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Limits Continuity and Differentiability question

2019 · 10 Apr · Shift 1 · Q39
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  5. /2019 · 10 Apr · Shift 1 · Q39

Limits Continuity and Differentiability question

2019 · 10 Apr · Shift 1 · Q39

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
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 :
  1. A
    differentiable if f '(c) = 0
  2. B
    differentiable if f '(c) eee 0
  3. C
    not differentiable
  4. D
    not differentiable if f '(c) = 0
View written solutionFree

Correct answer: A

  1. Given

    • f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R is differentiable at ccc.
    • f(c)=0f(c)=0f(c)=0.
    • g(x)=∣f(x)∣g(x)=|f(x)|g(x)=∣f(x)∣.

    We must determine when ggg is differentiable at x=cx=cx=c.

  2. Use the definition of derivative for ggg at ccc

    Since g(c)=∣f(c)∣=0g(c)=|f(c)|=0g(c)=∣f(c)∣=0, we have

    g′(c)=lim⁡x→cg(x)−g(c)x−c=lim⁡x→c∣f(x)∣x−c.g'(c)=\lim_{x\to c}\frac{g(x)-g(c)}{x-c} =\lim_{x\to c}\frac{|f(x)|}{x-c}.g′(c)=x→clim​x−cg(x)−g(c)​=x→clim​x−c∣f(x)∣​.

  3. Use differentiability of fff at ccc

    Because fff is differentiable at ccc and f(c)=0f(c)=0f(c)=0,

    f(x)=f(c)+f′(c)(x−c)+o(x−c)=f′(c)(x−c)+o(x−c).f(x)=f(c)+f'(c)(x-c)+o(x-c)=f'(c)(x-c)+o(x-c).f(x)=f(c)+f′(c)(x−c)+o(x−c)=f′(c)(x−c)+o(x−c).

    Therefore,

    ∣f(x)∣=∣f′(c)(x−c)+o(x−c)∣.|f(x)|=\left|f'(c)(x-c)+o(x-c)\right|.∣f(x)∣=∣f′(c)(x−c)+o(x−c)∣.

  4. Case 1: f′(c)=0f'(c)=0f′(c)=0

    Then

    f(x)=o(x−c).f(x)=o(x-c).f(x)=o(x−c).

    So,

    ∣f(x)∣=o(x−c).|f(x)|=o(x-c).∣f(x)∣=o(x−c).

    Hence,

    ∣f(x)∣x−c→0 as x→c.\frac{|f(x)|}{x-c}\to 0 \, \text{as } x\to c.x−c∣f(x)∣​→0as x→c.

    Thus g′(c)g'(c)g′(c) exists and equals 000.

    So, if f′(c)=0f'(c)=0f′(c)=0, then ggg is differentiable at ccc.

  5. Case 2: f′(c)≠0f'(c)\neq 0f′(c)=0

    Near x=cx=cx=c,

    f(x)≈f′(c)(x−c).f(x)\approx f'(c)(x-c).f(x)≈f′(c)(x−c).

    Hence,

    ∣f(x)∣≈∣f′(c)∣ ∣x−c∣.|f(x)|\approx |f'(c)|\,|x-c|.∣f(x)∣≈∣f′(c)∣∣x−c∣.

    Then

    ∣f(x)∣x−c≈∣f′(c)∣∣x−c∣x−c.\frac{|f(x)|}{x-c}\approx |f'(c)|\frac{|x-c|}{x-c}.x−c∣f(x)∣​≈∣f′(c)∣x−c∣x−c∣​.

    Now,

    ∣x−c∣x−c={1,x>c,−1,x<c.\frac{|x-c|}{x-c}= \begin{cases} 1, & x>c,\\ -1, & x<c. \end{cases}x−c∣x−c∣​={1,−1,​x>c,x<c.​

    Therefore,

    • Right-hand derivative =∣f′(c)∣=|f'(c)|=∣f′(c)∣
    • Left-hand derivative =−∣f′(c)∣=-|f'(c)|=−∣f′(c)∣

    These are unequal, so derivative does not exist.

    Thus, if f′(c)≠0f'(c)\neq 0f′(c)=0, then ggg is not differentiable at ccc.

  6. Conclusion

    g(x)=∣f(x)∣g(x)=|f(x)|g(x)=∣f(x)∣ is differentiable at x=cx=cx=c iff f′(c)=0f'(c)=0f′(c)=0.

    Therefore, the correct option is:

    A\boxed{\text{A}}A​

  7. Comparison with stored answer

    Stored correct answer: A

    Our derived answer is also A, so they agree.

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