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Integral Calculus: JEE Main Mathematics Question with Solution

\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}

be a continuous function satisfying

\int_0^{\pi/2} \mathrm{f}(\sin 2x) \cdot \sin x\,\mathrm{d}x + \alpha \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cdot \cos x\,\mathrm{d}x = 0,

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.

-\sqrt{3}

\sqrt{2}

\sqrt{3}

-\sqrt{2}

Step-by-step solutionView

Correct answer

By choosing the test function $\mathrm{f}(t) = 1$ or splitting and substituting within the integrals, we find $\alpha = -\sqrt{2}$ .

Option analysis

Why each option works or fails

A ·

-\sqrt{3}

Incorrectly evaluating $\int_0^{\pi/4} \cos x\,\mathrm{d}x$ as $\frac{\sqrt{3}}{2}$ instead of $\frac{1}{\sqrt{2}}$ . Recall that $\sin(\pi/4) = \frac{1}{\sqrt{2}}$ , not $\frac{\sqrt{3}}{2}$ .

B ·

\sqrt{2}

Moving a term to the opposite side of the equation without changing its sign, leading to $\alpha = +\sqrt{2}$ . When moving the first integral $I_1 = 1$ to the right-hand side in $I_1 + \alpha I_2 = 0$ , it becomes $\alpha I_2 = -1$ , yielding $\alpha = -\sqrt{2}$ .

C ·

\sqrt{3}

Combining a wrong evaluation of the trigonometric limit $\sin(\pi/4)$ as $\frac{\sqrt{3}}{2}$ with a sign error when solving for $\alpha$ . Ensure exact values for standard angles are used ( $\sin(\pi/4) = \frac{1}{\sqrt{2}}$ ) and track signs carefully across the equality.

D ·

-\sqrt{2}

Correct option. Evaluating with $\mathrm{f}(t) = 1$ gives $1 + \alpha \cdot \frac{1}{\sqrt{2}} = 0$ , which yields $\alpha = -\sqrt{2}$ .

Reviewed route

Solution

StepWorking

01given

Given the equation: $\int_0^{\pi/2} \mathrm{f}(\sin 2x) \sin x\,\mathrm{d}x + \alpha \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x = 0$

02goal

Find the constant $\alpha$ such that the equation holds for all continuous functions $\mathrm{f}$ .

03approach

Split the first integral at $x = \pi/4$ into $\int_0^{\pi/4}$ and $\int_{\pi/4}^{\pi/2}$ , then transform the domain of both to $[0, \pi/4]$ to express the entire equation in terms of $\int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x$ .

04execute

Split the first integral $I_1 = \int_0^{\pi/2} \mathrm{f}(\sin 2x) \sin x\,\mathrm{d}x$ : $I_1 = \int_0^{\pi/4} \mathrm{f}(\sin 2x) \sin x\,\mathrm{d}x + \int_{\pi/4}^{\pi/2} \mathrm{f}(\sin 2x) \sin x\,\mathrm{d}x$ For the first part, apply King's property ( $x \to \pi/4 - x$ ): $\int_0^{\pi/4} \mathrm{f}(\sin 2(\pi/4 - x)) \sin(\pi/4 - x)\,\mathrm{d}x = \int_0^{\pi/4} \mathrm{f}(\cos 2x) \sin\left(\frac{\pi}{4} - x\right)\,\mathrm{d}x$ For the second part, substitute $x - \pi/4 = t \implies x = \pi/4 + t$ with limits $t \in [0, \pi/4]$ : $\int_0^{\pi/4} \mathrm{f}(\sin 2(t + \pi/4)) \sin(t + \pi/4)\,\mathrm{dt} = \int_0^{\pi/4} \mathrm{f}(\cos 2t) \sin\left(t + \frac{\pi}{4}\right)\,\mathrm{dt}$

05execute

Combine the two sub-integrals: $I_1 = \int_0^{\pi/4} \mathrm{f}(\cos 2x) \left[ \sin\left(\frac{\pi}{4} - x\right) + \sin\left(\frac{\pi}{4} + x\right) \right] \mathrm{d}x$ Using the identity $\sin(A - B) + \sin(A + B) = 2\sin A \cos B$ : $\sin\left(\frac{\pi}{4} - x\right) + \sin\left(\frac{\pi}{4} + x\right) = 2\sin\frac{\pi}{4}\cos x = 2 \left(\frac{1}{\sqrt{2}}\right) \cos x = \sqrt{2}\cos x$ Thus, $I_1 = \sqrt{2}\int_0^{\pi/4} \mathrm{f}(\cos 2x)\cos x\,\mathrm{d}x$ .

