question_answer

                The area under the curve\[y=\left| \cos x-\sin x \right|,\]\[0\le x\le \frac{\pi }{2},\]and above \[x-\operatorname{axis}\] is:     JEE Main Online Paper ( Held On 23  April 2013 )

A)                  \[2\sqrt{2}\]                     

B)                                         \[2\sqrt{2}-2\]

C)                                         \[2\sqrt{2}+2\]

D)                                          0

Correct Answer: B

Solution :

                 \[y=|\cos x-\sin x|\] Required area\[=2\int\limits_{0}^{\pi /4}{(\cos x-\sin x)dx}\] \[=2[\sin x+\cos x]_{0}^{\pi /4}\] \[=2\left[ \frac{2}{\sqrt{2}}-1 \right]=(2\sqrt{2}-2)sq.\]units