Why did sin^(-1)√(3/7) + sin^(-1)(2/√(7)) equal π/2?

Because if sin α = √(3/7), then cos α = √(1 - 3/7) = √(4/7) = 2/√(7). Therefore, sin^(-1)(2/√(7)) = π/2 - α.

JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Area of Region Bounded by Circle, Parabola, and Vertical Line

Let Δ be the area of the region {(x, y) ∈ ℝ^2: x^2+y^2 ≤ 21, y^2 ≤ 4 x, x ≥ 1}. Then 1/2(Δ-21 sin ^(-1) 2/(√(7))) is equal to

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A2 √(3) - 2/3B√(3) - 4/3C√(3) - 2/3D2 √(3) - 1/3

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A · 2 √(3) - 2/3

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

B · √(3) - 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 $Δ = 2 \sqrt{3} - \frac{8}{3} + 21 \arcsin (2 / \sqrt{7})$ . Multiplying by $\frac{1}{2}$ after subtracting the inverse sine term yields $\sqrt{3} - \frac{4}{3}$ .

C · √(3) - 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 √(3) - 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 $\int_{1}^{3} 2 \sqrt{x} d x = {[\frac{4}{3} x^{3 / 2}]}_{1}^{3} = 4 \sqrt{3} - \frac{4}{3}$ .

Reviewed route

Solution

StepWorking

01Given

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

02Approach

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

03Execute

First integral: $2 \int_{1}^{3} 2 \sqrt{x} d x = 4 {[\frac{2}{3} x^{3 / 2}]}_{1}^{3} = \frac{8}{3} (3 \sqrt{3} - 1) = 8 \sqrt{3} - \frac{8}{3}$ .

04Execute

Second integral: $2 \int_{3}^{\sqrt{21}} \sqrt{21 - x^{2}} d x = 2 {[\frac{x}{2} \sqrt{21 - x^{2}} + \frac{21}{2} \sin^{- 1} (\frac{x}{\sqrt{21}})]}_{3}^{\sqrt{21}} = 21 \sin^{- 1} (1) - 3 \sqrt{12} - 21 \sin^{- 1} (\frac{3}{\sqrt{21}}) = \frac{21 π}{2} - 6 \sqrt{3} - 21 \sin^{- 1} (\sqrt{\frac{3}{7}})$ .

05Execute

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

✓Verify

Check that $\sin^{- 1} \sqrt{3 / 7} + \sin^{- 1} (2 / \sqrt{7}) = π / 2$ : let $θ = \sin^{- 1} (2 / \sqrt{7})$ , then $\cos θ = \sqrt{1 - 4 / 7} = \sqrt{3 / 7}$ , so $\sin^{- 1} \sqrt{3 / 7} = \frac{π}{2} - θ$ , 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 $y^{2} = 4 x$ ?
$x = 3, y = \pm 2 \sqrt{3}$
Using symmetry across the $x$ -axis, how should the total area $Δ$ be expressed as integrals over $x$ ?
$Δ = 2 [\int_{1}^{3} 2 \sqrt{x} d x + \int_{3}^{\sqrt{21}} \sqrt{21 - x^{2}} d x]$
What is the value of $\frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}})$ after evaluating the integrals?
$\sqrt{3} - \frac{4}{3}$

✓ 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 did

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

equal

π / 2

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

Answer

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

Why each option works or fails

A: 2 √(3) - 2/3 - The student forgets to multiply the entire expression by the outer factor of $\frac{1}{2}$ , computing $Δ - 21 \arcsin (2 / \sqrt{7})$ instead. Check the final expression requested in the stem and distribute the factor of $\frac{1}{2}$ to both terms.
B · correct: √(3) - 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 $Δ = 2 \sqrt{3} - \frac{8}{3} + 21 \arcsin (2 / \sqrt{7})$ . Multiplying by $\frac{1}{2}$ after subtracting the inverse sine term yields $\sqrt{3} - \frac{4}{3}$ .
C: √(3) - 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 √(3) - 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 $\int_{1}^{3} 2 \sqrt{x} d x = {[\frac{4}{3} x^{3 / 2}]}_{1}^{3} = 4 \sqrt{3} - \frac{4}{3}$ .

