Hint
The function is strictly increasing on its branch. At a common point (u,v), both f(u)=v and f(v)=u must hold.
Worked solution
(a)
Since the original x−1 is nonnegative, x−1=√(y−2). Exchanging variables gives 1+√(x−2), defined for x≥2.
Answer: \(f^{-1}(x)=1+\sqrt{x-2}\), \(x\ge2\).
(b)
The original branch starts at its vertex (1,2), so the inverse starts at (2,1). The point (4,11) reflects to (11,4).
Answer: Endpoint \((2,1)\); reflected point \((11,4)\).
(c)
If (u,v) is common, f(u)=v and f(v)=u. Strict increase rules out u<v and u>v, so u=v. A common point would therefore satisfy (u−1)²+2=u, or u²−3u+3=0. Its discriminant is 9−12=−3, so it has no real root and the graphs do not meet.
Answer: No common point.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Use the positive square-root branch.
- 1 mark: Obtain 1+√(x−2).
- 1 mark: State domain x≥2.
Alternative methods: A valid rearrangement with the branch condition is accepted.
Common error: Using the negative branch or the original domain as the inverse domain.
(b) · 2 marks
- 1 mark: Give endpoint (2,1).
- 1 mark: Give reflected point (11,4).
Alternative methods: Coordinate swapping with a clearly labelled endpoint is accepted.
Common error: Leaving the endpoint coordinates unchanged.
(c) · 3 marks
- 1 mark: Justify why a common point must have equal coordinates for this increasing branch.
- 1 mark: Obtain the fixed-point equation u²−3u+3=0 and its negative discriminant.
- 1 mark: Conclude that there is no common point.
Alternative methods: Showing f(x)>x for every permitted x also works: if f(u)=v then v>u, while f(v)=u would imply u>v, a contradiction.
Common error: Using the fixed-point test for every inverse pair without justification.
Total: 8 marks.