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 $f$.
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 M1AO 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
$f^{\prime \prime \prime}(x)=\frac{8\left(1-4 x^2\right)^{\frac{3}{2}}-8 x\left(\frac{3}{2}(-8 x)\left(1-4 x^2\right)^{\frac{1}{2}}\right)}{\left(1-4 x^2\right)^3}\left(=\frac{8\left(1-4 x^2\right)^{\frac{3}{2}}+96 x^2\left(1-4 x^2\right)^{\frac{1}{2}}}{\left(1-4 x^2\right)^3}\right) \quad$ A1
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
$$
f(x)=2 x+\frac{8 x^3}{6}+\ldots\left(=2 x+\frac{4 x^3}{3}+\ldots\right)
$$
(M1)A1
[8 marks]
b. METHOD 1
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\arcsin (2 x)-2 x}{(2 x)^3}=\lim _{x \rightarrow 0} \frac{2 x+\frac{4 x^3}{3}+\ldots-2 x}{8 x^3} \quad \text { M1 } \\
& =\lim _{x \rightarrow 0} \frac{\frac{4}{3}+\ldots \text { terms with } x}{8} \\
& \text { (M1) } \\
& =\frac{1}{6} \boldsymbol{A 1}
\end{aligned}
$$
(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
$$
=\lim _{x \rightarrow 0} \frac{\frac{2}{\sqrt{1-4 x^2}}-2}{24 x^2} \quad \text { M1 }
$$
$=\frac{0}{0}$ indeterminate form, using L’Hôpital’s rule again
$$
=\lim _{x \rightarrow 0} \frac{\frac{8 x}{\left(1-4 x^2\right)^{\frac{3}{2}}}}{48 x}\left(=\lim _{x \rightarrow 0} \frac{1}{6\left(1-4 x^2\right)^{\frac{3}{2}}}\right) \quad \boldsymbol{M 1}
$$
Note: Award $\boldsymbol{M 1}$ only if their previous expression is in indeterminate form.
$$
=\frac{1}{6} \quad \boldsymbol{A 1}
$$
Note: Award $\boldsymbol{F T}$ for use of their derivatives from part (a).
Question
This question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.
A power series in $x$ is defined as a function of the form $f(x)=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots$ where the $a_i \in \mathrm{R}$.
It can be considered as an infinite polynomial.
This is an example of a power series, but is only a finite power series, since only a finite number of the $a_i$ are non-zero.
We will now attempt to generalise further.
Suppose $(1+x)^q, q \in \mathrm{Q}$ can be written as the power series $a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots$
a. Expand $(1+x)^5$ using the Binomial Theorem.
b. Consider the power series $1-x+x^2-x^3+x^4-\ldots$
By considering the ratio of consecutive terms, explain why this series is equal to $(1+x)^{-1}$ and state the values of $x$ for which this equality is true.
c. Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for $(1+x)^{-2}$.
d. Repeat this process to find the first four terms in a power series for $(1+x)^{-3}$.
e. Hence, by recognising the pattern, deduce the first four terms in a power series for $(1+x)^{-n}, n \in \mathrm{Z}^{+}$.
f. By substituting $x=0$, find the value of $a_0$.
g. By differentiating both sides of the expression and then substituting $x=0$, find the value of $a_1$.
h. Repeat this procedure to find $a_2$ and $a_3$.
i. Hence, write down the first four terms in what is called the Extended Binomial Theorem for $(1+x)^q, q \in \mathrm{Q}$.
j. Write down the power series for $\frac{1}{1+x^2}$.
k. Hence, using integration, find the power series for $\arctan x$, giving the first four non-zero terms.
