Question
The function $f$ is defined by $f(x)=\arcsin (2 x)$, where $-\frac{1}{2} \leqslant x \leqslant \frac{1}{2}$.
a. By finding a suitable number of derivatives of $f$, find the first two non-zero terms in the Maclaurin series for $f$.
[8]
b. Hence or otherwise, find $\lim _{x \rightarrow 0} \frac{\arcsin (2 x)-2 x}{(2 x)^3}$.
▶️Answer/Explanation
Markscheme
a. $f(x)=\arcsin (2 x)$
$f^{\prime}(x)=\frac{2}{\sqrt{1-4 x^2}} \quad$ M1A1
Note: Award $\boldsymbol{M 1 A O}$ for $f^{\prime}(x)=\frac{1}{\sqrt{1-4 x^2}}$
$f^{\prime \prime}(x)=\frac{8 x}{\left(1-4 x^2\right)^{\frac{3}{2}}} \quad \boldsymbol{A 1}$
EITHER
$$
f^{\prime \prime \prime}(x)=\frac{8\left(1-4 x^2\right)^{\frac{3}{2}}-8 x\left(\frac{3}{2}(-8 x)\left(1-4 x^2\right)^{\frac{1}{2}}\right)}{\left(1-4 x^2\right)^3}\left(=\frac{8\left(1-4 x^2\right)^{\frac{3}{2}}+96 x^2\left(1-4 x^2\right)^{\frac{1}{2}}}{\left(1-4 x^2\right)^3}\right) \quad \boldsymbol{A 1}
$$
OR
$$
f^{\prime \prime \prime}(x)=8\left(1-4 x^2\right)^{-\frac{3}{2}}+8 x\left(-\frac{3}{2}\left(1-4 x^2\right)^{-\frac{5}{2}}\right)(-8 x)\left(=8\left(1-4 x^2\right)^{-\frac{3}{2}}+96 x^2\left(1-4 x^2\right)^{-\frac{5}{2}}\right) \quad \boldsymbol{A 1}
$$
THEN
substitute $x=0$ into $f$ or any of its derivatives
(M1)
$$
\begin{aligned}
& f(0)=0, f^{\prime}(0)=2 \text { and } f^{\prime \prime}(0)=0 \\
& f^{\prime \prime \prime}(0)=8
\end{aligned}
$$
A1
the Maclaurin series is
$$
f(x)=2 x+\frac{8 x^3}{6}+\ldots\left(=2 x+\frac{4 x^3}{3}+\ldots\right) \quad \text { (M1)A1 }
$$
[8 marks]
b. METHOD 1
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\arcsin (2 x)-2 x}{(2 x)^3}=\lim _{x \rightarrow 0} \frac{2 x+\frac{4 x^3}{3}+\ldots-2 x}{8 x^3} \quad \text { M1 } \\
& =\lim _{x \rightarrow 0} \frac{\frac{4}{3}+\ldots \text { terms with } x}{8} \\
& =\frac{1}{6} \text { A1 }
\end{aligned}
$$
(M1)
Note: Condone the omission of $+\ldots$ in their working.
METHOD 2
$\lim _{x \rightarrow 0} \frac{\arcsin (2 x)-2 x}{(2 x)^3}=\frac{0}{0}$ indeterminate form, using L’Hôpital’s rule
$=\lim _{x \rightarrow 0} \frac{\frac{2}{\sqrt{1-4 x^2}}-2}{24 x^2} \quad$ M1
$=\frac{0}{0}$ indeterminate form, using L’Hôpital’s rule again
$=\lim _{x \rightarrow 0} \frac{\frac{8 x}{\left(1-4 x^2\right)^{\frac{3}{2}}}}{48 x}\left(=\lim _{x \rightarrow 0} \frac{1}{6\left(1-4 x^2\right)^{\frac{3}{2}}}\right) \quad$ M1
Note: Award $\boldsymbol{M} 1$ only if their previous expression is in indeterminate form.
$$
=\frac{1}{6} \quad \text { A1 }
$$
Question
\(\text { Using L’Hôpital’s rule, find } \lim _{x \rightarrow 0}\left(\frac{\tan 3 x-3 \tan x}{\sin 3 x-3 \sin x}\right) \text {. }\)
▶️Answer/Explanation
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\tan 3 x-3 \tan x}{\sin 3 x-3 \sin x}\right) \\
& \lim _{x \rightarrow 0}\left(\frac{3 \sec ^2 3 x-3 \sec ^2 x}{3 \cos 3 x-3 \cos x}\right) \quad\left(=\lim _{x \rightarrow 0}\left(\frac{\sec ^2 3 x-\sec ^2 x}{\cos 3 x-\cos x}\right)\right) \quad \text { M1A1A1 }
\end{aligned}
$$
Note: Award $\boldsymbol{M} \boldsymbol{1}$ for attempt at differentiation using l’Hopital’s rule, $\boldsymbol{A} \boldsymbol{1}$ for numerator, $\boldsymbol{A} \boldsymbol{1}$ for denominator.
