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JEE Main 2019 · 9 Apr · Shift I

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

Let f(x)=15−∣x−10∣f(x) = 15 - |x - 10|; x∈Rx \in \mathbf{R}. Then, the set of all values of xx, at which the function, g(x)=f(f(x))g(x) = f(f(x)) is not differentiable, is
  1. A.
    {5,10,15,20}\{5, 10, 15, 20\}
  2. B.
    {5,10,15}\{5, 10, 15\}
  3. C.
    {10}\{10\}
  4. D.
    {10,15}\{10, 15\}
Show answer and solution

Correct option: B

Solution

f(x) = 15 - |x - 10|, so g(x) = f(f(x)) = 15 - |f(x) - 10| = 15 - |5 - |x - 10||.

g(x) is not differentiable where the inner expression changes its non-differentiable character. The absolute value |x - 10| creates a corner at x = 10. Then |5 - |x - 10|| creates corners where 5 - |x - 10| = 0, i.e., |x - 10| = 5, giving x = 5 and x = 15.

So the points of non-differentiability are x = 5, 10, 15 — the set {5, 10, 15}.

[JEE Main 2019 · 9 Apr · Shift I]

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