IB Maths AA SL 1 Number and Algebra Question Bank

Practise number and algebra across sequences, logarithms, financial models, proof, complex numbers, counting methods and exact symbolic manipulation at SL and HL.

Covered syllabus points

1.1 Number and algebra - SL content

[Maximum number: 4]

The diameter of a spherical planet is $6 \times 10^{4} \mathrm{~km}$ .

(a)

Write down the radius of the planet.

The volume of the planet can be expressed in the form $\pi\left(a \times 10^{k}\right) \mathrm{km}^{3}$ where $1 \leq a<10$ and $k \in \mathbb{Z}$ .

[ 1 ]

$3\times10^4$
or $30000\,\mathrm{km}$ . Accept $3\bullet10^4$ .

(b)

Find the value of a and the value of k.

[ 3 ]

Use the volume of a sphere with radius $3\times10^4$ :
$\frac43\pi(3\times10^4)^3=36\pi\times10^{12}=\pi(3.6\times10^{13}).$ $a=3.6,\qquad k=13$
A1, A1, A1

[Maximum number: 9]

Give your answers to parts (a)(ii), (c)(i) and (d) correct to two decimal places.
Daniela and Sorin have each recently received some money. Daniela won a cash prize and Sorin received an inheritance.
Daniela had two options to choose from to receive her winnings. In both options she receives a payment on the first day of each month for three years.
Option A Each payment is $ 5500.
Option B The first payment is $ 2000. In each month which follows, the payment is 6 % more than the previous month.

(a)

Find the total amount Daniela would receive if she chooses

[ 5 ]

(i)

Option A;

$5500 \times 36$
=($) 198000

(ii)

Option B.

Sorin received an inheritance of $ 120000. Sorin invested his inheritance in an account that pays a nominal annual interest rate of 4 % per annum, compounded monthly. The interest is added on the last day of each month.

[ 5 ]

recognizing sum of a geometric sequence is required
$\frac{2000\left(1-1.06^{36}\right)}{1-1.06}$ $=238241.7333 \ldots$
=($) 238241.73

(b)

Write down an expression for the value of Sorin's investment after n years.

Daniela chose Option B and received her first payment on $1^{\text {st }}$ January 2023. Sorin invested his inheritance on the same day.

[ 1 ]

Sorin's future value after n years $=120000\left(1+\frac{4}{100 \times 12}\right)^{12 n}$

(c)

Find the value of r if this investment grows to $ 53000 after six years.

[ 3 ]

EITHER
N=24
OR
N=6
$\mathrm{PV}=\mp 40000$ $\mathrm{PV}=\mp 40000$
PMT=0
PMT=0
$\mathrm{FV}= \pm 53000$ $\mathrm{FV}= \pm 53000$
P / Y=4
P / Y=1
C / Y=4
C / Y=4
(M1)(A1)
Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen, and award (A1) for all entries correct. PV and FV must have opposite signs.

OR
$40000\left(1+\frac{r}{100 \times 4}\right)^{6 \times 4}=53000$
(M1)(A1)

Note: Award (M1) for attempting to substitute into compound interest formula, award (A1) for correct equation.

THEN
$4.71781 \ldots$
( r= ) 4.72 (\%)

[Maximum number: 7]

(a)

Calculate the value of each of the following logarithms:

[ 7 ]

(i)

Calculate the value of each of the following logarithms:

$\quad \log _{2} \frac{1}{16}$ ;

valid approach to find the required logarithm

\begin{aligned} & 2^{x}=\frac{1}{16} \text { OR } 2^{x}=2^{-4} \text { OR } \frac{1}{16}=2^{-4} \text { OR } \log _{2} 1-\log _{2} 16 \\ & \log _{2} \frac{1}{16}=-4 \end{aligned}

(ii)

Calculate the value of each of the following logarithms:

$\quad \log _{9} 3$ ;

valid approach to find the required logarithm

\begin{aligned} & 9^{x}=3 \text { OR } 3^{2 x}=3 \text { OR } 3=9^{\frac{1}{2}} \text { OR } \frac{\log _{3} 3}{\log _{3} 9} \\ & \log _{9} 3=\frac{1}{2} \end{aligned}

(iii)

Calculate the value of each of the following logarithms:

$\quad \log _{\sqrt{3}} 81$ .

