If f : R \(\rightarrow\) R be a continuous function satisfying \(\int^{\frac{\pi}{2}}_0f(sin2x)sinxdx+\alpha\int^{\frac{\pi}{4}}_0f(cos2x)cosxdx=0\), then the value of \(\alpha\) is
- -√3
- √3
- -√2
- √2
The Correct Option is C
Solution and Explanation
We are given the integral equation:
\[ \int_{0}^{\frac{\pi}{2}} F(\sin^2x) \sin x \, dx + \alpha \int_{0}^{\frac{\pi}{2}} F(\cos^2x) \cos x \, dx = 0. \]
Step 1: Simplify the First Integral
Split the range of integration for the first term as follows:
\[ \int_{0}^{\frac{\pi}{2}} F(\sin^2x) \sin x \, dx = \int_{0}^{\frac{\pi}{4}} F(\sin^2x) \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} F(\sin^2x) \sin x \, dx. \]
Step 2: Use Symmetry for the Second Term
Rewriting the integral equation, we get:
\[ \int_{0}^{\frac{\pi}{4}} F(\sin^2x) \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} F(\sin^2x) \sin x \, dx + \alpha \int_{0}^{\frac{\pi}{2}} F(\cos^2x) \cos x \, dx = 0. \]
Step 3: Use the Property of Definite Integrals
Using the property of definite integrals:
\[ \int_{0}^{a} F(x) \, dx = \int_{0}^{a} F(a - x) \, dx, \]
we substitute \( x = t + \frac{\pi}{4} \) into the integrals. This simplifies to:
\[ \int_{0}^{\frac{\pi}{4}} F(\cos^2x) \sin\left(\frac{\pi}{4} - x\right) \, dx + \int_{0}^{\frac{\pi}{4}} F(\cos^2t) \sin\left(t + \frac{\pi}{4}\right) \, dt + \alpha \int_{0}^{\frac{\pi}{4}} F(\cos^2x) \cos x \, dx = 0. \]
Step 4: Simplify Using Trigonometric Identities
Using the addition formulas for sine:
\[ \sin\left(\frac{\pi}{4} - x\right) + \sin\left(x + \frac{\pi}{4}\right) = \sqrt{2} \cos x, \]
the equation becomes:
\[ \sqrt{2} \int_{0}^{\frac{\pi}{4}} F(\cos^2x) \cos x \, dx + \alpha \int_{0}^{\frac{\pi}{4}} F(\cos^2x) \cos x \, dx = 0. \]
Step 5: Factor Out Common Terms
Factoring out the common term \( \int_{0}^{\frac{\pi}{4}} F(\cos^2x) \cos x \, dx \), we get:
\[ \left(\sqrt{2} + \alpha\right) \int_{0}^{\frac{\pi}{4}} F(\cos^2x) \cos x \, dx = 0. \]
Since \( F(\cos^2x) \) and \( \cos x \) are non-zero in the interval \( [0, \frac{\pi}{4}] \), the term inside the parentheses must be zero:
\[ \sqrt{2} + \alpha = 0. \]
Step 6: Solve for \( \alpha \)
Rearranging, we find:
\[ \alpha = -\sqrt{2}. \]
Final Answer:
\( \alpha = -\sqrt{2} \).
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