(a)(i) Write down the binomial expansion of \((2 - \frac{1}{2}x)^{5}\) in ascending powers of x.
(ii) Using the expansion in (a)(i), find, correct to two decimal places, the value of \((1.99)^{5}\).
(b) The polynomial \(x^{3} + qx^{2} + rx + 9\), where q and r are constants, has (x + 1) as a factor and has a remainder -17 when divided by (x + 2). Find the values of q and r.
(a)(i) \((2 - \frac{1}{2}x)^{5} = ^{5}C_{5}(2)^{5}(-\frac{1}{2}x)^{0} + ^{5}C_{4}(2)^{4}(-\frac{1}{2}x)^{1} + ^{5}C_{3}(2)^{3}(-\frac{1}{2}x)^{2} + ^{5}C_{2}(2)^{2}(-\frac{1}{2}x)^{3} + ^{5}C_{1} (2)^{1}(-\frac{1}{2}x)^{4} + ^{5}C_{0} (2)^{0}(-\frac{1}{2}x)^{5}\)
= \(32 - 5(8x) + 10(2x^{2}) - 10(\frac{x^{3}}{2}) + 5(\frac{x^{4}}{8}) - \frac{x^{5}}{32}\)
= \(32 - 40x + 20x^{2} - 5x^{3} + \frac{5}{8}x^{4} - \frac{1}{32}x^{5}\)
(ii) \((1.99)^{5} = (2 - \frac{1}{2}x)^{5}\)
\(2 - \frac{1}{2}x = 1.99 \implies \frac{1}{2}x = 2 - 1.99 = 0.01\)
\(x = 0.01 \times 2 = 0.02\)
Put x = 0.02 in the equation, we have
\(32 - 40(0.02) + 20(0.02)^{2} - 5(0.02)^{3} + \frac{5}{8}(0.02)^{4} - \frac{1}{32}(0.02)^{5}\)
\(32 - 0.8 + 0.008 - ...\)
= \(31.208 \approxeq 31.21\)
(b) \(f(x) = x^{3} + qx^{2} + rx + 9\)
(x + 1) is a factor, hence, f(-1) = 0.
\(f(-1) = (-1)^{3} + q(-1)^{2} + r(-1) + 9\)
\(0 = -1 + q - r + 9\)
\(-8 = q - r ... (1)\)
(x + 2) has remainder -17, therefore f(-2) = -17.
\(-17 = (-2)^{3} + q(-2)^{2} + r(-2) + 9\)
\(-17 = -8 + 4q - 2r + 9\)
\(-18 = 4q - 2r \implies -9 = 2q - r ... (2)\)
From (1), r = q + 8.
\(\therefore 2q - q - 8 = -9\)
\(q = -1\)
\(r = -1 + 8 = 7\)
\((q, r) = (-1, 7)\)
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