Find the value of k .

The line \(y = p\) intersects the graph of f . Find all possible values of p .

valid approach     (M1)

e.g. \({b^2} - 4ac\) , \(\Delta = 0\) , \({( - 4k)^2} - 4(2k)(1)\)

correct equation     A1

e.g. \({( - 4k)^2} - 4(2k)(1) = 0\) , \(16{k^2} = 8k\) , \(2{k^2} - k = 0\)

correct manipulation     A1

e.g. \(8k(2k - 1)\) , \(\frac{{8 \pm \sqrt {64} }}{{32}}\)

\(k = \frac{1}{2}\)     A2     N3

[5 marks]

recognizing vertex is on the x-axis     M1

e.g. (1, 0) , sketch of parabola opening upward from the x-axis

\(p \ge 0\)     A1     N1

[2 marks]

Those who knew to set the discriminant to zero had little trouble completing part (a). Some knew that having two equal roots means the factors must be the same, and thus surmised that \(k = \frac{1}{2}\) will achieve \((x - 1)(x - 1)\) . This is a valid approach, provided the reasoning is completely communicated. Many candidates set \(f = 0\) and used the quadratic formula, which misses the approach entirely.

Part (b) proved challenging for most, and was often left blank. Those who considered a graphical interpretation and sketched the parabola found greater success.