JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Solving an Inverse Trigonometric Inequality and Determining Quadratic Coefficients via Domain Constraints

Q: Why does x have to be strictly equal to 3? Isn't it an equation with quadratic terms?

Because the argument of sin^(-1) and cos^(-1) must lie in [-1, 1]. Since (x - 3)^2 + 1 ≥ 1 for all real x, the only way it can be in [-1, 1] is if the argument equals 1, forcing x = 3.

Let (a, b) ⊂ (0, 2π) be the largest interval for which sin^(-1)(sin θ) - cos^(-1)(sin θ) > 0, θ ∈ (0, 2π), holds. If α x^2 + β x + sin^(-1)(x^2 - 6x + 10) + cos^(-1)(x^2 - 6x + 10) = 0 and α - β = b - a, then α is equal to:

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What feels right?

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Aπ/16Bπ/48Cπ/12Dπ/8

Step-by-step solutionView

Correct answer

By simplifying the inequality using $\sin^{- 1} t + \cos^{- 1} t = \frac{π}{2}$ , the largest interval is $(\frac{π}{4}, \frac{3 π}{4})$ , giving $b - a = \frac{π}{2}$ , which along with the domain constraint $x^{2} - 6 x + 10 = 1$ (forcing $x = 3$ ) gives $α = \frac{π}{12}$ .

Option analysis

Why each option works or fails

A · π/16

Believing that $x = 3$ yields the relation $9 α + 3 (α - \frac{π}{2}) + \frac{π}{2} = 0$ with a sign error on the linear term or evaluating $α - β$ as $\frac{π}{4}$ . Substitute $β = α - (b - a) = α - \frac{π}{2}$ into $9 α + 3 β + \frac{π}{2} = 0$ carefully: $9 α + 3 α - \frac{3 π}{2} + \frac{π}{2} = 0 ⟹ 12 α = π$ .

B · π/48

Considering the interval $(a, b)$ over $(0, \frac{π}{2})$ only, leading to $(a, b) = (\frac{π}{4}, \frac{π}{2})$ and length $b - a = \frac{π}{4}$ . Check all quadrants in $(0, 2 π)$ where $\sin θ > \frac{1}{\sqrt{2}}$ ; this holds for $θ \in (\frac{π}{4}, \frac{3 π}{4})$ , which has length $\frac{π}{2}$ .

C · π/12

This is the correct option. Correctly determined $(a, b) = (\frac{π}{4}, \frac{3 π}{4})$ , domain of $\sin^{- 1} (x^{2} - 6 x + 10)$ forcing $x = 3$ , and solved $12 α - π = 0$ to obtain $α = \frac{π}{12}$ .

D · π/8

Writing the quadratic in $x$ as having $x = 3$ but dropping the factor of 3 on $β$ , solving $9 α + β + \frac{π}{2} = 0$ . Ensure the term $β x$ is evaluated at $x = 3$ , giving $3 β$ , not $β$ .

Reviewed route

Solution

StepWorking

01Given

$\sin^{- 1} (\sin θ) - \cos^{- 1} (\sin θ) > 0$ on $(0, 2 π)$ defines the interval $(a, b)$ . The quadratic equation $α x^{2} + β x + \sin^{- 1} (x^{2} - 6 x + 10) + \cos^{- 1} (x^{2} - 6 x + 10) = 0$ holds with $α - β = b - a$ .

02Goal

Find the value of $α$ .

03Approach

1. Use identity $\sin^{- 1} (t) + \cos^{- 1} (t) = \frac{π}{2}$ to solve the inequality for $θ \in (0, 2 π)$ to find $a, b$ and thus $b - a$ . 2. Determine the domain of $\sin^{- 1} (x^{2} - 6 x + 10)$ to find the unique permissible value of $x$ . 3. Substitute $x$ into the equation to get a relation between $α$ and $β$ , then solve simultaneously with $α - β = b - a$ .

04Execute

Rewrite the inequality: $\sin^{- 1} (\sin θ) - (\frac{π}{2} - \sin^{- 1} (\sin θ)) > 0 ⟹ 2 \sin^{- 1} (\sin θ) > \frac{π}{2} ⟹ \sin^{- 1} (\sin θ) > \frac{π}{4}$ . Within $θ \in (0, 2 π)$ , this requires $\sin θ > \frac{1}{\sqrt{2}}$ , which gives $θ \in (\frac{π}{4}, \frac{3 π}{4})$ . Thus, $(a, b) = (\frac{π}{4}, \frac{3 π}{4})$ , giving $b - a = \frac{3 π}{4} - \frac{π}{4} = \frac{π}{2}$ .

