KVPY Sample Paper KVPY Stream-SX Model Paper-17

Let \[f:R\to R\] be a function such that \[\left| f(x) \right|\le {{x}^{2}}\], for all \[x\in R.\] Then, at \[x=0,f\]is:

A) Continuous but not differentiable.

B) Continuous as well as differentiable.

C) Neither continuous nor differentiable.

D) Differentiable but not continuous.

Correct Answer: B

Solution :

Let \[\left\| f(x) \right\|\le {{x}^{2}},\forall \,x\in R\]
Now, at \[x=0,\left\| f(0) \right\|\le 0\]
\[\Rightarrow \]\[f(0)=0\]
\[\therefore \]\[f'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h-0}\]
\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)}{h}\]	? (i)

Now, \[\left\| \frac{f(h)}{h} \right\|\le \left\| h \right\|\] \[(\because \left\| f(x) \right\|\le {{x}^{2}})\]
\[\Rightarrow -\left\| h \right\|\le \frac{f(h)}{h}\le \left\| h \right\|\]
\[\Rightarrow \]\[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)}{h}\to 0\]	? (ii)
(Using sandwich theorem)

\[\therefore \]from (1) and (2), we get \[f'(0)=0,\]

i.e.\[-f\left( x \right)\]is differentiability, at \[x=0\]

Since differentiability \[\Rightarrow \] continuity

\[\therefore \]\[\left| f(x) \right|\le {{x}^{2}},\]for all \[x\in R\] is continuous as well as differentiable at \[x=0\]