Numerical answer · +4 / −1
If f and g are both one-one functions, then fog is x (one-one/onto/bijective).
Answer and solution
Answer 1
Solution
Maths sir · 5 steps
Peel fog from the outside: f one-one strips f, then g one-one strips g, and x1 = x2 drops out.
- Reading the code. The choices are numbered in the order given: 1 for one-one, 2 for onto, 3 for bijective. So the task is to decide which property f ° g is guaranteed to have.
- Start from equal outputs. Suppose (f ° g)(x₁) = (f ° g)(x₂), that is, f(g(x₁)) = f(g(x₂)). To prove one-one, you must show this forces x₁ = x₂.
f(g(x₁)) = f(g(x₂)) ⇒ g(x₁) = g(x₂) ⇒ x₁ = x₂
- Why each arrow holds. The first arrow uses that f is one-one, so equal outputs of f come from equal inputs. The second arrow uses that g is one-one in the same way. So f ° g is one-one.
- Not onto in general. Take g(x) = eˣ and f(x) = x on ℝ. Both are one-one, but (f ° g)(x) = eˣ never takes negative values or zero, so it is not onto and therefore not bijective.
- So. The guaranteed property is one-one, which is choice 1.
The common slip: 3. Choosing bijective gives 3, but both functions being one-one says nothing about covering the whole codomain, so onto is not guaranteed.
The idea to keep. Composition carries a property through only when both functions have it: one-one with one-one gives one-one, onto with onto gives onto. To check one-one, start from equal outputs and peel off the outer function first.
Answer from the NTA final answer key. Solution by PrepWiser.