METHOD 1
using l’Hopital’s rule again
$$
=\lim _{x \rightarrow 0}\left(\frac{18 \sec ^2 3 x \tan 3 x-6 \sec ^2 x \tan x}{-9 \sin 3 x+3 \sin x}\right)\left(=\lim _{x \rightarrow 0}\left(\frac{6 \sec ^2 3 x \tan 3 x-2 \sec ^2 x \tan x}{-3 \sin 3 x+\sin x}\right)\right) \quad \text { A1A1 }
$$
EITHER
$$
=\lim _{x \rightarrow 0}\left(\frac{108 \sec ^2 3 x \tan ^2 3 x+54 \sec ^4 3 x-12 \sec ^2 x \tan ^2 x-6 \sec ^4 x}{-27 \cos 3 x+3 \cos x}\right) \quad \boldsymbol{A 1 A 1}
$$
Note: Not all terms in numerator need to be written in final fraction. Award $\boldsymbol{A} \boldsymbol{1}$ for $54 \sec ^4 3 x+\ldots-6 \sec ^4 x \ldots-$. However, if the terms are written, they must be correct to award $\mathrm{A} 1$.
attempt to substitute $x=0 \quad \boldsymbol{M 1}$
$$
=\frac{48}{-24}
$$
OR
$$
\begin{aligned}
& \left.\frac{\mathrm{d}}{\mathrm{d} x}\left(18 \sec ^2 3 x \tan 3 x-6 \sec ^2 x \tan x\right)\right|_{x=0}=48 \\
& \left.\frac{\mathrm{d}}{\mathrm{d} x}(-9 \sin 3 x+3 \sin x)\right|_{x=0}=-24 \quad \text { A1 }
\end{aligned}
$$
(M1)A1
THEN
$$
\left(\lim _{x \rightarrow 0}\left(\frac{\tan 3 x-3 \tan x}{\sin 3 x-3 \sin x}\right)\right)=-2 \quad \text { A1 }
$$
METHOD 2
$$
\begin{aligned}
& =\lim _{x \rightarrow 0}\left(\frac{\frac{3}{\cos ^2 3 x}-\frac{3}{\cos ^2 x}}{3 \cos 3 x-3 \cos x}\right) \quad \text { M1 } \\
& =\lim _{x \rightarrow 0}\left(\frac{\cos ^2 x-\cos ^2 3 x}{\cos ^2 3 x \cos ^2 x(\cos 3 x-\cos x)}\right) \\
& =\lim _{x \rightarrow 0}\left(\frac{\cos x+\cos 3 x}{-\cos ^2 3 x \cos ^2 x}\right) \quad \text { M1A1 }
\end{aligned}
$$
A1
attempt to substitute $x=0$
M1
$$
\begin{aligned}
& =\frac{2}{-1} \\
& =-2
\end{aligned}
$$
A1
[9 marks]
Question
a. Use L’Hôpital’s rule to determine the value of
$$
\lim _{x \rightarrow 0}\left(\frac{e^{-3 x^2}+3 \cos (2 x)-4}{3 x^2}\right)
$$
b.
Hence find $\lim _{x \rightarrow 0}\left(\frac{\int_0^x\left(e^{-3 t^2+3 \cos (2 t)-4}\right) \mathrm{d} t}{\int_0^x 3 t^2 \mathrm{~d} t}\right)$.
▶️Answer/Explanation
Markscheme
a. * This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\mathrm{e}^{-3 x^2}+3 \cos 2 x-4}{3 x^2}=\left(\begin{array}{l}
0 \\
0
\end{array}\right) \\
& =\lim _{x \rightarrow 0} \frac{-6 x e^{-3 x^2}-6 \sin 2 x}{6 x}=\left(\begin{array}{l}
0 \\
0
\end{array}\right) \quad \text { M1A1A1 } \\
& =\lim _{x \rightarrow 0} \frac{-6 \mathrm{e}^{-3 x^2}+36 x^2 e^{-3 x^2}-12 \cos 2 x}{6} \quad \text { A1 } \\
& =-3 \quad \text { A1 }
\end{aligned}
$$
[5 marks]
b.
$$
\lim _{x \rightarrow 0}\left(\frac{\int_0^x\left(\mathrm{e}^{-3 t^2}+3 \cos 2 t-4\right) \mathrm{d} t}{\int_0^x 3 t^2 \mathrm{~d} t}\right) \text { is of the form } \frac{0}{0}
$$
applying l’Hôpital’s rule
(M1)
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\mathrm{e}^{-3 x^2}+3 \cos 2 x-4}{3 x^2} \\
& =-3 \quad \text { A1 }
\end{aligned}
$$
[3 marks]
Question
Use l’Hôpital’s rule to determine the value of
$$
\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x \ln (1+x)} .
