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Let f(x) = \sin |x|+ |x|   \forall x\epsilon R  then at x= 0 , f(x) is 

  • Option 1)

    continous and diffrentiable both

  • 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 continuity 

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

\therefore f(x) is continous at x= 0

For diffrentiability 

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

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

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 }=1+1=2

LHD\neqRHD 

so non diffrentiable at x= 0

 

 

 

 

 

 


Option 1)

continous and diffrentiable both

Option 2)

continous but non diffrentiable 

Option 3)

neither continous nor diffrentiable 

Option 4)

discontinous but diffrentiable 

Posted by

Himanshu

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