IB Mathematics AA HL 5.1 Calculus Question Bank

Practise IB Mathematics AA HL 5.1 by solving extended differentiation, integration, differential-equation and optimisation problems with exact reasoning.

[Maximum number: 5]

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)

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: 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: 4]

Consider the function $f(x)=a x^{3}+b x^{2}+c x+d$ , where $x \in \mathbb{R}$ and $a, b, c, d \in \mathbb{R}$ .

Write down an expression for $f^{\prime}(x)$ .

$f^{\prime}(x)=3 a x^{2}+2 b x+c \quad \boldsymbol{A} \mathbf{1}$

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IB Mathematics AA HL 5.1 Calculus Question Bank

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5.1 Calculus - SL content question 1

Question (a)

Question (b)

5.1 Calculus - SL content question 2

Question (a)

Question (b)

Question (i)

Question (ii)

5.1 Calculus - SL content question 3