1.10: Past-paper question 6

4PM1/1/November/2024 — Question 6 · 12 marks

(a) Show that
a4+bx=a2(1+bx4)12whereaandb are positive integers.\frac{a}{\sqrt{4+bx}} = \frac{a}{2} \left( 1 + \frac{bx}{4} \right)^{-\frac{1}{2}} \mathrm{where} a \mathrm{and} b \text{ are positive integers.}
(2)
The expansion of a4+bx\displaystyle \frac{a}{\sqrt{4+bx}} in ascending powers of x\displaystyle x can be written as
P+Qx+Rx2+Sx3P + Qx + Rx^2 + Sx^3
where P\displaystyle P, Q\displaystyle Q, R\displaystyle R and S\displaystyle S are rational numbers.
(b) Show that Q=ab16\displaystyle Q = -\frac{ab}{16} and S=5ab32048\displaystyle S = -\frac{5ab^3}{2048}
and find P\displaystyle P and R\displaystyle R in terms of a\displaystyle a and b\displaystyle b, as fractions in their lowest terms.
(4)
Given that Q=1285S\displaystyle Q = \frac{128}{5}S and R=9256\displaystyle R = \frac{9}{256}
(c) show that a=3\displaystyle a = 3 and b=1\displaystyle b = 1
(3)
(d) Hence, using an appropriate value of x\displaystyle x, find, to 3 decimal places, an approximate
value for 62\displaystyle \frac{\sqrt{6}}{2}
(3)