A-LevelMathematicsFunctions9709/12 · May/June 2012 · Q1012 Marks

Functions f and g are defined by f: x → 2x + 5 for x ∈ R, g: x → 8/(x-3) for x ∈ R, x ≠ 3.

(i)Obtain expressions, in terms of x, for f⁻¹(x) and g⁻¹(x), stating the value of x for which g⁻¹(x) is not defined.[4]
(ii)Sketch the graphs of y = f(x) and y = f⁻¹(x) on the same diagram, making clear the relationship between the two graphs.[3]
(iii)Given that the equation fg(x) = 5 – kx, where k is a constant, has no solutions, find the set of possible values of k.[5]

Mark scheme

Marking points from the official mark scheme (May/June 2012). The mark or mark type is shown in brackets.

[B1] f⁻¹(x) = (x - 5)/2; co.
[M1 A1 B1] For g⁻¹(x): let y = 8/(x-3), rearrange to x = 8/y + 3. So g⁻¹(x) = (8 + 3x)/x. State x ≠ 0; Attempt at x the subject. co. Allow if a linear denominator.
[B1] Sketch of y = f(x) (a straight line with positive gradient 2, y-intercept 5); +ve gradient, +ve y intercept.
[B1] Sketch of y = f⁻¹(x) (a straight line with positive gradient 1/2, y-intercept -5/2); +ve gradient, +ve y intercept.
[B1] Graphs clearly show reflection symmetry about the line y = x (or stated); States, or shows the line y = x as a line of symmetry.
[B1] fg(x) = f(g(x)) = f(8/(x-3)) = 2(8/(x-3)) + 5 = 16/(x-3) + 5; co.
[M1] Forms equation: 16/(x-3) + 5 = 5 - kx; Must lead to a quadratic.
[A1] Rearranges to kx² - 3kx + 16 = 0; Correct quadratic.
[M1 A1] Uses discriminant b² - 4ac < 0: (-3k)² - 4(k)(16) < 0 → 9k² - 64k < 0; Use of b² - 4ac even if <0, >0. co Condone <.
[A1] Solves quadratic inequality for k: 0 < k < 64/9; co.

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