \implies x \in (-2, -\tfrac{1}{5}) \cup (\tfrac{1}{2}, 5)
 Step 3: Intersect
S=(−2,−12)∪(−13,−15)∪(12,5)S = (-2, -\tfrac{1}{2}) \cup (-\tfrac{1}{3}, -\tfrac{1}{5}) \cup (\tfrac{1}{2}, 5)
 The interval (1,4)(1, 4) lies inside (12,5)(\tfrac{1}{2}, 5), so SS contains (1,4)(1, 4). Answer: (1)
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