Question
Given that \({2^m} = 8\) and \({2^n} = 16\), write down the value of \(m\) and of \(n\).
Hence or otherwise solve \({8^{2x + 1}} = {16^{2x – 3}}\).
Answer/Explanation
Markscheme
\(m = 3,{\text{ }}n = 4\) A1A1 N2
[2 marks]
attempt to apply \({({2^a})^b} = {2^{ab}}\) (M1)
eg\(\;\;\;6x + 3,{\text{ }}4(2x – 3)\)
equating their powers of \(2\) (seen anywhere) M1
eg\(\;\;\;3(2x + 1) = 8x – 12\)
correct working A1
eg\(\;\;\;8x – 12 = 6x + 3,{\text{ }}2x = 15\)
\(x = \frac{{15}}{2}\;\;\;(7.5)\) A1 N2
[4 marks]
Total [6 marks]
Question
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 \(k\).
Answer/Explanation
Markscheme
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]