Question

Let $f(x) = {x^k},k \in \mathbb{R}$ . 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) $\dfrac{{3n - 2}}{3}$
(D) None of these

Answer 1

Hint : We need to first find the trend of differentiation of the given equation to an arbitrary number of times, say, $m$ . The differentiation of the given function is to be done by applying chain rule. By analysing the trend of the differentiated function we can determine the value of $k$ for which the function can be differentiated up to $(n - 1)$ times at $x = 0$ .

Complete step-by-step answer :
Given function is, $f(x) = {x^k}$ .
Now, differentiating the function $m$ -times, we get,
${f^m}(x) = k(k - 1)(k - 2)(k - 3).....(k - (m - 1)){x^{k - m}}$
Thus, we can say that, 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 $x = 0$ but not differentiable $n$ -times.
We must have $k \in \mathbb{R} - {\rm I}$ .
Now, by analysing the data we can see that,

Analysing option A,
$n$ is a positive integer.
But we must have valueof k as a real number excluding integers.
So, it doesn’t satisfy the condition for $k$ .
So, option A is incorrect.

Analysing option B,
Similarly, $(n - 1)$ is a positive integer.
But we must have valueof k as a real number excluding integers.
So, it doesn’t satisfy the condition for $k$ .
So, option B is incorrect.

Analysing option C,
$\dfrac{{3n - 2}}{3}$ is a positive real number but not an integer.
As, $\dfrac{{3n - 2}}{3} = n - \dfrac{2}{3}$ .
Also, we needed the value of k to be a real number excluding the integer set.
So, it satisfies the condition for $k$ .
So, option C is correct.
Therefore, the correct option is C.
So, the correct answer is “Option C”.

Note : Differentiable functions means that a function that can be approximated locally by a linear function. Now, there is a very basic relation between differentiability and continuity of function. That is, if $f:(a,b) \to \mathbb{R}$ is differentiable at $c \in (a,b)$ , then $f$ is continuous at $c$ . The conclusion that can be drawn from this is that, a continuous may or may not be differentiable, but a differentiable function will always be continuous.