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  • Please help! - Let and Then, in the interval, g is: - Limit , continuity and differentiability - JEE Main

Let f(x)=\left\{\begin{matrix} -1,\: \: \: -2\leq x< 0\\ x^{2}-1, \: \: \: 0\leq x\leq 2 \end{matrix}\right.   and  g(x)=\left | f(x) \right |+f(\left | x \right |).

Then, in the interval (-2,2), g is:

  • Option 1)

     

    not differentiable at two points

  • Option 2)

     

    differentiable at all points

  • Option 3)

     

    not differentiable at one point

  • Option 4)

     

    not continuous

Answers (1)

best_answer

 

Condition for differentiability -

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

-

 

 

Properties of differentiable functions -

At every corner point  f(x) is continuous but not differentiable.

ex:    | x - a |  is continuous but not differentiable at  x = a  for  a > 0 

- wherein

 

 

y=f(x)

only one nondifferential point at x=1

 

 


Option 1)

 

not differentiable at two points

Option 2)

 

differentiable at all points

Option 3)

 

not differentiable at one point

Option 4)

 

not continuous

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