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Trigonometry: JEE Main Mathematics Question with Solution

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

Because the argument of and must lie in . Since for all real , the only way it can be in is if the argument equals 1, forcing .

Let

(a, b) \subset (0, 2\pi)

be the largest interval for which

\sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0, \theta \in (0, 2\pi)

, holds. If

\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0

and

\alpha - \beta = b - a

, then

\alpha

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.

\frac{\pi}{16}

\frac{\pi}{48}

\frac{\pi}{12}

\frac{\pi}{8}

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

\frac{\pi}{16}

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

B ·

\frac{\pi}{48}

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

C ·

\frac{\pi}{12}

None. This is the correct option. Correctly determined $(a, b) = (\frac{\pi}{4}, \frac{3\pi}{4})$ , domain of $\sin^{-1}(x^2 - 6x + 10)$ forcing $x = 3$ , and solved $12\alpha - \pi = 0$ to obtain $\alpha = \frac{\pi}{12}$ .

D ·

\frac{\pi}{8}

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

Reviewed route

Solution

StepWorking

01given

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

02goal

Find the value of $\alpha$ .

03approach

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

04execute

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

05execute

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

06execute

Substitute $x = 3$ into the equation: $9\alpha + 3\beta + \frac{\pi}{2} = 0 \implies 3\alpha + \beta = -\frac{\pi}{6}$ . We also have $\alpha - \beta = b - a = \frac{\pi}{2}$ . Adding the two equations: $4\alpha = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \implies \alpha = \frac{\pi}{12}$ .

✓verify

Check: $\alpha = \frac{\pi}{12} \implies \beta = \alpha - \frac{\pi}{2} = \frac{\pi}{12} - \frac{6\pi}{12} = -\frac{5\pi}{12}$ . Then $3\alpha + \beta = \frac{3\pi}{12} - \frac{5\pi}{12} = -\frac{2\pi}{12} = -\frac{\pi}{6}$ , which matches.

Hints that build this answer step by step

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

✓ 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: 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 $(x - 3)^2 + 1 \ge 1$ 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 $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$ , the largest interval is $(\frac{\pi}{4}, \frac{3\pi}{4})$ , giving $b - a = \frac{\pi}{2}$ , which along with the domain constraint $x^2 - 6x + 10 = 1$ (forcing $x = 3$ ) gives $\alpha = \frac{\pi}{12}$ .

Why each option works or fails

A: $\frac{\pi}{16}$ - Believing that $x = 3$ yields the relation $9\alpha + 3(\alpha - \frac{\pi}{2}) + \frac{\pi}{2} = 0$ with a sign error on the linear term or evaluating $\alpha - \beta$ as $\frac{\pi}{4}$ . Substitute $\beta = \alpha - (b - a) = \alpha - \frac{\pi}{2}$ into $9\alpha + 3\beta + \frac{\pi}{2} = 0$ carefully: $9\alpha + 3\alpha - \frac{3\pi}{2} + \frac{\pi}{2} = 0 \implies 12\alpha = \pi$ .
B: $\frac{\pi}{48}$ - Considering the interval $(a, b)$ over $(0, \frac{\pi}{2})$ only, leading to $(a, b) = (\frac{\pi}{4}, \frac{\pi}{2})$ and length $b - a = \frac{\pi}{4}$ . Check all quadrants in $(0, 2\pi)$ where $\sin\theta > \frac{1}{\sqrt{2}}$ ; this holds for $(\frac{\pi}{4}, \frac{3\pi}{4})$ , which has length $\frac{\pi}{2}$ .
C · correct: $\frac{\pi}{12}$ - None. This is the correct option. Correctly determined $(a, b) = (\frac{\pi}{4}, \frac{3\pi}{4})$ , domain of $\sin^{-1}(x^2 - 6x + 10)$ forcing $x = 3$ , and solved $12\alpha - \pi = 0$ to obtain $\alpha = \frac{\pi}{12}$ .
D: $\frac{\pi}{8}$ - Writing the quadratic in $x$ as having $x = 3$ but dropping the factor of 3 on $\beta$ , solving $9\alpha + \beta + \frac{\pi}{2} = 0$ . Ensure the term $\beta x$ is evaluated at $x = 3$ , giving $3\beta$ , not $\beta$ .

Step-by-step solution

given: $\sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0$ on $(0, 2\pi)$ defines the interval $(a, b)$ . The quadratic equation $\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0$ holds with $\alpha - \beta = b - a$ .
goal: Find the value of $\alpha$ .
approach: 1. Use identity $\sin^{-1}(t) + \cos^{-1}(t) = \frac{\pi}{2}$ to solve the inequality for $\theta \in (0, 2\pi)$ to find $a, b$ and thus $b - a$ . 2. Determine the domain of $\sin^{-1}(x^2 - 6x + 10)$ to find the unique permissible value of $x$ . 3. Substitute $x$ into the equation to get a relation between $\alpha$ and $\beta$ , then solve simultaneously with $\alpha - \beta = b - a$ .
execute: Rewrite the inequality: $\sin^{-1}(\sin \theta) - \left(\frac{\pi}{2} - \sin^{-1}(\sin \theta)\right) > 0 \implies 2\sin^{-1}(\sin \theta) > \frac{\pi}{2} \implies \sin^{-1}(\sin \theta) > \frac{\pi}{4}$ . Within $\theta \in (0, 2\pi)$ , this requires $\sin \theta > \frac{1}{\sqrt{2}}$ , which gives $\theta \in \left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$ . Thus, $(a, b) = \left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$ , giving $b - a = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}$ .
execute: For $\sin^{-1}(x^2 - 6x + 10)$ and $\cos^{-1}(x^2 - 6x + 10)$ to be defined, $-1 \le x^2 - 6x + 10 \le 1$ . Since $x^2 - 6x + 10 = (x - 3)^2 + 1 \ge 1$ , the only real solution is when $(x - 3)^2 = 0 \implies x = 3$ . At $x = 3$ , $x^2 - 6x + 10 = 1$ , and $\sin^{-1}(1) + \cos^{-1}(1) = \frac{\pi}{2}$ .
execute: Substitute $x = 3$ into the equation: $9\alpha + 3\beta + \frac{\pi}{2} = 0 \implies 3\alpha + \beta = -\frac{\pi}{6}$ . We also have $\alpha - \beta = b - a = \frac{\pi}{2}$ . Adding the two equations: $4\alpha = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \implies \alpha = \frac{\pi}{12}$ .
verify: Check: $\alpha = \frac{\pi}{12} \implies \beta = \alpha - \frac{\pi}{2} = \frac{\pi}{12} - \frac{6\pi}{12} = -\frac{5\pi}{12}$ . Then $3\alpha + \beta = \frac{3\pi}{12} - \frac{5\pi}{12} = -\frac{2\pi}{12} = -\frac{\pi}{6}$ , which matches.

Trigonometry: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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