JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 9)
Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = $${\pi \over 2}$$ in the first quadrant. Then,
$${A_1}:{A_2} = 1:\sqrt 2 $$ and $${A_1} + {A_2} = 1$$
$${A_1} = {A_2}$$ and $${A_1} + {A_2} = \sqrt 2 $$
$$2{A_1} = {A_2}$$ and $${A_1} + {A_2} = 1 + \sqrt 2 $$
$${A_1}:{A_2} = 1:2$$ and $${A_1} + {A_2} = 1$$
Explanation
$${A_1} + {A_2} = \int\limits_0^{\pi /2} {\cos x.\,dx = \left. {\sin x} \right|_0^{\pi /2}} = 1$$
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$${A_1} = \left. {\int\limits_0^{\pi /4} {(\cos x - \sin x)dx = (\sin x + \cos x)} } \right|_0^{\pi /4} = \sqrt 2 - 1$$
$$ \therefore $$ $${A_2} = 1 - \left( {\sqrt 2 - 1} \right) = 2 - \sqrt 2 $$
$$ \therefore $$ $${{{A_1}} \over {{A_2}}} = {{\sqrt 2 - 1} \over {\sqrt 2 \left( {\sqrt 2 - 1} \right)}} = {1 \over {\sqrt 2 }}$$
$${A_1} = \left. {\int\limits_0^{\pi /4} {(\cos x - \sin x)dx = (\sin x + \cos x)} } \right|_0^{\pi /4} = \sqrt 2 - 1$$
$$ \therefore $$ $${A_2} = 1 - \left( {\sqrt 2 - 1} \right) = 2 - \sqrt 2 $$
$$ \therefore $$ $${{{A_1}} \over {{A_2}}} = {{\sqrt 2 - 1} \over {\sqrt 2 \left( {\sqrt 2 - 1} \right)}} = {1 \over {\sqrt 2 }}$$
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