JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 17)
Let f : R $$ \to $$ R be differentiable at c $$ \in $$ R and f(c) = 0. If g(x) = |f(x)| , then at x = c, g is :
differentiable if f '(c) = 0
differentiable if f '(c) $$ \ne $$ 0
not differentiable
not differentiable if f '(c) = 0
Explanation
$$g'(c) = \mathop {\lim }\limits_{x \to c} {{g(x) - g(c)} \over {x - c}}$$
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right| - \left| {f(c)} \right|} \over {x - c}}$$
$$ \therefore $$ f(c) = 0
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right|} \over {x - c}}$$
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{f(x)} \over {x - c}}$$ if f(x) > 0
and $$g'(c) = \mathop {\lim }\limits_{x \to c} {{ - f(x)} \over {x - c}}$$ if f(x) < 0
$$ \Rightarrow g'(c) = f'(c) = - f'(c)$$
$$ \Rightarrow $$ 2f'(c) = 0
$$ \Rightarrow $$ f'(c) = 0
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right| - \left| {f(c)} \right|} \over {x - c}}$$
$$ \therefore $$ f(c) = 0
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right|} \over {x - c}}$$
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{f(x)} \over {x - c}}$$ if f(x) > 0
and $$g'(c) = \mathop {\lim }\limits_{x \to c} {{ - f(x)} \over {x - c}}$$ if f(x) < 0
$$ \Rightarrow g'(c) = f'(c) = - f'(c)$$
$$ \Rightarrow $$ 2f'(c) = 0
$$ \Rightarrow $$ f'(c) = 0
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