IB Mathematics AA HL 5.2 Calculus Question Bank

[Maximum number: 6]

The general term of a sequence $\left\{a_{n}\right\}$ is given by the formula $a_{n}=\frac{\mathrm{e}^{n}+2^{n}}{2 \mathrm{e}^{n}}, n \in \mathbb{Z}^{+}$ .

(a)

Show that the sequence $\left\{a_{n}\right\}$ is convergent and find the limit L.

[ 2 ]

$\quad L=\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \frac{1}{2}+\frac{1}{2}\left(\frac{2}{\mathrm{e}}\right)^{n}=\frac{1}{2}+\frac{1}{2} \times 0=\frac{1}{2}$

(b)

Find the smallest value of $N \in \mathbb{Z}^{+}$ such that $\left|a_{n}-L\right|<0.001$ , for all $n \geq N$ .

[ 4 ]

$\quad\left|a_{n}-\frac{1}{2}\right|=\left|\frac{1}{2}+\frac{1}{2}\left(\frac{2}{\mathrm{e}}\right)^{n}-\frac{1}{2}\right|=\left|\frac{1}{2}\left(\frac{2}{\mathrm{e}}\right)^{n}\right|<\frac{1}{1000}$

EITHER
$\Rightarrow\left(\frac{\mathrm{e}}{2}\right)^{n}>500$ $\Rightarrow n>20.25 \ldots$

OR
$\Rightarrow\left(\frac{2}{\mathrm{e}}\right)^{n}<500$ $\Rightarrow n>20.25 \ldots$

Note: A1 for correct inequality;
A1 for correct value.

THEN
therefore N=21

[Maximum number: 6]

In this question you will investigate series of the form

\sum_{i=1}^{n} \boldsymbol{i}^{q}=1^{q}+2^{q}+3^{q}+\ldots+n^{q} \text { where } n, q \in \mathbb{Z}^{+}

and use various methods to find polynomials, in terms of n, for such series.
When q=1, the above series is arithmetic.

(a)

Show that $\lim _{x \rightarrow 1} f_{1}(x)$ is in indeterminate form.

[ 1 ]

$\quad \lim _{x \rightarrow 1} f_{1}(x)=\frac{n-(n+1)+1}{0}\left(=\frac{0}{0}\right)$

Note: Only award R1 for sufficient simplification of the numerator, for example, as shown above.

Do not award R1 if $\lim _{x \rightarrow 1}$ is not referred to or stated.

(b)

Hence, by applying l'Hôpital's rule, show that $\lim _{x \rightarrow 1} f_{1}(x)=\frac{1}{2} n(n+1)$ .

[ 5 ]

attempts to differentiate both the numerator and the denominator M1

\lim _{x \rightarrow 1} \frac{n(n+2) x^{n+1}-(n+1)^{2} x^{n}+1}{2(x-1)}

Note: Award A1 for $\left(\left(\lim _{x \rightarrow 1}\right) \frac{n(n+2) x^{n+1}-n(n+1) x^{n}-(n+1) x^{n}+1}{2(x-1)}\right)$ . This form can be used in subsequent work.
(l'Hôpital's rule applies again since)

\lim _{x \rightarrow 1} \frac{n(n+2) x^{n+1}-(n+1)^{2} x^{n}+1}{2(x-1)}=\frac{0}{0}

Note: Do not award R1 if $\lim _{x \rightarrow 1}$ is not referred to or stated.
Subsequent marks are independent of this R mark.
attempts to differentiate both the numerator and the denominator

\begin{aligned} & \lim _{x \rightarrow 1} \frac{n(n+2)(n+1) x^{n}-n(n+1)^{2} x^{n-1}}{2} \\ & =\frac{n(n+2)(n+1)-n(n+1)^{2}}{2}\left(=\frac{n^{3}+3 n^{2}+2 n-\left(n^{3}+2 n^{2}+n\right)}{2}\right) \\ & =\frac{n(n+1)((n+2)-(n+1))}{2}\left(=\frac{n^{2}+n}{2}\right) \\ & =\frac{1}{2} n(n+1) \end{aligned}

[Maximum number: 6]

Two boats A and B travel due north.
Initially, boat B is positioned 50 metres due east of boat A .
The distances travelled by boat A and boat B , after t seconds, are x metres and y metres respectively. The angle $\theta$ is the radian measure of the bearing of boat B from boat A . This information is shown on the following diagram.

