JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 8)
The area, enclosed by the curves $$y = \sin x + \cos x$$ and $$y = \left| {\cos x - \sin x} \right|$$ and the lines $$x = 0,x = {\pi \over 2}$$, is :
$$2\sqrt 2 (\sqrt 2 - 1)$$
$$2(\sqrt 2 + 1)$$
$$4(\sqrt 2 - 1)$$
$$2\sqrt 2 (\sqrt 2 + 1)$$
Explanation
$$A = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - \left| {\cos x - \sin x} \right|} \right)\,dx} $$
$$A = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\cos x - \sin x)} \right)\,dx} + \int_{{\pi \over 4}}^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\sin x - \cos x)} \right)\,dx} $$
$$A = 2\int_0^{{\pi \over 2}} {\sin x\,dx + 2\int_{{\pi \over 4}}^{{\pi \over 2}} {\cos x\,dx} } $$
$$A = - 2\left( {{1 \over {\sqrt 2 }} - 1} \right) + \left( {1 - {1 \over {\sqrt 2 }}} \right)$$
$$A = 4 - 2\sqrt 2 = 2\sqrt 2 (\sqrt 2 - 1)$$
Option (a)
$$A = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\cos x - \sin x)} \right)\,dx} + \int_{{\pi \over 4}}^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\sin x - \cos x)} \right)\,dx} $$
$$A = 2\int_0^{{\pi \over 2}} {\sin x\,dx + 2\int_{{\pi \over 4}}^{{\pi \over 2}} {\cos x\,dx} } $$
$$A = - 2\left( {{1 \over {\sqrt 2 }} - 1} \right) + \left( {1 - {1 \over {\sqrt 2 }}} \right)$$
$$A = 4 - 2\sqrt 2 = 2\sqrt 2 (\sqrt 2 - 1)$$
Option (a)
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