Step-by-step solution

Given: The region is defined by $x^{2} + y^{2} \leq 21$ , $y^{2} \leq 4 x$ , and $x \geq 1$ . We need to find $\frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}})$ , where $Δ$ is the area of this region.
Approach: Find intersection of circle $x^{2} + y^{2} = 21$ and parabola $y^{2} = 4 x$ : $x^{2} + 4 x - 21 = 0 ⟹ (x + 7) (x - 3) = 0$ , giving $x = 3$ since $x \geq 1$ . By symmetry across the $x$ -axis, the total area is $Δ = 2 [\int_{1}^{3} 2 \sqrt{x} d x + \int_{3}^{\sqrt{21}} \sqrt{21 - x^{2}} d x]$ . Evaluate the integrals and simplify the trigonometric inverse terms.
Execute: First integral: $2 \int_{1}^{3} 2 \sqrt{x} d x = 4 {[\frac{2}{3} x^{3 / 2}]}_{1}^{3} = \frac{8}{3} (3 \sqrt{3} - 1) = 8 \sqrt{3} - \frac{8}{3}$ .
Execute: Second integral: $2 \int_{3}^{\sqrt{21}} \sqrt{21 - x^{2}} d x = 2 {[\frac{x}{2} \sqrt{21 - x^{2}} + \frac{21}{2} \sin^{- 1} (\frac{x}{\sqrt{21}})]}_{3}^{\sqrt{21}} = 21 \sin^{- 1} (1) - 3 \sqrt{12} - 21 \sin^{- 1} (\frac{3}{\sqrt{21}}) = \frac{21 π}{2} - 6 \sqrt{3} - 21 \sin^{- 1} (\sqrt{\frac{3}{7}})$ .
Execute: Sum the areas: $Δ = (8 \sqrt{3} - \frac{8}{3}) + (\frac{21 π}{2} - 6 \sqrt{3} - 21 \sin^{- 1} \sqrt{\frac{3}{7}}) = 2 \sqrt{3} - \frac{8}{3} + \frac{21 π}{2} - 21 \sin^{- 1} \sqrt{\frac{3}{7}}$ . Substitute into the target: $\frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}) = \frac{1}{2} [2 \sqrt{3} - \frac{8}{3} + \frac{21 π}{2} - 21 (\sin^{- 1} \sqrt{\frac{3}{7}} + \sin^{- 1} \frac{2}{\sqrt{7}})]$ . Since ${\sqrt{\frac{3}{7}}}^{2} + {(\frac{2}{\sqrt{7}})}^{2} = \frac{3}{7} + \frac{4}{7} = 1$ , we have $\sin^{- 1} \sqrt{\frac{3}{7}} + \sin^{- 1} \frac{2}{\sqrt{7}} = \frac{π}{2}$ . Thus, the $π$ terms cancel: $\frac{21 π}{2} - 21 (\frac{π}{2}) = 0$ . Finally, the value is $\frac{1}{2} (2 \sqrt{3} - \frac{8}{3}) = \sqrt{3} - \frac{4}{3}$ .
Verify: Check that $\sin^{- 1} \sqrt{3 / 7} + \sin^{- 1} (2 / \sqrt{7}) = π / 2$ : let $θ = \sin^{- 1} (2 / \sqrt{7})$ , then $\cos θ = \sqrt{1 - 4 / 7} = \sqrt{3 / 7}$ , so $\sin^{- 1} \sqrt{3 / 7} = \frac{π}{2} - θ$ , which holds identically.

Area of Region Bounded by Circle, Parabola, and Vertical Line

What feels right?

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

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

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