Find the speed of boat A at time T.

attempt to differentiate with respect to t
$\frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{\mathrm{d} x}{\mathrm{~d} t}-50(\operatorname{cosec} \theta)^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} t}$
attempt to set speed of B equal to double the speed of A
$2 \frac{\mathrm{~d} x}{\mathrm{~d} t}=\frac{\mathrm{d} x}{\mathrm{~d} t}-50(\operatorname{cosec} \theta)^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} t}$ $\frac{\mathrm{d} x}{\mathrm{~d} t}=-50(\operatorname{cosec} \theta)^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} t}$ $\theta=\arctan 5(=1.373 \ldots=78.69 \ldots)$ OR $\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta=1+\left(\frac{1}{5}\right)^{2}=\frac{26}{25}$

Note: This A1 can be awarded independently of previous marks.
$\frac{\mathrm{d} x}{\mathrm{~d} t}=-50\left(\frac{26}{25}\right) \times-0.1$
So the speed of boat A is $5.2\left(\mathrm{~ms}^{-1}\right)$

Note: Accept 5.20 from the use of inexact values.

[Maximum number: 6]

A function f is defined by $f(x)=\arcsin \left(\frac{x^{2}-1}{x^{2}+1}\right), x \in \mathbb{R}$ .

Show that $f^{\prime}(x)=\frac{2 x}{\sqrt{x^{2}}\left(x^{2}+1\right)}$ for $x \in \mathbb{R}, x \neq 0$ .

attempting to use the quotient rule to find $\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{x^{2}-1}{x^{2}+1}\right)$ $\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x^{2}-1}{x^{2}+1}\right)=\frac{2 x\left(x^{2}+1\right)-2 x\left(x^{2}-1\right)}{\left(x^{2}+1\right)^{2}}\left(=\frac{4 x}{\left(x^{2}+1\right)^{2}}\right)$
attempting to use the chain rule to find $\frac{\mathrm{d}}{\mathrm{dx}}\left(\arcsin \left(\frac{x^{2}-1}{x^{2}+1}\right)\right)$
let $u=\frac{x^{2}-1}{x^{2}+1}$ and so $y=\arcsin u$ and $\frac{\mathrm{d} y}{\mathrm{~d} u}=\frac{1}{\sqrt{1-u^{2}}}$ $f^{\prime}(x)=\frac{1}{\sqrt{1-\left(\frac{x^{2}-1}{x^{2}+1}\right)^{2}}} \times \frac{4 x}{\left(x^{2}+1\right)^{2}}$ $=\frac{4 x}{\sqrt{\left(x^{2}+1\right)^{2}-\left(x^{2}-1\right)^{2}}} \times \frac{1}{\left(x^{2}+1\right)}$ $=\frac{4 x}{\sqrt{4 x^{2}}} \times \frac{1}{\left(x^{2}+1\right)}$ $=\frac{2 x}{\sqrt{x^{2}}\left(x^{2}+1\right)}$

[Maximum number: 4]

The following question uses Maclaurin series to investigate approximations of mathematical constants and the accuracies of such approximations.

By using integration by parts, show that $\int_{0}^{\frac{1}{\sqrt{3}}} \arctan x \mathrm{~d} x=\frac{\pi}{6 \sqrt{3}}-\frac{1}{2} \ln \frac{4}{3}$ .

