Let $f$, $\mathrm{g}:(1, \infty) \rightarrow \mathbb{R}$ be defined as $f(\mathrm{x})=\frac{2 x+3}{5 x+2}$ and $\mathrm{g}(\mathrm{x})=\frac{2-3 x}{1-x}$. If the range of the function $f \circ g:[2,4] \rightarrow \mathbb{R}$ is $[\alpha, \beta]$, then $\frac{1}{\beta-\alpha}$ is equal to

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Sets Relations and Functions

Let

f

g : (1, \infty) \to R

be defined as

f (x) = \frac{2 x + 3}{5 x + 2}

and

g (x) = \frac{2 - 3 x}{1 - x}

. If the range of the function

f \circ g : [2, 4] \to R

[α, β]

, then

\frac{1}{β - α}

is equal to

Answer: (D)

Composing,

f (g (x)) = \frac{2 g + 3}{5 g + 2}

with

g (x) = \frac{2 - 3 x}{1 - x}

simplifies to

\frac{7 - 9 x}{12 - 17 x}

. Its derivative is

\frac{11}{( 12 - 17 x ) ^{2}} > 0

, so it increases on

[2, 4]

; the endpoints give

\frac{7 - 18}{12 - 34} = \frac{1}{2}

and

\frac{7 - 36}{12 - 68} = \frac{29}{56}

. The range is

[\frac{1}{2}, \frac{29}{56}]

, so

β - α = \frac{29}{56} - \frac{28}{56} = \frac{1}{56}

and

\frac{1}{β - α} = 56

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option