Question

For a whole number n if $f\left(x\right)=x^{n-1}sin\left(\frac{1}{x}\right),x\ne 0andf\left(0\right)=0$ then in order that f is differentiable at all x, the value of n can be:

• A. 1
• B. 2
• C. 3

Answer 1

The correct option is C 3

$f^{\prime }\left(0\right)=\underset{x\to 0}{lim}\frac{h^{n-1}sin\left(\frac{1}{h}\right)-0}{h}=\underset{x\to 0}{lim}{h}^{n-2}sin\left(\frac{1}{h}\right)$
$\Rightarrow$ f' (0)exists and equal to 0 for n-2>0.
$\therefore n=3$.