▶️Answer/Explanation
Markscheme
a. $1+5 x+10 x^2+10 x^3+5 x^4+x^5 \quad$ M1A1
[2 marks]
b. It is an infinite GP with $a=1, r=-x \quad \operatorname{R1A1}$
$$
S_{\infty}=\frac{1}{1-(-x)}=\frac{1}{1+x}=(1+x)^{-1} \quad \text { M1A1AG }
$$
[4 marks]
$$
\begin{aligned}
& \text { c. }(1+x)^{-1}=1-x+x^2-x^3+x^4-\ldots \\
& \begin{array}{l}
-1(1+x)^{-2}=-1+2 x-3 x^2+4 x^3-\ldots \quad \text { A1 } \\
(1+x)^{-2}=1-2 x+3 x^2-4 x^3+\ldots \quad \text { A1 }
\end{array}
\end{aligned}
$$
d. $-2(1+x)^{-3}=-2+6 x-12 x^2+20 x^3 \ldots \quad$ A1
$(1+x)^{-3}=1-3 x+6 x^2-10 x^3 \ldots \quad$ A1
[2 marks]
e. $(1+x)^{-n}=1-n x+\frac{n(n+1)}{2 !} x^2-\frac{n(n+1)(n+2)}{3 !} x^3 \ldots \quad$ A1A1A1
[3 marks]
f. $1^q=a_0 \Rightarrow a_0=1$
$A 1$
[1 mark]
g. $q(1+x)^{q-1}=a_1+2 a_2 x+3 a_3 x^2+\ldots \quad \quad$ A1
$a_1=q \quad \boldsymbol{A 1}$
[2 marks]
h. $q(q-1)(1+x)^{q-2}=1 \times 2 a_2+2 \times 3 a_3 x+\ldots \quad$ A1
$a_2=\frac{q(q-1)}{2 !} \quad \boldsymbol{A 1}$
$q(q-1)(q-2)(1+x)^{q-3}=1 \times 2 \times 3 a_3+\ldots$
A1
$a_3=\frac{q(q-1)(q-2)}{3 !}$
A1
[4 marks]
i. $(1+x)^q=1+q x+\frac{q(q-1)}{2 !} x^2+\frac{q(q-1)(q-2)}{3 !} x^3 \ldots \quad \boldsymbol{A 1}$
k. $\arctan x+c=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots \quad$ M1A1
Putting $x=0 \Rightarrow c=0 \quad$ R1
So $\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$
A1
Question
The function $f$ is defined by $f(x)=\ln \left(1+x^2\right)$ where $-1<x<1$.
The seventh derivative of $f$ is given by $f^{(7)}(x)=\frac{1440 x\left(x^6-21 x^4+35 x^2-7\right)}{\left(1+x^2\right)^7}$.
a.i. Use the Maclaurin series for $\ln (1+x)$ to write down the first three non-zero terms of the Maclaurin series for $f(x)$.
a.ii.Hence find the first three non-zero terms of the Maclaurin series for $\frac{x}{1+x^2}$.
b. Use your answer to part (a)(i) to write down an estimate for $f(0.4)$.
c.i. Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating $f(0.4)$, using the first three nonzero terms of the Maclaurin series for $f(x)$.
c.ii.With reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for $f(0.4)$.
▶️Answer/Explanation
Markscheme
a.i.
substitution of $x^2$ in $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots$
(M1)
$x^2-\frac{x^4}{2}+\frac{x^6}{3}$
A1
[2 marks]
a.ii. $\frac{d}{d x}\left(\ln \left(1+x^2\right)\right)=\frac{2 x}{1+x^2}$
(M1)
Note: Award (M1) if this is seen in part (a)(i).
attempt to differentiate their answer in part (a)
(M1)
$\frac{2 x}{1+x^2}=2 x-\frac{4 x^3}{2}+\frac{6 x^5}{3} \quad$ M1
Note: Award $\boldsymbol{M} 1$ for equating their derivatives.
$\frac{x}{1+x^2}=x-x^3+x^5$
b. $f(0.4) \approx 0.149$
A1
Note: Accept an answer that rounds correct to 2 s.f. or better.
[1 mark]
c.i. attempt to find the maximum of $\left|f^{(7)}(c)\right|$ for $c \in[0,0.4]$
(M1)
maximum of $\left|f^{(7)}(c)\right|$ occurs at $c=0.199$
(A1)
$$
\left.\left|f^{(7)}(c)\right|<1232.97 \ldots \text { (for all } c \in\right] 0,0.4[\text { ) }
$$
use of $x=0.4$
(M1)
substitution of $n=6$ and $a=0$ and their value of $x$ and their value of $f^{(7)}(c)$ into Lagrange error term
(M1)
Note: Award (M1) for substitution of $n=3$ and $a=0$ and their value of $x$ and their value of $f^{(4)}(c)$ into Lagrange error term.
$$
\begin{aligned}
& \left|R_6(0.4)\right|<\frac{1232.97(0.4)^7}{7 !} \\
& \text { upper bound }=0.000401
\end{aligned}
$$
A1
Note: Accept an answer that rounds correct to 1 s.f or better.
c.ii. $.^{(7)}(c)<0$ (for all $\left.c \in\right] 0,0.4[$ ) $\quad \boldsymbol{R}$
Note: Accept $R_6(c)<0$ or “the error term is negative”.
the answer in (b) is an overestimate
A1
Note: The $\boldsymbol{A} \mathbf{1}$ is dependent on the $\boldsymbol{R} \mathbf{1}$.
