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  • Try this! - Limit , continuity and differentiability - JEE Main-2

Let f,g:R\rightarrow R Be two functions defined by

f(x)=\left \{ \right.and g(x)=xf(x)

Statement I : f is a continuous function at  x = 0.

Statement II : g is a differentiable function at x = 0.

 

  • Option 1)

     Both statements I and II are false.
     

     

  • Option 2)

     Both statements I and II are true.

  • Option 3)

    Statement I is true, statement II is false.

     

  • Option 4)

     Statement I is false, statement II is true.

 

Answers (1)

best_answer

As we learnt in

Condition for differentiable -

A function  f(x) is said to be differentiable at  x=x_{\circ }  if   Rf'(x_{\circ })\:\:and\:\:Lf'(x_{\circ })   both exist and are equal otherwise non differentiable

-

 

 f(x)=\left\{\begin{matrix} x \sin \left ( \frac{1}{x} \right )& x\neq 0\\ 0& x=0\end{matrix}\right. 

and     g(x)=xf(x)

\lim_{x \rightarrow0^{+}/0^{-}}\:\:\:\:\:\:x sin \frac{1}{x}=0\times finite =0

So f(x) is continuous at x = 0

 g(x)=\left\{\begin{matrix} x^{2} \sin \frac{1}{x} &x\neq0 \\ 0&x=0 \end{matrix}\right.

\lim_{h \rightarrow 0^{+}}\:\:\:\frac{h^{2}\sin \frac{1}{h}-0}{h}\:\:\:\:=h \sin \frac{1}{h}=0

\lim_{h \rightarrow 0^{-}}\:\:\:\frac{-h^{2}\sin \frac{1}{h}-0}{-h}\:\:\:\:=h \sin \frac{1}{h}=0

\therefore g'(0^{+})=g'(0^{-})

So, g(x) is differentiable at x=0

 

 

 


Option 1)

 Both statements I and II are false.
 

 

This option is incorrect.

Option 2)

 Both statements I and II are true.

This option is correct.

Option 3)

Statement I is true, statement II is false.

 

This option is incorrect.

Option 4)

 Statement I is false, statement II is true.

This option is incorrect.

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