Question

For a whole number $n$, if $f\left(x\right)=x^{n-1}\sin \left(1/x\right)$ for $x\ne 0$ and $f\left(0\right)=0$

then in order that $f$ is differentiable at all $x$, the value of $n$ can be?

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is C $3$

$f\left(x\right)=x^{n-1}\sin \left(\frac{1}{x}\right)$

$f^{\prime }\left(x\right)=x^{n-1}\cos \left(\frac{1}{x}\right)(-\frac{1}{x^2})+(n-1)x^{n-2}\sin \left(\frac{1}{x}\right)$

$=-x^{n-3}\cos \left(\frac{1}{x}\right)+(n-1)x^{n-2}\sin \left(\frac{1}{x}\right)$

For it to be differentiable everywhere $f^{\prime }\left(0\right)$ must be finite.

$\Rightarrow x^{n-3}=0$ at $x=0$

$\Rightarrow n-3\ge 0$

$\Rightarrow n\ge 3$