IB Maths AA HL 1.1 Number and Algebra Sl Content Questions

Practise IB Mathematics AA HL 1.1 by extending number-and-algebra models through sequences, series, proof, financial mathematics and complex symbolic reasoning.

Exam points

Model arithmetic and geometric sequences, finite or infinite sums, financial growth and decay, and solve inverse or threshold conditions with the correct restrictions.
Transform and solve exponential or logarithmic equations and models, preserving domain, base and parameter conditions.
Construct numerical and algebraic proofs involving identities, divisibility, parity, formulae or exceptional cases, with every step justified.
Use binomial coefficients and the general term to expand expressions, compare terms and determine restricted parameters in extended cases.

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: 11]

In this question, you will investigate the maximum product of positive real numbers with a given sum.
Consider the two numbers $x_{1}, x_{2} \in \mathbb{R}^{+}$ , such that $x_{1}+x_{2}=12$ .

(a)

Consider n positive real numbers, $x_{1}, x_{2}, \ldots, x_{n}$ .
The geometric mean is defined as $\left(x_{1} \times x_{2} \times \ldots \times x_{n}\right)^{\frac{1}{n}}$ . It is given that the geometric mean is always less than or equal to the arithmetic mean, so $\left(x_{1} \times x_{2} \times \ldots \times x_{n}\right)^{\frac{1}{n}} \leq \frac{\left(x_{1}+x_{2}+\ldots+x_{n}\right)}{n}$ .

[ 6 ]

(i)

Show that the geometric mean and arithmetic mean are equal when $x_{1}=x_{2}=\ldots=x_{n}$ .

[ 2 ]

let all $x_{i}$ be labelled as x (or $x_{1}$ or $x_{n}$ etc.)

Note: Do not accept use of a specific number for x or n.

\left(x^{n}\right)^{\frac{1}{n}}=x \text { and } \frac{n x}{n}=x

(ii)

Use this result to prove that $M_{n}(S)=\left(\frac{S}{n}\right)^{n}$ .

[ 4 ]

$x_{1}+x_{2}+\ldots+x_{n}=S$

\left(x_{1} \times x_{2} \times \ldots \times x_{n}\right)^{\frac{1}{n}} \leq \frac{S}{n}

Note: Award M1 for use of the inequality, which may be seen as an equality.
$x_{1} \times x_{2} \times \ldots \times x_{n} \leq\left(\frac{S}{n}\right)^{n}$ (as both sides are positive)

LHS and RHS are equal when all values of $x_{i}$ are equal (to $\frac{S}{n}$ )
$M_{n}(S)=\left(\frac{S}{n}\right)^{n}$

(b)

Verify that $g(x)=M_{x}(S)$ , when $x \in \mathbb{Z}^{+}$ .

[ 2 ]

$\quad \ln (g(x))=x \ln \left(\frac{S}{x}\right) \Rightarrow \ln (g(x))=\ln \left(\frac{S}{x}\right)^{x} \quad$ M1
$g(x)=\left(\frac{S}{x}\right)^{x}$ $=M_{x}(S)$ for $x \in \mathbb{Z}^{+}$
AG

(c)

Use your answer to part (h) to find the largest possible product of positive numbers whose sum is 100 . Give your answer in the form $a \times 10^{k}$ , where $1 \leq a<10$ and $k \in \mathbb{Z}^{+}$ .

[ 3 ]

$\frac{100}{\mathrm{e}}=36.8 \quad$ M1
$\left(\frac{100}{36}\right)^{36}=9.3996 \ldots \times 10^{15}$ AND $\left(\frac{100}{37}\right)^{37}=9.47406 \ldots \times 10^{15}$
largest possible product is $9.47 \times 10^{15}\left(9.47406 \ldots \times 10^{15}\right)$

Note: Award A1 independently of the R1 (but not independently of the M1).

[Maximum number: 7]

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 $\sum_{i=1}^{n} i=\frac{1}{2} n(n+1)$ .

Consider the case when q=2.

[ 1 ]

EITHER
$S_{n}=\frac{n}{2}(2 \times 1+(n-1) \times 1)$

OR
$u_{1}=1$ and either $u_{n}=n$ or d=1 stated explicitly

OR
$1+2+\ldots+n$ (or equivalent) stated explicitly

THEN
$S_{n}=\frac{n}{2}(1+n)$

Note: Award A0 for a numerical verification.

(b)

Using sigma notation, write down an expression for $f_{q}(1)$ .

[ 1 ]

$f_{q}(1)=1^{q}+2^{q}+3^{q}+\ldots+n^{q}$

=\sum_{i=1}^{n} i^{q}\left(=\sum_{i=1}^{n} 1^{i} i^{q}\right)

(c)

By considering $f(x)=1+x+x^{2}+\ldots+x^{n}$ as a geometric series, for $x \neq 1$ , show that $f(x)=\frac{x^{n+1}-1}{x-1}$ .

