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Integral Calculus: JEE Main Mathematics Question with Solution

Let

\Delta

be the area of the region

\left\{(x, y) \in \mathbb{R}^{2}: x^{2}+y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1\right\}

. Then

\frac{1}{2}\left(\Delta-21 \sin ^{-1} \frac{2}{\sqrt{7}}\right)

is equal to

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.

\sqrt{3}

\frac{2}{3}

\sqrt{3}

\frac{4}{3}

\sqrt{3}

\frac{2}{3}

\sqrt{3}

\frac{1}{3}

Step-by-step solutionView

Correct answer

The value of $\frac{1}{2}\left(\Delta-21 \sin^{-1} \frac{2}{\sqrt{7}}\right)$ is $\sqrt{3}-\frac{4}{3}$ .

Option analysis

Why each option works or fails

A · 2

\sqrt{3}

\frac{2}{3}

The student forgets to multiply the entire expression by the outer factor of $\frac{1}{2}$ , computing $\Delta - 21\arcsin(2/\sqrt{7})$ instead. Check the final expression requested in the stem and distribute the factor of $\frac{1}{2}$ to both terms.

B ·

\sqrt{3}

\frac{4}{3}

This is the correct option. The total area is decomposed by symmetry into upper and lower halves, evaluated via definite integrals of the parabola from $x=1$ to $x=3$ and the circle from $x=3$ to $x=\sqrt{21}$ , giving $\Delta = 2\sqrt{3} - \frac{8}{3} + 21\arcsin(2/\sqrt{7})$ . Multiplying by $\frac{1}{2}$ after subtracting the inverse sine term yields $\sqrt{3}-\frac{4}{3}$ .

C ·

\sqrt{3}

\frac{2}{3}

The student halves the polynomial term incorrectly, treating the parabolic contribution as $\frac{4}{3}$ instead of $\frac{8}{3}$ before applying the factor of $\frac{1}{2}$ . Ensure the parabolic integral is doubled for symmetry before subtracting and then halved at the very end.

D · 2

\sqrt{3}

\frac{1}{3}

The student makes an arithmetic slip when evaluating the limits of integration for the parabola, leading to an incorrect constant term. Carefully compute $\int_1^3 2\sqrt{x}\,dx = \left[\frac{4}{3}x^{3/2}\right]_1^3 = 4\sqrt{3} - \frac{4}{3}$ .

Reviewed route

Solution

StepWorking

01given

The region is defined by $x^2 + y^2 \leq 21$ , $y^2 \leq 4x$ , and $x \geq 1$ . We need to find $\frac{1}{2}\left(\Delta - 21 \sin^{-1} \frac{2}{\sqrt{7}}\right)$ , where $\Delta$ is the area of this region.

02approach

Find intersection of circle $x^2 + y^2 = 21$ and parabola $y^2 = 4x$ : $x^2 + 4x - 21 = 0 \implies (x+7)(x-3) = 0$ , giving $x = 3$ since $x \geq 1$ . By symmetry across the $x$ -axis, the total area is $\Delta = 2 \left[\int_{1}^{3} 2\sqrt{x}\,dx + \int_{3}^{\sqrt{21}} \sqrt{21-x^2}\,dx\right]$ . Evaluate the integrals and simplify the trigonometric inverse terms.

03execute

First integral: $2 \int_1^3 2\sqrt{x}\,dx = 4 \left[ \frac{2}{3}x^{3/2} \right]_1^3 = \frac{8}{3}(3\sqrt{3} - 1) = 8\sqrt{3} - \frac{8}{3}$ .

04execute

Second integral: $2 \int_3^{\sqrt{21}} \sqrt{21-x^2}\,dx = 2 \left[ \frac{x}{2}\sqrt{21-x^2} + \frac{21}{2}\sin^{-1}\left(\frac{x}{\sqrt{21}}\right) \right]_3^{\sqrt{21}} = 21\sin^{-1}(1) - 3\sqrt{12} - 21\sin^{-1}\left(\frac{3}{\sqrt{21}}\right) = \frac{21\pi}{2} - 6\sqrt{3} - 21\sin^{-1}\left(\sqrt{\frac{3}{7}}\right)$ .

