JEE Main 2021 · 20 July · morning · Single correct · +4 / −1
Let a function f : R → R be defined as f(x) = { sin x - eˣ, if, x ≤ 0; a + [ - x], if, 0 < x < 1; 2x - b, if, x ≥ 1
where [ x ] is the greatest integer less than or equal to x. If f is continuous on R, then (a + b) is equal to:
- (a) 4
- (b) 3
- (c) 2
- (d) 5
Answer and solution
Answer (b) 3
Solution
Maths sir · 4 steps
On 0 < x < 1 the value [-x] never changes, so match one constant to both neighbouring pieces.
- Middle piece. For 0 < x < 1, the number -x lies strictly between -1 and 0. So [-x] = -1, and the middle piece is the constant a - 1.
- Continuity at 0. f(0) = sin 0 - e⁰ = 0 - 1 = -1. The right-hand limit is a - 1. Setting a - 1 = -1 gives a = 0.
- Continuity at 1. As x → 1⁻, -x stays inside (-1, 0), so the left-hand limit is still a - 1 = -1. Also f(1) = 2(1) - b = 2 - b. Setting 2 - b = -1 gives b = 3.
- So. a + b = 0 + 3 = 3, which is option (b).
The idea to keep. Before taking a limit of [ · ], find which interval the inside lies in. If it stays between two integers, the greatest integer function is just a constant there.
Answer from the NTA final answer key. Solution by PrepWiser.