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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
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
    $\sqrt{3}$
  • ✓
    $-\sqrt{2}$
✓

Answer

Correct option: D.
$-\sqrt{2}$
d
$\text { Sol. } I=\int \limits_0^{\frac{\pi}{4}} f(\sin 2 x) \sin x d x+\int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x$

$+\alpha \int_0^{\frac{\pi}{4}} f (\cos 2 x) \cos x d x=0$

Apply king in first part and put $x -\frac{\pi}{4}= t$ in second part.

$I= \int \limits_0^{\frac{\pi}{4}} f (\cos 2 x ) \sin \left(\frac{\pi}{4}- x \right) dx +\int \limits_0^{\frac{\pi}{4}} f (\cos 2 t ) \sin \left(\frac{\pi}{4}+ t \right) dt$

$+\alpha \int \limits_0^{\frac{\pi}{4}} f (\cos 2 x ) \cos x d x=0$

$I =\int \limits_0^{\frac{\pi}{4}} f (\cos 2 x )\left[2 \sin \frac{\pi}{4} \cdot \cos x+\alpha \cos x\right] d x=0$

$I=(\alpha+\sqrt{2}) \int \limits_0^{\frac{\pi}{4}} f (\cos 2 x) \cos x d x=0$

$\therefore \alpha=-\sqrt{2}$

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