[ 2 ]

uses $S_{n}=\frac{u_{1}\left(r^{n}-1\right)}{r-1}$ with r=x and $u_{1}=1$
clear indication there are ( n+1 ) terms

f(x)=\frac{x^{n+1}-1}{x-1}

(d)

For $x \neq 1$ , show that $f_{1}(x)=\frac{n x^{n+2}-(n+1) x^{n+1}+x}{(x-1)^{2}}$ .

[ 3 ]

METHOD 1

\begin{aligned} & f_{1}(x)=x f^{\prime}(x) \\ & =x \frac{(x-1)(n+1) x^{n}-1 \times\left(x^{n+1}-1\right)}{(x-1)^{2}} \end{aligned}

Note: Award M1 for attempting to use the quotient or the product rule to find $f^{\prime}(x)$ .

=x \frac{(n x+x-n-1) x^{n}-\left(x^{n+1}-1\right)}{(x-1)^{2}}\left(=x \frac{n x^{n+1}-n x^{n}-x^{n}+1}{(x-1)^{2}}\right)

Note: Award A1 for any correct manipulation of the derivative that leads to the AG.

=\frac{n x^{n+2}-(n+1) x^{n+1}+x}{(x-1)^{2}}

METHOD 2
attempts to form $(x-1) f_{1}(x)$

\begin{aligned} & (x-1) f_{1}(x)=n x^{n+1}-\left(x+x^{2}+x^{3}+\ldots+x^{n}\right) \\ & f_{1}(x)=\frac{1}{x-1}\left(n x^{n+1}-\left(\frac{x^{n+1}-1}{x-1}-1\right)\right) \\ & f_{1}(x)=\frac{1}{x-1}\left(\frac{n x^{n+1}(x-1)-x^{n+1}+x}{x-1}\right)\left(=\frac{1}{x-1}\left(\frac{n x^{n+2}-n x^{n+1}-x^{n+1}+x}{x-1}\right)\right) \end{aligned}

Note: Award A1 for any correct manipulation of the derivative that leads to the AG.

f_{1}(x)=\frac{n x^{n+2}-(n+1) x^{n+1}+x}{(x-1)^{2}}

[Maximum number: 15]

Phil takes out a bank loan of $\$ 150000$ to buy a house, at an annual interest rate of 3.5 %. The interest is calculated at the end of each year and added to the amount outstanding.

(a)

Find the amount Phil would owe the bank after 20 years. Give your answer to the nearest dollar.

To pay off the loan, Phil makes annual deposits of $\$ P$ at the end of every year in a savings account, paying an annual interest rate of 2 %. He makes his first deposit at the end of the first year after taking out the loan.

[ 3 ]

$150000 \times 1.035^{20}$
= $298468

Note: Only accept answers to the nearest dollar. Accept $\$ 298469$ .

(b)

Show that the total value of Phil's savings after 20 years is $\frac{\left(1.02^{20}-1\right) P}{(1.02-1)}$ .

[ 3 ]

attempt to look for a pattern by considering 1 year, 2 years etc
recognising a geometric series with first term P and common ratio 1.02

EITHER

P+1.02 P+\ldots+1.02^{19} P\left(=P\left(1+1.02+\ldots+1.02^{19}\right)\right)

OR
explicitly identify $u_{1}=P, r=1.02$ and n=20 (may be seen as $S_{20}$ ).

THEN
$S_{20}=\frac{\left(1.02^{20}-1\right) P}{(1.02-1)}$

(c)

Given that Phil's aim is to own the house after 20 years, find the value for P to the nearest dollar.

David visits a different bank and makes a single deposit of $\$ Q$ , the annual interest rate being 2.8 %.

[ 3 ]

$\quad 24.297 \ldots P=298468$
P=12284

Note: Accept answers which round to 12284.

(d)

David wishes to withdraw $\$ 5000$ at the end of each year for a period of n years. Show that an expression for the minimum value of Q is

\frac{5000}{1.028}+\frac{5000}{1.028^{2}}+\ldots+\frac{5000}{1.028^{n}}

[ 3 ]

METHOD 1

$Q\left(1.028^{n}\right)=5000\left(1+1.028+1.028^{2}+1.028^{3}+\ldots+1.028^{n-1}\right)$
$Q=\frac{5000\left(1+1.028+1.028^{2}+1.028^{3}+\ldots+1.028^{n-1}\right)}{1.028^{n}}$
$=\frac{5000}{1.028}+\frac{5000}{1.028^{2}}+\ldots+\frac{5000}{1.028^{n}}$	AG

METHOD 2
the initial value of the first withdrawal is $\frac{5000}{1.028}$
the initial value of the second withdrawal is $\frac{5000}{1.028^{2}}$
the investment required for these two withdrawals is $\frac{5000}{1.028}+\frac{5000}{1.028^{2}}$ $Q=\frac{5000}{1.028}+\frac{5000}{1.028^{2}}+\ldots+\frac{5000}{1.028^{n}}$
AG

(e)

Hence or otherwise, find the minimum value of Q that would permit David to withdraw annual amounts of $\$ 5000$ indefinitely. Give your answer to the nearest dollar.

