A) \[\frac{\pi +1}{4}\]
B) \[\frac{1}{4}\]
C) \[\frac{\pi +2}{4}\]
D) \[\frac{\pi -1}{4}\]
Correct Answer: A
Solution :
[a] \[f'(x)={{\tan }^{-1}}\left( \frac{1+\sin x}{\cos x} \right)={{\tan }^{-1}}\left( \tan \left( \frac{\pi }{4}+\frac{x}{2} \right) \right)\] \[\because \,\,\,\,\,\,\,\,\,\,\frac{\pi }{4}+\frac{x}{2}\in \,\,\left( -\frac{\pi }{2},\frac{\pi }{2} \right)\] in the neighborhood of \[x=1\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,f'(x)=\frac{\pi }{4}+\frac{x}{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,f(x)=\frac{\pi }{4}x+\frac{{{x}^{2}}}{4}+c\] \[\because \,\,\,\,\,\,\,\,\,\,\,f(0)=0\] \[\Rightarrow \,\,\,\,c=0\] So \[f(1)=\frac{\pi }{4}+\frac{1}{4}=\frac{\pi +1}{4}\]
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