Q35 / 75 · JEE Main 2025 (4 Apr Shift 1) · Mathematics

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 $f \circ g:[2,4] \rightarrow \mathbb{R}$ is $[\alpha, \beta]$, then $\frac{1}{\beta-\alpha}$ is equal to

  • A
    68
  • B
    29
  • C
    2
  • Correct answer
    D
    56

Solution

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

Hence, the correct answer is option: 'D'.

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