JEE Mathematics 2025 Question with Solution

Given: $f(x) = \frac{2x + 3}{5x + 2}$, $g(x) = \frac{2 - 3x}{1 - x}$, and $x \in [2,4]$.

Find: $\frac{1}{\beta-\alpha}$ where the range of $f \circ g$ is $[\alpha,\beta]$.

First compute the composite function:

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

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

$$f(g(x))=\frac{2\left(\frac{2-3x}{1-x}\right)+3}{5\left(\frac{2-3x}{1-x}\right)+2}$$

Simplifying,

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

Now evaluate at the endpoints:

$$f(g(2))=\frac{7-18}{12-34}=\frac{-11}{-22}=\frac{1}{2}$$ $$f(g(4))=\frac{7-36}{12-68}=\frac{-29}{-56}=\frac{29}{56}$$

Also,

$$\frac{d}{dx}\left(\frac{7-9x}{12-17x}\right)=\frac{11}{(12-17x)^2}>0$$

so the function is increasing on $[2,4]$. Hence the range is $\left[\frac{1}{2},\frac{29}{56}\right]$. Therefore,

$$\beta-\alpha=\frac{29}{56}-\frac{1}{2}=\frac{1}{56}$$

Thus,

$$\frac{1}{\beta-\alpha}=56$$

Therefore, the correct option is D.

Given: $f(x) = \frac{2x + 3}{5x + 2}$ and $g(x) = \frac{2 - 3x}{1 - x}$.

Find: the value of $\frac{1}{\beta - \alpha}$ when the range of $f(g(x))$ on $[2,4]$ is $[\alpha,\beta]$.

Compute the composite function step by step:

$$f(g(x))=\frac{2g(x)+3}{5g(x)+2}$$ $$=\frac{2\left(\frac{2-3x}{1-x}\right)+3}{5\left(\frac{2-3x}{1-x}\right)+2}$$

Numerator:

$$2\left(\frac{2-3x}{1-x}\right)+3=\frac{4-6x+3(1-x)}{1-x}=\frac{7-9x}{1-x}$$

Denominator:

$$5\left(\frac{2-3x}{1-x}\right)+2=\frac{10-15x+2(1-x)}{1-x}=\frac{12-17x}{1-x}$$

Hence,

$$f(g(x))=\frac{\frac{7-9x}{1-x}}{\frac{12-17x}{1-x}}=\frac{7-9x}{12-17x}$$

Let

$$y=\frac{7-9x}{12-17x}$$

Now check monotonicity:

$$\frac{dy}{dx}=\frac{(-9)(12-17x)-(7-9x)(-17)}{(12-17x)^2}$$ $$=\frac{-108+153x+119-153x}{(12-17x)^2}=\frac{11}{(12-17x)^2}>0$$

So $f(g(x))$ is increasing on $[2,4]$.

Therefore, minimum value occurs at $x=2$ and maximum value occurs at $x=4$.

$$f(g(2))=\frac{7-18}{12-34}=\frac{1}{2}$$ $$f(g(4))=\frac{7-36}{12-68}=\frac{29}{56}$$

Thus,

$$[\alpha,\beta]=\left[\frac{1}{2},\frac{29}{56}\right]$$

Now,

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

Hence,

$$\frac{1}{\beta-\alpha}=56$$

Therefore, the correct option is D.