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Study 1.1 Number and algebra - SL content conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[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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IB Maths AA HL 1.1 Number and Algebra Sl Content Topic Practice

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