Question 1
In this question, you will be investigating the family of functions of the form $f(x) = x^{n}e^{-x}$.
Consider the family of functions $f_{n}(x) = x^{n}e^{-x}$, where $x \geq 0$ and $n \in \mathbb{Z}^{+}$.
When $n = 1$, the function $f_{1}(x) = xe^{-x}$ where $x \geq 0$.
Topic – SL:2.6
(a) Sketch the graph of $y = f_{1}(x)$, stating the coordinates of the local maximum point.
Topic – AHL:5.15
(b) Show that the area of the region bounded by the graph $y = f_{1}(x)$, the x-axis and the line $x = b$, where $b > 0$, is given by $\frac{e^{b}-b-1}{e^{b}}$.
You may assume that the total area, $A_{n}$, of the region between the graph $y = f_{n}(x)$ and the x-axis can be written as $A_{n} = \int_{0}^{\infty}f_{n}(x)dx$ and is given by $\lim_{b \rightarrow \infty} \int_{0}^{b}f_{n}(x)dx$.
Topic – AHL:5.17
(c) (i) Use l’Hôpital’s rule to find $\lim_{b \rightarrow \infty} \frac{e^{b}-b-1}{e^{b}}$. You may assume that the condition for applying l’Hôpital’s rule has been met.
Topic – AHL:5.15
(ii) Hence write down the value of $A_{1}$.
You are given that $A_{2} = 2$ and $A_{3} = 6$.
(d) Use your graphic display calculator, and an appropriate value for the upper limit, to determine the value of
Topic – SL:5.14
(i) $A_{4}$;
Topic – SL:5.14
(ii) $A_{5}$.
Topic – AHL:1.3
(e) Suggest an expression for $A_{n}$ in terms of $n$, where $n \in \mathbb{Z}^{+}$.
Topic – AHL:1.7
(f) Use mathematical induction to prove your conjecture from part (e). You may assume that, for any value of $m$, $\lim_{x \rightarrow \infty} x^{m}e^{-x} = 0$.
▶️Answer/Explanation
Solution: –
(a) 
A1 for (1, 0.368) or $\left(1, \frac{1}{e}\right)$ labelled at local maximum (accept correct coordinates written away from the graph)
A1 for graph clearly starting at, or passing through, the origin
A1 for correct domain
A1 for correct shape i.e.: single maximum, and asymptotic behaviour (equation not required) (or point of inflexion)
(b) $\int_{0}^{b}xe^{-x}dx$
Use of integration by parts
\[=[-xe^{-x}]_{0}^{b}+\int_{0}^{b}e^{-x}dx\]
\[=[-xe^{-x}]_{0}^{b}-[e^{-x}]_{0}^{b}\]
attempt to substitute limits
\[=-be^{-b}-e^{-b}+1\]
\[=\frac{e^{b}-b-1}{e^{b}}\]
(c) (i) $\lim_{b\to\infty}\frac{e^{b}-b-1}{e^{b}}=\lim_{b\to\infty}1-\frac{b}{e^{b}}-\frac{1}{e^{b}}$
(ii) $\int_{0}^{\infty}xe^{-x}dx=1$
(d) (i) correct integral
24
(ii) 120
(e) $A_{n}=n!$
(f) \[n=1\]
\[A_{1}=1\] and \(1!=1\)
so true for \(n=1\)
assume true for \(n=k, (A_{k}=\int_{0}^{\infty}x^{k}e^{-s}dx=k!)\)
when \(n=k+1\)
attempt to integrate by parts
Question 2
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 $x_{1} + x_{2} = 12$.
Topic – SL:2.5
(a) Find the product of $x_{1}$ and $x_{2}$ as a function, $f$, of $x_{1}$ only.
Topic – SL:6.2
(b) (i) Find the value of $x_{1}$ for which the function is maximum.
Topic – SL:6.2
(ii) Hence show that the maximum product of $x_{1}$ and $x_{2}$ is 36.
Consider $M_{n}(S)$ to be the maximum product of $n$ positive real numbers with a sum of $S$, where $n \in \mathbb{Z}^{+}$ and $S \in \mathbb{R}^{+}$.
For $n = 2$, the maximum product can be expressed as $M_{2}(S) = \left(\frac{S}{2}\right)^{2}$.
Topic – SL:2.5
(c) Verify that $M_{2}(S) = \left(\frac{S}{2}\right)^{2}$ is true for $S = 12$.
Consider $n$ positive real numbers, $x_{1}, x_{2}, …, x_{n}$.
The geometric mean is defined as $(x_{1} \times x_{2} \times … \times x_{n})^{\frac{1}{n}}$. It is given that the geometric mean is always less than or equal to the arithmetic mean, so $(x_{1} \times x_{2} \times … \times x_{n})^{\frac{1}{n}} \leq \frac{(x_{1} + x_{2} + … + x_{n})}{n}$.
Topic – AHL:1.9
(d) (i) Show that the geometric mean and arithmetic mean are equal when $x_{1}=x_{2}=…=x_{n}$.