$\quad u=\arctan x$ and $\frac{\mathrm{d} v}{\mathrm{~d} x}=1$ (or equivalent)

\int_{0}^{\frac{1}{\sqrt{3}}} \arctan x \mathrm{~d} x=[x \arctan x]_{0}^{\frac{1}{\sqrt{3}}}-\int_{0}^{\frac{1}{\sqrt{3}}} \frac{x}{1+x^{2}} \mathrm{~d} x

Note: Award A1 for each term.

\begin{aligned} & =[x \arctan x]_{0}^{\frac{1}{\sqrt{3}}}-\frac{1}{2}\left[\ln \left(1+x^{2}\right)\right]_{0}^{\frac{1}{\sqrt{3}}}\left(=\frac{1}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}}-\frac{1}{2} \ln \frac{4}{3}\right) \\ & \quad=\frac{\pi}{6 \sqrt{3}}-\frac{1}{2} \ln \frac{4}{3} \end{aligned}

[Maximum number: 6]

Consider the function defined by $f(x)=\frac{k x-5}{x-k}$ , where $x \in \mathbb{R} \backslash\{k\}$ and $k^{2} \neq 5$ .

The region bounded by the x-axis, the curve y=f(x), and the lines x=5 and x=7 is rotated through $2 \pi$ about the x-axis. Find the volume of the solid generated, giving your answer in the form $\pi(a+b \ln 2)$ , where $a, b \in \mathbb{Z}$ .

METHOD 1
volume $=\pi \int_{5}^{7}\left(\frac{3 x-5}{x-3}\right)^{2} d x$

EITHER
attempt to express $\frac{3 x-5}{x-3}$ in the form $p+\frac{q}{x-3}$ $\frac{3 x-5}{x-3}=3+\frac{4}{x-3}$

OR
attempt to expand $\left(\frac{3 x-5}{x-3}\right)^{2}$ or $(3 x-5)^{2}$ and divide out
$\left(\frac{3 x-5}{x-3}\right)^{2}=9+\frac{24 x-56}{(x-3)^{2}}$

THEN

\begin{array}{ll} \left(\frac{3 x-5}{x-3}\right)^{2}=9+\frac{24}{x-3}+\frac{16}{(x-3)^{2}} & \boldsymbol{A} \mathbf{1} \\ \text { volume }=\pi \int_{5}^{7}\left(9+\frac{24}{x-3}+\frac{16}{(x-3)^{2}}\right) \mathrm{d} x \\ =\pi\left[9 x+24 \ln (x-3)-\frac{16}{x-3}\right]_{5}^{7} \\ =\pi[(63+24 \ln 4-4)-(45+24 \ln 2-8)] & \boldsymbol{A} \mathbf{1} \\ =\pi(22+24 \ln 2) & \boldsymbol{A} \mathbf{1} \end{array}

METHOD 2

\begin{aligned} & \text { volume }=\pi \int_{5}^{7}\left(\frac{3 x-5}{x-3}\right)^{2} \mathrm{~d} x \\ & \text { substituting } u=x-3 \Rightarrow \frac{\mathrm{~d} u}{\mathrm{~d} x}=1 \\ & \begin{aligned} 3 x-5 & =3(u+3)-5=3 u+4 \\ \text { volume } & =\pi \int_{2}^{4}\left(\frac{3 u+4}{u}\right)^{2} \mathrm{~d} u \\ = & \pi \int_{2}^{4} 9+\frac{16}{u^{2}}+\frac{24}{u} \mathrm{~d} u \\ = & \pi\left[9 u-\frac{16}{u}+24 \ln u\right]_{2}^{4} \end{aligned} \end{aligned}

Note: Ignore absence of or incorrect limits seen up to this point.

[Maximum number: 9]

The following question explores features of a family of curves. The family is then linked to a homogeneous differential equation.
Consider the curve given by $y=\frac{x\left(x^{2}-16\right)}{x^{2}+16}$ .

(a)

Hence, show that a solution to the original differential equation may be expressed in the form $x^{2}=\left|\frac{A(x+y)}{x-y}\right|$ , where A is a positive constant.

Now consider only the case where $\frac{A(x+y)}{x-y}>0$ .

