July 22, 2025 · 6:43 am

Continuous Uniform Random Variable

My Year 12 Mathematics Methods students are doing continuous random variables at the moment and I thought it would be worthwhile deriving the mean and variance formulas for a uniform continuous random variable.

The probability density function for a uniform random variable is

    \begin{equation*}f(x)= \left \{ {\begin{matrix}\frac{1}{b-a} & a\le x\le b \\0 & \text {elsewhere}\end{matrix}}\end{equation}

and it looks like

Remember, the mean \mu or expected value E(X) of a continuous random variable is

(1)   \begin{equation*}E(X)=\int xp(x) dx\end{equation*}

and the variance \sigma^2 is

(2)   \begin{equation*}\sigma^2=\int (x-\mu)^2p(x) dx\end{equation*}

We are going to use equations 1 and 2 to find formulae for a uniform continuous random variable.

    \begin{equation*}\mu=\int_a^b x (\frac{1}{b-a}) dx\end{equation}

    \begin{equation*}\mu=\frac{x^2}{2(b-a)}|\begin{matrix}b\\a\end{matrix}\end{equation}

    \begin{equation*}\mu=\frac{b^2}{2(b-a)}-\frac{a^2}{2(b-a)}=\frac{b^2-a^2}{2(b-a)}\end{equation}

Factorise the numerator (using difference of squares)

    \begin{equation*}\mu=\frac{(b-a)(b+a)}{2(b-a)}\end{equation}

Hence,

    \begin{equation*}\mu=\frac{b+a}{2}\end{equation}

Now for the variance

    \begin{equation*}\sigma^2=\int_a^b (x-(\frac{a+b}{2}))^2(\frac{1}{b-a}) dx\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{(x-\frac{a+b}{2})^3}{3})|\begin{matrix}b\\a\end{matrix}\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}((\frac{(b-\frac{a+b}{2})^3}{3})-(\frac{(a-\frac{a+b}{2})^3}{3}))\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{-a^3}{12}+\frac{b^3}{12}+\frac{a^2b}{4}-\frac{ab^2}{4})\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{-a^3+b^3+3a^2b-3ab^2}{12})\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{b^3-3b^2a+3ba^2-a^3}{12})\end{equation}

From the binomial expansion theorem, we know

    \begin{equation*}b^3-3b^2a+3ba^2-a^3=(b-a)^3\end{equation}

Hence

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{(b-a)^3}{12}\end{equation}

and

    \begin{equation*}\sigma^2=\frac{(b-a)^2}{12}\end{equation}

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July 20, 2025 · 5:50 am

Integrating the Natural Log Function

One of the students asked me the other day how to integrate f(x)=ln(x) – it’s not part of their course, but I thought I would do it here.

We use integration by parts to integrate ln(x)

    \begin{equation*}\int u dv=uv-\int v du\end{equation}

    \begin{equation*}\int ln(x) dx\end{equation}

Let u=ln(x) and dv=1, then du=\frac{1}{x} and v=x

    \begin{equation*}\int ln(x) dx = xln(x)-\int x\times \frac{dx}{x}\end{equation}

    \begin{equation*}\int ln(x) dx = xln(x)-\int 1 dx\end{equation}

    \begin{equation*}\int ln(x) dx = xln(x)-x+c\end{equation}

What about \int 5ln(\sqrt{x}) dx

We can take advantage of log laws and the properties of integration.

    \begin{equation*}\int 5ln(\sqrt{x}) dx=5\int lnx^{\frac{1}{2}} dx=5\int \frac{1}{2}ln(x) dx=\frac{5}{2} \int ln(x) dx\end{equation}

    \begin{equation*}\frac{5}{2} \int ln(x) dx=\frac{5}{2}(xlnx-x)+c\end{equation}

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July 11, 2025 · 5:35 am

Simultaneous Equation (or is it?)

Solve simultaneously

X+Y+Z=10
XYZ=30
XY+YZ+XZ=31

We could attempt to solve this simultaneously, but I think the algebra would be tricky.

The three equations are related to the roots of a cubic polynomial.

