The function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}^{2}-6 \mathrm{x}-16}, \mathrm{x} \in \mathbb{R}-\{-2,8\}$

(B) decreases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$

(A) decreases in (-2, 8) and increases in $(-\infty,-2) \cup(8, \infty)$

(C) decreases in ( $-\infty,-2$ ) and increases in ( $8, \infty$ )

(D) increases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Application of Derivatives

The function

f (x) = \frac{x}{x ^{2} - 6 x - 16}, x \in R - {- 2, 8}

Answer: (B)

f (x) = \frac{x}{x ^{2} - 6 x - 16}

, where

x^{2} - 6 x - 16 = (x - 8) (x + 2)

f^{'} (x) = \frac{( x ^{2} - 6 x - 16 ) ( 1 ) - x ( 2 x - 6 )}{( x ^{2} - 6 x - 16 ) ^{2}} = \frac{x ^{2} - 6 x - 16 - 2 x ^{2} + 6 x}{( x ^{2} - 6 x - 16 ) ^{2}} = \frac{- ( x ^{2} + 16 )}{( x ^{2} - 6 x - 16 ) ^{2}}

.

Since

x^{2} + 16 > 0

always and the denominator is a square,

f^{'} (x) < 0

for every

x

in the domain.

So

f

decreases on each of

(- \infty, - 2)

(- 2, 8)

and

(8, \infty)

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option