ISI2014-DCG-29

Question

If $f(x) = \sin \bigg( \dfrac{1}{x^2+1} \bigg),$ then

• A. $f(x)$ is continuous at $x=0$, but not differentiable at $x=0$
• B. $f(x)$ is differentiable at $x=0$, and $f’(0) \neq 0$
• C. $f(x)$ is differentiable at $x=0$, and $f’(0) = 0$
• D. None of the above

Answer 1

Here, $\sin\left(\frac{1}{x^{2}+1}\right)$ is continuous and definitely differentiable.

$f(x)=\sin\left(\frac{1}{x^{2}+1}\right)\\ \Rightarrow f'(x)= \frac{-2x}{\left(x^{2}+1\right)^2}\cos\left(\frac{1}{x^{2}+1}\right)\\ \Rightarrow f'(0)=0$

 

So the correct answer is C.

 

Comments

Here's the graph of $\sin\left(\frac{1}{x^{2}+1}\right)$ below.

 

Answer 2

Few statements based on Continuity of functions:-

1) If $\lim_{x\to a }f(x) = b$ and $g(x)$ is continuous at $x=b$ then $\lim_{x\to a }g(f(x))  = g(b)$

2) If $\lim_{x\to a }f(x) = f(c)$ and $g(x)$ is continuous at $x=f(c) $ then $\lim_{x\to a }g(f(x))  = g(f(c))$ (in words, if f(x) is continuous at $x=c$

and g(x) is continuous at $x=f(c)$ then composition of $g(f(x))$ is also continuous at $x=c$.)

3) if $f(x)$ is continuous at $x=c$ then $\dfrac{1}{f}$ also continuous at $x=c$.(Provided that $f(c) \neq 0$).

now coming to question $f(x)= sin\left(\dfrac{1}{x^2+1}\right)$ ,

Let $g(x) = \dfrac{1}{x^2+1}$ , $h(x) = sin(x)$, now $f(x) $ is nothing but a compostion $h(g(x))$ ,  $x^2 +1$ is continuous everywhere because

 its polynomial and observe carefully it is always positive never equals to 0,  $\dfrac{1}{x^2+1}$ is also continuous everywhere (statement 3).

we all know that $sin(x)$ is continuous at every $x$, as $h(x) ,  g(x)$ continuous everywhere their composition $h(g(x))$ also continuous at every $x$.

so f(x) is continuous for all $x$.

$f'(x) = - 2xcos\left(\dfrac{1}{x^2+1}\right)\dfrac{1}{(x^2+1)^2}$,

now its trivial to answer the value of $f'(0)$ $= -2(0)cos(1)(1) = 0$

the answer is option C.