If the general equation of the polynomial is ax^3+bx^2+cx+d, then we know

The sum of the roots

\alpha+\beta+\gamma=\frac{-b}{a}

The product of the roots

\alpha\beta\gamma=\frac{-d}{a}

and

\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}

So from our three equations we have

(1)   \begin{equation*}X+Y+Z=10=\frac{-b}{a}\end{equation*}

(2)   \begin{equation*}XYZ=30=\frac{-d}{a}\end{equation*}

(3)   \begin{equation*}XY+YZ+XZ=31=\frac{c}{a}\end{equation*}

Let a=1, then b=-10, c=31, and d=-30

Our cubic is t^3-10t^2+31t-30 and we can try to solve it.

The roots will be factors of -30, so \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30

Try t=2

    \begin{equation*}(2)^3-10(2)^2+31(2)-30=0\end{equation}

Hence t=2 is a root.

Use synthetic division to find the quadratic factor

21-1031-30
2-1630
1-8150

The quadratic factor is x^2-8x+15, which factorises to (x-3)(x-5)

Hence the solutions are X=2, Y=3, and Z=5

We could assume the solutions are natural numbers, then we can look at factors of 30.

Factors of ThirtyX+Y+ZXY+YZ+XZ
1, 1, 301+1+30=321+30+30=61
1, 2, 151+2+15=182+15+30=47
1, 5, 61+5+6=125+6+30=41
2, 3, 52+3+5=106+10+15=31

Hence the solutions are X=2, Y=3, and Z=5

But with this approach we might not be able to find the solutions.

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July 8, 2025 · 6:17 am

Age Question (Year 8 equation solving)

Eight years ago my father was three times as old as I shall be in five years time. When I was born he was 41 years old. How old am I now?

I always find these age questions a bit weird – a bit of a riddle, and contrived (just so we can solve some equations)

Let x be my age now, and y be my fathers age now.

(1)   \begin{equation*}y=x+41\end{equation*}

Because my father was 41 when I was born.

(2)   \begin{equation*}y-8=3(x+5)\end{equation*}

y-8 for 8 years ago, and 3(x+5) for three times my age in 5 years.

Solve the equations simultaneously. Substitute y=x+41 into equation 2

    \begin{equation*}x+41-8=3(x+5)\end{equation}

    \begin{equation*}x-33=3x+15\end{equation}

    \begin{equation*}18=2x\end{equation}

    \begin{equation*}x=9\end{equation}

Hence my current age is 9.

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July 1, 2025 · 10:35 am

Synthetic Division (for factorising and/or solving polynomials)

I use synthetic division to factorise polynomials with a degree greater than 2. For example, f(x)=x^3+2x^2-5x-6

It works best with monic polynomials but can be adapted to non-monic ones (see example below).

The only problem is that you need to find a root to start.

Try the factors of -6 i.e. (-1, 1, 2, -2, 3, -3, 6, -6)

f(-1)=(-1)^3+2(-1)^2-5(-1)-6=0

Hence, x=-1 is a root and (x+1) is a factor of the polynomial.

Set up as follows

Bring the first number down

Multiply by the root and place under the second co-efficient

Add down

Repeat the process

The numbers at the bottom (1, 1, -6) are the coefficients of the polynomial factor.

We now know x^3+2x^2-5x-6=(x+1)(x^2+x-6).

We can factorise the quadratic in the usual way.

x^2+x-6=(x+3)(x-2)

Hence x^3+2x^2-5x-6=(x+1)(x+3)(x-2).

Let’s try a non-monic example

Factorise 6x^4+39x^3+91x^2+89x+30

I know -2 is a root. Otherwise I would try the factors of 30.

Use synthetic division

Because this was non-monic we need to divide our new co-efficients (6, 27, 37, 15) by 6 (the co-efficient of the x^4 term)

x^3+\frac{9}{2}x^2+\frac{37}{6}+\frac{5}{2}

We now need to go again. I know that \frac{-3}{2} is a root and (2x+3) is a factor.

Our quadratic factor is x^2+3x+5/3, which is 3x^2+9x+5.

The quadratic factor doesn’t have integer factors so,

6x^4+39x^3+91x^2+89x+30=(x+2)(2x+3)(3x^2+9x+5)

I think this is much quicker than polynomial long division.

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Filed under Algebra, Factorising, Factorising, Fractions, Polynomials, Quadratic, Simplifying fractions, Solving Equations

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June 26, 2025 · 7:22 am

Trigonometry Question

I don’t know where I found this question, but it does require algebra and problem solving (as well as right trig and Pythagoras)

From a point A, a lighthouse is on a bearing of 026^\circT and the top of the light house is at an angle of elevation of 20.25^\circ. From a point B, the lighthouse is on a bearing of 296^\cricT and the top of the lighthouse is at angle of elevation of 10.2^\circ. If A and B are 500 metres apart, find the height of the lighthouse.

