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  • Tell me? - Let S be the set of all points inat which the function,is not differentiable. Then S is a subset - Limit , continuity and differentiability - JEE Main

Let S be the set of all points in \left ( -\pi ,\pi \right ) at which the function, f(x)=min\left \{ \sin x,\cos x \right \} is not differentiable. Then S is a subset of which of the following ? 

 

  • Option 1)

    \left \{ -\frac{3\pi }{4},-\frac{\pi }{2},\frac{\pi }{2},\frac{3\pi }{4} \right \}

     

     

     

  • Option 2)

    \left \{ -\frac{3\pi }{4},-\frac{\pi }{4},\frac{3\pi }{4},\frac{\pi }{4} \right \}

  • Option 3)

    \left \{ -\frac{\pi }{4},0,\frac{\pi }{4} \right \}

  • Option 4)

    \left \{ -\frac{\pi }{2},-\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}\right \}

Answers (1)

best_answer

 

Differentiability -

Let  f(x) be a real valued function defined on an open interval (a, b) and  x\epsilon (a, b).Then  the function  f(x) is said to be differentiable at   x_{\circ }   if

\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}


or\:\:\:\lim_{x\rightarrow x_{0}}\:\frac{f(x)-f(x_{0})}{x-x_{0}}

-

Hence number of points f (x) is non-differentiable are 2 which are \frac{-3\pi}{4}  and  \frac{\pi}{4}.

 


Option 1)

\left \{ -\frac{3\pi }{4},-\frac{\pi }{2},\frac{\pi }{2},\frac{3\pi }{4} \right \}

 

 

 

Option 2)

\left \{ -\frac{3\pi }{4},-\frac{\pi }{4},\frac{3\pi }{4},\frac{\pi }{4} \right \}

Option 3)

\left \{ -\frac{\pi }{4},0,\frac{\pi }{4} \right \}

Option 4)

\left \{ -\frac{\pi }{2},-\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}\right \}

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