A) \[\frac{\pi }{2}\]
B) \[\pi \]
C) \[\frac{\pi }{4}\]
D) \[\frac{3\pi }{4}\]
E) \[\frac{\pi }{3}\]
Correct Answer: C
Solution :
Let\[I={{[\sin x+\cos x]}^{1+\sin 2x}}\] \[={{\left[ \sqrt{2}\sin \left( \frac{\pi }{4}+x \right) \right]}^{1+\sin 2x}}\] At \[x=\frac{\pi }{4},\] \[I={{\left[ \sqrt{2}\sin \left( \frac{\pi }{4}+\frac{\pi }{4} \right) \right]}^{1+\sin \frac{2\pi }{4}}}\] \[={{(\sqrt{2})}^{2}}=2\] Hence, option (c) is correct.
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