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 satisfying the inequality (34)6x+10−x2<2764\left(\dfrac{3}{4}\right)^{6x+10-x^2} < \dfrac{27}{64} is:
A66
B77correct
C88
Dinfinite
Solution
Step 1: Express both sides with the same base
(34)6x+10−x2<(34)3\left(\frac{3}{4}\right)^{6x+10-x^2} < \left(\frac{3}{4}\right)^3
 Step 2: Flip the inequality (base less than 1)
6x+10−x2>3  ⟹  x2−6x−7<0  ⟹  (x−7)(x+1)<0  ⟹  −1<x<76x+10-x^2 > 3 \implies x^2-6x-7 < 0 \implies (x-7)(x+1) < 0 \implies -1 < x < 7
 Step 3: Count integers The integers are 0,1,2,3,4,5,60, 1, 2, 3, 4, 5, 6, giving 7 values. Answer: (2)
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