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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 of one curvilinear triangle formed by $C_{1}$ and $C_{2}$ 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 enclosed by $C_{1}$ and $C_{2}$ in successive points of intersection is $2 \sqrt{2}$

Area of one curvilinear triangle formed by $C_{1}$ and $C_{2}$ is $5 - \sqrt{2}$

answer is A, B, D.

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Detailed Solution

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

B) R.A= $\int_{π / 4}^{5 π / 4} (\sin x - \cos x) d x$

$R_{1} = \int_{0}^{\frac{π}{4}} (\cos x - \sin x) dx = \sqrt{2} - 1 a n d R_{2} = \int_{0}^{\frac{π}{4}} \sin x + \int_{\frac{π}{4}}^{\frac{π}{2}} \cos xdx = 2 - \sqrt{2}$

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

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