JEE Mains · Maths · STD 12 - 7.2 definite integral
If \(f : R \rightarrow R\) be a continuous function satisfying \(\int \limits_0^{\pi / 2} f(\sin 2 x) \cdot \sin x d x+\alpha \int \limits_0^{\pi / 4} f(\cos 2 x) \cdot \cos x d x=0\)then \(\alpha\) is equal to
- A \(-\sqrt{3}\)
- B \(\sqrt{2}\)
- C \(\sqrt{3}\)
- D \(-\sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(-\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\text { Sol. } I=\int \limits_0^{\frac{\pi}{4}} f(\sin 2 x) \sin x d x+\int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x\) \(+\alpha \int_0^{\frac{\pi}{4}} f (\cos 2 x) \cos x d x=0\) Apply king in first part and put \(x -\frac{\pi}{4}= t\) in second part.…
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