If f: R R is a continuous function satisfying 0^ 2 f( 2x) x dx + 0^ 4 f( 2x) x dx = 0, then the value of is:

If f: R R is a continuous function satisfying 0^ 2 f( 2x) x dx +… — Solution | JEE Mathematics

MCQMediumJEE 2023Properties of Definite Integrals

If $f : \mathbb{R} \to \mathbb{R}$ is a continuous function satisfying

\int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int_{0}^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0,

then the value of $\alpha$ is:

Correct answer:C

Standard Method

Given:

\int_{0}^{\frac{\pi}{2}} F(\sin^2 x) \sin x \, dx + \alpha \int_{0}^{\frac{\pi}{4}} F(\cos^2 x) \cos x \, dx = 0

Find: The value of $\alpha$ .

From the solution working, split the first integral as

\int_{0}^{\frac{\pi}{2}} F(\sin^2 x) \sin x \, dx = \int_{0}^{\frac{\pi}{4}} F(\sin^2 x) \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} F(\sin^2 x) \sin x \, dx

so the equation becomes

\int_{0}^{\frac{\pi}{4}} F(\sin^2 x) \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} F(\sin^2 x) \sin x \, dx + \alpha \int_{0}^{\frac{\pi}{4}} F(\cos^2 x) \cos x \, dx = 0

Using symmetry and the substitution indicated in the solution, this is rewritten as

\int_{0}^{\frac{\pi}{4}} F(\cos^2 x) \sin\left(\frac{\pi}{4} - x\right) \, dx + \int_{0}^{\frac{\pi}{4}} F(\cos^2 t) \sin\left(t + \frac{\pi}{4}\right) \, dt + \alpha \int_{0}^{\frac{\pi}{4}} F(\cos^2 x) \cos x \, dx = 0

Now use

\sin\left(\frac{\pi}{4} - x\right) + \sin\left(x + \frac{\pi}{4}\right) = \sqrt{2} \cos x

Hence,

\sqrt{2} \int_{0}^{\frac{\pi}{4}} F(\cos^2 x) \cos x \, dx + \alpha \int_{0}^{\frac{\pi}{4}} F(\cos^2 x) \cos x \, dx = 0

Therefore,

(\sqrt{2} + \alpha) \int_{0}^{\frac{\pi}{4}} F(\cos^2 x) \cos x \, dx = 0

So,

\sqrt{2} + \alpha = 0

and therefore

\alpha = -\sqrt{2}

Thus, the correct option is C.

Using the given integrand without checking the transformation carefully. The working uses forms like $F(\sin^2 x)$ and $F(\cos^2 x)$ , so missing the intended substitution/symmetry step leads to the wrong comparison. Always rewrite both integrals into the same functional form before combining them.
Applying the identity for $\sin(A\pm B)$ incorrectly. If $\sin\left(\frac{\pi}{4}-x\right) + \sin\left(x+\frac{\pi}{4}\right)$ is simplified wrongly, the coefficient of the common integral is lost. Use the identity carefully to obtain $\sqrt{2}\cos x$ .
Forgetting to split the first integral over $\left[0, \frac{\pi}{2}\right]$ into two parts. Without splitting at $\frac{\pi}{4}$ , the symmetry argument used in the solution cannot be applied cleanly. First align the intervals, then combine the terms.

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