JEE Main 2022 · 26 June · evening · Single correct · +4 / −1
Let f : R → R be defined as f (x) = x - 1 and g : R - 1, -1 → R be defined as g(x) = x²/(x² - 1).
Then the function fog 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 (d) neither one-one nor onto
Solution
Maths sir · 4 steps
Simplify fog to 1/(x²-1) first. It is even and it never touches 0, so both properties fail.
- Build fog. Apply g first, then subtract 1. Over a common denominator the x² terms cancel.
(f° g)(x) = x²/(x²-1) - 1 = 1/(x²-1), x ≠ ± 1
- Test one-one. The formula depends only on x², so x and -x give the same output. For example, (f° g)(2) = (f° g)(-2) = 1/3. So fog is not one-one.
- Find the range. As x varies, x² - 1 takes every value in [-1, 0) ∪ (0, ∞). Its reciprocal then takes every value in (-∞, -1] ∪ (0, ∞). Values in (-1, 0], such as 0 or -1/2, are never reached, so fog is not onto ℝ.
- So. fog is neither one-one nor onto, which is option (d).
The idea to keep. Always simplify the composite before you test it. An even function on a symmetric domain is never one-one. A reciprocal 1/h(x) can never equal 0, so it is never onto ℝ.
Answer from the NTA final answer key. Solution by PrepWiser.