Further Mathematics Questions and Answers

\(f(x) = x^{2} + 1\)  and  \(g(x) = x - 2\)

\(f o g = f(g(x)) = f(x - 2) = (x - 2)^{2} + 1 \)

= \(x^{2} - 4x + 4 + 1 = x^{2} - 4x + 5\)

\(120 kmh^{-1} = \frac{120 \times 1000}{3600} = \frac{100}{3} = 33.3ms^{-1}\)

\(\frac{\mathrm d s(t)}{\mathrm d t} = v(t)\) and \(\frac{\mathrm d v(t)}{\mathrm d t} = a(t)\)

\(\therefore v(t) = \frac{\mathrm d (12t^{2} - 2t^{3})}{\mathrm d t} = 24t - 6t^{2}\)

\(\frac{\mathrm d (24t - 6t^{2})}{\mathrm d t} = 24 - 12t = a(t)\)

\(a(1) = 24 - 12(1) = 24 - 12 = 12ms^{-2}\)

Let the power of \(2x^{2}\) be t and the power of \(\frac{1}{x} \equiv x^{-1}\) = 9 - t.

\((2x^{2})^{t}(x^{-1})^{9 - t} = x^{0}\)

Dealing with x alone, we have

\((x^{2t})(x^{-9 + t}) = x^{0} \implies 2t - 9 + t = 0\)

\(3t - 9 = 0 \therefore t = 3\)

The binomial expansion is then,

\(^{9}C_{3} (2x^{2})^{3}(x^{-1})^{6} = \frac{9!}{(9-3)! 3!} \times 2^{3}\)

= 84 x 8

= 672

\(F_{1} = 7i + 0j\)

\(F_{2} = (4\cos\theta)i + (4\sin\theta)j\)

\(9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}\)

\(9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta\)

\(81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)\)

Recall, \(\cos^{2}\theta + \sin^{2}\theta = 1\)

\(81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta\)

\(16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857\)

\(\theta = \cos^{-1} 0.2857  = 73.40°\)