Application of IntegralshardPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Bijections and Bounded Area = (3−ln2)/(3ln2) | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let ee be the base of natural logarithm and let f:{1,2,3,4}→{1,e,e2,e3}f:\{1,2,3,4\}\to\{1,e,e^2,e^3\} and g:{1,e,e2,e3}→{1,12,13,14}g:\{1,e,e^2,e^3\}\to\left\{1,\dfrac12,\dfrac13,\dfrac14\right\} be two bijective functions such that ff is strictly decreasing and gg is strictly increasing. If ϕ(x)=[f−1{g−1(12)}]x\phi(x)=\left[f^{-1}\left\{g^{-1}\left(\dfrac12\right)\right\}\right]^x, then the area of the region R={(x,y):x2≤y≤ϕ(x), 0≤x≤1}R=\{(x,y):x^2\le y\le\phi(x),\ 0\le x\le1\} is
A3−log⁡e23log⁡e2\dfrac{3-\log_e 2}{3\log_e 2}correct
B13log⁡e2\dfrac{1}{3\log_e 2}
C3+log⁡e23+\log_e 2
D3+log⁡e22+log⁡e3\dfrac{3+\log_e 2}{2+\log_e 3}
Solution
Step 1: Domain of gg: 1<e<e2<e31<e<e^2<e^3; codomain: 14<13<12<1\dfrac14<\dfrac13<\dfrac12<1. gg strictly increasing:
g(1)=14, g(e)=13, g(e2)=12, g(e3)=1⇒g−1(12)=e2.g(1)=\dfrac14,\ g(e)=\dfrac13,\ g(e^2)=\dfrac12,\ g(e^3)=1\Rightarrow g^{-1}\left(\dfrac12\right)=e^2.
 Step 2: Domain of ff: 1<2<3<41<2<3<4; codomain 1<e<e2<e31<e<e^2<e^3. ff strictly decreasing:
f(1)=e3, f(2)=e2, f(3)=e, f(4)=1⇒f−1(e2)=2.f(1)=e^3,\ f(2)=e^2,\ f(3)=e,\ f(4)=1\Rightarrow f^{-1}(e^2)=2.
 Step 3: f−1{g−1(12)}=f−1(e2)=2⇒ϕ(x)=2xf^{-1}\big\{g^{-1}(\tfrac12)\big\}=f^{-1}(e^2)=2\Rightarrow\phi(x)=2^x. Step 4: On 0≤x≤10\le x\le1, 2x≥1≥x22^x\ge1\ge x^2:
R=∫01(2x−x2) dx=[2xln⁡2−x33]01=(2ln⁡2−13)−1ln⁡2=1ln⁡2−13.R=\int_0^1\big(2^x-x^2\big)\,dx=\left[\dfrac{2^x}{\ln2}-\dfrac{x^3}{3}\right]_0^1=\left(\dfrac{2}{\ln2}-\dfrac13\right)-\dfrac{1}{\ln2}=\dfrac{1}{\ln2}-\dfrac13.
 Step 5:
∴R=33ln⁡2−ln⁡23ln⁡2=3−ln⁡23ln⁡2=3−log⁡e23log⁡e2.\therefore R=\dfrac{3}{3\ln2}-\dfrac{\ln2}{3\ln2}=\dfrac{3-\ln2}{3\ln2}=\dfrac{3-\log_e2}{3\log_e2}.
 Correct answer: (1)
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