IB Maths AA HL 5 Calculus

[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}

Covered syllabus points

5 Calculus question 1

Question (a)

Question (b)

Question (c)

5 Calculus question 2

Question (a)

Question (b)

5 Calculus question 3

Question (a)

Question (b)

Question (i)

Question (ii)

5 Calculus question 4

Question (a)

Question (b)

Question (c)

Question (d)