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Let $A_{1}$ Be the area of the region bounded by the curves $y = \sin x, y = \cos x$ And y-axis in the first quadrant.Also, Let $A_{2}$ Be the area of the region bounded by the curves $y = \sin x, y = \cos x$ , x-axis and $x = \frac{π}{2}$ In the first quadrant. Then

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$2 A_{1} = A_{2} a n d A_{1} + A_{2} = 1 + \sqrt{2}$

$A_{1} : A_{2} = 1 : \sqrt{2} a n d A_{1} + A_{2} = 1$

$A_{1} = A_{2} a n d A_{1} + A_{2} = \sqrt{2}$

$A_{1} : A_{2} = 1 : 2 a n d A_{1} + A_{2} = 1$

answer is B.

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Detailed Solution

$A_{1} = \int_{0}^{\frac{π}{4}} (\cos x - \sin x) d x = \sin x {+ \cos x]}_{0}^{\frac{π}{4}} = \sqrt{2} - 1$

$A_{2} = \int_{0}^{\frac{π}{4}} \sin x d x + \int_{\frac{π}{4}}^{\frac{π}{2}} \cos x d x = 2 - \sqrt{2}_{1} : A_{2} = 1 : \sqrt{2}, A_{1} + A_{2} = 1.$

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