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  • Solve! - Limit , continuity and differentiability - JEE Main

Let f:R\rightarrow R be a function defined by f\left ( x \right )= min\left \{ x+1,\left | x \right |+1 \right \}

Then which of the following is true ?

  • Option 1)

    f\left ( x \right ) is differentiable everywhere

  • Option 2)

    f\left ( x \right ) is not differentiable at x= 0

  • Option 3)

    f\left ( x \right ) \geq 1\: for \: all\: x\in R

  • Option 4)

    f\left ( x \right ) is not differentiable at x= 1

 

Answers (1)

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)=\:min\begin{Bmatrix} x+1,\left | x \right |+1 \end{Bmatrix}

f(x)= x+1, x\:\epsilon R

So f(x) is differentiable every where.


Option 1)

f\left ( x \right ) is differentiable everywhere

Correct

Option 2)

f\left ( x \right ) is not differentiable at x= 0

Incorrect

Option 3)

f\left ( x \right ) \geq 1\: for \: all\: x\in R

Incorrect

Option 4)

f\left ( x \right ) is not differentiable at x= 1

Incorrect

Posted by

Sabhrant Ambastha

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