A) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\]does not exist
B) \[f(x)\]is not continuous at\[x=0\]
C) \[f(x)\]is not differentiable at\[x=0\]
D) \[f'(0)=0\]
Correct Answer: D
Solution :
\[f(x)={{(\tan x+\sin x)}^{2}}\] \[\because \]tan x & sin x are both continuous & differentiable at x = 0 \[\Rightarrow \]\[f(x)\]is cont. & diff at x = 0 \[f'(x)=2(\tan x+\sin x)[{{\sec }^{2}}x+\cos x],\]\[f'(0)=0\]
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