Q53 / 90 · KVPY 2012 (SB/SX) · Mathematics

The value Figure

  • A
    $\frac{\sqrt{2}+1}{\sqrt{2}-1}$
  • B
    $\frac{\sqrt{2}-1}{\sqrt{2}+1}$
  • C
    $\frac{\sqrt{2}+1}{\sqrt{2}}$
  • Correct answer
    D
    $2-\sqrt{2}$

Solution

$\begin{array}{l} I_{1}=\int_{0}^{\pi / 2}(\sin x)^{\sqrt{2}} \cdot \sin x d x \quad I_{2}=\int_{0}^{\pi / 2}(\sin x)^{\sqrt{2}-1} d x \\ I_{1}=\left((\sin x)^{\sqrt{2}} \int \sin x d x\right)_{0}^{\pi / 2}-\int_{0}^{\pi / 2}\left(\sqrt{2}(\sin x)^{\sqrt{2}-1} \cos x \int \sin x d x\right) \\ =-\left(\cos x(\sin x)^{\sqrt{2}}\right)_{0}^{\pi / 2}+\sqrt{2} \int_{0}^{\pi / 2}(\sin x)^{\sqrt{2}-1}\left(1-\sin ^{2} x\right) d x \\ \frac{I_{1}}{I_{2}}=\frac{\sqrt{2}}{1+\sqrt{2}} \times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}=2-\sqrt{2} \end{array}$

Hence, the correct answer is option: 'D'.

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