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Mathematics · Functions

JEE Main 2026 — 6 April, Morning Shift — Question 40

Let e be the base of natural logarithm and let f:{1,2,3,4}→{1,e,e2,e3}f:\{1,2,3,4\} \rightarrow\left\{1, \mathrm{e}, \mathrm{e}^{2}, \mathrm{e}^{3}\right\} and g:{1,e,e2,e3}→{1,12,13,14}\mathrm{g}:\left\{1, \mathrm{e}, \mathrm{e}^{2}, \mathrm{e}^{3}\right\} \rightarrow\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\} be two bijective functions such that ff is strictly decreasing and g is strictly increasing. If ϕ(x)=[f−1{ g−1(12)}]x\phi(\mathrm{x})=\left[f^{-1}\left\{\mathrm{~g}^{-1}\left(\frac{1}{2}\right)\right\}\right]^{\mathrm{x}}, then the area of the region R={(x,y):x2≤y≤ϕ(x)\mathrm{R}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2} \leq \mathrm{y} \leq \phi(\mathrm{x})\right., 0≤x≤1}0 \leq \mathrm{x} \leq 1\} is :

  1. Option A:

    3−log⁡e(2)3log⁡e(2)\frac{3-\log _{e}(2)}{3 \log _{e}(2)}

    Correct
  2. Option B:

    13log⁡e(2)\frac{1}{3 \log _{e}(2)}

  3. Option C:

    3+log⁡e(2)3+\log _{e}(2)

  4. Option D:

    3+log⁡e(2)2+log⁡e(3)\frac{3+\log _{e}(2)}{2+\log _{e}(3)}

Answer: A

Step-by-step solution

f(x)=ex−1f(x)=e^{x-1} g(x)=1log⁡ex+1\mathrm{g}(\mathrm{x})=\frac{1}{\log _{\mathrm{e}} \mathrm{x}+1} f−1( g−1(1/2))=f−1(e)=2\mathrm{f}^{-1}\left(\mathrm{~g}^{-1}(1 / 2)\right)=\mathrm{f}^{-1}(\mathrm{e})=2 ϕ(x)=2x\phi(\mathrm{x})=2^{\mathrm{x}} ∴∫01(2x−x2)dx=(2xln⁡2−x33)01=1ln⁡2−13\therefore \int_{0}^{1}\left(2^{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx}=\left(\frac{2^{\mathrm{x}}}{\ln 2}-\frac{\mathrm{x}^{3}}{3}\right)_{0}^{1}=\frac{1}{\ln 2}-\frac{1}{3} =3−ℓn23ℓn2=\frac{3-\ell \mathrm{n} 2}{3 \ell \mathrm{n} 2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Mathematics
Chapter
Functions
Topic
Functional Equations

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