JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
The function
f(x) = { π/4 + tan ⁻¹x,, | x | ≤ 1; 1/2( | x | - 1 ),, | x | > 1 is :
- (a) continuous on R–{–1} and differentiable on R–{–1, 1}
- (b) both continuous and differentiable on R–{1}
- (c) both continuous and differentiable on R–{–1}
- (d) continuous on R–{1} and differentiable on R–{–1, 1}
Answer and solution
Answer (d) continuous on R–{1} and differentiable on R–{–1, 1}
Solution
Maths sir · 3 steps
Only x = ± 1 can cause trouble. Match the values there first, then compare the slopes on each side.
- Where to look. Inside each piece, f is smooth. tan⁻¹ x is differentiable everywhere, and 1/2(|x|-1) only has a kink at x = 0, which lies outside its piece. So only the joins at x = -1 and x = 1 need checking.
- Slopes near the joins. For |x| < 1, f'(x) = 1/(1+x²), which tends to 1/2 at both joins. For x < -1, |x| = -x, so f(x) = -1/2(x+1) with slope -1/2.
Point Value of f Limit from the outer piece Slopes (left, right) Verdict x = -1 π/4 + tan⁻¹(-1) = 0 1/2(1 - 1) = 0 -1/2, 1/2 continuous, not differentiable (corner) x = 1 π/4 + π/4 = π/2 1/2(1 - 1) = 0 not needed jump, so neither - So. f fails continuity only at x = 1, and fails differentiability at both x = 1 and x = -1. It is continuous on ℝ - {1} and differentiable on ℝ - {-1, 1}, which is option (d).
The idea to keep. For a piecewise function, check continuity at each join before anything else. A jump kills differentiability at once. If the values match, compare the left and right slopes.
Answer from the NTA final answer key. Solution by PrepWiser.