[ 5 ]

attempt to separate variables and integrate

\begin{aligned} & \int \frac{\mathrm{d} x}{x}=\int \frac{\mathrm{d} v}{1-v^{2}} \\ & \ln |x|=\frac{1}{2} \ln \left|\frac{A(1+v)}{1-v}\right| \end{aligned}

attempt to use power rule of logs

\begin{aligned} & \ln |x|=\ln \left|\frac{A(1+v)}{1-v}\right|^{\frac{1}{2}} \text { OR } \ln |x|^{2}=\ln \left|\frac{A(1+v)}{1-v}\right| \\ & |x|=\left|\frac{A(1+v)}{1-v}\right|^{\frac{1}{2}} \text { OR } x^{2}=\left|\frac{A(1+v)}{1-v}\right| \\ & x^{2}=\left|\frac{A\left(1+\frac{y}{x}\right)}{1-\frac{y}{x}}\right| \\ & x^{2}=\left|\frac{A(x+y)}{x-y}\right| \end{aligned}

Note: Award at most (M1)A1(M1)A1A0 for incorrect use or omission of moduli.
Note: The substitution of $\frac{y}{x}$ may be seen earlier.

(b)

Show that a solution to the original differential equation is $y=\frac{x\left(x^{2}-A\right)}{x^{2}+A}$ .

[ 4 ]

$x^{2}=\frac{A(x+y)}{x-y}$
attempt to clear denominator
$x^{2}(x-y)=A(x+y)$ $x^{3}-x^{2} y=A x+A y$
attempt to isolate y
$x^{3}-A x=x^{2} y+A y$ $x\left(x^{2}-A\right)=y\left(x^{2}+A\right)\left(x^{3}-A x=y\left(x^{2}+A\right)\right)$ $y=\frac{x\left(x^{2}-A\right)}{x^{2}+A}$

Note: The y must be factored out for the final A1.

[Maximum number: 11]

The following question uses Maclaurin series to investigate approximations of mathematical constants and the accuracies of such approximations.

(a)

Using $x=\frac{1}{\sqrt{3}}$ and the first three (non-zero) terms of the Maclaurin series of $\arctan x$ , find an approximation for $\pi$ to three decimal places.

The Maclaurin series of $\arctan x$ is an example of an alternating series, ie a series where consecutive terms are positive and negative. Consider the following theorem.

Theorem: For alternating series with terms of decreasing magnitude, the error obtained in using a finite number of terms is less than or equal to the absolute value of the next term in the sequence.

Using the theorem, the maximum error in using the first three (non-zero) terms as an approximation to $\arctan x$ is given by $\left|-\frac{x^{7}}{7}\right|$ . In other words, $\left|\arctan x-\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{5}\right)\right| \leq\left|-\frac{x^{7}}{7}\right|$ .

[ 3 ]

$\quad \arctan \frac{1}{\sqrt{3}}=\frac{\pi}{6}\left(\approx \frac{1}{\sqrt{3}}-\frac{1}{3}\left(\frac{1}{\sqrt{3}}\right)^{3}+\frac{1}{5}\left(\frac{1}{\sqrt{3}}\right)^{5}\right)$
attempt to evaluate $\frac{1}{\sqrt{3}}-\frac{1}{3}\left(\frac{1}{\sqrt{3}}\right)^{3}+\frac{1}{5}\left(\frac{1}{\sqrt{3}}\right)^{5}$ $\pi \approx 3.156$

(b)

Determine how many (non-zero) terms of the series would need to be used, such that the error in approximating $\arctan \left(\frac{1}{\sqrt{3}}\right)$ is less than 0.0001 .

[ 3 ]

METHOD 1
recognition that in the Maclaurin expansion of $\arctan x$ , the error term is $\frac{x^{2 m-1}}{2 m-1}$
attempt to solve $\frac{1}{2 m-1} \times\left(\frac{1}{\sqrt{3}}\right)^{2 m-1}<0.0001$ OR $m=6.60583 \ldots(=7)$

6 (non-zero) terms are needed

Note: Accept the error term in an absolute value.
Give marks as above for candidates who use 2 m+1 instead of 2 m-1.
This gives $m=5.60583 \ldots(=6)$ and leads to needing 6 terms.
Give marks as above for candidates who use n as the exponent and divisor. This gives $n=12.211675 \ldots$ and leads to needing 6 terms.

