A) A minimum but no maximum
B) A maximum but no minimum
C) Neither maximum nor minimum
D) Both maximum and minimum
Correct Answer: C
Solution :
\[f(x)=x+\sin x\] Þ \[{f}'(x)=1+\cos x\] Now \[{f}'(x)=0\Rightarrow 1+\cos x=0\Rightarrow \cos x=-1\Rightarrow x=\pi \] Now \[{f}''(x)=-\sin x\],\[{f}''(\pi )=0\],\[f'''(x)=-\cos x\], \[{f}'''(\pi )=1\ne 0\] \ Neither maximum nor minimum.You need to login to perform this action.
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