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Basics & Logarithms: Find Number Negative Integers Satisfying

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

Question

Find the number of negative integers satisfying

-1 \leq \dfrac{x^2-5x+4}{x^2-4} \leq 1

Solution

Answer: 0

Domain:

x \neq \pm 2

. Left inequality:

\dfrac{x^2-5x+4}{x^2-4} + 1 \geq 0 \implies \dfrac{2x^2-5x}{x^2-4} \geq 0 \implies \dfrac{x(2x-5)}{(x-2)(x+2)} \geq 0

Sign analysis with critical points

-2, 0, 2, \frac{5}{2}

: Solution:

x \in (-\infty,-2) \cup [0,2) \cup [\frac{5}{2},+\infty)

. Right inequality:

\dfrac{x^2-5x+4}{x^2-4} - 1 \leq 0 \implies \dfrac{8-5x}{x^2-4} \leq 0 \implies \dfrac{5x-8}{(x-2)(x+2)} \geq 0

Sign analysis with critical points

-2, \frac{8}{5}, 2

: Solution:

x \in (-2, \frac{8}{5}] \cup (2,+\infty)

. Intersection:

[0, \frac{8}{5}] \cup [\frac{5}{2}, +\infty)

. Negative integers in

[0, \frac{8}{5}] \cup [\frac{5}{2},+\infty)

: none. Answer: 0

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