where g(x) is an even function differentiable at x = 0, passing through the origin. Then f' (0)
A) is equal to 1
B) is equal to 0
C) is equal to 2
D) does not exist
Correct Answer: C
Solution :
[c] :\[f'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{g(0+h)cos(1/h)-0}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{g(h)cos(1/h)}{h}=\underset{h\to 0}{\mathop{\lim }}\,g'(0)cos(1/h)=0\] \[[\because g'(-x)=-g'(x)\Rightarrow g'(0)=0]\]
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