$$
▶️Answer/Explanation
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to use l’Hôpital’s rule, $M 1$
limit $=\lim _{x \rightarrow 0} \frac{2 \sin x \cos x}{\ln (1+x)+\frac{x}{1+x}}$ or $\frac{\sin 2 x}{\ln (1+x)+\frac{x}{1+x}} \quad$ A1A1
Note: Award $\boldsymbol{A} \boldsymbol{1}$ for numerator $\boldsymbol{A} \boldsymbol{1}$ for denominator.
this gives $0 / 0$ so use the rule again
(M1)
$$
=\lim _{x \rightarrow 0} \frac{2 \cos ^2 x-2 \sin ^2 x}{\frac{1}{1+x}+\frac{1+x-x}{(1+x)^2}} \text { or } \frac{2 \cos 2 x}{\frac{2+x}{(1+x)^2}}
$$
A1A1
Note: Award $\boldsymbol{A} \boldsymbol{1}$ for numerator $\boldsymbol{A} \boldsymbol{1}$ for denominator.
$$
=1 \quad \text { A }
$$
Note: This $\boldsymbol{A 1}$ is dependent on all previous marks being awarded, except when the first application of L’Hopital’s does not lead to 0/0, when it should be awarded for the correct limit of their derived function.
[7 marks]
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 $A B C$ 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 $A B C$ of side length, $x$ units, inscribed in a circle of radius 1 unit and centre $O$ as shown in the following diagram.
The equilateral triangle $A B C$ 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 $A B C$ 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
[3] 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.
[2]
d. Show that $P_i(n)=2 n \sin \left(\frac{\pi}{n}\right)$.
[3]
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 (f) to determine an inequality for the value of $\pi$ in terms of $n$.
[2]
i. The inequality found in part (h) can be used to determine lower and upper bound approximations for the value of $\pi$.
[3]
Determine the least value for $n$ such that the lower bound and upper bound approximations are both within 0.005 of $\pi$.
▶️Answer/Explanation
Markscheme
a. $M E T H O D 1$
consider right-angled triangle $\mathrm{OCX}$ where $\mathrm{CX}=\frac{x}{2}$
$$
\begin{aligned}
& \sin \frac{\pi}{3}=\frac{\frac{x}{2}}{1} \quad \text { MIA1 } \\
& \Rightarrow \frac{x}{2}=\frac{\sqrt{3}}{2} \Rightarrow x=\sqrt{3} \quad \text { A1 } \\
& P_i=3 \times x=3 \sqrt{3} \quad \text { AG }
\end{aligned}
$$
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
$P_i=3 \times x=3 \sqrt{3} \quad \boldsymbol{A G}$
Note: Accept use of sine rule.
[3 marks]
b. $\sin _{\frac{\pi}{4}}^{\frac{\pi}{4}}=\frac{1}{x}$ where $x=$ side of square
M1
$x=\sqrt{2} \quad$ A1
$P_i=4 \sqrt{2} \quad$ A1
[3 marks]
c. 6 equilateral triangles $\Rightarrow x=1$
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
$\Rightarrow x=2 \sin \left(\frac{\pi}{n}\right) \quad$ A1
$P_i=n \times x$
$P_i=n \times 2 \sin \left(\frac{\pi}{n}\right) \quad \boldsymbol{M} 1$
$P_i=2 n \sin \left(\frac{\pi}{n}\right) \quad$ AG
[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
$$
\begin{aligned}
& 2 n \sin \left(\frac{\pi}{n}\right)=2 n\left(\frac{\pi}{n}-\frac{\pi^3}{6 n^3}+\frac{\pi^5}{120 n^5}-\ldots\right) \\
& =2\left(\pi-\frac{\pi^3}{6 n^2}+\frac{\pi^5}{120 n^4}-\ldots\right) \quad \boldsymbol{A 1} \\
& \Rightarrow \lim _{n \rightarrow \infty} 2 n \sin \left(\frac{\pi}{n}\right)=2 \pi \quad \boldsymbol{A 1}
\end{aligned}
$$
(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.
