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 :
\[f(x)={{(\tan x+\sin x)}^{2}}\] \[\because \,\,\tan x\And \sin x\] 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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