Basics & LogarithmseasyFree

Basics & Logarithms — JEE Maths practice question

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

Question
Number of integral values of xx which satisfies the inequality (x+6)2(x+5)(x+1)2(x−3)≤0\dfrac{(x+6)^2(x+5)(x+1)^2}{(x-3)} \leq 0 is:
A1010
B66
C88
D99correct
Solution
Step 1: Rewrite the inequality Since (x+6)2≥0(x+6)^2 \geq 0 and (x+1)2≥0(x+1)^2 \geq 0, the sign of the expression (away from x=−6,−1x = -6, -1) is governed by x+5x−3\dfrac{x+5}{x-3}, with equality at x=−6,−5,−1x = -6, -5, -1. Step 2: Solve the reduced inequality
x+5x−3≤0  ⟹  −5≤x<3\frac{x+5}{x-3} \leq 0 \implies -5 \leq x < 3
 Including the isolated zero x=−6x = -6:
x∈[−5,3)∪{−6}x \in [-5, 3) \cup \{-6\}
 Step 3: Count integers The integers are −6,−5,−4,−3,−2,−1,0,1,2-6, -5, -4, -3, -2, -1, 0, 1, 2, giving 9 values. Answer: (4)
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