Question

Let $f\left(x\right)=x^k,k\in R$.The value of k so that f is differentiable $(n-1)$ times at $x=0$, but not differentiable n-times at $x=0$ is

• A. n
• B. $n-1$
• C. $\frac{3n-2}{3}$
• D. None of these

Answer 1

The correct option is C $\frac{3n-2}{3}$

$f^k\left(x\right)=k(k-1)(k-2)...(k-(m-1))x^{k-m}$.
Thus if $k$ is positive integer then $f$ is differentiable $n-$times for every $n$ and for all $x.$
Hence if f is differentiable $(n-1)$ times at $x=0$ bu8t not
differentiable $n-$times
We must have $k\in R\sim I.$
The only such value is $\frac{(3n-2)}{3}$