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.