JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Area Under Curves

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 :

Answer: (A)

Solution

Since

f

is a strictly decreasing bijection,

f (1) = e^{3}, f (2) = e^{2}, f (3) = e, f (4) = 1

, and since

g

is strictly increasing,

g (1) = \frac{1}{4}, g (e) = \frac{1}{3}, g (e^{2}) = \frac{1}{2}, g (e^{3}) = 1

. Thus

g^{- 1} (\frac{1}{2}) = e^{2}

and

f^{- 1} (e^{2}) = 2

, so

ϕ (x) = 2^{x}

. On

[0, 1]

2^{x} \geq x^{2}

, so the area is

\int_{0}^{1} (2^{x} - x^{2}) d x = [\frac{2 ^{x}}{l n 2} - \frac{x ^{3}}{3}]_{0}^{1} = \frac{1}{l n 2} - \frac{1}{3} = \frac{3 - l n 2}{3 l n 2}

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option