Question
The function $f$ is defined by $f(x)=\ln \left(1+x^2\right)$ where $-1<x<1$.
The seventh derivative of $f$ is given by $f^{(7)}(x)=\frac{1440 x\left(x^6-21 x^4+35 x^2-7\right)}{\left(1+x^2\right)^7}$.
a.i.Use the Maclaurin series for $\ln (1+x)$ to write down the first three non-zero terms of the Maclaurin series for $f(x)$.
a.ii.Hence find the first three non-zero terms of the Maclaurin series for $\frac{x}{1+x^2}$.
b. Use your answer to part
▶️Answer/Explanation
(a)(i) to write down an estimate for $f(0.4)$.
c.i. Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating $f(0.4)$, using the first three nonzero terms of the Maclaurin series for $f(x)$.
c.ii.With reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for $f(0.4)$.a.i.
substitution of $x^2$ in $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots$
(M1)
$x^2-\frac{x^4}{2}+\frac{x^6}{3} \quad$ A1
[2 marks]
a.ii. $\frac{\mathrm{d}}{\mathrm{d} x}\left(\ln \left(1+x^2\right)\right)=\frac{2 x}{1+x^2}$
(M1)
Note: Award (M1) if this is seen in part (a)(i).
attempt to differentiate their answer in part (a)
(M1)
$\frac{2 x}{1+x^2}=2 x-\frac{4 x^3}{2}+\frac{6 x^5}{3} \quad$ M1
Note: Award $\boldsymbol{M} 1$ for equating their derivatives.
$\frac{x}{1+x^2}=x-x^3+x^5 \quad$ A1
[4 marks]
b. $f(0.4) \approx 0.149$
A1
Note: Accept an answer that rounds correct to 2 s.f. or better.
[1 mark]
c.i. attempt to find the maximum of $\left|f^{(7)}(c)\right|$ for $c \in[0,0.4] \quad$ (M1)
maximum of $\left|f^{(7)}(c)\right|$ occurs at $c=0.199$
(A1)
$\left|f^{(7)}(c)\right|<1232.97 \ldots \quad$ (for all $\left.c \in\right] 0,0.4[$ )
(A1)
use of $x=0.4$
(M1)
substitution of $n=6$ and $a=0$ and their value of $x$ and their value of $f^{(7)}(c)$ into Lagrange error term
(M1)
Note: Award (M1) for substitution of $n=3$ and $a=0$ and their value of $x$ and their value of $f^{(4)}(c)$ into Lagrange error term.
$$
\begin{aligned}
& \left|R_6(0.4)\right|<\frac{1232.97(0.4)^7}{7 !} \\
& \text { upper bound }=0.000401
\end{aligned}
$$
A1
Note: Accept an answer that rounds correct to 1 s.f or better.
c.ii. $f^{(7)}(c)<0$ (for all $\left.c \in\right] 0,0.4[) \quad R 1$
Note: Accept $R_6(c)<0$ or “the error term is negative”.
the answer in (b) is an overestimate
A1
Note: The $\boldsymbol{A} 1$ is dependent on the $\boldsymbol{R 1 .}$
Question
The function f is defined by
\[f(x) = \ln \left( {\frac{1}{{1 – x}}} \right).\]
(a) Write down the value of the constant term in the Maclaurin series for \(f(x)\) .
(b) Find the first three derivatives of \(f(x)\) and hence show that the Maclaurin series for \(f(x)\) up to and including the \({x^3}\) term is \(x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\).
(c) Use this series to find an approximate value for ln 2 .
(d) Use the Lagrange form of the remainder to find an upper bound for the error in this approximation.
(e) How good is this upper bound as an estimate for the actual error?
