IB Maths AA HL 5 Calculus Questions

Practise IB Maths AA HL calculus through shared-core and HL limits, differentiation, integration, differential equations and proof with exact reasoning.

Exam points

interpret derivatives as gradients and rates of change, including increasing/decreasing behaviour and graph features
differentiate powers and standard functions, then form tangents, normals and related rates where required
integrate functions and use boundary conditions, definite integrals and areas under or between curves
solve optimisation, kinematics and area problems by combining derivative tests, graph interpretation and integrals
apply advanced differentiation, limits, implicit methods and L’Hôpital’s rule to constrained functions
use substitution, parts, partial fractions, differential equations and Maclaurin series in supported models
calculate areas or volumes of revolution and interpret constants, domains and approximation error

Covered syllabus points

5.1 Calculus - SL content 5.2 Calculus - AHL content

[Maximum number: 11]

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

(a)

By considering limits, show that the graph of y=f(x) has a horizontal asymptote and state its equation.

[ 2 ]

as $x \rightarrow \pm \infty, f(x) \rightarrow \arcsin 1\left(\rightarrow \frac{\pi}{2}\right)$
so the horizontal asymptote is $y=\frac{\pi}{2}$

(b)

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

[ 6 ]

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 $\frac{\mathrm{d} y}{\mathrm{~d} u}=\frac{1}{\sqrt{1-u^{2}}}$ $f^{\prime}(x)=\frac{1}{\sqrt{1-\left(\frac{x^{2}-1}{x^{2}+1}\right)^{2}}} \times \frac{4 x}{\left(x^{2}+1\right)^{2}}$ $=\frac{4 x}{\sqrt{\left(x^{2}+1\right)^{2}-\left(x^{2}-1\right)^{2}}} \times \frac{1}{\left(x^{2}+1\right)}$ $=\frac{4 x}{\sqrt{4 x^{2}}} \times \frac{1}{\left(x^{2}+1\right)}$ $=\frac{2 x}{\sqrt{x^{2}}\left(x^{2}+1\right)}$

(c)

By using the expression for $f^{\prime}(x)$ and the result $\sqrt{x^{2}}=|x|$ , show that f is decreasing for x<0.

A function g is defined by $g(x)=\arcsin \left(\frac{x^{2}-1}{x^{2}+1}\right), x \in \mathbb{R}, x \geq 0$ .

[ 3 ]

$f^{\prime}(x)=\frac{2 x}{|x|\left(x^{2}+1\right)}$
EITHER
for x<0,|x|=-x
so $f^{\prime}(x)=-\frac{2}{x^{2}+1}$
OR
|x|>0 and $x^{2}+1>0$
2 x<0, x<0
THEN
$f^{\prime}(x)<0$
Note: Award R1 for stating that in $f^{\prime}(x)$ , the numerator is negative, and the denominator is positive.
so f is decreasing for x<0
Note: Do not accept a graphical solution.

[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: 12]

The functions f and g are defined by

\begin{aligned} & f(x)=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}, x \in \mathbb{R} \\ & g(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}, x \in \mathbb{R} \end{aligned}

(a)

Use the substitution $u=\mathrm{e}^{x}$ to find $\int_{0}^{\ln 3} \frac{1}{4 f(x)-2 g(x)} \mathrm{d} x$ . Give your answer in the form $\frac{\pi \sqrt{a}}{b}$ where $a, b \in \mathbb{Z}^{+}$ .

[ 6 ]

$\quad u=\mathrm{e}^{x} \Rightarrow \mathrm{~d} u=\mathrm{e}^{x} \mathrm{~d} x$

\int \frac{\mathrm{e}^{x}}{\mathrm{e}^{2 x}+3} \mathrm{~d} x=\int \frac{1}{u^{2}+3} \mathrm{~d} u

(when x=0, u=1 and when $x=\ln 3, u=3$ )

\begin{aligned} & \int_{1}^{3} \frac{1}{u^{2}+3} \mathrm{~d} u=\left[\frac{1}{\sqrt{3}} \arctan \left(\frac{u}{\sqrt{3}}\right)\right]_{1}^{3} \\ & \left.=\left[\frac{1}{\sqrt{3}} \arctan \left(\frac{\mathrm{e}^{x}}{\sqrt{3}}\right)\right]_{0}^{\ln 3}\right) \\ & =\frac{\pi \sqrt{3}}{9}-\frac{\pi \sqrt{3}}{18} \\ & =\frac{\pi \sqrt{3}}{18} \end{aligned}

(b)

Let $t(x)=\frac{g(x)}{f(x)}$ .

[ 6 ]

(i)

Show that $t^{\prime}(x)=\frac{[f(x)]^{2}-[g(x)]^{2}}{[f(x)]^{2}}$ for $x \in \mathbb{R}$ .