Let’s draw a diagram.

Let the height of the lighthouse be h

We can find the angle between A, the lighthouse, and B by using the base triangle

The red line from L is parallel to the two north lines. Hence \theta=26^\circ+64^\circ=90^\circ (Alternate angles in parallel lines are congruent)

It’s a right triangle so we know

(1)   \begin{equation*}500^2=(AL)^2+(BL)^2\end{equation*}

We are going to use the other two triangles to find AL and BL


tan(20.25)=\frac{h}{AL}
AL=\frac{h}{tan(20.25)}
tan(10.2)=\frac{h}{BL}
BL=\frac{h}{tan(10.2)}

Substitute AL and BL into equation 1

    \begin{equation*}500^2=(\frac{h}{tan(20.25)})^2+(\frac{h}{tan(10.2)})^2\end{equation}

Solve for h.

    \begin{equation*}500^2=7.35h^2+30.89h^2=38.24h^2\end{equation}

    \begin{equation*}h^2=6538.3\end{equation}

    \begin{equation*}h=80.9m\end{equation}

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June 19, 2025 · 7:18 am

Converting recurring (non-terminating) decimals to fractions

The easiest approach is to jump right in with some examples.

Example 1

Convert 0.\overline{5} to a fraction.

Let x=0.\overline{5}

(1)   \begin{equation*}x=0.\overline{5}\end{equation*}

(2)   \begin{equation*}10x=5.\overline{5}\end{equation*}

Subtract equation 1 from equation 2

    \begin{equation*}9x=5\end{equation}

Hence x=\frac{5}{9} so 0.\overline{5}=\frac{5}{9}

Example 2

Convert 0.\overline{12} to a fraction.

Let x=0.\overline{12}

(3)   \begin{equation*}x=0.\overline{12}\end{equation*}

(4)   \begin{equation*}100x=12.\overline{12}\end{equation*}

Subtract equation 3 from equation 4.

    \begin{equation*}99x=12\end{equation}

    \begin{equation*}x=\frac{12}{99}=\frac{4}{33}\end{equation}

Example 3

Convert 0.1\overline{23} to a fraction

Let x=0.1\overline{23}

(5)   \begin{equation*}x=0.1\overline{23}\end{equation*}

(6)   \begin{equation*}10x=1.\overline{23}\end{equation*}

(7)   \begin{equation*}1000x=123.\overline{23}\end{equation*}

Subtract equation 6 from equation 7

    \begin{equation*}990x=122\end{equation}

    \begin{equation*}x=\frac{122}{990}=\frac{61}{495}\end{equation}

Our aim is to manipulate the recurring decimal to create two numbers each which have only the repeated digits after the decimal point.

One more example.

Example 4

Convert 3.4\overline{56} to a fraction

Let x=3.4\overline{56}

If I multiply by 10, then I will have 34.\overline{56} – only repeated digits after the decimal point.

If I multiply by 1000, then I will have 3456.\overline{56}– only repeated digits after the decimal point.

So I get,

    \begin{equation*}990x=3422\end{equation}

    \begin{equation*}x=\frac{3422}{990}=3\frac{226}{495}\end{equation}

You can also use your Casio classpad to do the conversion. Although I think it is easier just to do it yourself.

Let’s think about example 4,

3.4\overline{56}=3.4+\frac{56}{1000}+\frac{56}{100000}+\frac{56}{10000000}+...

Which is

3.4+\frac{56}{1000\times 100^0}+\frac{56}{1000\times100^1}+\frac{56}{1000\times 100^2}...

3.4+\Sigma_{x=0}^\infty(\frac{56}{1000\times 100^x})

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June 10, 2025 · 1:54 am

Square Root Puzzle

Is it possible to find three numbers, a, b, c, none of which is zero or a perfect square, such that
\sqrt{a}+\sqrt{b}=\sqrt{c}

Can You Solve These – David Wells

As a, b and c can’t be perfect squares, let a=d\times e^2, b=f\times g^2 and c=h\times k^2 where d, e, f, g, h and k are real numbers.