▶️Answer/Explanation
Markscheme
(a) Constant term = 0 A1
[1 mark]
(b) \(f'(x) = \frac{1}{{1 – x}}\) A1
\(f”(x) = \frac{1}{{{{(1 – x)}^2}}}\) A1
\(f”'(x) = \frac{2}{{{{(1 – x)}^3}}}\) A1
\(f'(0) = 1;{\text{ }}f”(0) = 1;{\text{ }}f”'(0) = 2\) A1
Note: Allow FT on their derivatives.
\(f(x) = 0 + \frac{{1 \times x}}{{1!}} + \frac{{1 \times {x^2}}}{{2!}} + \frac{{2 \times {x^3}}}{{3!}} + …\) M1A1
\( = x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\) AG
[6 marks]
(c) \(\frac{1}{{1 – x}} = 2 \Rightarrow x = \frac{1}{2}\) (A1)
\(\ln 2 \approx \frac{1}{2} + \frac{1}{8} + \frac{1}{{24}}\) M1
\( = \frac{2}{3}{\text{ (0.667)}}\) A1
[3 marks]
(d) Lagrange error \({\text{ = }}\frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}} \times {\left( {\frac{1}{2}} \right)^{n + 1}}\) (M1)
\( = \frac{6}{{{{(1 – c)}^4}}} \times \frac{1}{{24}} \times {\left( {\frac{1}{2}} \right)^4}\) A1
\( < \frac{6}{{{{\left( {1 – \frac{1}{2}} \right)}^4}}} \times \frac{1}{{24}} \times \frac{1}{{16}}\) A2
giving an upper bound of 0.25. A1
[5 marks]
(e) Actual error \( = \ln 2 – \frac{2}{3} = 0.0265\) A1
The upper bound calculated is much larger that the actual error therefore cannot be considered a good estimate. R1
[2 marks]
Total [17 marks]
Examiners report
In (a), some candidates appeared not to understand the term ‘constant term’. In (b), many candidates found the differentiation beyond them with only a handful realising that the best way to proceed was to rewrite the function as \(f(x) = – \ln (1 – x)\). In (d), many candidates were unable to use the Lagrange formula for the upper bound so that (e) became inaccessible.
Question
(a) Given that \(y = \ln \cos x\) , show that the first two non-zero terms of the Maclaurin series for y are \( – \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}}\).
(b) Use this series to find an approximation in terms of \(\pi {\text{ for }}\ln 2\) .
▶️Answer/Explanation
Markscheme
(a) \(f(x) = \ln \cos x\)
\(f'(x) = \frac{{ – \sin x}}{{\cos x}} = – \tan x\) M1A1
\(f”(x) = – {\sec ^2}x\) M1
\(f”'(x) = – 2\sec x\sec x\tan x\) A1
\({f^{iv}}(x) = – 2{\sec ^2}x({\sec ^2}x) – 2\tan x(2{\sec ^2}x\tan x)\)
\( = – 2{\sec ^4}x – 4{\sec ^2}x{\tan ^2}x\) A1
\(f(x) = f(0) + xf'(0) + \frac{{{x^2}}}{{2!}}f”(0) + \frac{{{x^3}}}{{3!}}f”'(0) + \frac{{{x^4}}}{{4!}}{f^{iv}}(0) + …\)
\(f(0) = 0,\) M1
\(f'(0) = 0,\)
\(f”(0) = – 1,\)
\(f”'(0) = 0,\)
\({f^{iv}}(0) = – 2,\) A1
Notes: Award the A1 if all the substitutions are correct.
Allow FT from their derivatives.
