IBDP Maths analysis and approaches Topic: AHL 5.12 Informal ideas of limit, continuity and convergence HL Paper 1

Solution:-

(a)  n=1:

$LHS=f^{(1)}(x)=-\frac{1}{2}x-a(1-ax)^{\frac{3}{2}}(-\frac{a}{2}(1-ax)^{\frac{3}{2}})$
$RHS=\frac{a(1)(1-ax)^{\frac{1}{2}}}{2^{1}(0)!}$ therefore true for $n=1$

assume that the result is true for $n=k$

$f^{(k)}(x)=\frac{a^{k}(2k-1)!(1-ax)^{\frac{2k+1}{2}}}{2^{2k-1}(k-1)!}$
attempt to differentiate the right-hand side with respect to x:

$f^{(k+1)}(x)=\frac{d}{dx}(f^{(k)}(x))$
$=-\frac{(2k+1)\times-a}{2}\times\frac{a^{k}(2k-1)!(1-ax)^{\frac{2k+1}{2}-1}}{2^{2k-1}(k-1)!}$ (or equivalent)

attempt to multiply top and bottom by 2k

$=\frac{(2k+1)}{2}\times\frac{a^{k+1}(2k-1)!(1-ax)^{\frac{2k+1}{2}-1}}{2^{2k-1}(k-1)!}\times\frac{2k}{2k}$
$=\frac{a^{k+1}(2k+1)!(1-ax)^{\frac{2k+1}{2}}}{2^{2k}(k)!}$

hence if the result is true for $n=k$ then it is true for $n=k+1$ and as it is true for $n=1$
is true for all $n \in \mathbb{Z}^+$

(b)

$f(x)=f(0)+xf'(0)+\frac{x^{2}}{2}f”(0)+….$

$f'(x)=\frac{a^{2}(3)(1-ax)^{-\frac{5}{2}}}{2^{2}(1!)}$ OR $\frac{3}{4}a^{2}(1-ax)^{-\frac{5}{2}}$ OR $f'(0)=\frac{a^{2}(3)}{2^{2}}$

$f(0)=1, f'(0)=\frac{a}{2}, f”(0)=\frac{6a^{2}}{8}$

$f(x)=1+\frac{1}{2}ax+\frac{3}{8}a^{2}x^{2}+….$

(c) attempt to use a=2 or a=4 in the expansion

$(1-2x)^{\frac{1}{2}}=1+\frac{2x}{2}+\frac{3\times4x^2}{8}(=1+x+\frac{3}{2}x^2+…)$
$(1-4x)^{\frac{1}{3}}=1+\frac{4x}{2}+\frac{3\times16x^2}{8}(=1+2x+6x^2+…)$

attempt to multiply their two expansions together

$(1+x+\frac{3}{2}x^2+…)(1+2x+6x^2+…)=1+2x+6x^2+x+2x^2+\frac{3}{2}x^2+…$
$=1+3x+\frac{19}{2}x^2+…$ OR $\frac{2+4x+12x^2+2x+4x^2+3x^2+…}{2}$

$(1-2x)^{\frac{1}{2}}(1-4x)^{\frac{1}{3}}\approx\frac{2+6x+19x^2}{2}$

(d) $|x|<\frac{1}{4}$

(e)
$(\frac{2+6x+19x^{2}}{2})=\frac{2+0.6+0.19}{2}(=\frac{279}{200})$ (or equivalent)

attempt to substitute $x=\frac{1}{10}$ into $(1-2x)^{-\frac{1}{2}}(1-4x)^{-\frac{1}{2}}$

$((1-2x)^{-\frac{1}{2}}(1-4x)^{-\frac{1}{2}}=\frac{10}{\sqrt{48}}$ (or equivalent)

$\frac{10}{4\sqrt{3}}=\frac{279}{200}$ (or equivalent in terms of $\sqrt{3}$)

$\frac{1}{\sqrt{3}}=\frac{279}{500}$

$\sqrt{3}=\frac{500}{279}$