Question

Let $f\left(x\right)=\{\begin{array}{c}g\left(x\right).\cos \frac{1}{x}\text{if}x\ne 0\\ 0\text{if}x=0\end{array}$

where $g\left(x\right)$ is an even function differentiable at $x=0$, passing through the origin. Then $f^{\prime }\left(0\right)$

• A. is equal to $1$
• B. is equal to $0$
• C. is equal to $2$
• D. does not exist

Answer 1

Answer: (B)

is equal to $0$

$f(-x)=g(-x).\cos (-\frac{1}{x})$ if $x\ne 0$

$=g\left(x\right)\cos \left(\frac{1}{x}\right)$ ...... (As $g\left(x\right)$ and $\cos x$ are even functions)

$=f\left(x\right)$

So, $f\left(x\right)$ is an even function.

Clearly then if it passes through origin i.e at $x=0$, then the origin must be a point of minima

Hence $f^{\prime }\left(0\right)=0$