IB Mathematics AA HL 5.1 Calculus Question Bank

[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: 1]

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 $f^{\prime}(x)$ .

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

[Maximum number: 2]

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

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

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

[Maximum number: 3]

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}

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

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

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: 6]

Consider the function f defined by $f(x)=90 \mathrm{e}^{-0.5 x}$ for $x \in \mathbb{R}^{+}$ .
The graph of f and the line y=x intersect at point P .

(a) attempt to find the point of intersection of the graph of f and the line y=x (M1)
$x=5.56619 \ldots$
=5.57

(a)

Find the exact coordinates of Q.

[ 4 ]

$\quad f^{\prime}(x)=-45 \mathrm{e}^{-0.5 x}$
attempt to set the gradient of f equal to -1
$-45 \mathrm{e}^{-0.5 x}=-1$
Q has coordinates $(2 \ln 45,2)$ (accept $\left(-2 \ln \frac{1}{45}, 2\right)$ )

Note: Award A1 for each value, even if the answer is not given as a coordinate pair.

Do not accept $\frac{\ln \frac{1}{45}}{-0.5}$ or $\frac{\ln 45}{0.5}$ as a final value for x. Do not accept 2.0 or 2.00 as a final value for y.

(b)

Show that the equation of L is $y=-x+2 \ln 45+2$ .

The shaded region A is enclosed by the graph of f and the lines y=x and L.

Figure for Question (b) — IB Maths AA HL

[ 2 ]

attempt to substitute coordinates of Q ( in any order)
into an appropriate equation
$y-2=-(x-2 \ln 45)$ OR $2=-2 \ln 45+c$
equation of L is $y=-x+2 \ln 45+2$

[Maximum number: 3]

Consider the differential equation $y \frac{\mathrm{~d} y}{\mathrm{~d} x}=\cos 2 x$ .

For which value of the constant of integration does your solution coincide with the function given in part (i)?

$\quad \sin 2 x+A \equiv(\cos x+\sin x)^{2}$ $(\cos x+\sin x)^{2}=\cos ^{2} x+2 \sin x \cos x+\sin ^{2} x$
use of $\sin 2 x \equiv 2 \sin x \cos x$ .
A=1

[Maximum number: 5]

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)

Given $f(x)=\frac{x\left(x^{2}-A\right)}{x^{2}+A}$ , prove that $f^{\prime}(\sqrt{A})$ is independent of A.

[ 4 ]

attempt to use quotient rule or product rule
$f^{\prime}(x)=\frac{\left(x^{2}+A\right)\left(3 x^{2}-A\right)-2 x\left(x^{3}-A x\right)}{\left(x^{2}+A\right)^{2}} \quad\left(f^{\prime}(x)=\frac{2 x^{2}+\left(x^{2}-A\right)}{x^{2}+A}-\frac{2 x^{2}\left(x^{2}-A\right)}{\left(x^{2}+A\right)^{2}}\right)$
(or equivalent)

Note: Award A1 for numerator, A1 for denominator.
Note: If product rule used, award A1 for each correct term.
Note: Award (M1) for an attempt to use quotient rule or product rule with A=16 or any other constant. Award no further marks.

f^{\prime}(\sqrt{A})\left(=\frac{(2 A)(2 A)-0}{(2 A)^{2}}\right)=1

hence independent of A

(b)

Write down the coordinates of a point on the curve where the oblique asymptote is parallel to the tangent to the curve at that point.

Now consider the differential equation $x^{2} \frac{\mathrm{~d} y}{\mathrm{~d} x}=x(x+y)-y^{2}$ , where $x \neq 0, y \neq \pm x$ .
Using the substitution y=v x, the differential equation can be written as $x \frac{\mathrm{~d} v}{\mathrm{~d} x}=1-v^{2}$ .

