Let \(f(x) = k{x^2} + kx\) and \(g(x) = x - 0.8\). The graphs of \(f\) and \(g\) intersect at two distinct points.

Find the possible values of \(k\).

attempt to set up equation     (M1)

eg\(\;\;\;f = g,{\text{ }}k{x^2} + kx = x - 0.8\)

rearranging their equation to equal zero     M1

eg\(\;\;\;k{x^2} + kx - x + 0.8 = 0,{\text{ }}k{x^2} + x(k - 1) + 0.8 = 0\)

evidence of discriminant (if seen explicitly, not just in quadratic formula)     (M1)

eg\(\;\;\;{b^2} - 4ac,{\text{ }}\Delta  = {(k - 1)^2} - 4k \times 0.8,{\text{ }}D = 0\)

correct discriminant     (A1)

eg\(\;\;\;{(k - 1)^2} - 4k \times 0.8,{\text{ }}{k^2} - 5.2k + 1\)

evidence of correct discriminant greater than zero     R1

eg\(\;\;\;{k^2} - 5.2k + 1 > 0,{\text{ }}{(k - 1)^2} - 4k \times 0.8 > 0\), correct answer

both correct values     (A1)

eg\(\;\;\;0.2,{\text{ }}5\)

correct answer     A2     N3

eg\(\;\;\;k < 0.2,{\text{ }}k \ne 0,{\text{ }}k > 5\)

[8 marks]

Many candidates knew to set the equations equal, and then some knew to manipulate the equation such that it is equal to zero. Those who recognized the discriminant in this equation earned further marks, although few set a correct discriminant greater than zero. Even in such cases, finding both inequalities proved elusive for most.

An alternative method was to graph each function and find where the line intersects the parabola in exactly one and in two places. Few could carry this approach to adequate completion, often neglecting a second inequality for \(k\).