05execute

Sum the areas: $\Delta = \left(8\sqrt{3} - \frac{8}{3}\right) + \left(\frac{21\pi}{2} - 6\sqrt{3} - 21\sin^{-1}\sqrt{\frac{3}{7}}\right) = 2\sqrt{3} - \frac{8}{3} + \frac{21\pi}{2} - 21\sin^{-1}\sqrt{\frac{3}{7}}$ . Substitute into the target: $\frac{1}{2}\left(\Delta - 21\sin^{-1}\frac{2}{\sqrt{7}}\right) = \frac{1}{2}\left[2\sqrt{3} - \frac{8}{3} + \frac{21\pi}{2} - 21\left(\sin^{-1}\sqrt{\frac{3}{7}} + \sin^{-1}\frac{2}{\sqrt{7}}\right)\right]$ . Since $\sqrt{\frac{3}{7}}^2 + \left(\frac{2}{\sqrt{7}}\right)^2 = \frac{3}{7} + \frac{4}{7} = 1$ , we have $\sin^{-1}\sqrt{\frac{3}{7}} + \sin^{-1}\frac{2}{\sqrt{7}} = \frac{\pi}{2}$ . Thus, the $\pi$ terms cancel: $\frac{21\pi}{2} - 21\left(\frac{\pi}{2}\right) = 0$ . Finally, the value is $\frac{1}{2}\left(2\sqrt{3} - \frac{8}{3}\right) = \sqrt{3} - \frac{4}{3}$ .

✓verify

Check that $\sin^{-1}\sqrt{3/7} + \sin^{-1}(2/\sqrt{7}) = \pi/2$ : let $\theta = \sin^{-1}(2/\sqrt{7})$ , then $\cos\theta = \sqrt{1 - 4/7} = \sqrt{3/7}$ , so $\sin^{-1}\sqrt{3/7} = \frac{\pi}{2} - \theta$ , which holds identically.

Hints that build this answer step by step

What are the points of intersection between the circle $x^2 + y^2 = 21$ and the parabola $y^2 = 4x$ ?
$x = 3, y = \pm 2\sqrt{3}$
Using symmetry across the $x$ -axis, how should the total area $\Delta$ be expressed as integrals over $x$ ?
$\Delta = 2\left[\int_{1}^{3} 2\sqrt{x}\,dx + \int_{3}^{\sqrt{21}} \sqrt{21-x^2}\,dx\right]$
What is the value of $\frac{1}{2}\left(\Delta-21 \sin ^{-1} \frac{2}{\sqrt{7}}\right)$ after evaluating the integrals?
$\sqrt{3} - \frac{4}{3}$

✓ 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 did

\sin^{-1}\sqrt{3/7} + \sin^{-1}(2/\sqrt{7})

equal

\pi/2

Because if $\sin \alpha = \sqrt{3/7}$ , then $\cos \alpha = \sqrt{1 - 3/7} = \sqrt{4/7} = 2/\sqrt{7}$ . Therefore, $\sin^{-1}(2/\sqrt{7}) = \frac{\pi}{2} - \alpha$ .

Answer

The value of $\frac{1}{2}\left(\Delta-21 \sin^{-1} \frac{2}{\sqrt{7}}\right)$ is $\sqrt{3}-\frac{4}{3}$ .

Why each option works or fails

A: 2 $\sqrt{3}$ - $\frac{2}{3}$ - The student forgets to multiply the entire expression by the outer factor of $\frac{1}{2}$ , computing $\Delta - 21\arcsin(2/\sqrt{7})$ instead. Check the final expression requested in the stem and distribute the factor of $\frac{1}{2}$ to both terms.
B · correct: $\sqrt{3}$ - $\frac{4}{3}$ - This is the correct option. The total area is decomposed by symmetry into upper and lower halves, evaluated via definite integrals of the parabola from $x=1$ to $x=3$ and the circle from $x=3$ to $x=\sqrt{21}$ , giving $\Delta = 2\sqrt{3} - \frac{8}{3} + 21\arcsin(2/\sqrt{7})$ . Multiplying by $\frac{1}{2}$ after subtracting the inverse sine term yields $\sqrt{3}-\frac{4}{3}$ .
C: $\sqrt{3}$ - $\frac{2}{3}$ - The student halves the polynomial term incorrectly, treating the parabolic contribution as $\frac{4}{3}$ instead of $\frac{8}{3}$ before applying the factor of $\frac{1}{2}$ . Ensure the parabolic integral is doubled for symmetry before subtracting and then halved at the very end.
D: 2 $\sqrt{3}$ - $\frac{1}{3}$ - The student makes an arithmetic slip when evaluating the limits of integration for the parabola, leading to an incorrect constant term. Carefully compute $\int_1^3 2\sqrt{x}\,dx = \left[\frac{4}{3}x^{3/2}\right]_1^3 = 4\sqrt{3} - \frac{4}{3}$ .

