A) \[4(\sqrt{2}-1)\]
B) \[2\sqrt{2}(\sqrt{2}-1)\]
C) \[2(\sqrt{2}+1)\]
D) \[2\sqrt{2}(\sqrt{2}+1)\]
Correct Answer: B
Solution :
[b] Since \[\sin x\] and \[\cos x>0\] for \[x\in [0,\pi /2]\], the graph of \[y=\sin x+\cos x\]always lies above the graph of \[y=\left| \cos x-\sin x \right|\] Also \[\cos x>\sin x\] for \[x\in [0,\pi /4]\] and \[\sin x>\cos x\] for \[x\in [\pi /4,\pi /2]\] \[\Rightarrow \] Area \[=\int\limits_{0}^{\pi /4}{((sinx+cosx)-(cosx-sinx))dx}\] \[+\int\limits_{\pi /4}^{\pi /2}{((sinx+cosx)-(sinx-\operatorname{cosx}))dx}\] \[=4-2\sqrt{2}\]
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