\(\ln (\cos x) \approx – \frac{{{x^2}}}{{2!}} – \frac{{2{x^4}}}{{4!}}\) A1
\( = – \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}}\) AG
[8 marks]
(b) Some consideration of the manipulation of ln 2 (M1)
Attempt to find an angle (M1)
EITHER
Taking \(x = \frac{\pi }{3}\) A1
\(\ln \frac{1}{2} \approx – \frac{{{{\left( {\frac{\pi }{3}} \right)}^2}}}{{2!}} – \frac{{2{{\left( {\frac{\pi }{3}} \right)}^4}}}{{4!}}\) A1
\( – \ln 2 \approx – \frac{{\frac{{{\pi ^2}}}{9}}}{{2!}} – \frac{{2\frac{{{\pi ^4}}}{{81}}}}{{4!}}\) A1
\(\ln 2 \approx \frac{{{\pi ^2}}}{{18}} + \frac{{{\pi ^4}}}{{972}} = \frac{{{\pi ^2}}}{9}\left( {\frac{1}{2} + \frac{{{\pi ^2}}}{{108}}} \right)\) A1
OR
Taking \(x = \frac{\pi }{4}\) A1
\(\ln \frac{1}{{\sqrt 2 }} \approx – \frac{{{{\left( {\frac{\pi }{4}} \right)}^2}}}{{2!}} – \frac{{2{{\left( {\frac{\pi }{4}} \right)}^4}}}{{4!}}\) A1
\( – \frac{1}{2}\ln 2 \approx – \frac{{\frac{{{\pi ^2}}}{{16}}}}{{2!}} – \frac{{2\frac{{{\pi ^4}}}{{256}}}}{{4!}}\) A1
\(\ln 2 \approx \frac{{{\pi ^2}}}{{16}} + \frac{{{\pi ^4}}}{{1536}} = \frac{{{\pi ^2}}}{8}\left( {\frac{1}{2} + \frac{{{\pi ^2}}}{{192}}} \right)\) A1
[6 marks]
Total [14 marks]
Examiners report
Some candidates had difficulty organizing the derivatives but most were successful in getting the series. Using the series to find the approximation for \(\ln 2\) in terms of \(\pi \) was another story and it was rare to see a good solution.
Question
The variables x and y are related by \(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\tan x = \cos x\) .
(a) Find the Maclaurin series for y up to and including the term in \({x^2}\) given that
\(y = – \frac{\pi }{2}\) when x = 0 .
(b) Solve the differential equation given that y = 0 when \(x = \pi \) . Give the solution in the form \(y = f(x)\) .
▶️Answer/Explanation
Markscheme
(a) from \(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\tan x + \cos x\) , \(f'(0) = 1\) A1
now \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = y{\sec ^2}x + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x – \sin x\) M1A1A1A1
Note: Award A1 for each term on RHS.
\( \Rightarrow f”(0) = – \frac{\pi }{2}\) A1
\( \Rightarrow y = – \frac{\pi }{2} + x – \frac{{\pi {x^2}}}{4}\) A1
[7 marks]
(b) recognition of integrating factor (M1)
integrating factor is \({{\text{e}}^{\int { – \tan x{\text{d}}x} }}\)
\( = {{\text{e}}^{\ln \cos x}}\) (A1)
\( = \cos x\) (A1)
\( \Rightarrow y\cos x = \int {{{\cos }^2}x{\text{d}}x} \) M1
\( \Rightarrow y\cos x = \frac{1}{2}\int {(1 + \cos 2x){\text{d}}x} \) A1
\( \Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} + k\) A1
when \(x = \pi ,{\text{ }}y = 0 \Rightarrow k = – \frac{\pi }{2}\) M1A1
\( \Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} – \frac{\pi }{2}\) (A1)
\( \Rightarrow y = \sec x\left( {\frac{x}{2} + \frac{{\sin 2x}}{4} – \frac{\pi }{2}} \right)\) A1
[10 marks]
Total [17 marks]
Examiners report
Part (a) of the question was set up in an unusual way, which caused a problem for a number of candidates as they tried to do part (b) first and then find the Maclaurin series by a standard method. Few were successful as they were usually weaker candidates and made errors in finding the solution \(y = f(x)\) . The majority of candidates knew how to start part (b) and recognised the need to use an integrating factor, but a number failed because they missed out the negative sign on the integrating factor, did not realise that \({{\text{e}}^{\ln \cos x}} = \cos x\) or were unable to integrate \({{{\cos }^2}x}\) . Having said this, a number of candidates succeeded in gaining full marks on this question.
