The function f is defined as follows: f(x) = √x−1 for x > 1. The diagram shows the graph of y = g(x) where g(x) = 1/(x²+2) for x ∈ R. The function h is defined by h(x) = 1/(x²+2) for x ≥ 0. [Figure 4.1]
- (a)Find an expression for f¯¹(x).[1]
- (b)State the range of g and explain whether g¯¹ exists.[2]
- (c)Solve the equation hf(x) = f (25/16). Give your answer in the form a+b√c, where a, b and c are integers.[4]
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Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [B1] [f⁻¹(x) =](x+1)² ISW Condone 'y = '.
- [B1] 0 < g(x) ≤ 1/2 or g(x) > 0 and g(x) ≤ 1/2 or (0, 1/2] Do not allow g(x)>0,g(x)<1/2. Condone g or y in place of g(x).
- [B1] g⁻¹ does not exist because it is one to many or g⁻¹ does not exist because it is not one to one. Or g⁻¹ does not exist because g is not one to one or g⁻¹ does not exist because g is many to one or g⁻¹ does not exist because g fails the horizontal line test.
- [B1] f(25/16) = 1/4 SOI
- [M1] Equating 1/((√x-1)²+2) = 1/4, or their 'simplified' version, to their f(25/16)
- [A1] [(√x-1)²+2 = 4 leading to] √x-1 = √2 leading to x = (1±√2)² Or [x-2√x+1+2 = 4 leading to] x- 2√x-1 = 0 leading to x = (1+√2)² Or [x-1=2√x leading to] x²-6x+1=0 leading to x = (6±√36-4) / 2. Allow just + in the results because - can be disregarded at this stage.
- [A1] 3+2√2. Must discount the solution 3 –2√2.
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