$$
\begin{aligned}
& \text { consider } \lim _{n \rightarrow \infty} 2 n \tan \left(\frac{\pi}{n}\right)=\lim _{n \rightarrow \infty}\left(\frac{2 \tan \left(\frac{\pi}{n}\right)}{\frac{1}{n}}\right) \\
& =\lim _{n \rightarrow \infty}\left(\frac{2 \tan \left(\frac{\pi}{n}\right)}{\frac{1}{n}}\right)=\frac{0}{0}
\end{aligned}
$$
attempt to use L’Hopital’s rule M1
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left(\frac{-\frac{2 \pi}{n^2} \sec ^2\left(\frac{\pi}{n}\right)}{-\frac{1}{n^2}}\right) \quad \text { A1A1 } \\
& =2 \pi
\end{aligned}
$$
[5 marks]
h. $P_i<2 \pi<P_c$
$$
\begin{aligned}
& 2 n \sin \left(\frac{\pi}{n}\right)<2 \pi<2 n \tan \left(\frac{\pi}{n}\right) \quad \text { M1 } \\
& n \sin \left(\frac{\pi}{n}\right)<\pi<n \tan \left(\frac{\pi}{n}\right) \quad \text { A1 } \\
& {[2 \text { marks] }}
\end{aligned}
$$
i. attempt to find the lower bound and upper bound approximations within 0.005 of $\pi$
(M1)
$$
n=46 \quad \text { A2 }
$$
[3 marks]
Question
Use l’Hôpital’s rule to find
$$
\lim _{x \rightarrow 1} \frac{\cos \left(x^2-1\right)-1}{\mathrm{e}^{x-1}-x}
$$
▶️Answer/Explanation
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to use l’Hôpital’s rule $\quad M 1$
$$
=\lim _{x \rightarrow 1} \frac{-2 x \sin \left(x^2-1\right)}{\mathrm{e}^{x-1}-1}
$$
A1A1
Note: Award $\boldsymbol{A} \mathbf{1}$ for the numerator and $\boldsymbol{A} \mathbf{1}$ for the denominator.
substitution of 1 into their expression
(M1)
$=\frac{0}{0}$ hence use l’Hôpital’s rule again
Note: If the first use of l’Hôpital’s rule results in an expression which is not in indeterminate form, do not award any further marks.
attempt to use product rule in numerator
M1
$$
\begin{aligned}
& =\lim _{x \rightarrow 1} \frac{-4 x^2 \cos \left(x^2-1\right)-2 \sin \left(x^2-1\right)}{\mathrm{e}^{x-1}} \\
& =-4 \text { A1 }
\end{aligned}
$$
A1
[7 marks]
Question
The function f is defined by \(f(x) = {{\text{e}}^{({{\text{e}}^x} – 1)}}\) .
(a) Assuming the Maclaurin series for \({{\text{e}}^x}\) , show that the Maclaurin series for \(f(x)\)
is \(1 + x + {x^2} + \frac{5}{6}{x^3} + \ldots {\text{ .}}\)
(b) Hence or otherwise find the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{f(x) – 1}}{{f'(x) – 1}}\) .
▶️Answer/Explanation
Markscheme
(a) \({{\text{e}}^x} – 1 = x + \frac{{{x^2}}}{2} + \frac{{{x^2}}}{6} + \ldots \) A1
\({{\text{e}}^{{{\text{e}}^x} – 1}} = 1 + \left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right) + \frac{{{{\left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)}^2}}}{2} + \frac{{{{\left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)}^3}}}{6} + \ldots \) M1A1
\( = 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^3}}}{6} + \ldots \) M1A1
\( = 1 + x + {x^2} + \frac{5}{6}{x^3} + \ldots \) AG
[5 marks]
(b) EITHER
\(f'(x) = 1 + 2x + \frac{{5{x^2}}}{2} + \ldots \) A1
\(\frac{{f(x) – 1}}{{f'(x) – 1}} = \frac{{x + {x^2} + 5{x^3}/6 + \ldots }}{{2x + 5{x^2}/2 + \ldots }}\) M1A1
\( = \frac{{1 + x + \ldots }}{{2 + 5x/2 + \ldots }}\) A1
\( \to \frac{1}{2}{\text{ as }}x \to 0\) A1
[5 marks]
OR
using l’Hopital’s rule, M1
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1}}{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1′ – 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1}}{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}} – 1}}\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}}}}{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}} \times ({{\text{e}}^x} + 1)}}\) A1
\( = \frac{1}{2}\) A1
[5 marks]
Total [10 marks]
Examiners report
Many candidates obtained the required series by finding the values of successive derivatives at x = 0 , failing to realise that the intention was to start with the exponential series and replace x by the series for \({{\text{e}}^x} – 1\). Candidates who did this were given partial credit for using this method. Part (b) was reasonably well answered using a variety of methods.
Question
(a) Using the Maclaurin series for \({(1 + x)^n}\), write down and simplify the Maclaurin series approximation for \({(1 – {x^2})^{ – \frac{1}{2}}}\) as far as the term in \({x^4}\)
(b) Use your result to show that a series approximation for arccos x is
\[\arccos x \approx \frac{\pi }{2} – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5}.\]
(c) Evaluate \(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} – \arccos ({x^2}) – {x^2}}}{{{x^6}}}\).
