(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?
Explanation
(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)