JEE Main 2026 Apr 6 Shift 1, Mathematics Q20: Area as Definite Integral

JEE Main2026Apr 6, Shift 1Mathematics

Let $e$ be the base of natural logarithm and let $f : {1, 2, 3, 4} \to {1, e, e^{2}, e^{3}}$ and $g : {1, e, e^{2}, e^{3}} \to {1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $ϕ (x) = [f^{- 1} {g^{- 1} (\frac{1}{2})}]^{x}$ , then the area of the region $R = {(x, y) : x^{2} \leq y \leq ϕ (x), 0 \leq x \leq 1}$ is:

A
$\frac{3 - lo g _{e} ( 2 )}{3 lo g _{e} ( 2 )}$
B
$\frac{1}{3 lo g _{e} ( 2 )}$
C
$3 + lo g_{e} (2)$
D
$\frac{3 + lo g _{e} ( 2 )}{2 + lo g _{e} ( 3 )}$

Solution

Since $f$ is strictly decreasing and bijective: $f (1) = e^{3}, f (2) = e^{2}, f (3) = e, f (4) = 1$ . Since $g$ is strictly increasing and bijective: $g (1) = 1/4, g (e) = 1/3, g (e^{2}) = 1/2, g (e^{3}) = 1$ .

So $g^{- 1} (1/2) = e^{2}$ , and $f^{- 1} (e^{2}) = 2$ . Hence $ϕ (x) = 2^{x}$ .

On $[0, 1]$ , $2^{x} \geq x^{2}$ , so

$A = \int_{0}^{1} (2^{x} - x^{2}) d x = [\frac{2 ^{x}}{ln 2} - \frac{x ^{3}}{3}]_{0}^{1} = \frac{2}{ln 2} - \frac{1}{3} - \frac{1}{ln 2} = \frac{1}{ln 2} - \frac{1}{3} = \frac{3 - ln 2}{3 ln 2}$ .

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q20: Area as Definite Integral

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