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JEE Mains · Maths · STD 12 - 7.2 definite integral

જો \(f : R \rightarrow R\) એ \(\int \limits_0^{\pi / 2} f(\sin 2 x) \cdot \sin x d x+\alpha \int \limits_0^{\pi / 4} f(\cos 2 x) \cdot \cos x d x=0\)નું સમાધાન કરતુ સતત વિધેય હોય,તો \(\alpha\)નું મૂલ્ય \(............\) છે.

  1. A \(-\sqrt{3}\)
  2. B \(\sqrt{2}\)
  3. C \(\sqrt{3}\)
  4. D \(-\sqrt{2}\)
Verified Solution

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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