Further Mathematics Questions and Answers

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

\(2 ( 1 - \cos^{2} \theta) = 1 + \cos \theta\)

\(2 - 2\cos^{2} \theta = 1 + \cos \theta\)

\(0 = 1 - 2 + \cos \theta + 2\cos^{2} \theta\)

\(2\cos^{2} \theta + \cos \theta - 1 = 0\)

Factorizing, we have

\((\cos \theta + 1)(2\cos \theta - 1) = 0\)

Note: In the range, \(0° \leq \theta \leq 90°\), all trig functions are positive, so we consider

\(2\cos \theta = 1 \implies \cos \theta = \frac{1}{2}\)

\(\theta = 60°\).

\(^{n}C_{r} = \frac{n!}{(n-r)! r!}\)

\(^{n}C_{r - 1} = \frac{n!}{(n - (r - 1))! (r - 1)!}\)

\(^{n}C_{r} ÷ ^{n}C_{r - 1} = \frac{n!}{(n - r)! r!} ÷ \frac{n!}{(n-(r-1))!(r-1)!}\)

= \(\frac{n!}{(n-r)! r!} \times \frac{(n-(r-1)! (r-1)!}{n!}\)

= \(\frac{(n + 1 - r)! (r - 1)!}{(n - r)! r!}\)

= \(\frac{(n+1-r)(n-r)! (r-1)!}{(n-r)! r (r - 1)!}\)

= \(\frac{n + 1 - r}{r}\)

Let the sum of the 8 numbers = y.

\(\frac{y}{8} = 20 \implies y = 20 \times 8 = 160\)

\(160 - 17 = 143 (\text{the sum of the other 7 numbers})\)

\(mean = \frac{143 + 25}{8} = \frac{168}{8}\)

= 21

Let \(p(head) = p = \frac{1}{2}\) and \(p(tail) = q = \frac{1}{2}\)

\((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}\)

The probability of equal heads and tails = \(6p^{2}q^{2} = 6(\frac{1}{2}^{2})(\frac{1}{2}^{2})\)

= \(\frac{6}{16} = \frac{3}{8}\).

\(y = 4 - 9x\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{\Delta y}{\Delta x} = -9\)

\(\frac{\Delta y}{0.1} = -9 \implies \Delta y = -9 \times 0.1 = -0.9\)