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}}

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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)