JEE Main 2025 Apr 4 Shift 1, Mathematics Q1: Invertible And Composite Functions

Substitute $g(x)$ into $f$:

$f(g(x))=\frac{2\left(\frac{2-3x}{1-x}\right)+3}{5\left(\frac{2-3x}{1-x}\right)+2}=\frac{(4-6x)+3(1-x)}{(10-15x)+2(1-x)}=\frac{7-9x}{12-17x}$.

This function is monotonic on any interval not containing $x=\frac{12}{17}$. The interval $[2,4]$ lies entirely to the right of $\frac{12}{17}$, so $f\circ g$ is monotonic there; the extreme values occur at the endpoints.

$f\circ g(2)=\frac{7-18}{12-34}=\frac{-11}{-22}=\frac{1}{2}$.

$f\circ g(4)=\frac{7-36}{12-68}=\frac{-29}{-56}=\frac{29}{56}$.

Since $\frac{1}{2}=\frac{28}{56}<\frac{29}{56}$, the range is $[\alpha ,\beta ]=\left[\frac{1}{2},\frac{29}{56}\right]$.

$\beta -\alpha =\frac{29}{56}-\frac{28}{56}=\frac{1}{56}$, hence $\frac{1}{\beta -\alpha }=56$.