JEE Main 2023 · 29 January · morning · Single correct · +4 / −1
Let f:R → R be a function such that f(x) = (x² + 2x + 1)/(x² + 1). Then
- (a) f(x) is many-one in ( - ∞ , - 1)
- (b) f(x) is one-one in ( - ∞ ,∞ )
- (c) f(x) is one-one in [1,∞ ) but not in ( - ∞ ,∞ )
- (d) f(x) is many-one in (1,∞ )
Answer and solution
Answer (c) f(x) is one-one in [1,∞ ) but not in ( - ∞ ,∞ )
Solution
Maths sir · 4 steps
Split off the 1, find where f' changes sign, and the one-one pieces show up.
- Rewrite f. The numerator is x² + 1 + 2x, so f(x) = 1 + 2x/(x² + 1). Now only the fraction decides the behaviour.
f'(x) = (2(x²+1) - 2x · 2x)/(x²+1)² = 2(1 - x²)/(x²+1)²
- Read the sign. The denominator is always positive, so the sign of f' is the sign of 1 - x². f is increasing on (-1, 1) and strictly decreasing on (-∞, -1) and on (1, ∞). A strictly monotonic function is one-one, so f is one-one on [1, ∞) and on (-∞, -1).
- Not one-one on R. Since 2x/(x²+1) is unchanged when x becomes 1/x, two inputs give the same value: f(2) = 1 + 4/5 = 9/5 and f(1/2) = 1 + 1/(5/4) = 9/5.
- So. f is one-one on [1, ∞) but not on (-∞, ∞), which is option (c).
The idea to keep. To test one-one on an interval, check the sign of f'. If it never changes sign there, the function is one-one. A sign change anywhere in the domain means it is many-one over the whole domain.
Answer from the NTA final answer key. Solution by PrepWiser.