IB Maths AA SL 1.1 Number and Algebra Sl Content Questions

Practise IB Mathematics AA SL 1.1 by working with number systems, sequences, arithmetic and geometric structure, financial applications and technology-supported algebra.

Exam points

Use arithmetic or geometric terms and sums to model sequences, financial growth or decay, then solve inverse, threshold or infinite-sum conditions.
Apply exponent and logarithm laws to solve equations or determine an unknown parameter, checking base, argument and domain restrictions.
Write a numerical or algebraic proof using equality steps, integer representation, divisibility or parity, and justify each implication.
Expand a binomial or use its general term to find a coefficient, power or unknown parameter, checking all allowed values.

Covered syllabus points

SL 1.1—Scientific notation SL 1.2—Arithmetic sequences and series SL 1.3—Geometric sequences and series SL 1.4—Financial applications SL 1.5—Integer exponents and logarithms SL 1.6—Deductive proof SL 1.7—Rational exponents and logarithm laws SL 1.8—Infinite geometric series SL 1.9—Binomial theorem

[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)

The volume of the planet can be expressed in the form $\pi(a\times10^k)\ \mathrm{km}^3$ , where $1\leq a<10$ and $k\in\mathbb Z$ .

Find the value of a and the value of k.

[ 3 ]

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

[Maximum number: 15]

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;

[ 2 ]

$5500 \times 36$
=($) 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.

[ 3 ]

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 total value of Daniela's winnings and Sorin's investment on the last day of the sixth month.

[ 3 ]

Sorin's total $=120000\left(1+\frac{4}{100 \times 12}\right)^{6}(=122420.09)$

Daniela's total $=\frac{2000\left(1-1.06^{6}\right)}{1-1.06}(=13950.64)$
total value =($) 136370.73

(d)

Find the minimum number of complete months before the total value of Daniela's winnings and Sorin's investment is at least $\$ 250000$ .

At the end of the three years, Daniela invested $40000 for a further six years in a second account that pays a nominal interest rate of r % per annum compounded quarterly.

[ 3 ]

EITHER (finding number of months, m )
$120000\left(1+\frac{4}{100 \times 12}\right)^{m}+\frac{2000\left(1-1.06^{m}\right)}{1-1.06}(\geq 250000)$ $m \geq 26.0905 \ldots$ OR $(m=26 \Rightarrow) 249157 \ldots$ AND $(m=27 \Rightarrow) 258692 \ldots$

Note: Condone use of an equation or strict inequality.
OR (finding number of years, n )
$120000\left(1+\frac{4}{100 \times 12}\right)^{12 \times n}+\frac{2000\left(1-1.06^{12 \times n}\right)}{1-1.06}(\geq 250000)$ $n \geq 2.17421 \ldots$ (years)

Note: Condone use of an equation or strict inequality.

THEN
m=27 (months)

(e)

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
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
$40000\left(1+\frac{r}{100 \times 4}\right)^{6 \times 4}=53000$
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: 11]

(a)

Calculate the value of each of the following logarithms:

[ 7 ]

(i)

Calculate the value of each of the following logarithms:

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

[ 1 ]

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:

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

[ 1 ]

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:

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

[ 5 ]

$(\sqrt{3})^{x}=81$ 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 $a, b \in \mathbb{R}^{+}, a b \neq 1$ .

[ 4 ]

(i)

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

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

[ 4 ]

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}

[Maximum number: 2]

Find the total amount Daniela would receive if she chooses

Option A;

$5500 \times 36$
=($) 198000

[Maximum number: 1]

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}

[Maximum number: 2]

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 $x^{2}+2 x+2>0$

[Maximum number: 7]

(a)

Consider an infinite geometric sequence with $u_{1}=40$ and $r=\frac{1}{2}$ .

[ 2 ]

(i)

Find the sum of the infinite sequence.

[ 2 ]

correct substitution into formula for infinite sum
e.g. $S_{\infty}=\frac{40}{1-0.5}, S_{\infty}=\frac{40}{0.5}$

S_{\infty}=80

(b)

The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n.

[ 5 ]

multiplying $S_{n}$ (AP) by 2 or dividing S (infinite GP) by 2
e.g. $2 S_{n}, \frac{S_{\infty}}{2}, 40$
evidence of substituting into $2 S_{n}=S_{\infty}$
e.g. $2 n^{2}-38 n=40,4 n^{2}-76 n-80(=0)$
attempt to solve their quadratic (equation)
e.g. intersection of graphs, formula

n=20

9. Note: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their 3 sf answers in subsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT.

[Maximum number: 5]

Consider the expansion of $(x+3)^{10}$ .

(a)

Write down the number of terms in this expansion.

[ 1 ]

(b)

Find the term containing $x^{3}$ .

[ 4 ]

evidence of binomial expansion
eg $\quad\binom{n}{r} a^{n-r} b^{r}$ , attempt to expand
evidence of choosing correct term
eg $\quad 8^{\text {th }}$ term, $r=7,\binom{10}{7},(x)^{3}(3)^{7}$
correct working
eg $\quad\binom{10}{7}(x)^{3}(3)^{7},\binom{10}{3}(x)^{3}(3)^{7}$ ,
$262440 x^{3}$ (accept $262000 x^{3}$ )

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

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Question (i)

Question (ii)

Question (b)

Question (c)

Question (d)

Question (e)

Question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

Question (i)

Question 4

Question 5

Question 6

Question 7

Question (a)

Question (i)

Question (b)

Question 8

Question (a)

Question (b)