Let f, ~g:(1, ∞) arrow ℝ be defined as f(x)=(2 x+3)/(5 x+2) and g(x)=(2-3 x)/(1-x) . If the range of the function f g:[2,4] arrow ℝ is [α, β] , then (1)/(β-α) is equal to

Correct answer: D. 56 aligned & fog(x)=f(g(x)) & =f((2-3 x)/(1-x))=(2((2-3 x)/(1-x))+3)/(5((2-3 x)/(1-x))+2) & =(4-6 x+3-3 x)/(10-15 x+2-2 x)=((7-9 x)/(12-17 x)) & [ array c 12-7 x ≠ 0 x ≠ (12)/(17) array . & [ array l fog(2)=(7-9(2))/(12-17(2))=(-11)/(-22)=(1)/(2) fog(4)=(7-9(4))/(12-17(4))=(-29)/(-56)=(29)/(56) array . & Range of fog : [α, β]=[(1)/(2), (29)/(56)] & (β-α)=(29)/(56)-(1)/(2)=(29-28)/(56)=(1)/(56) & (1)/((β-α))=56 aligned

JEE Main 20254 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Functions

Let f, ~g:(1, ∞) arrow ℝ be defined as f(x)=(2 x+3)/(5 x+2) and...

Let

f, \mathrm{~g}:(1, \infty) \rightarrow \mathbb{R}

be defined as

f(\mathrm{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] \rightarrow \mathbb{R}

[\alpha, \beta]

, then

\frac{1}{\beta-\alpha}

is equal to

Options

Correct answer

D. 56

Step-by-step solution

\begin{aligned}& \operatorname{fog}(x)=f(g(x)) \\ & =f\left(\frac{2-3 x}{1-x}\right)=\frac{2\left(\frac{2-3 x}{1-x}\right)+3}{5\left(\frac{2-3 x}{1-x}\right)+2} \\ & =\frac{4-6 x+3-3 x}{10-15 x+2-2 x}=\left(\frac{7-9 x}{12-17 x}\right) \\ & \therefore\left[\begin{array}{c}12-7 x \neq 0 \\ x \neq \frac{12}{17}\end{array}\right. \\ & {\left[\begin{\text{ array }}{l}\operatorname{fog}(2)=\frac{7-9(2)}{12-17(2)}=\frac{-11}{-22}=\frac{1}{2} \\ \operatorname{fog}(4)=\frac{7-9(4)}{12-17(4)}=\frac{-29}{-56}=\frac{29}{56}\end{\text{ array }}\right.} \\ & \text{ Range of fog : }[\alpha, \beta]=\left[\frac{1}{2}, \frac{29}{56}\right] \\ & \therefore(\beta-\alpha)=\frac{29}{56}-\frac{1}{2}=\frac{29-28}{56}=\frac{1}{56} \\ & \frac{1}{(\beta-\alpha)}=56\end{aligned}

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