IB Mathematics AHL 5.19 Maclaurin series AA HL Paper 3
Question
If two functions $f(x)$ and $g(x)$ are differentiable, then their product is differentiable and the two functions satisfy the product rule: $(f(x)g(x))’ = f'(x)g(x) + g'(x)f'(x)$.
In this question, you will meet examples of pairs of differentiable functions, $f(x)$ and $g(x)$, that also satisfy $(f(x)g(x))’ = f'(x)g'(x)$.
In part (a), consider
$$f(x) = \frac{1}{(2-x)^2}$$
where $x \in \mathbb{R}$, $x \neq 2$, and $g(x) = x^2$ where $x \in \mathbb{R}$.
(a) (i) Find an expression for $f'(x)$.
(ii) Show that
$$f'(x)g'(x) = \frac{4x}{(2-x)^3}$$
(iii) Show that
$$f(x)g'(x) + g(x)f'(x) = \frac{4x}{(2-x)^3}$$
In parts (b) and (c), consider two non-constant functions, $f(x)$ and $g(x)$, where $f(x) > 0$ and $g(x) \neq g'(x)$.
(b) By rearranging the equation $f(x)g'(x) + g(x)f'(x) = f'(x)g'(x)$, show that
$$\frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x)-g(x)}$$
(c) Hence, by integrating both sides of $\frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x)-g(x)}$, show that
$$f(x) = Ae^{\int \frac{g'(x)}{g'(x)-g(x)}dx}$$
where A is an arbitrary positive constant.
The result from part (c) can be used to find pairs of functions, $f(x)$ and $g(x)$, which satisfy both of the following:
$$(f(x)g(x))’ = f(x)g'(x) + g(x)f'(x) \text{ and } (f(x)g(x))’ = f'(x)g'(x).$$
In parts (d) and (e), use the result in part (c) with A = 1.
(d) Consider $g(x) = xe^x$.
Find $f(x)$ such that $f(x)$ and $g(x)$ satisfy the above two equations.
(e) Consider $g(x) = \sin x + \cos x$.
Find $f(x)$ such that $f(x)$ and $g(x)$ satisfy the above two equations over the domain $0 < x < \pi$.
Give your answer in the form $f(x) = \sqrt{e^{h(x)}}$, where $h(x)$ is a function to be determined.
▶️Answer/Explanation
Detailed Solution
Part (a) (i) Given \(f(x) = \frac{1}{(2 – x)^2}\) where \(x \neq 2\),
OR \[ f(x) = (2 – x)^{-2} \]
Using the chain rule, let \(u = 2 – x\), so \(f(x) = u^{-2}\) and \(\frac{du}{dx} = -1\). The derivative is:
\[ f'(x) = \frac{d}{dx} [u^{-2}] = -2 u^{-3} \cdot \frac{du}{dx} = -2 (2 – x)^{-3} \cdot (-1) = 2 (2 – x)^{-3} \]
Alternatively, express it as:
\[ f'(x) = \frac{2}{(2 – x)^3} \]……………..(eqn 1)
(ii) Given \(g(x) = x^2\), finding its derivative:
\[ g'(x) = 2x \]……………..(eqn 2)
Now, taking the product \(f'(x)g'(x)\) from equation 1 and 2:
\[ f'(x)g'(x) = \frac{2}{(2 – x)^3} \cdot 2x = \frac{4x}{(2 – x)^3} \]
\[ f'(x)g'(x) = \frac{4x}{(2 – x)^3} \]
(iii) To show \(f(x)g'(x) + g(x)f'(x) = \frac{4x}{(2 – x)^3}\)
Using the product rule, compute \((f(x)g(x))’\):
\[ (f(x)g(x))’ = f(x)g'(x) + g(x)f'(x) \]
Substitute the known functions and derivatives:
\(f(x) = \frac{1}{(2 – x)^2}\)
\(g(x) = x^2\)
\(f'(x) = \frac{2}{(2 – x)^3}\)
\(g'(x) = 2x\)
First term:
\[ f(x)g'(x) = \frac{1}{(2 – x)^2} \cdot 2x = \frac{2x}{(2 – x)^2} \]
Second term:
\[ g(x)f'(x) = x^2 \cdot \frac{2}{(2 – x)^3} = \frac{2x^2}{(2 – x)^3} \]