(d) Use the series approximation for \(\arccos x\) to find an approximate value for
\[\int_0^{0.2} {\arccos \left( {\sqrt x } \right){\text{d}}x} ,\]
giving your answer to 5 decimal places. Does your answer give the actual value of the integral to 5 decimal places?
▶️Answer/Explanation
Markscheme
(a) using or obtaining \({(1 + x)^n} = 1 + nx + \frac{{n(n – 1)}}{2}{x^2} + \ldots \) (M1)
\({(1 – {n^2})^{ – \frac{1}{2}}} = 1 + ( – {x^2}) \times \left( { – \frac{1}{2}} \right) + \frac{{{{( – {x^2})}^2}}}{2} \times \left( { – \frac{1}{2}} \right) \times \left( { – \frac{3}{2}} \right) + \ldots \) (A1)
\( = 1 + \frac{1}{2}{x^2} + \frac{3}{8}{x^4} + \ldots \) A1
[3 marks]
(b) integrating, and changing sign
\(\arccos x = – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5} + C + \ldots \) M1A1
put x = 0,
\(\frac{\pi }{2} = C\) M1
\(\left( {\arccos x \approx \frac{\pi }{2} – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5}} \right)\) AG
[3 marks]
(c) EITHER
using \(\arccos {x^2} \approx \frac{\pi }{2} – {x^2} – \frac{1}{6}{x^6} – \frac{3}{{40}}{x^{10}}\) M1A1
\(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} – \arccos {x^2} – {x^2}}}{{{x^6}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{x^6}}}{6} + {\text{higher powers}}}}{{{x^6}}}\) M1A1
\( = \frac{1}{6}\) A1
OR
using l’Hôpital’s Rule M1
\({\text{limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 – {x^4}} }} \times 2x – 2x}}{{6{x^5}}}\) M1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 – {x^4}} }} – 1}}{{3{x^4}}}\) A1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{ – \frac{1}{2} \times \frac{1}{{{{(1 – {x^4})}^{3/2}}}} \times – 4{x^3}}}{{12{x^3}}}\) M1
\( = \frac{1}{6}\) A1
[5 marks]
(d) \(\int_0^{0.2} {\arccos \sqrt x {\text{d}}x \approx \int_0^{0.2} {\left( {\frac{\pi }{2} – {x^{\frac{1}{2}}} – \frac{1}{6}{x^{\frac{3}{2}}} – \frac{3}{{40}}{x^{\frac{5}{2}}}} \right){\text{d}}x} } \) M1
\( = \left[ {\frac{\pi }{2}x – \frac{2}{3}{x^{\frac{3}{2}}} – \frac{1}{{15}}{x^{\frac{5}{2}}} – \frac{3}{{140}}{x^{\frac{7}{2}}}} \right]_0^{0.2}\) (A1)
\( = \frac{\pi }{2} \times 0.2 – \frac{2}{3} \times {0.2^{\frac{3}{2}}} – \frac{1}{{15}} \times {0.2^{\frac{5}{2}}} – \frac{3}{{140}} \times {0.2^{\frac{7}{2}}}\) (A1)
= 0.25326 (to 5 decimal places) A1
Note: Accept integration of the series approximation using a GDC.
using a GDC, the actual value is 0.25325 A1
so the approximation is not correct to 5 decimal places R1
[6 marks]
Total [17 marks]
Examiners report
Many candidates ignored the instruction in the question to use the series for \({(1 + x)^n}\) to deduce the series for \({(1 – {x^2})^{ – 1/2}}\) and attempted instead to obtain it by successive differentiation. It was decided at the standardisation meeting to award full credit for this method although in the event the algebra proved to be too difficult for many. Many candidates used l’Hopital’s Rule in (c) – this was much more difficult algebraically than using the series and it usually ended unsuccessfully. Candidates should realise that if a question on evaluating an indeterminate limit follows the determination of a Maclaurin series then it is likely that the series will be helpful in evaluating the limit. Part (d) caused problems for many candidates with algebraic errors being common. Many candidates failed to realise that the best way to find the exact value of the integral was to use the calculator.
Question
Find \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 – \cos {x^6}}}{{{x^{12}}}}} \right)\).
