Let be a function defined by . If be the set of all points at which is not differentiable, then has exactly
- A.three elements
- B.five elements
- C.two elements
- D.one element
Show answer and solution
Correct option: A
Solution
f(x) = max{-|x|, -√(1-x²)}.
Both -|x| and -√(1-x²) are negative. We need to find where each is larger (less negative).
-|x| ≥ -√(1-x²) ⟺ √(1-x²) ≥ |x| ⟺ 1-x² ≥ x² ⟺ x² ≤ 1/2 ⟺ |x| ≤ 1/√2.
So f(x) = -|x| for |x| ≤ 1/√2, and f(x) = -√(1-x²) for |x| ≥ 1/√2.
The three points where f is not differentiable are:
- x = 0: f(x) = -|x| has a corner (non-differentiable).
- x = 1/√2 and x = -1/√2: the two pieces -|x| and -√(1-x²) meet, but their derivatives differ there (slope of -|x| is ∓1, slope of -√(1-x²) is x/√(1-x²) = ±1 at |x|=1/√2, but left and right derivatives don't match smoothly — actually the derivatives match in magnitude but this is the junction of two different functional forms, and checking carefully: derivative of -|x| at x=1/√2⁻ is -1, derivative of -√(1-x²) at x=1/√2⁺ is x/√(1-
Both -|x| and -√(1-x²) are negative. We need to find where each is larger (less negative).
-|x| ≥ -√(1-x²) ⟺ √(1-x²) ≥ |x| ⟺ 1-x² ≥ x² ⟺ x² ≤ 1/2 ⟺ |x| ≤ 1/√2.
So f(x) = -|x| for |x| ≤ 1/√2, and f(x) = -√(1-x²) for |x| ≥ 1/√2.
The three points where f is not differentiable are:
- x = 0: f(x) = -|x| has a corner (non-differentiable).
- x = 1/√2 and x = -1/√2: the two pieces -|x| and -√(1-x²) meet, but their derivatives differ there (slope of -|x| is ∓1, slope of -√(1-x²) is x/√(1-x²) = ±1 at |x|=1/√2, but left and right derivatives don't match smoothly — actually the derivatives match in magnitude but this is the junction of two different functional forms, and checking carefully: derivative of -|x| at x=1/√2⁻ is -1, derivative of -√(1-x²) at x=1/√2⁺ is x/√(1-
[JEE Main 2019 · 10 Jan · Shift II]