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Consider the curve $C_{1} : y = \cos x$ and the curve $C_{2} : y = \sin x$ then

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Area enclosed $C_{1}$ and $C_{2}$ in successive points of intersections is $2 \sqrt{2}$

Ratio of the area enclosed by $C_{1}$ in $[0, \frac{π}{2}]$ divided by the curve $C_{2}$ is $1 : \sqrt{2}$

Area bounded by one arc of $C_{1}$ is equal to one arc of $C_{2}$

Area of one curvilinear triangle formed by $C_{1}$ and $C_{2}$ is $2 - \sqrt{2}$

answer is A, B, C, D.

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Detailed Solution

$a) R . A = \int_{0}^{\frac{π}{4}} \sin xdx + \int_{\frac{π}{4}}^{\frac{π}{2}} \cos xdx = 2 - \sqrt{2}$

$b) R \cdot A = \int_{\frac{π}{4}}^{\frac{5 π}{4}} (\sin x - \cos x) dx$

$c) R_{2} = \int_{0}^{\frac{π}{4}} \sin x + \int_{\frac{π}{4}}^{\frac{π}{2}} \cos xdx = 2 - \sqrt{2}$ ;

$R_{1} = \int_{0}^{\frac{π}{4}} (\cos x - \sin x) dx = \sqrt{2} - 1$ ;

Required to $= R_{1}; R_{2}$

$d) A_{1} : A_{2} = 2 R_{1} : R_{2} = 1 : \sqrt{2}$

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