06execute

Substitute $I_1$ into the original equation: $\sqrt{2} \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x + \alpha \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x = 0$ $(\alpha + \sqrt{2}) \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x = 0 \implies \alpha = -\sqrt{2}$

✓verify

Test with a test function $\mathrm{f}(u) = 1$ : $\int_0^{\pi/2} \sin x\,\mathrm{d}x = [-\cos x]_0^{\pi/2} = 1$ . $\int_0^{\pi/4} \cos x\,\mathrm{d}x = [\sin x]_0^{\pi/4} = \frac{1}{\sqrt{2}}$ . Then $1 + \alpha \left(\frac{1}{\sqrt{2}}\right) = 0 \implies \alpha = -\sqrt{2}$ . Matches.

✓ 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: 7 September 2026

Quick checks

Students also ask

Why did we split the first integral at pi/4?

Because the second integral has upper limit pi/4 and contains f(cos 2x). Splitting at pi/4 allows transforming both parts into intervals [0, pi/4] where sin(2(pi/4 - x)) = cos(2x) and sin(2(pi/4 + t)) = cos(2t).

Answer

By choosing the test function $\mathrm{f}(t) = 1$ or splitting and substituting within the integrals, we find $\alpha = -\sqrt{2}$ .

Why each option works or fails

A: $-\sqrt{3}$ - Incorrectly evaluating $\int_0^{\pi/4} \cos x\,\mathrm{d}x$ as $\frac{\sqrt{3}}{2}$ instead of $\frac{1}{\sqrt{2}}$ . Recall that $\sin(\pi/4) = \frac{1}{\sqrt{2}}$ , not $\frac{\sqrt{3}}{2}$ .
B: $\sqrt{2}$ - Moving a term to the opposite side of the equation without changing its sign, leading to $\alpha = +\sqrt{2}$ . When moving the first integral $I_1 = 1$ to the right-hand side in $I_1 + \alpha I_2 = 0$ , it becomes $\alpha I_2 = -1$ , yielding $-\sqrt{3}$ .
C: $\sqrt{3}$ - Combining a wrong evaluation of the trigonometric limit $\sin(\pi/4)$ as $\frac{\sqrt{3}}{2}$ with a sign error when solving for $\alpha$ . Ensure exact values for standard angles are used ( $\sin(\pi/4) = \frac{1}{\sqrt{2}}$ ) and track signs carefully across the equality.
D · correct: $-\sqrt{2}$ - Correct option. Evaluating with $\mathrm{f}(t) = 1$ gives $1 + \alpha \cdot \frac{1}{\sqrt{2}} = 0$ , which yields $\alpha = -\sqrt{2}$ .

