Verified Solution • 2025–26 Session moderate

QJEEM19Madfaa: JEE Main 2019 Mathematics - Class 12 JEEM19Madfaa Solution: Let f : R R be differentiable at c R and f(c) = 0. If g(x) =…

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 :

Multiple Choice Options

  • A differentiable if f '(c) = 0
  • B differentiable if f '(c) ≠\ne 0
  • C not differentiable
  • D not differentiable if f '(c) = 0

Step-by-Step Verified Solution

g′(c)=lim⁡x→cg(x)−g(c)x−cg'(c) = \mathop {\lim }\limits_{x \to c} {{g(x) - g(c)} \over {x - c}}

⇒g′(c)=lim⁡x→c∣f(x)∣−∣f(c)∣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

⇒g′(c)=lim⁡x→c∣f(x)∣x−c\Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right|} \over {x - c}}

⇒g′(c)=lim⁡x→cf(x)x−c\Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{f(x)} \over {x - c}} if f(x) > 0

and g′(c)=lim⁡x→c−f(x)x−cg'(c) = \mathop {\lim }\limits_{x \to c} {{ - f(x)} \over {x - c}} if f(x) < 0

⇒g′(c)=f′(c)=−f′(c)\Rightarrow g'(c) = f'(c) = - f'(c)

⇒\Rightarrow 2f'(c) = 0

⇒\Rightarrow f'(c) = 0

Related Learning Resources & Exam Preparation