[ 3 ]

sum to infinity is $\frac{\frac{5000}{1.028}}{1-\frac{1}{1.028}}$ $=178571.428 \ldots$
so minimum amount is $\$ 178572$
A1
Note: Accept answers which round to $\$ 178571$ or $\$ 178572$ .

[Maximum number: 3]

In this question, you will investigate the maximum product of positive real numbers with a given sum.
Consider the two numbers $x_{1}, x_{2} \in \mathbb{R}^{+}$ , such that $x_{1}+x_{2}=12$ .

[Maximum number: 2]

It is given that $\log _{10} a=\frac{1}{3}$ , where a>0.
Find the value of

$\quad \log _{10}\left(\frac{1}{a}\right)$ ;

$\log _{10} 1-\log _{10} a$ OR $\log _{10} a^{-1}=-\log _{10} a$ OR $\log _{10} 10^{-\frac{1}{3}}$ OR $10^{x}=\frac{1}{10^{\frac{1}{3}}}$ $=-\frac{1}{3}$

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

Show that $x-\frac{2 A x}{x^{2}+A} \equiv \frac{x\left(x^{2}-A\right)}{x^{2}+A}$ .

METHOD 1

\begin{aligned} & x-\frac{2 A x}{x^{2}+A} \equiv \frac{x\left(x^{2}+A\right)-2 A x}{x^{2}+A}\left(\equiv \frac{x^{3}+A x-2 A x}{x^{2}+A}\right) \\ & \equiv \frac{x^{3}-A x}{x^{2}+A}\left(\equiv \frac{x\left(x^{2}+A-2 A\right)}{x^{2}+A}\right) \\ & x-\frac{2 A x}{x^{2}+A} \equiv \frac{x\left(x^{2}-A\right)}{x^{2}+A} \end{aligned}

Note: Award no marks for numerical approaches.

METHOD 2

EITHER
attempt to do long division on $\frac{x\left(x^{2}-A\right)}{x^{2}+A}$ or $\frac{x^{3}-A x}{x^{2}+A}$
quotient of x and remainder of -2 A x
$x-\frac{2 A x}{x^{2}+A} \equiv \frac{x\left(x^{2}-A\right)}{x^{2}+A}$

OR
$\frac{x^{3}}{x^{2}+A}-\frac{A x}{x^{2}+A} \equiv \frac{x^{3}+A x}{x^{2}+A}-\frac{A x+A x}{x^{2}+A}$ $\equiv \frac{x\left(x^{2}+A\right)}{\left(x^{2}+A\right)}-\frac{2 A x}{\left(x^{2}+A\right)}$ $\equiv x-\frac{2 A x}{x^{2}+A} \equiv \frac{x\left(x^{2}-A\right)}{x^{2}+A}$

Note: Award no marks for numerical approaches.

[Maximum number: 3]

It is given that $\log _{10} a=\frac{1}{3}$ , where a>0.
Find the value of

$\log _{1000} a$ .

$\frac{\log _{10} a}{\log _{10} 1000}$ OR $\frac{1}{3} \log _{1000} 10$ OR $\log _{1000} \sqrt[3]{1000^{\frac{1}{3}}}$ OR $10^{\frac{1}{3}}=1000^{x}\left(=\left(10^{3}\right)^{x}\right)$ $\frac{\log _{10} a}{3}$ OR $\frac{1}{3} \log _{1000} 1000^{\frac{1}{3}}$ OR $\log _{1000} 1000^{\frac{1}{9}}$ OR $3 x=\frac{1}{3}$ $=\frac{1}{9}$

[Maximum number: 2]

The following question uses Maclaurin series to investigate approximations of mathematical constants and the accuracies of such approximations.

Given |x|<1, find the sum to infinity of the geometric series $1-x^{2}+x^{4}-x^{6}+\ldots$ .

attempt to use $S_{\infty}=\frac{u_{1}}{1-r}$

Note: Allow working with $u_{1}=1$ and $r=-x^{2}$ or $u_{1}=1-x^{2}$ and $r=x^{4}$ .

=\frac{1}{1+x^{2}}

[Maximum number: 3]

Consider the expression $\frac{1}{1+x^{2}}$ , where |x|<1.

Show that the first four terms in the binomial expansion of $\frac{1}{1+x^{2}}$
are $1-x^{2}+x^{4}-x^{6}$ .

attempt to create binomial expansion to four terms with n=-1

\begin{aligned} & 1+\binom{-1}{1} x^{2}+\binom{-1}{2}\left(x^{2}\right)^{2}+\binom{-1}{3}\left(x^{2}\right)^{2}+\ldots \\ & 1-x^{2}+\frac{(-1)(-2)}{2}\left(x^{2}\right)^{2}+\frac{(-1)(-2)(-3)}{6}\left(x^{2}\right)^{3} \end{aligned}

Note: Award A1 for the third term and A1 for the fourth term.

1-x^{2}+x^{4}-x^{6}

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9