Add them:
\[ f(x)g'(x) + g(x)f'(x) = \frac{2x}{(2 – x)^2} + \frac{2x^2}{(2 – x)^3} \]
To combine, use a common denominator \((2 – x)^3\):
Rewrite the first term: \(\frac{2x}{(2 – x)^2} = \frac{2x (2 – x)}{(2 – x)^3} = \frac{4x – 2x^2}{(2 – x)^3}\)
Second term: \(\frac{2x^2}{(2 – x)^3}\)
\[ \frac{4x – 2x^2}{(2 – x)^3} + \frac{2x^2}{(2 – x)^3} = \frac{4x – 2x^2 + 2x^2}{(2 – x)^3} = \frac{4x}{(2 – x)^3} \] Hence proved
Part (b) To rearrange:
\[ f(x)g'(x) + g(x)f'(x) = f'(x)g'(x) \]
Move all terms to one side:
\[ f(x)g'(x) + g(x)f'(x) – f'(x)g'(x) = 0 \]
Factorize:
\[ f(x)g'(x) + f'(x)(g(x) – g'(x)) = 0 \]
\[ f'(x)(g(x) – g'(x)) = -f(x)g'(x) \]
Since \(g(x) \neq g'(x)\), divide both sides by \(f(x)(g'(x) – g(x))\) (noting \(f(x) > 0\)):
\[ \frac{f'(x)}{f(x)} = \frac{-f(x)g'(x)}{f(x)(g'(x) – g(x))} = \frac{-g'(x)}{g'(x) – g(x)} = \frac{g'(x)}{g(x) – g'(x)} \]
Thus:
\[ \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x) – g(x)} \]
Part (c) Integrate both sides of:
\[ \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x) – g(x)} \]
Left side:
\[ \int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C_1 \]
Since \(f(x) > 0\), this is:
\[ \ln f(x) + C_1 \]
Right side:
This simplifies directly to:
\[ \int \frac{g'(x)}{g'(x) – g(x)} \, dx \]
Equate and solve:
\[ \ln f(x) + C_1 = \int \frac{g'(x)}{g'(x) – g(x)} \, dx + C_2 \]
\[ \ln f(x) = \int \frac{g'(x)}{g'(x) – g(x)} \, dx + C_2 – C_1 \]
\[ f(x) = e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx + C} = e^C \cdot e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \]
Since \(f(x) > 0\), let \(A = e^C > 0\):
\[ f(x) = A e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \] …………………….(eqn 3)
Part (d) Given \(g(x) = x e^x\) and \(A = 1\), find \(f(x)\):
\[ g'(x) = e^x + x e^x = e^x (1 + x) \]
\[ g'(x) – g(x) = e^x (1 + x) – x e^x = e^x (1 + x – x) = e^x \]
\[ \frac{g'(x)}{g'(x) – g(x)} = \frac{e^x (1 + x)}{e^x} = 1 + x \]
\[ \int (1 + x) \, dx = x + \frac{x^2}{2} + C \]
From equation 3, \[ f(x) = 1 \cdot e^{x + \frac{x^2}{2} + C} = e^C e^{x + \frac{x^2}{2}} \]
Let \(e^C = k > 0\), but since \(A = 1\), adjust \(C = 0\):
\[ f(x) = e^{x + \frac{x^2}{2}} \]
Part (e)
Given \(g(x) = \sin x + \cos x\) over \(0 < x < \pi\), and \(f(x) = \sqrt{e^{h(x)}}\):
\[ g'(x) = \cos x – \sin x \]
\[ g'(x) – g(x) = (\cos x – \sin x) – (\sin x + \cos x) = 2 \cos x – 2 \sin x = 2 (\cos x – \sin x) \]
\[ \frac{g'(x)}{g'(x) – g(x)} = \frac{\cos x – \sin x}{2 (\cos x – \sin x)} = \frac{1}{2} \quad (\text{for } \cos x \neq \sin x) \]
At \(x = \frac{\pi}{4}\), \(\cos x = \sin x\), so adjust domain consideration. Generally:
\[ \int \frac{1}{2} \, dx = \frac{x}{2} + C \]
From equation 3 \[ f(x) = e^{\frac{x}{2} + C} = k e^{\frac{x}{2}}, \quad k = e^C \]
\[ f(x) = \sqrt{e^{\frac{x}{2} + C}} = e^{\frac{1}{2} (\frac{x}{2} + C)} \]
With \(A = 1\), \(C = 0\):
\[ f(x) = \sqrt{e^{\frac{x}{2}}} \]
So, \(h(x) = \frac{x}{2}\).
