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My attempt:

I have to choose one from option A and option D. Option B can be eliminated by taking $m=1, n=2$. option C can also be eliminated by taking $m=4, n=3$. Please help me to choose one from A and D.

Anne Bauval
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ketan
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2 Answers2

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I'll assume $n\ge1$, otherwise the problem is trivial.

Use the definition of derivative: the derivative exists if and only if $$ f'(0)=\lim_{x\to 0}x^{m-1}\sin\frac{1}{x^n} $$ exists and is finite.

If $m>1$, this limit exists and is zero, because, for $x\ne0$, $$ -|x^{m-1}|\le x^{m-1}\sin\frac{1}{x^n}\le |x^{m-1}| $$ and the squeeze theorem applies, so $f'(0)=0$. If $m\le1$ the limit does not exist.

No hypothesis whatsoever is needed on $n$. So, with this interpretation, A and D are true.

The case would be much different if the question is “does the function have continuous derivative at $0$?” But “differentiable at $0$” usually means “the function has a derivative at $0$”.

If continuous differentiability is required, then we know that $m>1$ and $f'(0)=0$, in this case. Moreover, for $x\ne0$, $$ f'(x)=mx^{m-1}\sin\frac{1}{x^n}-nx^{m-n-1}\cos\frac{1}{x^n} $$ and this is continuous everywhere, except possibly at $0$.

The limit of this function at $0$ should be $0$. Since $m>1$, we already know that $$ \lim_{x\to0}x^{m-1}\sin\frac{1}{x^n}=0 $$ so we also need that $$ \lim_{x\to0}x^{m-n-1}\cos\frac{1}{x^n}=0 $$ Note that if $m-n-1\le0$, the limit doesn't exist. Instead, if $m-n-1>0$, the limit is $0$ with the same argument as before.

So the condition for continuous differentiability at $0$ is $m>n+1$.

For example, if $m=2$ and $n=1$, the derivative is $$ f'(x)=\begin{cases} 2x\sin\dfrac{1}{x}-\cos\dfrac{1}{x} & \text{if $x\ne0$}\\ 0 & \text{if $x=0$} \end{cases} $$ and this function is not continuous at $0$.

So, if we consider this as the question, we have to choose non differentiability, so either C or D.

egreg
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  • is A correct or D? – ketan Mar 08 '15 at 04:38
  • @ketan If $m>1$ then the function is differentiable at $0$. Suppose $m>n>0$: then $m>1$ and the function is differentiable at $0$; on the other hand, if $n=0$ the function is differentiable. Hence A is correct. But also D is correct, because for $m=1$ and $n>1$ the function is not differentiable at $0$. – egreg Mar 08 '15 at 09:39
  • So (to make sure and improve my english): D must be interpreted as "not [diff for each m<n]" and not as "[not diff] for each m<n"? – Anne Bauval Jan 05 '24 at 17:04
  • @egreg the condition for $m,n$ is already given that they belongs to natural numbers so further assumption for them to be greater than $0$ is not needed. So that makes option d also true, but official answer key to that question says only a option is correct check question 28 answer key – Gajjze Jan 05 '24 at 19:07
  • @DeepakKumar (context=my previous comment) Interpreting rather D as "[not diff] for each m<n" makes it false, which agrees with your answer key. – Anne Bauval Jan 05 '24 at 19:28
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    Yes, that makes sense. Thankyou ✨ – Gajjze Jan 05 '24 at 19:32
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$f'(x)=mx^{m-1}sin(1/x^n)-nx^{m-n-1}cos(1/x^n)$ (for $x \neq 0$). The sine and cosine function are continuous; therefore differentiability arises if there are factors $x^a$ with $a \geq 0$. Otherwise the derivative of $f(x)$ would have a jump at $x=0$.

kryomaxim
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