Limits, Continuity & Differentiability Question - None

Question

If f(x) = |1 − x|, then the points where sin⁻¹(f(|x|)) is non-differentiable, are

Accepted Answer

Step 1: Given f(x) = |1 − x|

Step 2: f(|x|) = |1 − |x|| = \begin{cases} 1 - x, & x \geq 1 \ 1 - x, & 0 \leq x < 1 \ 1 + x, & -1 < x < 0 \ -1 - x, & x \leq -1 \end{cases}

Step 3: For sin⁻¹(f(|x|)) to be defined: −1 ≤ f(|x|) ≤ 1, which gives |1 − |x|| ≤ 1

Step 4: This means −1 ≤ 1 − |x| ≤ 1, so −2 ≤ −|x| ≤ 0, thus |x| ≤ 2, i.e., x ∈ [−2, 2]

Step 5: sin⁻¹(f(|x|)) is non-differentiable at points where f(|x|) has corners or cusps: x = 0, 1, −1

∴ Answer is (c).

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