————Markscheme—————–
Solution: –
(a) (i) attempts chain rule differentiation to find f(x)
$$f'(x)=\frac{2}{(2-x)^3}=(-1)(-2)(2-x)^{-3}$$
(ii)
$$g'(x)=2x$$
$$f'(x)g'(x)=(2(2-x)^{-3})(2x)\left [ =\frac{2(2x)}{(2-x)^3}\right ] (or equivalent)$$
$$=\frac{4x}{(2-x)^3} $$
(iii)
substitution f(x),g(x) and their g'(x),f'(x) into the given expression
$EITHER$
$$f(x)g'(x)+g(x)f'(x)=2x(2-x)^{-2}+2x^2(2-x)^{-3}$$
attempts to factorise their expression
$$=2x(2-x)^{-3}((2-x)+x)$$
OR
$$f(x)g'(x)+g(x)f'(x)=\frac{2x}{\left ( 2-x \right )^{2}}+\frac{2x^{2}}{\left ( 2-x \right )^{2}}$$
attempts to form an expression with a common denominator
$$=\frac{2x(2-x)}{(2-x)^3}+\frac{2x^2}{(2-x)^3} = \left(\frac{4x-2x^2+2x^2}{(2-x)^3}\right)$$
THEN
$$=\frac{4x}{(2-x)^3}$$
(b) METHOD 1
$$f'(x)g'(x)-g(x)f'(x)=f(x)g'(x)$$
$$f'(x)g'(x)-g(x)f'(x)-f(x)g'(x)=0$$
$$f'(x)(g'(x)-g(x))=f(x)g'(x)$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)}{g'(x)-g(x)}$$
METHOD 2
$$g'(x)=\frac{f'(x)g'(x)}{f(x)}-\frac{g(x)f(x)}{f(x)}$$
$$g'(x)=\frac{f'(x)}{f(x)}(g(x)-g(x))$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)}{g'(x)-g(x)}$$
METHOD 3
$$g'(x)=\frac{f(x)g'(x)}{f'(x)}+g(x)$$
$$\frac{f(x)}{f'(x)}=\frac{g'(x)-g(x)}{g'(x)}$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)}{g'(x)-g(x)}$$
METHOD 4
$$\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}=1$$
$$\frac{f'(x)g(x)+g'(x)f(x)}{f(x)g(x)}=1$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)g(x)-g(x)}{g'(x)}$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)}{g'(x)-g(x)}$$
(c) METHOD 1
OR
$$\ln f(x)+C=\int \frac{g'(x)}{g'(x)-g(x)} dx$$
$$f(x)=e\left ( ^{\int \frac{g'(x)}{g'(x)-g(x)} dx}e^-C \right )\left [f(x)e^{c}=e\left ( ^{\int \frac{g'(x)}{g'(x)-g(x)} dx} \right ),f(x)=e^{\left ( \int \frac{g'(x)}{g'(x)-g(x)}dx-c \right )} \right ]$$
THEN
$$f(x)=Ae^{\int \frac{g'(x)}{g'(x)-g(x)}dx}$$
METHOD 2
$$f'(x)-\frac{g'(x)}{g'(x)-g(x)}f(x)=0$$
Integrating factor:$$ e^{-\int \frac{g'(x)}{g'(x)-g(x)}dx}$$
$$\frac{d}{dx}\left [f(x)e^{-\int \frac{g'(x)}{g'(x)-g(x)}dx}\right ]=0$$
$$f(x)e^\left ( {-\int \frac{g'(x)}{g'(x)-g(x)}dx} \right )=A$$
$$f(x)=Ae^\left ({\int \frac{g'(x)}{g'(x)-g(x)}dx}\right )$$
(d) $g'(x) = xe^x + e^x \text{ (seen anywhere)}$
$\text{attempts to find an expression for } \frac{g'(x)}{g'(x) – g(x)}$
$=\frac{xe^x + e^x}{e^x}\left [ = \frac{e^x(x+1)}{e^x}\right ] =x+1(as e^{x}\neq 0)$
$\text{attempts to integrate their } \frac{g'(x)}{g'(x) – g(x)}$
$\int (x+1) dx = \frac{1}{2}x^2 + x \left ( + C \right )$
$f(x) = e^{\frac{1}{2}x^2 + x }$
(e) $g'(x) = \cos x – \sin x$ (seen anywhere)
$\text{attempts to find an expression for } \frac{g'(x)}{g'(x) – g(x)}$
$= \frac{\cos x – \sin x}{\cos x – \sin x- \sin x – \cos x} \left [= \frac{\sin x – \cos x}{2 \sin x}\right ]$
$= \frac{1}{2} – \frac{1}{2} \cot x \text{ (as } \sin x \neq 0 \text{)} \text{ OR }= \frac{1}{2}-\frac{1}{2} \frac{\cos x}{\sin x} \text{ (as } \sin x \neq 0 \text{)}$