Step-by-step solution

given: The region is defined by $x^2 + y^2 \leq 21$ , $y^2 \leq 4x$ , and $x \geq 1$ . We need to find $\frac{1}{2}\left(\Delta - 21 \sin^{-1} \frac{2}{\sqrt{7}}\right)$ , where $\Delta$ is the area of this region.
approach: Find intersection of circle $x^2 + y^2 = 21$ and parabola $y^2 = 4x$ : $x^2 + 4x - 21 = 0 \implies (x+7)(x-3) = 0$ , giving $x = 3$ since $x \geq 1$ . By symmetry across the $x$ -axis, the total area is $\Delta = 2 \left[\int_{1}^{3} 2\sqrt{x}\,dx + \int_{3}^{\sqrt{21}} \sqrt{21-x^2}\,dx\right]$ . Evaluate the integrals and simplify the trigonometric inverse terms.
execute: First integral: $\sqrt{3}$ .
execute: Second integral: $2 \int_3^{\sqrt{21}} \sqrt{21-x^2}\,dx = 2 \left[ \frac{x}{2}\sqrt{21-x^2} + \frac{21}{2}\sin^{-1}\left(\frac{x}{\sqrt{21}}\right) \right]_3^{\sqrt{21}} = 21\sin^{-1}(1) - 3\sqrt{12} - 21\sin^{-1}\left(\frac{3}{\sqrt{21}}\right) = \frac{21\pi}{2} - 6\sqrt{3} - 21\sin^{-1}\left(\sqrt{\frac{3}{7}}\right)$ .
execute: Sum the areas: $\Delta = \left(8\sqrt{3} - \frac{8}{3}\right) + \left(\frac{21\pi}{2} - 6\sqrt{3} - 21\sin^{-1}\sqrt{\frac{3}{7}}\right) = 2\sqrt{3} - \frac{8}{3} + \frac{21\pi}{2} - 21\sin^{-1}\sqrt{\frac{3}{7}}$ . Substitute into the target: $\frac{1}{2}\left(\Delta - 21\sin^{-1}\frac{2}{\sqrt{7}}\right) = \frac{1}{2}\left[2\sqrt{3} - \frac{8}{3} + \frac{21\pi}{2} - 21\left(\sin^{-1}\sqrt{\frac{3}{7}} + \sin^{-1}\frac{2}{\sqrt{7}}\right)\right]$ . Since $\sqrt{\frac{3}{7}}^2 + \left(\frac{2}{\sqrt{7}}\right)^2 = \frac{3}{7} + \frac{4}{7} = 1$ , we have $\sin^{-1}\sqrt{\frac{3}{7}} + \sin^{-1}\frac{2}{\sqrt{7}} = \frac{\pi}{2}$ . Thus, the $\pi$ terms cancel: $\frac{21\pi}{2} - 21\left(\frac{\pi}{2}\right) = 0$ . Finally, the value is $\frac{1}{2}\left(2\sqrt{3} - \frac{8}{3}\right) = \sqrt{3} - \frac{4}{3}$ .
verify: Check that $\sin^{-1}\sqrt{3/7} + \sin^{-1}(2/\sqrt{7}) = \pi/2$ : let $\theta = \sin^{-1}(2/\sqrt{7})$ , then $\cos\theta = \sqrt{1 - 4/7} = \sqrt{3/7}$ , so $\sin^{-1}\sqrt{3/7} = \frac{\pi}{2} - \theta$ , which holds identically.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The value of $\frac{1}{2}\left(\Delta-21 \sin^{-1} \frac{2}{\sqrt{7}}\right)$ is $\sqrt{3}-\frac{4}{3}$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

✓verify

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