▶️Answer/Explanation
Markscheme
METHOD 1
\(f(0) = \frac{0}{0}\), hence using l’Hôpital’s Rule, (M1)
\(g(x) = 1 – \cos ({x^6}),{\text{ }}h(x) = {x^{12}};{\text{ }}\frac{{g'(x)}}{{h'(x)}} = \frac{{6{x^5}\sin ({x^6})}}{{12{x^{11}}}} = \frac{{\sin ({x^6})}}{{2{x^6}}}\) A1A1
EITHER
\(\frac{{g'(0)}}{{h'(0)}} = \frac{0}{0}\), using l’Hôpital’s Rule again, (M1)
\(\frac{{g”(x)}}{{h”(x)}} = \frac{{6{x^5}\cos ({x^6})}}{{12{x^5}}} = \frac{{\cos ({x^6})}}{2}\) A1A1
\(\frac{{g”(0)}}{{h”(0)}} = \frac{1}{2}\), hence the limit is \(\frac{1}{2}\) A1
OR
So \(\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos {x^6}}}{{{x^{12}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}}\) A1
\( = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}}\) A1
\( = \frac{1}{2}{\text{ since }}\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}} = 1\) A1 (R1)
METHOD 2
substituting \({{x^6}}\) for x in the expansion \(\cos x = 1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} \ldots \) (M1)
\(\frac{{1 – \cos {x^6}}}{{{x^{12}}}} = \frac{{1 – \left( {1 – \frac{{{x^{12}}}}{2} + \frac{{{x^{24}}}}{{24}}} \right) \ldots }}{{{x^{12}}}}\) M1A1
\( = \frac{1}{2} – \frac{{{x^{12}}}}{{24}} + …\) A1A1
\(\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos {x^6}}}{{{x^{12}}}} = \frac{1}{2}\) M1A1
Note: Accept solutions using Maclaurin expansions.
[7 marks]
Examiners report
Surprisingly, some weaker candidates were more successful in answering this question than stronger candidates. If candidates failed to simplify the expression after the first application of L’Hôpital’s rule, they generally were not successful in correctly differentiating the expression a \({2^{{\text{nd}}}}\) time, hence could not achieve the final three A marks.
Question
Find \(\mathop {\lim }\limits_{x \to \frac{1}{2}} \left( {\frac{{\left( {\frac{1}{4} – {x^2}} \right)}}{{\cot \pi x}}} \right)\).
▶️Answer/Explanation
Markscheme
using l’Hôpital’s Rule (M1)
\(\mathop {\lim }\limits_{x \to \frac{1}{2}} \left( {\frac{{\left( {\frac{1}{4} – {x^2}} \right)}}{{\cot \pi x}}} \right) = \mathop {\lim }\limits_{x \to \frac{1}{2}} \left[ {\frac{{ – 2}}{{ – \pi {\text{cose}}{{\text{c}}^2}\pi x}}} \right]\) A1A1
\( = \frac{{ – 1}}{{ – \pi {\text{cose}}{{\text{c}}^2}\frac{\pi }{2}}} = \frac{1}{\pi }\) (M1)A1
[5 marks]
Examiners report
This question was accessible to the vast majority of candidates, who recognised that L’Hôpital’s rule was required. However, some candidates omitted the factor \(\pi \) in the differentiation of \({\cot \pi x}\). Some candidates replaced \({\cot \pi x}\) by \(\cos \pi x{\text{/}}\sin \pi x\), which is a valid method but the extra algebra involved often led to an incorrect answer. Many fully correct solutions were seen.
Question
Use L’Hôpital’s Rule to find \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} – 1 – x\cos x}}{{{{\sin }^2}x}}\) .
▶️Answer/Explanation
Markscheme
apply l’Hôpital’s Rule to a \(0/0\) type limit
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} – 1 – x\cos x}}{{{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} – \cos x + x\sin x}}{{2\sin x\cos x}}\) M1A1
noting this is also a \(0/0\) type limit, apply l’Hôpital’s Rule again (M1)
obtain \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} + \sin x + x\cos x + \sin x}}{{2\cos 2x}}\) A1
substitution of x = 0 (M1)
= 0.5 A1
[6 marks]
Examiners report
The vast majority of candidates were familiar with L’Hôpitals rule and were also able to apply the technique twice as required by the problem. The errors that occurred were mostly due to difficulty in applying the differentiation rules correctly or errors in algebra. A small minority of candidates tried to use the quotient rule but it seemed that most candidates had a good understanding of L’Hôpital’s rule and its application to finding a limit.
Question
(a) Using l’Hopital’s Rule, show that \(\mathop {\lim }\limits_{x \to \infty } x{{\text{e}}^{ – x}} = 0\) .
(b) Determine \(\int_0^a {x{{\text{e}}^{ – x}}{\text{d}}x} \) .
(c) Show that the integral \(\int_0^\infty {x{{\text{e}}^{ – x}}{\text{d}}x} \) is convergent and find its value.
▶️Answer/Explanation
Markscheme
(a) \(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{{\text{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{{\text{e}}^x}}}\) M1A1
= 0 AG
[2 marks]
(b) Using integration by parts M1
\(\int_0^a {x{{\text{e}}^{ – x}}{\text{d}}x} = \left[ { – x{{\text{e}}^{ – x}}} \right]_0^a + \int_0^a {{{\text{e}}^{ – x}}{\text{d}}x} \) A1A1
\( = – a{{\text{e}}^{ – a}} – \left[ {{e^{ – x}}} \right]_0^a\) A1
\( = 1 – a{{\text{e}}^{ – a}} – {{\text{e}}^{ – a}}\) A1
[5 marks]
(c) Since \({{\text{e}}^{ – a}}\) and \(a{{\text{e}}^{ – a}}\) are both convergent (to zero), the integral is convergent. R1
Its value is 1. A1
[2 marks]
Total [9 marks]
Examiners report
Most candidates made a reasonable attempt at (a). In (b), however, it was disappointing to note that some candidates were unable to use integration by parts to perform the integration. In (c), while many candidates obtained the correct value of the integral, proof of its convergence was often unconvincing.
