JEE Main 2022 · 28 June · morning · Single correct · +4 / −1
Let f : R → R be defined as
f(x) = [ [eˣ],, x < 0; aeˣ + [x - 1],, 0 ≤ x < 1; b + [sin (π x)],, 1 ≤ x < 2; [e⁻ˣ] - c,, x ≥ 2
where a, b, c ∈ R and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true?
- (a) There exists a, b, c ∈ R such that f is continuous on R.
- (b) If f is discontinuous at exactly one point, then a + b + c = 1
- (c) If f is discontinuous at exactly one point, then a + b + c ≠ 1
- (d) f is discontinuous at at least two points, for any values of a, b and c
Answer and solution
Answer (c) If f is discontinuous at exactly one point, then a + b + c ≠ 1
Solution
Maths sir · 3 steps
Remove every bracket first. Then x=1 breaks on its own, whatever a, b and c are.
Interval Value inside the bracket f(x) x<0 0<eˣ<1, so [eˣ]=0 0 0≤ x<1 -1≤ x-1<0, so [x-1]=-1 aeˣ-1 x=1 sin π=0, so the bracket is 0 b 1<x<2 -1≤sin(π x)<0, so the bracket is -1 b-1 x≥ 2 0<e⁻ˣ≤ e⁻², so [e⁻ˣ]=0 -c - Always broken at 1. f(1)=b, but the right-hand limit is b-1. These never match, so f is always discontinuous at x=1. Option (a) is therefore false.
- Fixing 0 and 2. At x=0 the left limit is 0 and f(0)=a-1, so continuity needs a=1. At x=2 the left limit is b-1 and f(2)=-c, so continuity needs b+c=1.
- So. Discontinuity at exactly one point is possible, so (d) is false. It happens only when a=1 and b+c=1, which gives a+b+c=2≠ 1. The answer is (c).
The idea to keep. With greatest integer pieces, work out the bracket's value on each interval first. Then check the endpoints where the inside lands exactly on an integer, like sin π x at x=1.
Answer from the NTA final answer key. Solution by PrepWiser.