Maths Q20 – JEE-MAIN 2026

Home Maths Relations and FunctionsQuestion #375

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 {\frac{1}{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:

Clue 1 of 3: Function Composition and Inverse Functions

First, determine the values of $g^{- 1} (\frac{1}{2})$ and then $f^{- 1}$ of that result, using the properties of strictly increasing and decreasing bijective functions.

🥷

Ninja Strategy•Magnitude Estimation

Estimate the approximate value of the integral by evaluating the bounds of the integrand, then compare with the approximate values of the options.

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● Video Walkthrough

Step 1: Determine

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

and

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

✦ Active

Given $g : {1, e, e^{2}, e^{3}} \to {1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}}$ is strictly increasing. Sorting the codomain elements in increasing order gives ${\frac{1}{4}, \frac{1}{3}, \frac{1}{2}, 1}$ . Thus, $g (1) = \frac{1}{4}$ , $g (e) = \frac{1}{3}$ , $g (e^{2}) = \frac{1}{2}$ , $g (e^{3}) = 1$ . Therefore, $g^{- 1} (\frac{1}{2}) = e^{2}$ .

Given $f : {1, 2, 3, 4} \to {1, e, e^{2}, e^{3}}$ is strictly decreasing. Sorting the codomain elements in decreasing order gives ${e^{3}, e^{2}, e, 1}$ . Thus, $f (1) = e^{3}$ , $f (2) = e^{2}$ , $f (3) = e$ , $f (4) = 1$ . Therefore, $f^{- 1} (e^{2}) = 2$ .

Step 2: Determine the function

ϕ (x)

○ Expand

Substitute the result from Step 1 into the definition of $ϕ (x)$ .

ϕ (x) = {[f^{- 1} {g^{- 1} (\frac{1}{2})}]}^{x} = [f^{- 1} (e^{2})]^{x} = [2]^{x} = 2^{x}

Step 3: Calculate the area of the region

R

○ Expand

The region is defined by $x^{2} \leq y \leq ϕ (x)$ for $0 \leq x \leq 1$ . In the interval $[0, 1]$ , $2^{x} \geq x^{2}$ . The area is given by the integral of the difference between the upper and lower functions.

Area = \int_{0}^{1} (ϕ (x) - x^{2}) d x = \int_{0}^{1} (2^{x} - x^{2}) d x

Area = {[\frac{2^{x}}{\log_{e} (2)} - \frac{x^{3}}{3}]}_{0}^{1}

Area = (\frac{2^{1}}{\log_{e} (2)} - \frac{1^{3}}{3}) - (\frac{2^{0}}{\log_{e} (2)} - \frac{0^{3}}{3}) = (\frac{2}{\log_{e} (2)} - \frac{1}{3}) - (\frac{1}{\log_{e} (2)} - 0)

Area = \frac{1}{\log_{e} (2)} - \frac{1}{3} = \frac{3 - \log_{e} (2)}{3 \log_{e} (2)}

This matches option 1.

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Maths Question 20 – JEE-MAIN 2026

Clue 1 of 3: Function Composition and Inverse Functions

🧠 Full Solution Path