O-LevelAdditional MathematicsFunctions4037/11 · May/June 2011 · Q1113 Marks

(a) The function f is such that f(x) = 2x² – 8x + 5.

(a)(i)Show that f(x) = 2(x + a)² + b, where a and b are to be found.[2]
(a)(ii)Hence, or otherwise, write down a suitable domain for f so that f⁻¹ exists.[1]
(b)(i)Write down the range of g and of h⁻¹.[2]
(b)(ii)On the axes below, sketch the graphs of y = g(x) and y = g⁻¹(x), showing the coordinates of any points where the curves meet the coordinate axes. [Figure 11.1][4]
(b)(iii)Find the value of x for which gh(x) = 85.[4]

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Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

[1] B1 for finding a = -2.
[1] B1 for finding b = -3. (from f(x) = 2(x-2)² - 3).
[1] √B1 for x ≥ 2 or equivalent (following their 'a' value).
[1] B1 for Range of g: g(x) ≥ 4.
[1] B1 for Range of h⁻¹: h⁻¹(x) ≥ 0.
[1] B1 for correct sketch of y = g(x) (parabola, vertex at (0,4), restricted to x≥0).
[1] B1 for correct sketch of y = g⁻¹(x) (reflection of g(x) in y=x, starting at (4,0)).
[1] B1 for showing points where curves meet axes (e.g., (0,4) for g(x), (4,0) for g⁻¹(x)).
[1] M1 for correctly forming the composite function gh(x) = g(4x-25).
[1] DM1 for setting (4x-25)² + 4 = 85 and attempting to solve the quadratic for x.
[1] A1 for obtaining both solutions x = 17/2 and x = 4.
[1] B1 for discarding x = 4 (because h(x) domain is x≥0 and 4x-25 must be ≥0 for g(h(x))). Or more precisely, h(x) must be in the domain of g, which is x≥0, so h(x) = 4x-25 ≥ 0 implies x ≥ 25/4 = 6.25).

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