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  • How to solve this problem- - Limit , continuity and differentiability - JEE Main-10

Let f(x) = \sin |x| -|x| ,   then at x= 0  , f(x) is 

  • Option 1)

    Both continous and diffrentiable 

  • Option 2)

     continous but non diffrentiable 

  • Option 3)

    neither continous nor diffrentiable 

  • Option 4)

    discontinous but  diffrentiable 

 

Answers (1)

best_answer

As we have learned 

 

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

-

 

 for continuty \rightarrow

\lim_{x\rightarrow 0^{-}} \sin |x|-|x| = \lim_{x\rightarrow 0^{+}}\sin |x| -|x| = f(0)=0

f(x) is continous at x=0 

 For diffrentiability 

LHD= \lim_{h\rightarrow 0^{+}}\frac{f(0-h)-f(0)}{-h} =\lim_{h\rightarrow 0^{+}}\frac{\sin |h|-|h|-0}{-h }

=-\lim_{h\rightarrow 0^{+}}\frac{\sin h-h-0}{h }= -\lim_{h\rightarrow 0^{+}}\left ( \frac{\sin h}{h}-\frac{h}{h} \right )= -(1-1)=0

 

RHD= \lim_{h\rightarrow 0^{+}}\frac{f(0-h)-f(0)}{h} =\lim_{h\rightarrow 0^{+}}\frac{\sin |h|-|h|-0}{h }

 

=-\lim_{h\rightarrow 0^{+}}\frac{\sin h-h-0}{h }= \lim_{h\rightarrow 0^{+}}\left ( \frac{\sin h}{h}-\frac{h}{h} \right )= (1-1)=0

LHD=RHD so f(x) is diffrentiable at x =0

 

 


Option 1)

Both continous and diffrentiable 

Option 2)

 continous but non diffrentiable 

Option 3)

neither continous nor diffrentiable 

Option 4)

discontinous but  diffrentiable 

Posted by

Himanshu

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