Hence \sqrt{a}=e\sqrt{d}, \sqrt{b}=g\sqrt{f} and \sqrt{c}=k\sqrt{h}.

    \begin{equation*}\sqrt{a}+\sqrt{b}=\sqrt{c}\end{equation}

    \begin{equation*}e\sqrt{d}+g\sqrt{f}=k\sqrt{h}\end{equation}

For the above equation to be possible d, f and h must simplify to the same surd. Because we are looking for one set of numbers, let d=f=h.

    \begin{equation*}e\sqrt{d}+g\sqrt{d}=k\sqrt{d}\end{equation}

    \begin{equation*}e+g=k\end{equation}

Let’s think of some numbers that might work…

1+2=3 or 2+3=5, etc.

Let’s try e=1, g=2, and k=3

We now have a=d, b=4d, and c=9d

As a can’t be a square number, d can’t be a square number.

Try d=2

    \begin{equation*}\sqrt{2}+\sqrt{8}=\sqrt{18}\end{equation}

LHS=\sqrt{2}+2\sqrt{2}

LHS=3\sqrt{2}

LHS=\sqrt{9\times 2}

LHS=\sqrt{18}

LHS=RHS

One set of possible numbers are 2, 8,and 18.

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June 4, 2025 · 5:30 am

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

At 1pm, object H travelling with constant velocity \begin{pmatrix}200\\10\end{pmatrix}km/h is sighted at the point with position vector \begin{pmatrix}-90\\-100\end{pmatrix}km. At 2pm object J travelling with constant velocity \begin{pmatrix}100\\-100\end{pmatrix}km/h is sighted at the point with position vector \begin{pmatrix}20\\-120\end{pmatrix}km. Determine the minimum distance between H and J and when this occurs.

OT Lee Mathematics Specialist Year 11 Unit 1 and 2 Exercise 10.1 Question 6.

(1)   \begin{equation*}\mathbf{r_H}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}\end{equation*}

(2)   \begin{equation*}\mathbf{r_J}=\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix}\end{equation*}

\begin{pmatrix}-80\\-20\end{pmatrix} is the position vector of J at 1pm.

Find the relative displacement of H to J

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\mathbf{r_H}-\mathbf{r_J}\end{equation}

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}-(\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix})\end{equation}

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix}\end{equation}

Find the relative velocity of H to J

    \begin{equation*}\mathbf{_H}\mathbf{v_J}=\begin{pmatrix}100\\110\end{pmatrix}\end{equation}

The relative displacement is perpendicular to the relative velocity at the closest approach.

That is

(3)   \begin{equation*}\mathbf{_H}\mathbf{r_J}\cdot\mathbf{_H}\mathbf{v_J}=0\end{equation*}

    \begin{equation*}(\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix})\cdot(\begin{pmatrix}100\\110\end{pmatrix})=0\end{equation}

    \begin{equation*}(-10+100t)(100)+(-80+110t)(110)=0\end{equation}

    \begin{equation*}-1000+10 000t-8800+12100t=0\end{equation}

    \begin{equation*}22100t=9800\end{equation}

    \begin{equation*}t=\frac{98}{221}\end{equation}

Substitute t=\frac{98}{221} into the relative displacement and find the magnitude.

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+\frac{98}{221}\begin{pmatrix}100\\110\end{pmatrix}\end{equation}

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\end{equation}

    \begin{equation*}\|\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\|=46.4\end{equation}

The closest objects H and J get to each other is 46.4km at 1:27pm.

I have made an e-activity for this.

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May 28, 2025 · 1:33 pm

A Counting Question

How many three digit numbers can you form from the digits 1, 2, 3, 4 and 5 if

(a) the digits must occur in increasing order?

(b) adjacent digits differ by 2?

Cambridge Year 11 Specialist Mathematics Skill Sheet 1A

(a) There are 5\times 4\times 3=60 permutations of three digits from the five digits, but how many of those are in the right order?

Each set of 3 digits has 6 arrangements (3\times 2\times 1=6).

For example, if the set is {1, 2, 3}, then the possible arrangements are:

123, 132, 213, 231, 312, and 321.

Only one of those arrangements is in numerical order.

Hence what we want is \begin{pmatrix}5\\3\end{pmatrix}=10

(b) I think this one is more about creating a list. I shall start with 1

135, 531, 131, 242, 353, 424, 535, 315

There are 8 possibilities.

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