$f(x) = e^{\int (\frac{1}{2} – \frac{1}{2} \cot x) dx}$
$\text{attempts to find the indefinite integral of } (\pm k) \cot x \text{ OR }(\pm k) \frac{\cos x}{\sin x}$
$\int \left( \frac{1}{2} – \frac{1}{2} \cot x \right) dx = \frac{x}{2} – \frac{1}{2} \ln |\sin x|( + C)\left [ =\frac{1}{2}\left ( x-1n\left|sin x \right|(+c) \right ) \right ]$
$f(x) = e^{\frac{x}{2}} e^{-\frac{1}{2} \ln |\sin x|}( e^C)$
$=e^{\frac{x}{2}}e^{1n\sqrt{\frac{1}{sin x}}}(e^{c})\left [ =e^{\frac{x}{2}}e^{\frac{1}{2}1n(\frac{1}{sin x})} (e^{c}),=\sqrt{e^{x-1n(sin x)}}(e^{c}) \right ]$
$=e^{\frac{x}{2}}\sqrt{\frac{1}{sin x}}$
$= \sqrt{e^x \csc x}\left [ =\sqrt{\frac{e^{x}}{sin x}} \right ]\text{ (where } h(x) = \frac{1}{\sin x} \text{)}$
Question
This question asks you to investigate regular $n$-sided polygons inscribed and circumscribed in a circle, and the perimeter of these as $n$ tends to infinity, to make an approximation for $\pi$.
Let $P_i(n)$ represent the perimeter of any $n$-sided regular polygon inscribed in a circle of radius 1 unit.
Consider an equilateral triangle ABC of side length, $x$ units, circumscribed about a circle of radius 1 unit and centre $\mathrm{O}$ as shown in the following diagram.
Let $P_c(n)$ represent the perimeter of any $n$-sided regular polygon circumscribed about a circle of radius 1 unit.
a. Consider an equilateral triangle $\mathrm{ABC}$ of side length, $x$ units, inscribed in a circle of radius 1 unit and centre $\mathrm{O}$ as shown in the following diagram.
The equilateral triangle $\mathrm{ABC}$ can be divided into three smaller isosceles triangles, each subtending an angle of $\frac{2 \pi}{3}$ at $\mathrm{O}$, as shown in the following diagram.
Using right-angled trigonometry or otherwise, show that the perimeter of the equilateral triangle $\mathrm{ABC}$ is equal to $3 \sqrt{3}$ units.
b. Consider a square of side length, $x$ units, inscribed in a circle of radius 1 unit. By dividing the inscribed square into four isosceles triangles, find the exact perimeter of the inscribed square.
c. Find the perimeter of a regular hexagon, of side length, $x$ units, inscribed in a circle of radius 1 unit.
d. Show that $P_i(n)=2 n \sin \left(\frac{\pi}{n}\right)$.
e. Use an appropriate Maclaurin series expansion to find $\lim _{n \rightarrow \infty} P_i(n)$ and interpret this result geometrically.
f. Show that $P_c(n)=2 n \tan \left(\frac{\pi}{n}\right)$.
g. By writing $P_c(n)$ in the form $\frac{2 \tan \left(\frac{\pi}{n}\right)}{\frac{1}{n}}$, find $\lim _{n \rightarrow \infty} P_c(n)$.
h. Use the results from part (d) and part ( $\mathrm{f}$ ) to determine an inequality for the value of $\pi$ in terms of $n$.
i. The inequality found in part (h) can be used to determine lower and upper bound approximations for the value of $\pi$.
Determine the least value for $n$ such that the lower bound and upper bound approximations are both within 0.005 of $\pi$.
