(a) Simplify : \((\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1})\)
(b) A point P is 40km from Q on a bearing 061°. Calculate, correct to one decimal place, the distance of P to (i) north of Q ; (ii) east of Q.
(c) A man left N5,720 to be shared among his son and three daughters. Each daughter's share was \(\frac{3}{4}\) of the son's share. How much did the son receive?
(a) \((\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1})\)
\(\frac{x^{2}}{2} - x + \frac{1}{2} = \frac{x^{2} - 2x + 1}{2}\)
= \(\frac{(x - 1)^{2}}{2}\)
\(\therefore (\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1}) = (\frac{(x - 1)^{2}}{2})(\frac{1}{x - 1})\)
= \(\frac{x - 1}{2}\)
(b) 
(i) TQ = PR (North of Q)
\(\implies \sin 29 = \frac{PR}{40}\)
\(PR = 40 \sin 29 = 19.39 km\)
(ii) QR = East of Q
\(\frac{QR}{40} = \cos 29\)
\(QR = 40 \cos 29\)
= 34.98km
\(\approxeq\) 35km.
(c) Let the son's share = x.
Each daughter's share = \(\frac{3}{4}x\)
For the three daughters = \(3 \times \frac{3}{4} = \frac{9}{4}\)
\(x + \frac{9}{4}x = 5720 \implies \frac{13}{4}x = 5720\)
\(x = \frac{5720 \times 4}{13} = N1760\)
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