Let \(f(x) = 3{\tan ^4}x + 2k\) and \(g(x) =  - {\tan ^4}x + 8k{\tan ^2}x + k\), for \(0 \leqslant x \leqslant 1\), where \(0 < k < 1\). The graphs of \(f\) and \(g\) intersect at exactly one point. Find the value of \(k\).

discriminant \( = 0\) (seen anywhere)     M1

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f = g,{\text{ }}3{\tan ^4}x + 2k =  - {\tan ^4}x + 8k{\tan ^2}x + k\)

rearranging their equation (to equal zero)     (M1)

eg\(\,\,\,\,\,\)\(4{\tan ^4}x - 8k{\tan ^2}x + k = 0,{\text{ }}4{\tan ^4}x - 8k{\tan ^2}x + k\)

recognizing LHS is quadratic     (M1)

eg\(\,\,\,\,\,\)\(4{({\tan ^2}x)^2} - 8k{\tan ^2}x + k = 0,{\text{ }}4{m^2} - 8km + k\)

correct substitution into discriminant     A1

eg\(\,\,\,\,\,\)\({( - 8k)^2} - 4(4)(k)\)

correct working to find discriminant or solve discriminant \( = 0\)     (A1)

eg\(\,\,\,\,\,\)\(64{k^2} - 16k,{\text{ }}\frac{{ - ( - 16) \pm \sqrt {{{16}^2}} }}{{2 \times 64}}\)

correct simplification     (A1)

egx\(\,\,\,\,\,\)\(16k(4k - 1),{\text{ }}\frac{{32}}{{2 \times 64}}\)

\(k = \frac{1}{4}\)     A1     N2

[8 marks]

There was a minor issue with the domain of the function, but this did not affect any candidate. The question was amended for publication.

Most candidates recognized the need to set the functions equal to each other and many rearranged the equation to equal zero. Few students then recognized the quadratic form and the need to find the discriminant. Those who did use the discriminant generally completed it correctly.