A-LevelMathematicsFunctions9709/12 · Oct/Nov 2022 · Q98 Marks
Functions f and g are defined by f(x) = x + 1/x for x > 0, g(x) = ax + 1 for x ∈ R, where a is a constant.
- (a)Find an expression for gf(x).[1]
- (b)Given that gf(2) = 11, find the value of a.[2]
- (c)Given that the graph of y = f(x) has a minimum point when x = 1, explain whether or not f has an inverse.[1]
- (d)It is given instead that a = 5. Find and simplify an expression for g⁻¹f(x).[3]
- (e)Explain why the composite function fg cannot be formed.[1]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.
- B1 ISW. a(x + 1/x) + 1
- M1 Substitute x = 2 into their expression from (a) and equate to 11. This may be done in 2 stages: f(2)=2.5,g(2.5)=11. a(2 + 1/2) + 1 = 11
- A1 [a =] 4
- B1 Or other suitable explanation that may include one to many or many to one. No, [because it is] not one-one
- B1 Condone use of a instead of 5. [g⁻¹(x)] = (x-1)/5 WWW
- M1 Correct combination of their g⁻¹(x) with given f(x) Condone use of a instead of 5. [g⁻¹f(x) =] (x + 1/x - 1)/5 OE
- A1 Must not contain unresolved fractions e.g. (x + x⁻¹ - 1)/5. (x² - x + 1)/(5x) or 1/5(x + x⁻¹ - 1) or 1/5(x - 1 + 1/x) OE ISW
- B1 Accept an answer that includes an example outside the domain of f, e.g. g(-1) = - 4 but for f, x>0. The domain of f does not include the whole of the range of g. Or The range of g does not lie in the domain of f.
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