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  • Please help! - Limit , continuity and differentiability - JEE Main-19

Let f:R\rightarrow R be differentiable at c\epsilon R and f(c) = 0.

If g(x)=\left | f(x) \right |, then at x=c , g is :

  • Option 1)

    not differentiable if f{}'(c)=0

  • Option 2)

    differentiable if f{}'(c)\neq0

  • Option 3)

    differentiable if f{}'(c)=0

  • Option 4)

    not differentiable 

 

Answers (1)

Given g(x) = | f(x) | and also given f(c)=0

g'(c^{+})=\lim_{x\rightarrow c^{+}}\frac{|f(x)|-f(c)}{x-c}

              =\lim_{x\rightarrow c^{+}}\frac{\pm f(x)}{x-c}

              =\pm f'(c)

g'(c^{-})=\lim_{x\rightarrow c^{-}}\frac{|f(x)|-f(c)}{x-c}

              =\pm f'(c)

For  g(x)  to be differentiable at C    

f'(c)=0

correct option is (3)


Option 1)

not differentiable if f{}'(c)=0

Option 2)

differentiable if f{}'(c)\neq0

Option 3)

differentiable if f{}'(c)=0

Option 4)

not differentiable 

Posted by

Vakul

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