JEE MainMathematics

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Definite Integral Relation Involving an Arbitrary Continuous Function

If f:ℝ→ℝ be a continuous function satisfying integral from 0 to π/2 of f(sin 2x) · sin x dx + α integral from 0 to π/4 of f(cos 2x) · cos x dx = 0, then α is equal to

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A-√(3)B√(2)C√(3)D-√(2)

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A · -√(3)

Incorrectly evaluating $\int_{0}^{π / 4} \cos x d x$ as $\frac{\sqrt{3}}{2}$ instead of $\frac{1}{\sqrt{2}}$ . Recall that $\sin (π / 4) = \frac{1}{\sqrt{2}}$ , not $\frac{\sqrt{3}}{2}$ .

B · √(2)

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

C · √(3)

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

D · -√(2)

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

Reviewed route

Solution

StepWorking

01Given

Given the equation: $\int_{0}^{π / 2} f (\sin 2 x) \sin x d x + α \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0$

02Goal

Find the constant $α$ such that the equation holds for all continuous functions $f$ .

03Approach

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

04Execute

Split the first integral $I_{1} = \int_{0}^{π / 2} f (\sin 2 x) \sin x d x$ : $I_{1} = \int_{0}^{π / 4} f (\sin 2 x) \sin x d x + \int_{π / 4}^{π / 2} f (\sin 2 x) \sin x d x$ For the first part, apply King's property ( $x \to π / 4 - x$ ): $\int_{0}^{π / 4} f (\sin 2 (π / 4 - x)) \sin (π / 4 - x) d x = \int_{0}^{π / 4} f (\cos 2 x) \sin (\frac{π}{4} - x) d x$ For the second part, substitute $x - π / 4 = t ⟹ x = π / 4 + t$ with limits $t \in [0, π / 4]$ : $\int_{0}^{π / 4} f (\sin 2 (t + π / 4)) \sin (t + π / 4) d t = \int_{0}^{π / 4} f (\cos 2 t) \sin (t + \frac{π}{4}) d t$

05Execute

Combine the two sub-integrals: $I_{1} = \int_{0}^{π / 4} f (\cos 2 x) [\sin (\frac{π}{4} - x) + \sin (\frac{π}{4} + x)] d x$ Using the identity $\sin (A - B) + \sin (A + B) = 2 \sin A \cos B$ : $\sin (\frac{π}{4} - x) + \sin (\frac{π}{4} + x) = 2 \sin \frac{π}{4} \cos x = 2 (\frac{1}{\sqrt{2}}) \cos x = \sqrt{2} \cos x$ Thus, $I_{1} = \sqrt{2} \int_{0}^{π / 4} f (\cos 2 x) \cos x d x$ .

06Execute

Substitute $I_{1}$ into the original equation: $\sqrt{2} \int_{0}^{π / 4} f (\cos 2 x) \cos x d x + α \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0$ $(α + \sqrt{2}) \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0 ⟹ α = - \sqrt{2}$

✓Verify

Test with a test function $f (u) = 1$ : $\int_{0}^{π / 2} \sin x d x = [- \cos x]_{0}^{π / 2} = 1$ . $\int_{0}^{π / 4} \cos x d x = [\sin x]_{0}^{π / 4} = \frac{1}{\sqrt{2}}$ . Then $1 + α (\frac{1}{\sqrt{2}}) = 0 ⟹ α = - \sqrt{2}$ . Matches.

✓ 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: 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 $f (t) = 1$ or splitting and substituting within the integrals, we find $α = - \sqrt{2}$ .