[ 7 ]

$(\sqrt{3})^{x}=81$ OR $\frac{\log _{3} 81}{\log _{3} \sqrt{3}}$
(3) $\frac{x}{2}=3^{4}$ OR $\frac{x}{2}=4$ OR $\frac{4}{\frac{1}{2}}$
x=8

(b)

It is given that $\log _{a b} a=3$ , where $a, b \in \mathbb{R}^{+}, a b \neq 1$ .

[ 8 ]

(i)

It is given that $\log _{a b} a=3$ , where $a, b \in \mathbb{R}^{+}, a b \neq 1$ .

Show that $\log _{a b} b=-2$ .

Note: There are many valid approaches to the question, and the steps may be seen in different ways. Some possible methods are given here, but candidates may use a combination of one or more of these methods.

In all methods, the final A mark is awarded for working which leads directly to the AG.

METHOD 1

(a b)^{3}=a

attempt to isolate b or a power of b
correct working

\begin{aligned} & b=\frac{a}{a^{3} b^{2}} \text { OR } b^{3}=a^{-2} \text { OR } b^{-1}=(a b)^{2} \text { OR } b^{3}=\frac{1}{a^{2}} \\ & b=\frac{1}{a^{2} b^{2}} \text { OR } b=(a b)^{-2} \text { OR } 3 \log _{a b} b=-2 \log _{a b} a \text { OR }-\log _{a b} b=2 \log _{a b} a b \\ & \log _{a b} b=-2 \end{aligned}

METHOD 2

(a b)^{3}=a

taking logarithm to base a b on both sides

\log _{a b}(a b)^{3}=\log _{a b} a \quad \text { OR } \quad \log _{a b} a^{3} b^{3}=\log _{a b} a

correct application of log rules leading to equation in terms of $\log _{a b}$

\begin{aligned} & 3 \log _{a b} a+3 \log _{a b} b=\log _{a b} a \quad \text { OR } \quad 3 \log _{a b} b=-2 \log _{a b} a \quad \text { OR } \log _{a b} b^{3}=\log _{a b} a^{-2} \\ & \log _{a b} b=\log _{a b} a^{-\frac{2}{3}} \quad \text { OR } \quad \log _{a b} b=-\frac{2}{3} \log _{a b} a \quad \text { OR } \quad \log _{a b} b=-\frac{2}{3}(3) \\ & \log _{a b} b=-2 \end{aligned}

Note: Candidates may substitute $\log _{a b} a=3$ at any point in their working.

METHOD 3

\log _{a b} a=3

writing in terms of base a

\frac{\log _{a} a}{\log _{a} a b}(=3)

correct application of log rules

\begin{aligned} & \frac{\log _{a} a}{\log _{a} a+\log _{a} b}(=3) \quad \mathrm{OR} \frac{1}{1+\log _{a} b}(=3) \quad \mathrm{OR} 3 \log _{a} b=-2 \quad \mathrm{OR} \\ & \log _{a} b=-\frac{2}{3} \end{aligned}

writing $\log _{a b} b$ in terms of base a

\log _{a b} b=\frac{\log _{a} b}{\log _{a} a+\log _{a} b}

correct working

\log _{a b} b=\frac{-\frac{2}{3}}{1-\frac{2}{3}} \text { OR } \frac{\left(-\frac{2}{3}\right)}{\left(\frac{1}{3}\right)}

\log _{a b} b=-2

METHOD 4

\begin{aligned} & \log _{a b} a b=1 \\ & \log _{a b} a+\log _{a b} b=1 \\ & 3+\log _{a b} b=1 \\ & \log _{a b} b=-2 \end{aligned}

The derivative of a function f is given by $f^{\prime}(x)=\frac{2 x+2}{x^{2}+2 x+2}$ , for $x \in \mathbb{R}$ .

Show that $x^{2}+2 x+2>0$ for all values of x.

METHOD 1
$x^{2}+2 x+2=(x+1)^{2}+1$ $(x+1)^{2} \geq 0$ $(x+1)^{2}+1>0$

METHOD 2
discriminant $\Delta=4-8(=-4)$ $\Delta<0$ and concave up OR $\Delta<0$ and coefficient of $x^{2}>0$ (may be seen in diagram)
hence $x^{2}+2 x+2>0$

All question bank results loaded

Covered syllabus points

1 Number and algebra question 1

Question (a)

Question (b)

1 Number and algebra question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

1 Number and algebra question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

Question (i)

1 Number and algebra question 4