Logo
Jeeify
MCQMediumJEE 2025Composition & Inverse Functions

JEE Mathematics 2025 Question with Solution

Let f,g:(1,∞)→Rf, g: (1, \infty) \rightarrow \mathbb{R} be defined as f(x)=2x+35x+2f(x) = \frac{2x + 3}{5x + 2} and g(x)=2−3x1−xg(x) = \frac{2 - 3x}{1 - x}. If the range of the function fog:[2,4]→Rfog: [2, 4] \rightarrow \mathbb{R} is [α,β][\alpha, \beta], then 1β−α\frac{1}{\beta - \alpha} is equal to

  • A

    6868

  • B

    2929

  • C

    22

  • D

    5656

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: f(x)=2x+35x+2f(x) = \frac{2x + 3}{5x + 2}, g(x)=2−3x1−xg(x) = \frac{2 - 3x}{1 - x}, and x∈[2,4]x \in [2,4].

Find: 1β−α\frac{1}{\beta-\alpha} where the range of f∘gf \circ g is [α,β][\alpha,\beta].

First compute the composite function:

f(g(x))=2g(x)+35g(x)+2f(g(x))=\frac{2g(x)+3}{5g(x)+2}

Substituting g(x)=2−3x1−xg(x)=\frac{2-3x}{1-x},

f(g(x))=2(2−3x1−x)+35(2−3x1−x)+2f(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))=4−6x+3−3x1−x10−15x+2−2x1−x=7−9x12−17xf(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))=7−1812−34=−11−22=12f(g(2))=\frac{7-18}{12-34}=\frac{-11}{-22}=\frac{1}{2} f(g(4))=7−3612−68=−29−56=2956f(g(4))=\frac{7-36}{12-68}=\frac{-29}{-56}=\frac{29}{56}

Also,

ddx(7−9x12−17x)=11(12−17x)2>0\frac{d}{dx}\left(\frac{7-9x}{12-17x}\right)=\frac{11}{(12-17x)^2}>0

so the function is increasing on [2,4][2,4]. Hence the range is [12,2956]\left[\frac{1}{2},\frac{29}{56}\right]. Therefore,

β−α=2956−12=156\beta-\alpha=\frac{29}{56}-\frac{1}{2}=\frac{1}{56}

Thus,

1β−α=56\frac{1}{\beta-\alpha}=56

Therefore, the correct option is D.

Endpoint Evaluation with Monotonicity Check

Given: f(x)=2x+35x+2f(x) = \frac{2x + 3}{5x + 2} and g(x)=2−3x1−xg(x) = \frac{2 - 3x}{1 - x}.

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

Compute the composite function step by step:

f(g(x))=2g(x)+35g(x)+2f(g(x))=\frac{2g(x)+3}{5g(x)+2} =2(2−3x1−x)+35(2−3x1−x)+2=\frac{2\left(\frac{2-3x}{1-x}\right)+3}{5\left(\frac{2-3x}{1-x}\right)+2}

Numerator:

2(2−3x1−x)+3=4−6x+3(1−x)1−x=7−9x1−x2\left(\frac{2-3x}{1-x}\right)+3=\frac{4-6x+3(1-x)}{1-x}=\frac{7-9x}{1-x}

Denominator:

5(2−3x1−x)+2=10−15x+2(1−x)1−x=12−17x1−x5\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))=7−9x1−x12−17x1−x=7−9x12−17xf(g(x))=\frac{\frac{7-9x}{1-x}}{\frac{12-17x}{1-x}}=\frac{7-9x}{12-17x}

Let

y=7−9x12−17xy=\frac{7-9x}{12-17x}

Now check monotonicity:

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

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

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

f(g(2))=7−1812−34=12f(g(2))=\frac{7-18}{12-34}=\frac{1}{2} f(g(4))=7−3612−68=2956f(g(4))=\frac{7-36}{12-68}=\frac{29}{56}

Thus,

[α,β]=[12,2956][\alpha,\beta]=\left[\frac{1}{2},\frac{29}{56}\right]

Now,

β−α=2956−2856=156\beta-\alpha=\frac{29}{56}-\frac{28}{56}=\frac{1}{56}

Hence,

1β−α=56\frac{1}{\beta-\alpha}=56

Therefore, the correct option is D.

Common mistakes

  • Evaluating only the composite function and not checking whether it is increasing or decreasing on [2,4][2,4]. This is wrong because endpoint values give the range only after confirming monotonicity. Instead, verify monotonicity using the derivative or rational-function behavior before assigning α\alpha and β\beta.

  • Making an algebraic mistake while simplifying f(g(x))f(g(x)). This is wrong because an incorrect numerator or denominator changes the range completely. Instead, substitute g(x)g(x) carefully into both 2g(x)+32g(x)+3 and 5g(x)+25g(x)+2 and simplify step by step.

  • Reversing α\alpha and β\beta and computing β−α\beta-\alpha with the wrong order. This is wrong because α\alpha is the lower endpoint and β\beta is the upper endpoint of the range. Instead, identify the smaller and larger values correctly before subtraction.

Practice more Composition & Inverse Functions questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Composition & Inverse Functions

Related questions