Why each option works or fails

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

Step-by-step solution

Given: Given the equation: $\int_{0}^{π / 2} f (\sin 2 x) \sin x d x + α \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0$
Goal: Find the constant $α$ such that the equation holds for all continuous functions $f$ .
Approach: Split the first integral at $x = π / 4$ into $\int_{0}^{π / 4}$ and $\int_{π / 4}^{π / 2}$ , then transform the domain of both to $[0, π / 4]$ to express the entire equation in terms of $\int_{0}^{π / 4} f (\cos 2 x) \cos x d x$ .
Execute: Split the first integral $I_{1} = \int_{0}^{π / 2} f (\sin 2 x) \sin x d x$ : $I_{1} = \int_{0}^{π / 4} f (\sin 2 x) \sin x d x + \int_{π / 4}^{π / 2} f (\sin 2 x) \sin x d x$ For the first part, apply King's property ( $x \to π / 4 - x$ ): $\int_{0}^{π / 4} f (\sin 2 (π / 4 - x)) \sin (π / 4 - x) d x = \int_{0}^{π / 4} f (\cos 2 x) \sin (\frac{π}{4} - x) d x$ For the second part, substitute $x - π / 4 = t ⟹ x = π / 4 + t$ with limits $t \in [0, π / 4]$ : $\int_{0}^{π / 4} f (\sin 2 (t + π / 4)) \sin (t + π / 4) d t = \int_{0}^{π / 4} f (\cos 2 t) \sin (t + \frac{π}{4}) d t$
Execute: Combine the two sub-integrals: $I_{1} = \int_{0}^{π / 4} f (\cos 2 x) [\sin (\frac{π}{4} - x) + \sin (\frac{π}{4} + x)] d x$ Using the identity $\sin (A - B) + \sin (A + B) = 2 \sin A \cos B$ : $\sin (\frac{π}{4} - x) + \sin (\frac{π}{4} + x) = 2 \sin \frac{π}{4} \cos x = 2 (\frac{1}{\sqrt{2}}) \cos x = \sqrt{2} \cos x$ Thus, $I_{1} = \sqrt{2} \int_{0}^{π / 4} f (\cos 2 x) \cos x d x$ .
Execute: Substitute $I_{1}$ into the original equation: $\sqrt{2} \int_{0}^{π / 4} f (\cos 2 x) \cos x d x + α \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0$ $(α + \sqrt{2}) \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0 ⟹ α = - \sqrt{2}$
Verify: Test with a test function $f (u) = 1$ : $\int_{0}^{π / 2} \sin x d x = [- \cos x]_{0}^{π / 2} = 1$ . $\int_{0}^{π / 4} \cos x d x = [\sin x]_{0}^{π / 4} = \frac{1}{\sqrt{2}}$ . Then $1 + α (\frac{1}{\sqrt{2}}) = 0 ⟹ α = - \sqrt{2}$ . 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 $f : R \to R$ .

Goal: Determine the value of the parameter $α$ .

Approach: Choose the simplest continuous function $f (u) = 1$ to quickly evaluate both integrals.

Execute: Let $f (u) = 1$ : $\int_{0}^{π / 2} \sin x d x + α \int_{0}^{π / 4} \cos x d x = 0$ $[- \cos x]_{0}^{π / 2} + α [\sin x]_{0}^{π / 4} = 0$ $1 + α (\frac{1}{\sqrt{2}}) = 0 ⟹ α = - \sqrt{2}$

Verify: Choose another test function $f (u) = u$ to cross-check: $\int_{0}^{π / 2} \sin 2 x \sin x d x = 2 \int_{0}^{π / 2} \sin^{2} x \cos x d x = 2 [\frac{\sin^{3} x}{3}]_{0}^{π / 2} = \frac{2}{3}$ . $\int_{0}^{π / 4} \cos 2 x \cos x d x = \int_{0}^{π / 4} (1 - 2 \sin^{2} x) \cos x d x = [\sin x - \frac{2 \sin^{3} x}{3}]_{0}^{π / 4} = \frac{1}{\sqrt{2}} - \frac{2}{3 \cdot 2 \sqrt{2}} = \frac{2}{3 \sqrt{2}}$ . Then $\frac{2}{3} + α \frac{2}{3 \sqrt{2}} = 0 ⟹ α = - \sqrt{2}$ . Fully consistent.

Definite Integral Relation Involving an Arbitrary Continuous Function

What feels right?

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

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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