Topic – AHL:1.9
(ii) Use this result to prove that $M_{n}(S)=\left(\frac{S}{n}\right)^{n}$.
(e) Hence determine the value of
Topic – SL:1.3
(i) $M_{3}(12)$;
Topic – SL:1.3
(ii) $M_{4}(12)$;
Topic – SL:1.3
(iii) $M_{5}(12)$.
For $n\in\mathbb{Z}^{+}$ let $P(S)$ denote the maximum value of $M_{n}(S)$ across all possible values of $n$.
Topic – SL:1.3
(f) Write down the value of $P(12)$ and the value of $n$ at which it occurs.
Topic – SL:1.3
(g) Determine the value of $P(20)$ and the value of $n$ at which it occurs.
Consider the function $g$, defined by $\ln(g(x)) = x\ln\left(\frac{S}{x}\right)$, where $x \in \mathbb{R}^{+}$.
A sketch of the graph of $y = g(x)$ is shown in the following diagram. Point A is the maximum point on this graph.
Topic – AHL:6.10
(h) Find, in terms of $S$, the $x$-coordinate of point A.
Topic – AHL:2.10
(i) Verify that $g(x) = M_{n}(S)$, when $x \in \mathbb{Z}^{+}$.
Topic – AHL:2.10 & Topic – AHL:6.10
(j) 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 $k \in \mathbb{Z}$.
▶️Answer/Explanation
Solution: –
(a) \(x_{2}=12-x_{1}\)
\[f(x)=x_{1}(12-x_{1})\]
(b) (i) \((x_{1}=)6\)
(ii) \(f(6)\) OR \(6^{2}\) OR graph with maximum at (6, 36) = 36
(c) \[M_{2}(12)=(\frac{12}{2})^{2}=36\]
which is the maximum product (from \((b)(ii))\)
(d) (i) let all \(x_{i}\) be labelled as x (or x_{1} or \(x_{n}\) etc.)
\[(x^{n})^{\frac{1}{n}}=x\] and \[\frac{nx}{n}=x\]
(ii) \(x_{1}+x_{2}+…+x_{n}=S\)
\[(x_{1}x_{2}…x_{n})^{\frac{1}{n}}\le\frac{S}{n}\]
\(x_{1}x_{2}…x_{n}\le\left(\frac{S}{n}\right)^{n}\) (as both sides are positive)
LHS and RHS are equal when all values of \(x_{i}\) are equal (to \(\frac{S}{n}\))
\[M_{n}(S)=\left(\frac{S}{n}\right)^{n}\]
\[x_{1}x_{2}…x_{n}\le\left(\frac{S}{n}\right)^{n}\]
(e) (i) \(M_{3}(12)=4^{3}=64\)
(ii) \(M_{4}(12)=3^{4}=81\)
(iii) \(M_{5}(12)=2.4^{5}=79.6\) (79.6262…)
(f) considering \(M_{n}(12)\) for higher values of n
\[P(12)=81\]
\[n=4\]
(g) Consideration of graph or table of \((\frac{20}{n})^{n}\) including values either side of 7
Maximum occurs when \(n=7\)
\[P(20)=(\frac{20}{7})^{7}=1550(1554.260…)\]
(h) EITHER
\[ln(g(x))=x(ln(S)-ln~x)\]
attempt to use implicit differentiation and product rule
\[\frac{g^{\prime}(x)}{g(x)}=ln~S-ln~x-x\frac{1}{x}\]
OR
attempt to use implicit differentiation, product rule and chain rule
\[\frac{g^{\prime}(x)}{g(x)}=ln~\frac{S}{x}+(x\frac{x}{S}\times\frac{-S}{x^{2}})\]
OR
attempt to make equation explicit to
\[g(x)=e^{x~ln(\frac{S}{x})}\]
attempt to use product rule and chain rule
\[g^{\prime}(x)=e^{x~ln(\frac{S}{x})}[x\times\frac{x}{S}\times(-Sx^{-2})+ln(\frac{S}{x})]\]
\[=e^{x~ln(\frac{S}{x})}[ln(\frac{S}{x})-1]\]
THEN
\[g^{\prime}(x)=\left(ln\frac{S}{x}-1\right)g(x)\]
\[g(x)\ne0\]
\[g^{\prime}(x)=0\Rightarrow ln\frac{S}{x}-1=0\]
\[x=\frac{S}{e}(0.368S, 0.36789…S)\]
(i) \[ln(g(x))=x~ln(\frac{S}{x})\Rightarrow g(x)=\left(\frac{S}{x}\right)^{x}\]
\[g(x)=\left(\frac{S}{x}\right)^{x}\]
\[=M_{x}(S)\] for \(x\in\mathbb{Z}^{+}\)
(ii) \[\frac{100}{e}=36.8\]
\[\left(\frac{100}{36}\right)^{36}=9.3996…x10^{15}\] AND \[\left(\frac{100}{37}\right)^{37}=9.47406…x10^{15}\]
largest possible product is \(9.47×10^{15}\) \(9.47406…x10^{15}\)