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
a.Given that \(f(x) = \ln x\), use the mean value theorem to show that, for \(0 < a < b\), \(\frac{{b – a}}{b} < \ln \frac{b}{a} < \frac{{b – a}}{a}\).[7]
b.Hence show that \(\ln (1.2)\) lies between \(\frac{1}{m}\) and \(\frac{1}{n}\), where \(m\), \(n\) are consecutive positive integers to be determined.[2]
Answer/Explanation
Markscheme
\(f'(x) = \frac{1}{x}\) (A1)
using the MVT \(f'(c) = \frac{{f(b) – f(a)}}{{b – a}}\) (where \(c\) lies between \(a\) and \(b\)) (M1)
\(f'(c) = \frac{{\ln b – \ln a}}{{b – a}}\) A1
\(\ln \frac{b}{a} = \ln b – \ln a\) (M1)
\(f'(c) = \frac{{\ln \frac{b}{a}}}{{b – a}}\)
since \(f'(x)\) is a decreasing function or \(a < c < b \Rightarrow \frac{1}{b} < \frac{1}{c} < \frac{1}{a}\) R1
\(f'(b) < f'(c) < f'(a)\) (M1)
\(\frac{1}{b} < \frac{{\ln \frac{b}{a}}}{{b – a}} < \frac{1}{a}\) A1
\(\frac{{b – a}}{b} < \ln \frac{b}{a} < \frac{{b – a}}{a}\) AG
[7 marks]
putting \(b = 1.2,{\text{ }}a = 1\), or equivalent M1
\(\frac{1}{6} < \ln 1.2 < \frac{1}{5}\) A1
\((m = 6,{\text{ }}n = 5)\)
[2 marks]
Examiners report
Although many candidates achieved at least a few marks in this question, the answers revealed difficulties in setting up a proof. The Mean value theorem was poorly quoted and steps were often skipped. The conditions under which the Mean value theorem is valid were largely ignored, as were the reasoned steps towards the answer.
There were inequalities everywhere, without a great deal of meaning or showing progress. A number of candidates attempted to work backwards and presented the work in a way that made it difficult to follow their reasoning; in part (b) many candidates ignored the instruction ‘hence’ and just used GDC to find the required values; candidates that did notice the link to part a) answered this question well in general. A number of candidates guessed the answer and did not present an analytical derivation as required.
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 $f$.
b. Hence or otherwise, find $\lim _{x \rightarrow 0} \frac{\arcsin (2 x)-2 x}{(2 x)^3}$.
▶️Answer/Explanation
a. $f(x)=\arcsin (2 x)$
$f^{\prime}(x)=\frac{2}{\sqrt{1-4 x^2}} \quad$ M1A1
Note: Award M1AO for $f^{\prime}(x)=\frac{1}{\sqrt{1-4 x^2}}$
$f^{\prime \prime}(x)=\frac{8 x}{\left(1-4 x^2\right)^{\frac{3}{2}}}$
A1
EITHER
$f^{\prime \prime \prime}(x)=\frac{8\left(1-4 x^2\right)^{\frac{3}{2}}-8 x\left(\frac{3}{2}(-8 x)\left(1-4 x^2\right)^{\frac{1}{2}}\right)}{\left(1-4 x^2\right)^3}\left(=\frac{8\left(1-4 x^2\right)^{\frac{3}{2}}+96 x^2\left(1-4 x^2\right)^{\frac{1}{2}}}{\left(1-4 x^2\right)^3}\right)$
A1
OR
$$
f^{\prime \prime \prime}(x)=8\left(1-4 x^2\right)^{-\frac{3}{2}}+8 x\left(-\frac{3}{2}\left(1-4 x^2\right)^{-\frac{5}{2}}\right)(-8 x)\left(=8\left(1-4 x^2\right)^{-\frac{3}{2}}+96 x^2\left(1-4 x^2\right)^{-\frac{5}{2}}\right)
$$
THEN
substitute $x=0$ into $f$ or any of its derivatives
(M1)
$f(0)=0, f^{\prime}(0)=2$ and $f^{\prime \prime}(0)=0$
A1
$$
f^{\prime \prime \prime}(0)=8
$$
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
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\arcsin (2 x)-2 x}{(2 x)^3}=\lim _{x \rightarrow 0} \frac{2 x+\frac{4 x^3}{3}+\ldots-2 x}{8 x^3} \quad \text { M1 } \\
& =\lim _{x \rightarrow 0} \frac{\frac{4}{3}+\ldots \text { terms with } x}{8} \quad \text { (M1) } \\
& =\frac{1}{6} \quad \text { A1 }
\end{aligned}
$$
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.
$=\frac{1}{6}$
A1
Note: Award $\boldsymbol{F T}$ for use of their derivatives from part (a).
[3 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 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)
$$
\begin{aligned}
& n=46 \quad \text { A2 } \\
& \text { [3 marks] }
\end{aligned}
$$