Question
(a) Show that the solution of the differential equation
\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = \cos x{\cos ^2}y{\text{,}}\]
given that \(y = \frac{\pi }{4}{\text{ when }}x = \pi {\text{, is }}y = \arctan (1 + \sin x){\text{.}}\)
(b) Determine the value of the constant a for which the following limit exists
\[\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\arctan (1 + \sin x) – a}}{{{{\left( {x – \frac{\pi }{2}} \right)}^2}}}\]
and evaluate that limit.
▶️Answer/Explanation
Markscheme
(a) this separable equation has general solution
\(\int {{{\sec }^2}y{\text{d}}y = \int {\cos x{\text{d}}x} } \) (M1)(A1)
\(\tan y = \sin x + c\) A1
the condition gives
\(\tan \frac{\pi }{4} = \sin \pi + c \Rightarrow c = 1\) M1
the solution is \(\tan y = 1 + \sin x\) A1
\(y = \arctan (1 + \sin x)\) AG
[5 marks]
(b) the limit cannot exist unless \(a = \arctan \left( {1 + \sin \frac{\pi }{2}} \right) = \arctan 2\) R1A1
in that case the limit can be evaluated using l’Hopital’s rule (twice) limit is
\(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\left( {\arctan (1 + \sin x)} \right)}^\prime }}}{{2\left( {x – \frac{\pi }{2}} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y’}}{{2\left( {x – \frac{\pi }{2}} \right)}}\) M1A1
where y is the solution of the differential equation
the numerator has zero limit (from the factor \(\cos x\) in the differential equation) R1
so required limit is
\(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y”}}{2}\) M1A1
finally,
\(y” = – \sin x{\cos ^2}y – 2\cos x\cos y\sin y \times y'(x)\) M1A1
since \(\cos y\left( {\frac{\pi }{2}} \right) = \frac{1}{{\sqrt 5 }}\) A1
\(y” = – \frac{1}{5}{\text{ at }}x = \frac{\pi }{2}\) A1
the required limit is \( – \frac{1}{{10}}\) A1
[12 marks]
Total [17 marks]
Examiners report
Many candidates successfully obtained the displayed solution of the differential equation in part(a). Few complete solutions to part(b) were seen which used the result in part(a). The problem can, however, be solved by direct differentiation although this is algebraically more complicated. Some successful solutions using this method were seen.
Question
The function f is defined by \(f(x) = {{\text{e}}^{({{\text{e}}^x} – 1)}}\) .
(a) Assuming the Maclaurin series for \({{\text{e}}^x}\) , show that the Maclaurin series for \(f(x)\)
is \(1 + x + {x^2} + \frac{5}{6}{x^3} + \ldots {\text{ .}}\)
(b) Hence or otherwise find the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{f(x) – 1}}{{f'(x) – 1}}\) .
▶️Answer/Explanation
Markscheme
(a) \({{\text{e}}^x} – 1 = x + \frac{{{x^2}}}{2} + \frac{{{x^2}}}{6} + \ldots \) A1
\({{\text{e}}^{{{\text{e}}^x} – 1}} = 1 + \left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right) + \frac{{{{\left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)}^2}}}{2} + \frac{{{{\left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)}^3}}}{6} + \ldots \) M1A1
\( = 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^3}}}{6} + \ldots \) M1A1
\( = 1 + x + {x^2} + \frac{5}{6}{x^3} + \ldots \) AG
[5 marks]
(b) EITHER
\(f'(x) = 1 + 2x + \frac{{5{x^2}}}{2} + \ldots \) A1
\(\frac{{f(x) – 1}}{{f'(x) – 1}} = \frac{{x + {x^2} + 5{x^3}/6 + \ldots }}{{2x + 5{x^2}/2 + \ldots }}\) M1A1
\( = \frac{{1 + x + \ldots }}{{2 + 5x/2 + \ldots }}\) A1
\( \to \frac{1}{2}{\text{ as }}x \to 0\) A1
[5 marks]
OR
using l’Hopital’s rule, M1
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1}}{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1′ – 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1}}{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}} – 1}}\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}}}}{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}} \times ({{\text{e}}^x} + 1)}}\) A1
\( = \frac{1}{2}\) A1
[5 marks]
Total [10 marks]
Examiners report
Many candidates obtained the required series by finding the values of successive derivatives at x = 0 , failing to realise that the intention was to start with the exponential series and replace x by the series for \({{\text{e}}^x} – 1\). Candidates who did this were given partial credit for using this method. Part (b) was reasonably well answered using a variety of methods.
