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Integral Calculus: JEE Main Mathematics Question with Solution

Let

e

be the base of natural logarithm and let

f : \{1, 2, 3, 4\} \rightarrow \{1, e, e^2, e^3\}

and

g : \{1, e, e^2, e^3\} \rightarrow \left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}

be two bijective functions such that

f

is strictly decreasing and

g

is strictly increasing. If

\phi(x) = \left[ f^{-1} \left\{ g^{-1} \left( \frac{1}{2} \right) \right\} \right]^x

, then the area of the region

R = \{(x, y) : x^2 \le y \le \phi(x), \ 0 \le x \le 1\}

is :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A

\frac{3 - \log_e(2)}{3 \log_e(2)}

B

\frac{1}{3 \log_e(2)}

C

3 + \log_e(2)

D

\frac{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 $\phi(x) - x^2 = 2^x - x^2$ over $[0, 1]$ yields $\frac{3 - \log_e(2)}{3 \log_e(2)}$ .

Option analysis

Why each option works or fails

A ·

\frac{3 - \log_e(2)}{3 \log_e(2)}

None. This is the correct value obtained from evaluating $\int_0^1 (2^x - x^2)\,dx = \frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3\ln 2}$ . Correctly compute the base of $\phi(x)$ as $2$ and integrate $\int_0^1 (2^x - x^2)\,dx$ .

B ·

\frac{1}{3 \log_e(2)}

Forgetting to subtract the area under the lower curve $y = x^2$ , integrating only $\phi(x) = 2^x$ , or mishandling the subtraction of $\frac{1}{3}$ . Remember that the area between two curves $y_2 \ge y_1$ is $\int (y_2 - y_1)\,dx$ , so subtract $\int_0^1 x^2\,dx = \frac{1}{3}$ from $\int_0^1 2^x\,dx$ .

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 ·

\frac{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. $\phi(x) = \left[ f^{-1} \left( g^{-1}(1/2) \right) \right]^x$ . Region $R = \{(x, y) : x^2 \le y \le \phi(x), \ 0 \le x \le 1\}$ .

02goal

Find the area of the region $R$ , which is given by $\int_0^1 (\phi(x) - x^2) \, dx$ .

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 $\phi(x)$ , and integrate $\phi(x) - x^2$ 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 $\left\{\frac{1}{4}, \frac{1}{3}, \frac{1}{2}, 1\right\}$ . 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 $\phi(x) = 2^x$ .

05execute

Compute the area: $\text{Area} = \int_0^1 (2^x - x^2) \, dx = \left[ \frac{2^x}{\ln 2} - \frac{x^3}{3} \right]_0^1 = \left( \frac{2}{\ln 2} - \frac{1}{3} \right) - \left( \frac{1}{\ln 2} - 0 \right) = \frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$ .

✓verify

At $x=0$ , $2^0 - 0^2 = 1 > 0$ ; at $x=1$ , $2^1 - 1^2 = 1 > 0$ . Over $[0, 1]$ , $2^x \ge x^2$ 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 $g : \{1, e, e^2, e^3\} \to \left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$ is strictly increasing, what is the value of $g^{-1}\left(\frac{1}{2}\right)$ ?
$g^{-1}\left(\frac{1}{2}\right) = e^2$
Given that $f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$ is strictly decreasing, what is $\phi(x) = \left[ f^{-1}(e^2) \right]^x$ ?
$\phi(x) = 2^x$
What is the area of the region $R = \{(x, y) : x^2 \le y \le 2^x, \ 0 \le x \le 1\}$ ?
$\frac{3 - \log_e(2)}{3 \log_e(2)}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $\phi(x) - x^2 = 2^x - x^2$ over $[0, 1]$ yields $e$ .

Why each option works or fails

A · correct: $\frac{3 - \log_e(2)}{3 \log_e(2)}$ - None. This is the correct value obtained from evaluating $\int_0^1 (2^x - x^2)\,dx = \frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3\ln 2}$ . Correctly compute the base of $\phi(x)$ as $2$ and integrate $\int_0^1 (2^x - x^2)\,dx$ .
B: $\frac{1}{3 \log_e(2)}$ - Forgetting to subtract the area under the lower curve $y = x^2$ , integrating only $\phi(x) = 2^x$ , or mishandling the subtraction of $\frac{1}{3}$ . Remember that the area between two curves $y_2 \ge y_1$ is $\int (y_2 - y_1)\,dx$ , so subtract $\int_0^1 x^2\,dx = \frac{1}{3}$ from $\int_0^1 2^x\,dx$ .
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: $\frac{3 + \log_e(2)}{2 + \log_e(3)}$ - 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. $\phi(x) = \left[ f^{-1} \left( g^{-1}(1/2) \right) \right]^x$ . Region $R = \{(x, y) : x^2 \le y \le \phi(x), \ 0 \le x \le 1\}$ .
goal: Find the area of the region $R$ , which is given by $\int_0^1 (\phi(x) - x^2) \, dx$ .
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 $\phi(x)$ , and integrate $\phi(x) - x^2$ 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 $\left\{\frac{1}{4}, \frac{1}{3}, \frac{1}{2}, 1\right\}$ . 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 $\phi(x) = 2^x$ .
execute: Compute the area: $\text{Area} = \int_0^1 (2^x - x^2) \, dx = \left[ \frac{2^x}{\ln 2} - \frac{x^3}{3} \right]_0^1 = \left( \frac{2}{\ln 2} - \frac{1}{3} \right) - \left( \frac{1}{\ln 2} - 0 \right) = \frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$ .
verify: At $x=0$ , $2^0 - 0^2 = 1 > 0$ ; at $x=1$ , $2^1 - 1^2 = 1 > 0$ . Over $[0, 1]$ , $2^x \ge x^2$ holds strictly, so the integrand is non-negative throughout, and the value $\approx 1.4427 - 0.3333 = 1.1094 > 0$ is consistent.