05Execute

For $\sin^{- 1} (x^{2} - 6 x + 10)$ and $\cos^{- 1} (x^{2} - 6 x + 10)$ to be defined, $- 1 \leq x^{2} - 6 x + 10 \leq 1$ . Since $x^{2} - 6 x + 10 = (x - 3)^{2} + 1 \geq 1$ , the only real solution is when $(x - 3)^{2} = 0 ⟹ x = 3$ . At $x = 3$ , $x^{2} - 6 x + 10 = 1$ , and $\sin^{- 1} (1) + \cos^{- 1} (1) = \frac{π}{2}$ .

06Execute

Substitute $x = 3$ into the equation: $9 α + 3 β + \frac{π}{2} = 0 ⟹ 3 α + β = - \frac{π}{6}$ . We also have $α - β = b - a = \frac{π}{2}$ . Adding the two equations: $4 α = \frac{π}{2} - \frac{π}{6} = \frac{π}{3} ⟹ α = \frac{π}{12}$ .

✓Verify

Check: $α = \frac{π}{12} ⟹ β = α - \frac{π}{2} = \frac{π}{12} - \frac{6 π}{12} = - \frac{5 π}{12}$ . Then $3 α + β = \frac{3 π}{12} - \frac{5 π}{12} = - \frac{2 π}{12} = - \frac{π}{6}$ , which matches.

Hints that build this answer step by step

Using the identity $\cos^{- 1} t = \frac{π}{2} - \sin^{- 1} t$ , what condition on $\sin θ$ does the inequality $\sin^{- 1} (\sin θ) - \cos^{- 1} (\sin θ) > 0$ simplify to?
$\sin θ > \frac{1}{\sqrt{2}}$
What is the largest interval $(a, b) \subset (0, 2 π)$ satisfying $\sin θ > \frac{1}{\sqrt{2}}$ , and what is $b - a$ ?
$(a, b) = (\frac{π}{4}, \frac{3 π}{4})$ , so $b - a = \frac{π}{2}$
For which real value(s) of $x$ is the expression $\sin^{- 1} (x^{2} - 6 x + 10) + \cos^{- 1} (x^{2} - 6 x + 10)$ defined?
$x = 3$ only
Substituting $x = 3$ and $b - a = \frac{π}{2}$ into $α x^{2} + β x + \frac{π}{2} = 0$ with $β = α - \frac{π}{2}$ , what is $α$ ?
$α = \frac{π}{12}$

✓ 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: 9 September 2026

Quick checks

Students also ask

Why does

x

have to be strictly equal to 3? Isn't it an equation with quadratic terms?

Because the argument of $\sin^{- 1}$ and $\cos^{- 1}$ must lie in $[- 1, 1]$ . Since $(x - 3)^{2} + 1 \geq 1$ for all real $x$ , the only way it can be in $[- 1, 1]$ is if the argument equals 1, forcing $x = 3$ .

Answer

By simplifying the inequality using $\sin^{- 1} t + \cos^{- 1} t = \frac{π}{2}$ , the largest interval is $(\frac{π}{4}, \frac{3 π}{4})$ , giving $b - a = \frac{π}{2}$ , which along with the domain constraint $x^{2} - 6 x + 10 = 1$ (forcing $x = 3$ ) gives $α = \frac{π}{12}$ .