[ 3 ]

METHOD 1

\begin{aligned} & t(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{\mathrm{e}^{x}+\mathrm{e}^{-x}} \\ & t^{\prime}(x)=\frac{\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)^{2}-\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)^{2}}{\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)^{2}} \\ & t^{\prime}(x)=\frac{\left(\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}\right)^{2}-\left(\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}\right)^{2}}{\left(\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}\right)^{2}} \\ & =\frac{[f(x)]^{2}-[g(x)]^{2}}{[f(x)]^{2}} \end{aligned}

METHOD 2

\begin{aligned} & t^{\prime}(x)=\frac{f(x) g^{\prime}(x)-g(x) f^{\prime}(x)}{[f(x)]^{2}} \\ & g^{\prime}(x)=f(x) \text { and } f^{\prime}(x)=g(x) \\ & =\frac{[f(x)]^{2}-[g(x)]^{2}}{[f(x)]^{2}} \end{aligned}

METHOD 3
$t(x)=\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)^{-1}$ $t^{\prime}(x)=1-\frac{\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)^{2}}{\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)^{2}}$ $=1-\frac{[g(x)]^{2}}{[f(x)]^{2}}$ $=\frac{[f(x)]^{2}-[g(x)]^{2}}{[f(x)]^{2}}$
METHOD 4
$t^{\prime}(x)=\frac{g^{\prime}(x)}{f(x)}-\frac{g(x) f^{\prime}(x)}{[f(x)]^{2}}$ $g^{\prime}(x)=f(x)$ and $f^{\prime}(x)=g(x)$ gives $t^{\prime}(x)=1-\frac{[g(x)]^{2}}{[f(x)]^{2}}$ $=\frac{[f(x)]^{2}-[g(x)]^{2}}{[f(x)]^{2}}$

(ii)

Hence show that $t^{\prime}(x)>0$ for $x \in \mathbb{R}$ .

[ 3 ]

METHOD 1
$[f(x)]^{2}>[g(x)]^{2}$ (or equivalent)
$[f(x)]^{2}>0$
hence $t^{\prime}(x)>0, x \in \mathbb{R}$
Note: Award as above for use of either $f(x)=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}$ and $g(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}$ or $\mathrm{e}^{x}+\mathrm{e}^{-x}$ and $\mathrm{e}^{x}-\mathrm{e}^{-x}$ .
METHOD 2

\begin{array}{lr} {[f(x)]^{2}-[g(x)]^{2}=1 \text { (or equivalent) }} & \text { M1A1 } \\ {[f(x)]^{2}>0} & \text { R1 } \\ \text { hence } t^{\prime}(x)>0, x \in \mathbb{R} & \text { AG } \end{array}

Note: Award as above for use of either $f(x)=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}$ and $g(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}$ or $\mathrm{e}^{x}+\mathrm{e}^{-x}$ and $\mathrm{e}^{x}-\mathrm{e}^{-x}$ .

METHOD 3

[Maximum number: 9]

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 $x f^{\prime}(x)=x+2 x^{2}+3 x^{3}+\ldots+n x^{n}$ .

Let $f_{1}(x)=x f^{\prime}(x)$ and consider the following family of functions:

\begin{gathered} f_{2}(x)=x f_{1}^{\prime}(x) \\ f_{3}(x)=x f_{2}^{\prime}(x) \\ f_{4}(x)=x f_{3}^{\prime}(x) \\ \ldots \\ f_{q}(x)=x f_{q-1}^{\prime}(x) \end{gathered}

[ 1 ]

$f^{\prime}(x)=1+2 x+3 x^{2}+\ldots+n x^{n-1}$

Note: Award A1 for $f^{\prime}(x)=\sum_{i=1}^{n} i x^{i-1}$ .

\Rightarrow x f^{\prime}(x)=x+2 x^{2}+3 x^{3}+\ldots+n x^{n}

(b)

Show that $f_{2}(x)=\sum_{i=1}^{n} i^{2} x^{i}$ .

[ 2 ]

METHOD 1

\begin{aligned} & f_{2}(x)=x f_{1}^{\prime}(x) \\ & f_{1}^{\prime}(x)=1^{2}+2^{2} x+\left(3^{2} x^{2}\right)+\ldots+n^{2} x^{n-1}\left(=1+4 x+\left(9 x^{2}\right)+\ldots+n^{2} x^{n-1}\right) \end{aligned}

Note: Award A1 for

x f_{1}^{\prime}(x)=x\left(1^{2}+2^{2} x+\left(3^{2} x^{2}\right)+\ldots+n^{2} x^{n-1}\right)\left(=x\left(1+4 x+\left(9 x^{2}\right)+\ldots+n^{2} x^{n-1}\right)\right)

x f_{1}^{\prime}(x)=1^{2} x+2^{2} x^{2}+\left(3^{2} x^{3}\right)+\ldots+n^{2} x^{n}\left(=x+4 x^{2}+\left(9 x^{3}\right)+\ldots+n^{2} x^{n}\right)

Note: Award A1 for $f_{1}^{\prime}(x)=\sum_{i=1}^{n} i^{2} x^{i-1}$ and A1 for $x f_{1}^{\prime}(x)=x \sum_{i=1}^{n} i^{2} x^{i-1}$ .
The second A1 is dependent on the first A1.
Award a maximum of A0A1 if a general term is not considered.

=\sum_{i=1}^{n} i^{2} x^{i}

METHOD 2

\begin{aligned} & f_{2}(x)=x \frac{\mathrm{~d}}{\mathrm{~d} x}\left(x f^{\prime}(x)\right) \\ & =x\left(f^{\prime}(x)+x f^{\prime \prime}(x)\right)\left(=x f^{\prime}(x)+x^{2} f^{\prime \prime}(x)\right) \\ & =x \sum_{i=1}^{n} i x^{i-1}+x^{2} \sum_{i=1}^{n} i(i-1) x^{i-2} \\ & =\sum_{i=1}^{n} i x^{i}+\sum_{i=1}^{n} i(i-1) x^{i}\left(=\sum_{i=1}^{n}\left(i+i^{2}-i\right) x^{i}\right) \\ & =\sum_{i=1}^{n} i^{2} x^{i} \end{aligned}

(c)

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.

(d)

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}

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (i)

Question (ii)

Question 4

Question (a)

Question (b)

Question (c)

Question (d)