[ 1 ]

$(\sqrt{A}, 0) \mathrm{OR}(-\sqrt{A}, 0)$

[Maximum number: 3]

This question investigates a ratio of lengths found from the line passing through the points of inflexion of a quartic curve of the form $y=x^{4}-m x^{3}+n x$ .
The curve $y=x^{4}-3 x^{3}+3 x$ has points of inflexion at B and C . The line passing through B and C intersects the curve again at points A and D . This is shown in the following graph.

Find $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}$ .

$\frac{\mathrm{d} y}{\mathrm{~d} x}=4 x^{3}-9 x^{2}+3$
(M1)(A1)

Note: Award M1 for at least two correct terms.

\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=12 x^{2}-18 x

[Maximum number: 2]

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 $y=\frac{x\left(x^{2}-16\right)}{x^{2}+16}$ .

State the coordinates of the local maximum point and the coordinates of the local minimum point.

Max $(-1.94,1.20)(-1.94347 \ldots, 1.20113 \ldots)$

Min $(1.94,-1.20)(1.94347 \ldots,-1.20113 \ldots)$

[Maximum number: 12]

A particle P moves along the x-axis. The velocity of P is $v \mathrm{~m} \mathrm{~s}^{-1}$ at time t seconds, where $v(t)=4+4 t-3 t^{2}$ for $0 \leq t \leq 3$ . When t=0, P is at the origin O .

(a)

Find the value of t when P reaches its maximum velocity.

[ 2 ]

valid approach to find turning point $\left(v^{\prime}=0,-\frac{b}{2 a}\right.$ , average of roots)
4-6 t=0 OR $-\frac{4}{2(-3)}$ OR $\frac{-\frac{2}{3}+2}{2}$ $t=\frac{2}{3}(\mathrm{~s})$

(b)

Show that the distance of P from O at this time is $\frac{88}{27}$ metres.

[ 5 ]

attempt to integrate v
$\int v \mathrm{~d} t=\int\left(4+4 t-3 t^{2}\right) \mathrm{d} t=4 t+2 t^{2}-t^{3}(+c)$
Note: Award A1 for $4 t+2 t^{2}, \boldsymbol{A} \mathbf{1}$ for $-t^{3}$ .
attempt to substitute their t into their solution for the integral
distance $=4\left(\frac{2}{3}\right)+2\left(\frac{2}{3}\right)^{2}-\left(\frac{2}{3}\right)^{3}$ $=\frac{8}{3}+\frac{8}{9}-\frac{8}{27}$ (or equivalent)
$=\frac{88}{27}(\mathrm{~m})$

(c)

Find the total distance travelled by P.

[ 5 ]

recognising to integrate between 0 and 2, or 2 and 3 OR $\int_{0}^{3}\left|4+4 t-3 t^{2}\right| \mathrm{d} t$ $\int_{0}^{2}\left(4+4 t-3 t^{2}\right) d t$
=8
$\int_{2}^{3}\left(4+4 t-3 t^{2}\right) d t$
=-5
valid approach to sum the two areas (seen anywhere)
$\int_{0}^{2} v \mathrm{~d} t-\int_{2}^{3} v \mathrm{~d} t \quad$ OR $\int_{0}^{2} v \mathrm{~d} t+\left|\int_{2}^{3} v \mathrm{~d} t\right|$
total distance travelled $=13(\mathrm{~m})$

[Maximum number: 6]

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}

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 $a, b \in \mathbb{Z}^{+}$ .

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

\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}

[Maximum number: 3]

Consider the differential equation $y \frac{\mathrm{~d} y}{\mathrm{~d} x}=\cos 2 x$ .

A different solution of the differential equation, satisfying y=2 when $x=\frac{\pi}{4}$ ,
defines a curve C.

Find the area of R.

$\quad \int_{0}^{\frac{\pi}{2}}(\sin 2 x+3)^{\frac{1}{2}} \mathrm{~d} x$
=2.99

IB Maths AA HL 5.1 Calculus Sl Content Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question 3

Question 4

Question 5

Question 6

Question (a)

Question (b)

Question 7

Question 8

Question (a)

Question (b)

Question 9

Question 10

Question 11

Question (a)

Question (b)

Question (c)

Question 12

Question 13