A) \[\left( -\frac{\pi }{4},\frac{\pi }{4} \right)\]
B) \[\left( 0,\frac{\pi }{2} \right)\]
C) \[\left( -\frac{\pi }{2},-\frac{\pi }{4} \right)\]
D) \[\left( \frac{\pi }{4},\frac{\pi }{2} \right)\]
Correct Answer: A
Solution :
Let \[y=f(x)={{\tan }^{-1}}\,(\sin \,x+\cos x)\] \[\Rightarrow \] \[y={{\tan }^{-1}}\left[ \sqrt{2}\left( \frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x \right) \right]\] \[\Rightarrow \] \[y={{\tan }^{-1}}\left[ \sqrt{2}\,\sin \,\left( x+\frac{\pi }{4} \right) \right]\] \[\Rightarrow \] \[\tan y=\sqrt{2}\,\sin \left( x+\frac{\pi }{4} \right)\] On differentiating w. r. t. x, we get \[{{\sec }^{2}}y\,\frac{dy}{dx}=\sqrt{2}\,\cos \,\left( x+\frac{\pi }{4} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{\sqrt{2}\,\cos \,\left( x+\frac{\pi }{4} \right)}{1+{{\tan }^{2}}y}=\frac{\sqrt{2}\,\cos \,\left( x+\frac{\pi }{4} \right)}{2+\sin \,2x}\] For increasing, \[\frac{dy}{dx}\ge 0\] \[\Rightarrow \] \[\cos \,\left( x+\frac{\pi }{4} \right)\ge 0\] [\[\because \] DR?s is positive] \[\therefore \] x is increasing in \[\left( -\frac{\pi }{4},\frac{\pi }{4} \right).\]
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