A) \[\sqrt{3}\]
B) 2
C) \[\sqrt{2}\]
D) \[-\sqrt{2}\]
Correct Answer: C
Solution :
\[y=\sin x-\cos x,\]for maximum or minimum value of y \[\frac{dy}{dx}=\cos x-\sin x=0\] \[\Rightarrow \] \[\cos x=\sin x=\cos \left( \frac{\pi }{2}-x \right)\] \[\Rightarrow \] \[x=\frac{\pi }{2}-x\] \[\therefore \] \[x=\frac{\pi }{4}\] Again,\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\sin x-\cos x\] \[=-\,\,(\sin \,x+\cos x)\] \[=-\,\](maximum) Maximum value of \[\sin x+\cos x=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}\] \[=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\]
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