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  • Need explanation for: - Limit , continuity and differentiability - JEE Main-3

Let S={t \epsilon R : f (x)=| x-\pi|⋅ (e^{\left | x \right |} - 1 )\sin \left | x \right |)  is not differentiable at t}. Then the set S is equal to :

  • Option 1)

    \left \{ 0,\pi \right \}}

  • Option 2)

    \varnothing

  • Option 3)

    \left \{0 \right \}

  • Option 4)

    \left \{\pi \right \}14306

 

Answers (2)

best_answer

We have to check Differntiability of s at x = 0,\pi

f(x)= \left | x-\pi \left ( e^{\left | n \right |}-1(\sin x \right )) \right |

In he vincity of x=0 ; 

f(0^{+})= \left | 0-\pi \left ( e^{\left | n \right |}-1(\sin h\right )) \right |

f(0^{-})= \left | 0-\pi \left ( e^{\left | -h \right |}-1(\sin (-h)\right ) \right |

Thus differentiable in vthe vincity of x= \pi

\left | x-\pi \right | \sin \left | x \right |

no sharp corners at x= \pi

Thus , Differentialble

 

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

 

 


Option 1)

\left \{ 0,\pi \right \}}

Option 2)

\varnothing

Option 3)

\left \{0 \right \}

Option 4)

\left \{\pi \right \}14306

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Aadil

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