JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Area Between Curves via Composite Inverses of Monotonic Functions

Q: Why is g(e^2) = 1/2 and not g(e) = 1/2?

The domain elements ordered ascending are 1 < e < e^2 < e^3. The codomain elements ordered ascending are 1/4 < 1/3 < 1/2 < 1. Since g is strictly increasing, 3rd domain element maps to 3rd codomain element, so g(e^2) = 1/2.

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

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A(3 - log_e(2))/(3 log_e(2))B1/(3 log_e(2))C3 + log_e(2)D(3 + log_e(2))/(2 + log_e(3))

Step-by-step solutionView

Correct answer

The base of the exponential function is found to be 2 by tracking the strictly monotonic bijections, and integrating $ϕ (x) - x^{2} = 2^{x} - x^{2}$ over $[0, 1]$ yields $\frac{3 - \log_{e} (2)}{3 \log_{e} (2)}$ .

Option analysis

Why each option works or fails

A · (3 - log_e(2))/(3 log_e(2))

This is the correct value obtained from evaluating $\int_{0}^{1} (2^{x} - x^{2}) d x = \frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$ . Correctly compute the base of $ϕ (x)$ as $2$ and integrate $\int_{0}^{1} (2^{x} - x^{2}) d x$ .

B · 1/(3 log_e(2))

Forgetting to subtract the area under the lower curve $y = x^{2}$ , integrating only $ϕ (x) = 2^{x}$ , or mishandling the subtraction of $\frac{1}{3}$ . Remember that the area between two curves $y_{2} \geq y_{1}$ is $\int (y_{2} - y_{1}) d x$ , so subtract $\int_{0}^{1} x^{2} d x = \frac{1}{3}$ from $\int_{0}^{1} 2^{x} d x$ .

C · 3 + log_e(2)

Inverting the division when integrating $2^{x}$ (using $2^{x} \ln 2$ instead of $\frac{2^{x}}{\ln 2}$ ) and making algebraic sign errors when combining terms. Recall that the antiderivative of $a^{x}$ with respect to $x$ is $\frac{a^{x}}{\ln a}$ , not $a^{x} \ln a$ .

D · (3 + log_e(2))/(2 + log_e(3))

Mismatcing the order of domains/codomains when determining $g^{- 1} (1 / 2)$ and $f^{- 1}$ , leading to an incorrect base such as $3 / 2$ or combining unrelated logarithm arguments. Order the elements of the domain and codomain strictly to determine the unique strictly increasing or decreasing bijections.

Reviewed route

Solution

StepWorking

01Given

$f : {1, 2, 3, 4} \to {1, e, e^{2}, e^{3}}$ is strictly decreasing and bijective. $g : {1, e, e^{2}, e^{3}} \to {1, 1 / 2, 1 / 3, 1 / 4}$ is strictly increasing and bijective. $ϕ (x) = {[f^{- 1} (g^{- 1} (1 / 2))]}^{x}$ . Region $R = {(x, y) : x^{2} \leq y \leq ϕ (x), 0 \leq x \leq 1}$ .

02Goal

Find the area of the region $R$ , which is given by $\int_{0}^{1} (ϕ (x) - x^{2}) d x$ .

03Approach

Use the monotonic bijection properties to determine the values $g^{- 1} (1 / 2)$ and $f^{- 1} (g^{- 1} (1 / 2))$ , evaluate the base of $ϕ (x)$ , and integrate $ϕ (x) - x^{2}$ from $x = 0$ to $x = 1$ .

04Execute

Since $g$ is strictly increasing: $g (1) < g (e) < g (e^{2}) < g (e^{3})$ , and the target set sorted in ascending order is ${\frac{1}{4}, \frac{1}{3}, \frac{1}{2}, 1}$ . Thus, $g (1) = 1 / 4$ , $g (e) = 1 / 3$ , $g (e^{2}) = 1 / 2$ , $g (e^{3}) = 1$ . Therefore, $g^{- 1} (1 / 2) = e^{2}$ . Since $f$ is strictly decreasing: $f (1) > f (2) > f (3) > f (4)$ , matching the target set sorted in descending order ${e^{3}, e^{2}, e, 1}$ . Thus, $f (1) = e^{3}$ , $f (2) = e^{2}$ , $f (3) = e$ , $f (4) = 1$ . Therefore, $f^{- 1} (e^{2}) = 2$ . Hence, the base is $f^{- 1} (g^{- 1} (1 / 2)) = f^{- 1} (e^{2}) = 2$ , which gives $ϕ (x) = 2^{x}$ .

05Execute

Compute the area: $Area = \int_{0}^{1} (2^{x} - x^{2}) d x = {[\frac{2^{x}}{\ln 2} - \frac{x^{3}}{3}]}_{0}^{1} = (\frac{2}{\ln 2} - \frac{1}{3}) - (\frac{1}{\ln 2} - 0) = \frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$ .

✓Verify

At $x = 0$ , $2^{0} - 0^{2} = 1 > 0$ ; at $x = 1$ , $2^{1} - 1^{2} = 1 > 0$ . Over $[0, 1]$ , $2^{x} \geq x^{2}$ holds strictly, so the integrand is non-negative throughout, and the value $\approx 1.4427 - 0.3333 = 1.1094 > 0$ is consistent.

