JEE Main 2025 · 28 January · evening · Single correct · +4 / −1
Let f:[0,3] → A be defined by f(x)=2 x³-15 x²+36 x+7 and g:[0, ∞) → B be defined by g(x)=x²⁰²⁵/(x²⁰²⁵+1), If both the functions are onto and S={ x ∈ Z ; x ∈ A or x ∈ B }, then n(S) is equal to :
- (a) 29
- (b) 31
- (c) 30
- (d) 36
Answer and solution
Answer (c) 30
Solution
Maths sir · 4 steps
Onto means A and B are exactly the ranges, so find each range and count the integers in them.
- Critical points of f. f'(x) = 6x² - 30x + 36 = 6(x-2)(x-3). So f increases on [0,2] and decreases on [2,3].
- Range of f. f(0) = 7, f(2) = 16 - 60 + 72 + 7 = 35 and f(3) = 54 - 135 + 108 + 7 = 34. So A = [7, 35], which holds the 29 integers from 7 to 35.
g(x) = x²⁰²⁵/(x²⁰²⁵+1) = 1 - 1/(x²⁰²⁵+1)
- Range of g. At x = 0, g = 0. As x grows, g increases towards 1 but never reaches it. So B = [0, 1), and its only integer is 0.
- So. A and B do not overlap, so n(S) = 29 + 1 = 30, option (c).
The idea to keep. For a cubic on a closed interval, check the values at the endpoints and at the critical points inside it. For a function like t/(t+1), watch which end is reached and which is only approached.
Answer from the NTA final answer key. Solution by PrepWiser.