JEE Main 2025 · 24 January · evening · Single correct · +4 / −1
The function f:(-∞, ∞) →(-∞, 1), defined by f(x)=(2ˣ-2⁻ˣ)/(2ˣ+2⁻ˣ) is :
- (a) One-one but not onto
- (b) Onto but not one-one
- (c) Both one-one and onto
- (d) Neither one-one nor onto
Answer and solution
Answer (a) One-one but not onto
Solution
Maths sir · 5 steps
Multiply top and bottom by 2ˣ, write it as 1 - 2/(4ˣ+1), and both questions answer themselves.
- Rewrite the function. Multiply numerator and denominator by 2ˣ to get f(x) = (4ˣ - 1)/(4ˣ + 1). Splitting this up gives a form where x appears only once.
f(x) = 1 - 2/(4ˣ + 1)
- Check one-one. As x increases, 4ˣ increases, so 2/(4ˣ+1) decreases and f(x) increases. A strictly increasing function never takes the same value twice, so f is one-one.
- Find the range. As x → -∞, 4ˣ → 0 and f(x) → 1 - 2 = -1. As x → ∞, 4ˣ → ∞ and f(x) → 1. Neither end value is reached, so the range is (-1, 1).
- Compare with codomain. The codomain is (-∞, 1), but values such as -1 or -5 are never produced. The range is smaller than the codomain, so f is not onto.
- So. f is one-one but not onto, which is option (a).
The idea to keep. For onto, always find the actual range and compare it with the given codomain. A strictly increasing or decreasing function is automatically one-one.
Answer from the NTA final answer key. Solution by PrepWiser.