METHOD 2
recognition that in the Maclaurin expansion of $\arctan x$ , the error term is $\frac{x^{2 m-1}}{2 m-1}$
attempt to find the value of the error term for at least two consecutive values of m
error terms are
$\frac{\left(\frac{1}{\sqrt{3}}\right)^{7}}{7}=0.003054 \ldots, \frac{\left(\frac{1}{\sqrt{3}}\right)^{9}}{9}=0.0007919 \ldots, \frac{\left(\frac{1}{\sqrt{3}}\right)^{11}}{11}=0.0002159 \ldots, \frac{\left(\frac{1}{\sqrt{3}}\right)^{13}}{13}=0.00006092 \ldots$

6 (non-zero) terms are needed

Note: Accept the error term in an absolute value.

Accept 2 m+1 or n in place of 2 m-1.

METHOD 3
attempt to find the actual error using $\arctan x-$ (first m non-zero terms)
attempt to find the error for at least two consecutive values of m
$(m=4 \Rightarrow 0.000623, m=5 \Rightarrow 0.000169, m=6 \Rightarrow 0.0000473)$

6 (non-zero) terms are needed

(c)

Determine the smallest number of (non-zero) terms of the Maclaurin series for $\arctan x$ that should be used.

[ 5 ]

METHOD 1
recognition error is of the form $\int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{n}}{n}\right) \mathrm{d} x$

Note: Condone absence of limits of integration for (M1)

\frac{\left(\frac{1}{\sqrt{3}}\right)^{n+1}}{n(n+1)}<1 \times 10^{-6}

attempt to solve using GDC
$n=14.3336 \ldots$ $\Rightarrow n=15$
(hence $8^{\text {th }}$ term is the error term so) we require 7 (non-zero) terms

METHOD 2
recognition error is of the form $\int_{0}^{\frac{1}{\sqrt{3}}} \frac{x^{2 m-1}}{2 m-1} d x \operatorname{OR} \int_{0}^{\frac{1}{\sqrt{3}}} \frac{x^{2 m+1}}{2 m+1} d x$

Note: Condone absence of limits of integration for (M1)

\frac{\left(\frac{1}{\sqrt{3}}\right)^{2 m}}{2 m(2 m-1)}<1 \times 10^{-6} \text { OR } \frac{\left(\frac{1}{\sqrt{3}}\right)^{2 m+2}}{(2 m+1)(2 m+2)}<1 \times 10^{-6}

attempt to solve using GDC

\Rightarrow m=7.66681 \ldots \text { OR } \Rightarrow m=6.66681 \ldots

(hence $8^{\text {th }}$ term is the error term so) we require 7 (non-zero) terms

METHOD 3
recognition error is of the form $\int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{n}}{n}\right) \mathrm{d} x$

Note: Condone absence of limits of integration for (M1)

\int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{13}}{13}\right) \mathrm{d} x=2.51234 \ldots \times 10^{-6} \text { OR } \int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{15}}{15}\right) \mathrm{d} x=6.35065 \ldots \times 10^{-7}

attempt to find the value of the integral for at least two consecutive odd values of n, n>9
$\int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{13}}{13}\right) \mathrm{d} x=2.51234 \ldots \times 10^{-6}, \int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{15}}{15}\right) \mathrm{d} x=6.35065 \ldots \times 10^{-7}$

Note: Award this A1 for the second correct value.
(hence $8^{\text {th }}$ term is the error term so) we require 7 (non-zero) terms

METHOD 4
attempt to find the actual error using $\int_{0}^{\frac{1}{\sqrt{3}}} \arctan x d x-$ (first m non-zero terms)
error with 6 terms is $2.00830 \ldots \times 10^{-6} \mathbf{OR}$ error with 7 terms is $5.04041 \ldots \times 10^{-7}$
attempt to find the error for at least two consecutive values of m
error with 6 terms is $2.00830 \ldots \times 10^{-6}$ and error with 7 terms is $5.04041 \ldots \times 10^{-7}$

Note: Award this A1 for the second correct value.
we require 7 (non-zero) terms

IB Maths AA HL 5.2 Calculus Ahl Content Questions

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