▶️Answer/Explanation
a. METHOD 1
consider right-angled triangle $\mathrm{OCX}$ where $\mathrm{CX}=\frac{x}{2}$
$$
\begin{aligned}
& \sin \frac{\pi}{3}=\frac{\frac{x}{2}}{1} \quad \text { M1A1 } \\
& \Rightarrow \frac{x}{2}=\frac{\sqrt{3}}{2} \Rightarrow x=\sqrt{3} \quad \text { A1 } \\
& P_i=3 \times x=3 \sqrt{3} \quad \text { AG }
\end{aligned}
$$
A1
METHOD 2
eg use of the cosine rule $x^2=1^2+1^2-2(1)(1) \cos \frac{2 \pi}{3} \quad$ M1A1
$x=\sqrt{3} \quad$ A1
$P_i=3 \times x=3 \sqrt{3} \quad A G$
Note: Accept use of sine rule.
[3 marks]
b. $\sin \frac{\pi}{4}=\frac{1}{x}$ where $x=$ side of square $\quad M 1$
$x=\sqrt{2} \quad$ A1
$P_i=4 \sqrt{2}$
A1
[3 marks]
c. 6 equilateral triangles $\Rightarrow x=1 \quad$ A1
$P_i=6$
A1
[2 marks]
d. in right-angled triangle $\sin \left(\frac{\pi}{n}\right)=\frac{\frac{x}{2}}{1} \quad$ M1
$$
\begin{aligned}
& \Rightarrow x=2 \sin \left(\frac{\pi}{n}\right) \quad \text { A1 } \\
& P_i=n \times x \\
& P_i=n \times 2 \sin \left(\frac{\pi}{n}\right) \quad \text { M1 } \\
& P_i=2 n \sin \left(\frac{\pi}{n}\right) \quad \text { AG }
\end{aligned}
$$
A1
[3 marks]
e. consider $\lim _{n \rightarrow \infty} 2 n \sin \left(\frac{\pi}{n}\right)$
use of $\sin x=x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\ldots \quad$ M1
$2 n \sin \left(\frac{\pi}{n}\right)=2 n\left(\frac{\pi}{n}-\frac{\pi^3}{6 n^3}+\frac{\pi^5}{120 n^5}-\ldots\right)$
$=2\left(\pi-\frac{\pi^3}{6 n^2}+\frac{\pi^5}{120 n^4}-\ldots\right) \quad \boldsymbol{A 1}$
$\Rightarrow \lim _{n \rightarrow \infty} 2 n \sin \left(\frac{\pi}{n}\right)=2 \pi \quad$ A1
as $n \rightarrow \infty$ polygon becomes a circle of radius 1 and $P_i=2 \pi$
R1
[5 marks]
f. consider an $n$-sided polygon of side length $x$
$2 n$ right-angled triangles with angle $\frac{2 \pi}{2 n}=\frac{\pi}{n}$ at centre
M1A1
opposite side $\frac{x}{2}=\tan \left(\frac{\pi}{n}\right) \Rightarrow x=2 \tan \left(\frac{\pi}{n}\right) \quad$ M1A1
Perimeter $P_c=2 n \tan \left(\frac{\pi}{n}\right) \quad$ AG
[4 marks]
g. consider $\lim _{n \rightarrow \infty} 2 n \tan \left(\frac{\pi}{n}\right)=\lim _{n \rightarrow \infty}\left(\frac{2 \tan \left(\frac{\pi}{n}\right)}{\frac{1}{n}}\right)$
$$
=\lim _{n \rightarrow \infty}\left(\frac{2 \tan \left(\frac{\pi}{n}\right)}{\frac{1}{n}}\right)=\frac{0}{0}
$$
attempt to use L’Hopital’s rule $\quad M 1$
$=\lim _{n \rightarrow \infty}\left(\frac{-\frac{2 \pi}{n^2} \sec ^2\left(\frac{\pi}{n}\right)}{-\frac{1}{n^2}}\right) \quad$ A1A1
$=2 \pi \quad$ A1
[5 marks]
h. $P_i<2 \pi<P_c$
$$
\begin{aligned}
& 2 n \sin \left(\frac{\pi}{n}\right)<2 \pi<2 n \tan \left(\frac{\pi}{n}\right) \quad \text { M1 } \\
& n \sin \left(\frac{\pi}{n}\right)<\pi<n \tan \left(\frac{\pi}{n}\right) \quad \text { A1 }
\end{aligned}
$$
A1
[2 marks]
i. attempt to find the lower bound and upper bound approximations within 0.005 of $\pi \quad$ (M1)
$$
\begin{aligned}
& n=46 \quad \text { A2 } \\
& \text { [3 marks] }
\end{aligned}
$$