Question
(a) Using the Maclaurin series for \({(1 + x)^n}\), write down and simplify the Maclaurin series approximation for \({(1 – {x^2})^{ – \frac{1}{2}}}\) as far as the term in \({x^4}\)
(b) Use your result to show that a series approximation for arccos x is
\[\arccos x \approx \frac{\pi }{2} – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5}.\]
(c) Evaluate \(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} – \arccos ({x^2}) – {x^2}}}{{{x^6}}}\).
(d) Use the series approximation for \(\arccos x\) to find an approximate value for
\[\int_0^{0.2} {\arccos \left( {\sqrt x } \right){\text{d}}x} ,\]
giving your answer to 5 decimal places. Does your answer give the actual value of the integral to 5 decimal places?
▶️Answer/Explanation
Markscheme
(a) using or obtaining \({(1 + x)^n} = 1 + nx + \frac{{n(n – 1)}}{2}{x^2} + \ldots \) (M1)
\({(1 – {n^2})^{ – \frac{1}{2}}} = 1 + ( – {x^2}) \times \left( { – \frac{1}{2}} \right) + \frac{{{{( – {x^2})}^2}}}{2} \times \left( { – \frac{1}{2}} \right) \times \left( { – \frac{3}{2}} \right) + \ldots \) (A1)
\( = 1 + \frac{1}{2}{x^2} + \frac{3}{8}{x^4} + \ldots \) A1
[3 marks]
(b) integrating, and changing sign
\(\arccos x = – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5} + C + \ldots \) M1A1
put x = 0,
\(\frac{\pi }{2} = C\) M1
\(\left( {\arccos x \approx \frac{\pi }{2} – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5}} \right)\) AG
[3 marks]
(c) EITHER
using \(\arccos {x^2} \approx \frac{\pi }{2} – {x^2} – \frac{1}{6}{x^6} – \frac{3}{{40}}{x^{10}}\) M1A1
\(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} – \arccos {x^2} – {x^2}}}{{{x^6}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{x^6}}}{6} + {\text{higher powers}}}}{{{x^6}}}\) M1A1
\( = \frac{1}{6}\) A1
OR
using l’Hôpital’s Rule M1
\({\text{limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 – {x^4}} }} \times 2x – 2x}}{{6{x^5}}}\) M1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 – {x^4}} }} – 1}}{{3{x^4}}}\) A1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{ – \frac{1}{2} \times \frac{1}{{{{(1 – {x^4})}^{3/2}}}} \times – 4{x^3}}}{{12{x^3}}}\) M1
\( = \frac{1}{6}\) A1
[5 marks]
(d) \(\int_0^{0.2} {\arccos \sqrt x {\text{d}}x \approx \int_0^{0.2} {\left( {\frac{\pi }{2} – {x^{\frac{1}{2}}} – \frac{1}{6}{x^{\frac{3}{2}}} – \frac{3}{{40}}{x^{\frac{5}{2}}}} \right){\text{d}}x} } \) M1
\( = \left[ {\frac{\pi }{2}x – \frac{2}{3}{x^{\frac{3}{2}}} – \frac{1}{{15}}{x^{\frac{5}{2}}} – \frac{3}{{140}}{x^{\frac{7}{2}}}} \right]_0^{0.2}\) (A1)
\( = \frac{\pi }{2} \times 0.2 – \frac{2}{3} \times {0.2^{\frac{3}{2}}} – \frac{1}{{15}} \times {0.2^{\frac{5}{2}}} – \frac{3}{{140}} \times {0.2^{\frac{7}{2}}}\) (A1)
= 0.25326 (to 5 decimal places) A1
Note: Accept integration of the series approximation using a GDC.
using a GDC, the actual value is 0.25325 A1
so the approximation is not correct to 5 decimal places R1
[6 marks]
Total [17 marks]
Examiners report
Many candidates ignored the instruction in the question to use the series for \({(1 + x)^n}\) to deduce the series for \({(1 – {x^2})^{ – 1/2}}\) and attempted instead to obtain it by successive differentiation. It was decided at the standardisation meeting to award full credit for this method although in the event the algebra proved to be too difficult for many. Many candidates used l’Hopital’s Rule in (c) – this was much more difficult algebraically than using the series and it usually ended unsuccessfully. Candidates should realise that if a question on evaluating an indeterminate limit follows the determination of a Maclaurin series then it is likely that the series will be helpful in evaluating the limit. Part (d) caused problems for many candidates with algebraic errors being common. Many candidates failed to realise that the best way to find the exact value of the integral was to use the calculator.