Hints that build this answer step by step

Given that $g : {1, e, e^{2}, e^{3}} \to {1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}}$ is strictly increasing, what is the value of $g^{- 1} (\frac{1}{2})$ ?
$g^{- 1} (\frac{1}{2}) = e^{2}$
Given that $f : {1, 2, 3, 4} \to {1, e, e^{2}, e^{3}}$ is strictly decreasing, what is $ϕ (x) = {[f^{- 1} (e^{2})]}^{x}$ ?
$ϕ (x) = 2^{x}$
What is the area of the region $R = {(x, y) : x^{2} \leq y \leq 2^{x}, 0 \leq x \leq 1}$ ?
$\frac{3 - \log_{e} (2)}{3 \log_{e} (2)}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is

g (e^{2}) = 1 / 2

and not

g (e) = 1 / 2

The domain elements ordered ascending are $1 < e < e^{2} < e^{3}$ . The codomain elements ordered ascending are $1 / 4 < 1 / 3 < 1 / 2 < 1$ . Since $g$ is strictly increasing, 3rd domain element maps to 3rd codomain element, so $g (e^{2}) = 1 / 2$ .

Answer

The base of the exponential function is found to be 2 by tracking the strictly monotonic bijections, and integrating $ϕ (x) - x^{2} = 2^{x} - x^{2}$ over $[0, 1]$ yields $\frac{3 - \log_{e} (2)}{3 \log_{e} (2)}$ .

Why each option works or fails

A · correct: (3 - log_e(2))/(3 log_e(2)) - This is the correct value obtained from evaluating $\int_{0}^{1} (2^{x} - x^{2}) d x = \frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$ . Correctly compute the base of $ϕ (x)$ as $2$ and integrate $\int_{0}^{1} (2^{x} - x^{2}) d x$ .
B: 1/(3 log_e(2)) - Forgetting to subtract the area under the lower curve $y = x^{2}$ , integrating only $ϕ (x) = 2^{x}$ , or mishandling the subtraction of $\frac{1}{3}$ . Remember that the area between two curves $y_{2} \geq y_{1}$ is $\int (y_{2} - y_{1}) d x$ , so subtract $\int_{0}^{1} x^{2} d x = \frac{1}{3}$ from $\int_{0}^{1} 2^{x} d x$ .
C: 3 + log_e(2) - Inverting the division when integrating $2^{x}$ (using $2^{x} \ln 2$ instead of $\frac{2^{x}}{\ln 2}$ ) and making algebraic sign errors when combining terms. Recall that the antiderivative of $a^{x}$ with respect to $x$ is $\frac{a^{x}}{\ln a}$ , not $a^{x} \ln a$ .
D: (3 + log_e(2))/(2 + log_e(3)) - Mismatcing the order of domains/codomains when determining $g^{- 1} (1 / 2)$ and $f^{- 1}$ , leading to an incorrect base such as $3 / 2$ or combining unrelated logarithm arguments. Order the elements of the domain and codomain strictly to determine the unique strictly increasing or decreasing bijections.

Step-by-step solution

Given: $f : {1, 2, 3, 4} \to {1, e, e^{2}, e^{3}}$ is strictly decreasing and bijective. $g : {1, e, e^{2}, e^{3}} \to {1, 1 / 2, 1 / 3, 1 / 4}$ is strictly increasing and bijective. $ϕ (x) = {[f^{- 1} (g^{- 1} (1 / 2))]}^{x}$ . Region $R = {(x, y) : x^{2} \leq y \leq ϕ (x), 0 \leq x \leq 1}$ .
Goal: Find the area of the region $\frac{a^{x}}{\ln a}$ , which is given by $\int_{0}^{1} (ϕ (x) - x^{2}) d x$ .
Approach: Use the monotonic bijection properties to determine the values $g^{- 1} (1 / 2)$ and $f^{- 1} (g^{- 1} (1 / 2))$ , evaluate the base of $ϕ (x)$ , and integrate $ϕ (x) - x^{2}$ from $x = 0$ to $x = 1$ .
Execute: Since $g$ is strictly increasing: $g (1) < g (e) < g (e^{2}) < g (e^{3})$ , and the target set sorted in ascending order is ${\frac{1}{4}, \frac{1}{3}, \frac{1}{2}, 1}$ . Thus, $g (1) = 1 / 4$ , $g (e) = 1 / 3$ , $g (e^{2}) = 1 / 2$ , $g (e^{3}) = 1$ . Therefore, $g^{- 1} (1 / 2) = e^{2}$ . Since $f$ is strictly decreasing: $f (1) > f (2) > f (3) > f (4)$ , matching the target set sorted in descending order ${e^{3}, e^{2}, e, 1}$ . Thus, $f (1) = e^{3}$ , $f (2) = e^{2}$ , $f (3) = e$ , $f (4) = 1$ . Therefore, $f^{- 1} (e^{2}) = 2$ . Hence, the base is $f^{- 1} (g^{- 1} (1 / 2)) = f^{- 1} (e^{2}) = 2$ , which gives $ϕ (x) = 2^{x}$ .
Execute: Compute the area: $Area = \int_{0}^{1} (2^{x} - x^{2}) d x = {[\frac{2^{x}}{\ln 2} - \frac{x^{3}}{3}]}_{0}^{1} = (\frac{2}{\ln 2} - \frac{1}{3}) - (\frac{1}{\ln 2} - 0) = \frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$ .
Verify: At $x = 0$ , $2^{0} - 0^{2} = 1 > 0$ ; at $x = 1$ , $2^{1} - 1^{2} = 1 > 0$ . Over $[0, 1]$ , $2^{x} \geq x^{2}$ holds strictly, so the integrand is non-negative throughout, and the value $\approx 1.4427 - 0.3333 = 1.1094 > 0$ is consistent.

Area Between Curves via Composite Inverses of Monotonic Functions

What feels right?

The base of the exponential function is found to be 2 by tracking the strictly monotonic bijections, and integrating $ϕ (x) - x^{2} = 2^{x} - x^{2}$ over $[0, 1]$ yields $\frac{3 - \log_{e} (2)}{3 \log_{e} (2)}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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