A) \[2(\sqrt{2}-1)\,sq\]units
B) \[(\sqrt{3}+1)\,sq\] units
C) \[2(\sqrt{3}-1)\,sq\]units
D) None of these
Correct Answer: A
Solution :
We know that, \[\sin \,x\ge \,\cos \,x\]for \[x\in \,\left[ \frac{x}{4},\,\frac{\pi }{2} \right]\] and \[\sin \,x\le \cos \,x\] for \[x\in \,\left[ 0,\,\frac{\pi }{4} \right]\]. Therefore, the required area \[=\,\int_{0}^{\pi /4}{(\cos \,x-\sin \,x)\,dx}\] \[+\,\int_{\pi /4}^{\pi /2}{(\sin \,x-\cos \,x)dx}\] \[=\,[\sin \,x+\cos \,x]_{0}^{\pi /4}+[-\cos \,x-\sin \,x]_{\pi /4}^{\pi /2}\] \[=(\sqrt{2}-1)+\sqrt{2}-1\] \[=2\,(\sqrt{2}-1)\,sq\] unitsYou need to login to perform this action.
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