Why each option works or fails

A: π/16 - Believing that $x = 3$ yields the relation $9 α + 3 (α - \frac{π}{2}) + \frac{π}{2} = 0$ with a sign error on the linear term or evaluating $α - β$ as $\frac{π}{4}$ . Substitute $β = α - (b - a) = α - \frac{π}{2}$ into $9 α + 3 β + \frac{π}{2} = 0$ carefully: $9 α + 3 α - \frac{3 π}{2} + \frac{π}{2} = 0 ⟹ 12 α = π$ .
B: π/48 - Considering the interval $(a, b)$ over $(0, \frac{π}{2})$ only, leading to $(a, b) = (\frac{π}{4}, \frac{π}{2})$ and length $b - a = \frac{π}{4}$ . Check all quadrants in $(0, 2 π)$ where $\sin θ > \frac{1}{\sqrt{2}}$ ; this holds for $θ \in (\frac{π}{4}, \frac{3 π}{4})$ , which has length $\frac{π}{2}$ .
C · correct: π/12 - This is the correct option. Correctly determined $(a, b) = (\frac{π}{4}, \frac{3 π}{4})$ , domain of $\sin^{- 1} (x^{2} - 6 x + 10)$ forcing $(a, b)$ , and solved $12 α - π = 0$ to obtain $α = \frac{π}{12}$ .
D: π/8 - Writing the quadratic in $x$ as having $x = 3$ but dropping the factor of 3 on $β$ , solving $9 α + β + \frac{π}{2} = 0$ . Ensure the term $β x$ is evaluated at $x = 3$ , giving $3 β$ , not $β$ .

Step-by-step solution

Given: $\sin^{- 1} (\sin θ) - \cos^{- 1} (\sin θ) > 0$ on $(0, 2 π)$ defines the interval $(a, b)$ . The quadratic equation $α x^{2} + β x + \sin^{- 1} (x^{2} - 6 x + 10) + \cos^{- 1} (x^{2} - 6 x + 10) = 0$ holds with $α - β = b - a$ .
Goal: Find the value of $α = \frac{π}{12}$ .
Approach: 1. Use identity $\sin^{- 1} (t) + \cos^{- 1} (t) = \frac{π}{2}$ to solve the inequality for $θ \in (0, 2 π)$ to find $a, b$ and thus $b - a$ . 2. Determine the domain of $\sin^{- 1} (x^{2} - 6 x + 10)$ to find the unique permissible value of $x$ . 3. Substitute $x$ into the equation to get a relation between $α$ and $β$ , then solve simultaneously with $α - β = b - a$ .
Execute: Rewrite the inequality: $(a, b)$ . Within $θ \in (0, 2 π)$ , this requires $\sin θ > \frac{1}{\sqrt{2}}$ , which gives $θ \in (\frac{π}{4}, \frac{3 π}{4})$ . Thus, $(a, b) = (\frac{π}{4}, \frac{3 π}{4})$ , giving $b - a = \frac{3 π}{4} - \frac{π}{4} = \frac{π}{2}$ .
Execute: For $\sin^{- 1} (x^{2} - 6 x + 10)$ and $\cos^{- 1} (x^{2} - 6 x + 10)$ to be defined, $- 1 \leq x^{2} - 6 x + 10 \leq 1$ . Since $x^{2} - 6 x + 10 = (x - 3)^{2} + 1 \geq 1$ , the only real solution is when $θ \in (\frac{π}{4}, \frac{3 π}{4})$ . At $x = 3$ , $x^{2} - 6 x + 10 = 1$ , and $\sin^{- 1} (1) + \cos^{- 1} (1) = \frac{π}{2}$ .
Execute: Substitute $x = 3$ into the equation: $9 α + 3 β + \frac{π}{2} = 0 ⟹ 3 α + β = - \frac{π}{6}$ . We also have $α - β = b - a = \frac{π}{2}$ . Adding the two equations: $4 α = \frac{π}{2} - \frac{π}{6} = \frac{π}{3} ⟹ α = \frac{π}{12}$ .
Verify: Check: $α = \frac{π}{12} ⟹ β = α - \frac{π}{2} = \frac{π}{12} - \frac{6 π}{12} = - \frac{5 π}{12}$ . Then $3 α + β = \frac{3 π}{12} - \frac{5 π}{12} = - \frac{2 π}{12} = - \frac{π}{6}$ , which matches.

Solving an Inverse Trigonometric Inequality and Determining Quadratic Coefficients via Domain Constraints

What feels right?

By simplifying the inequality using $\sin^{- 1} t + \cos^{- 1} t = \frac{π}{2}$ , the largest interval is $(\frac{π}{4}, \frac{3 π}{4})$ , giving $b - a = \frac{π}{2}$ , which along with the domain constraint $x^{2} - 6 x + 10 = 1$ (forcing $x = 3$ ) gives $α = \frac{π}{12}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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