Basics & LogarithmsmediumFree

Basics & Logarithms — JEE Maths practice question

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

Question
The solution set of the inequality 3x−4xx2−3x−4≥0\dfrac{3^x-4^x}{x^2-3x-4} \geq 0 is:
A(−∞,0](-\infty, 0]
B[0,∞)[0, \infty)
C(−∞,−1)∪[0,4)(-\infty, -1) \cup [0, 4)correct
D(−∞,1)∪[0,4)(-\infty, 1) \cup [0, 4)
Solution
Step 1: Analyze numerator and denominator Numerator: 3x−4x>03^x - 4^x > 0 for x<0x < 0, =0= 0 at x=0x = 0, <0< 0 for x>0x > 0. Denominator: x2−3x−4=(x−4)(x+1)x^2 - 3x - 4 = (x-4)(x+1), with critical points −1,4-1, 4. Step 2: Sign analysis at critical points −1,0,4-1, 0, 4 The expression is positive on (−∞,−1)(-\infty, -1), zero at x=0x = 0, and positive on (0,4)(0, 4):
x∈(−∞,−1)∪[0,4)x \in (-\infty, -1) \cup [0, 4)
 Answer: (3)
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