Step-by-step solution

given: Given the equation: $\int_0^{\pi/2} \mathrm{f}(\sin 2x) \sin x\,\mathrm{d}x + \alpha \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x = 0$
goal: Find the constant $\alpha$ such that the equation holds for all continuous functions $\mathrm{f}$ .
approach: Split the first integral at $x = \pi/4$ into $\int_0^{\pi/4}$ and $\int_{\pi/4}^{\pi/2}$ , then transform the domain of both to $[0, \pi/4]$ to express the entire equation in terms of $\int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x$ .
execute: Split the first integral $I_1 = \int_0^{\pi/2} \mathrm{f}(\sin 2x) \sin x\,\mathrm{d}x$ : $I_1 = \int_0^{\pi/4} \mathrm{f}(\sin 2x) \sin x\,\mathrm{d}x + \int_{\pi/4}^{\pi/2} \mathrm{f}(\sin 2x) \sin x\,\mathrm{d}x$ For the first part, apply King's property ( $x \to \pi/4 - x$ ): $\int_0^{\pi/4} \mathrm{f}(\sin 2(\pi/4 - x)) \sin(\pi/4 - x)\,\mathrm{d}x = \int_0^{\pi/4} \mathrm{f}(\cos 2x) \sin\left(\frac{\pi}{4} - x\right)\,\mathrm{d}x$ For the second part, substitute $x - \pi/4 = t \implies x = \pi/4 + t$ with limits $t \in [0, \pi/4]$ : $\int_0^{\pi/4} \mathrm{f}(\sin 2(t + \pi/4)) \sin(t + \pi/4)\,\mathrm{dt} = \int_0^{\pi/4} \mathrm{f}(\cos 2t) \sin\left(t + \frac{\pi}{4}\right)\,\mathrm{dt}$
execute: Combine the two sub-integrals: $I_1 = \int_0^{\pi/4} \mathrm{f}(\cos 2x) \left[ \sin\left(\frac{\pi}{4} - x\right) + \sin\left(\frac{\pi}{4} + x\right) \right] \mathrm{d}x$ Using the identity $\sin(A - B) + \sin(A + B) = 2\sin A \cos B$ : $\sin\left(\frac{\pi}{4} - x\right) + \sin\left(\frac{\pi}{4} + x\right) = 2\sin\frac{\pi}{4}\cos x = 2 \left(\frac{1}{\sqrt{2}}\right) \cos x = \sqrt{2}\cos x$ Thus, $I_1 = \sqrt{2}\int_0^{\pi/4} \mathrm{f}(\cos 2x)\cos x\,\mathrm{d}x$ .
execute: Substitute $I_1$ into the original equation: $\sqrt{2} \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x + \alpha \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x = 0$ $(\alpha + \sqrt{2}) \int_0^{\pi/4} \mathrm{f}(\cos 2x) \cos x\,\mathrm{d}x = 0 \implies \alpha = -\sqrt{2}$
verify: Test with a test function $\mathrm{f}(u) = 1$ : $\int_0^{\pi/2} \sin x\,\mathrm{d}x = [-\cos x]_0^{\pi/2} = 1$ . $\int_0^{\pi/4} \cos x\,\mathrm{d}x = [\sin x]_0^{\pi/4} = \frac{1}{\sqrt{2}}$ . Then $1 + \alpha \left(\frac{1}{\sqrt{2}}\right) = 0 \implies \alpha = -\sqrt{2}$ . Matches.

Shortcut:

When to use it: When the relation must hold for all continuous functions f, choose the simplest constant function

given: The relation holds for any continuous function $\mathrm{f}:\mathbb{R}\to\mathbb{R}$ .

goal: Determine the value of the parameter $\alpha$ .

approach: Choose the simplest continuous function $\mathrm{f}(u) = 1$ to quickly evaluate both integrals.

execute: Let $\mathrm{f}(u) = 1$ : $\int_0^{\pi/2} \sin x\,\mathrm{d}x + \alpha \int_0^{\pi/4} \cos x\,\mathrm{d}x = 0$ $[-\cos x]_0^{\pi/2} + \alpha [\sin x]_0^{\pi/4} = 0$ $1 + \alpha \left(\frac{1}{\sqrt{2}}\right) = 0 \implies \alpha = -\sqrt{2}$

verify: Choose another test function $\mathrm{f}(u) = u$ to cross-check: $\int_0^{\pi/2} \sin 2x \sin x\,\mathrm{d}x = 2\int_0^{\pi/2} \sin^2 x \cos x\,\mathrm{d}x = 2 [\frac{\sin^3 x}{3}]_0^{\pi/2} = \frac{2}{3}$ . $\int_0^{\pi/4} \cos 2x \cos x\,\mathrm{d}x = \int_0^{\pi/4} (1 - 2\sin^2 x)\cos x\,\mathrm{d}x = [\sin x - \frac{2\sin^3 x}{3}]_0^{\pi/4} = \frac{1}{\sqrt{2}} - \frac{2}{3 \cdot 2\sqrt{2}} = \frac{2}{3\sqrt{2}}$ . Then $\frac{2}{3} + \alpha \frac{2}{3\sqrt{2}} = 0 \implies \alpha = -\sqrt{2}$ . Fully consistent.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

By choosing the test function $\mathrm{f}(t) = 1$ or splitting and substituting within the integrals, we find $\alpha = -\sqrt{2}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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