March 9, 2025 · 11:26 am

Circle Geometry Question

In the above diagram O is the centre of the larger circle. A, B,D and E are points on the circumference of the larger circle. A, C, E and 0 are points on the circumference of the smaller circle. Show that \angle{CAB}=\angle{ABC}. AB, AC and BC are straight lines.

AO=OB (radii of the larger circle)

At a line from O to E (it is also a radius of the larger circle)

Let \angle{CAB}=\alpha.

ACEO is a cyclic quadrilateral.

Hence, \angle{CED}=180-\alpha (AECO is a cyclic quadrilateral)

As CB is a straight line \angle{OEB}=180-(180-\alpha)=\alpha.

\Delta OEB is an isosceles triangle.

Therefore, \angle{ABC}=\alpha

Therefore \angle{ABC}=\angle{CAB}

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Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

March 5, 2025 · 2:00 am

Binomial Expansion Theorem

My Year 11 Mathematics Methods students are working on the Binomial Expansion Theorem.

But before we get onto that, remember Pascal’s triangle

First 8 rows of Pascal’s triangle

Now we can use combinations to find the numbers in each row. For example, 1 4 6 4 1 is \begin{pmatrix}4\\0\end{pmatrix}=1, \begin{pmatrix}4\\1\end{pmatrix}=4, \begin{pmatrix}4\\2\end{pmatrix}=6, \begin{pmatrix}4\\3\end{pmatrix}=4, \begin{pmatrix}4\\4\end{pmatrix}=1

ExpressionExpansionCo-efficients
(x+y)^2x^2+2xy+y^21, 2, 1
(x+y)^3x^3+3x^2y+3xy^2+y^31, 3, 3, 1
(x+y)^4x^4+4x^3y+6x^2y^2+4xy^3+y^41, 4, 6, 4, 1

As you can see, the coefficients are the row of pascal’s triangle corresponding to the power. So (x+y)^6 would have co-efficients from the sixth row of the table 1, 6, 15, 20, 15, 6, 1.

To generalise

(x+y)^n=\begin{pmatrix}n\\0\end{pmatrix}x^ny^0+\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}y^1+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}y^2+ ...+\begin{pmatrix}n\\n-1\end{pmatrix}x^1{y^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}x^0y^n

Which we can condense to

(x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i

Worked Examples

(1) Expand (2x-3)^4

(2x-3)^4=\begin{pmatrix}4\\0\end{pmatrix}(2x)^4(-3)^0+\begin{pmatrix}4\\1\end{pmatrix}(2x)^3(-3)^1+\begin{pmatrix}4\\2\end{pmatrix}(2x)^2(-3)^2+\begin{pmatrix}4\\3\end{pmatrix}(2x)^1(-3)^3+\begin{pmatrix}4\\4\end{pmatrix}(2x)^0(-3)^4
(2x-3)^4=16x^4-96x^3+216x^2-216x+81

(2) Find the co-efficient of the x^3 term in the expansion of (2-5x)^5.

Remember (x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i, the x^3 is when i=3
\begin{pmatrix}5\\3\end{pmatrix}(2)^2(-5)^3=10\times 2\times -125=-5000

(3) Find the constant term in the expansion of (x^2+\frac{3}{x^4})^6

We need to find the term where the x‘s cancel out. Each term is \begin{pmatrix}6\\i\end{pmatrix}(x^2)^{6-i}(\frac{3}{x^4})^i.
\begin{pmatrix}6\\i\end{pmatrix}(x^{12-2i})(3^ix^{-4i}).
We need 12-2i-4i=0, hence i=2
Therefore, the co-efficient is \begin{pmatrix}6\\2\end{pmatrix}\times3^2=135

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Filed under Algebra, Binomial Expansion Theorem, Counting Techniques, Year 11 Mathematical Methods

March 3, 2025 · 2:51 am

Derangements

From Wikipedia

In combinatorial mathematics, a derangement is a permutation of the elements of a set in which no element appears in its original position.

For example, if we have the set {A, B, C}, there are 6 permutations

ABC, ACB, BAC, BCA, CAB, CBA

But only 2 of them are derangements – BCA and CAB

I did a practical question on this here. In that question I used a tree diagram, but there must be a way to determine the number of derangements given n elements.

We know 3 elements has 2 derangements, what about 4 elements? We know there are 4!=24 permutations

ABCDBACDCABDDABC
ACBDBADCCADBDACB
ACDBBCADCBADDBAC
ABDCBCDACBDADBCA
ADBCBDACCDABDCAB
ADCBBDCACDBADCBA

I have highlighted the derangements. when n=4 there are 9 derangements.

Let’s try to generalise.

  • In how many ways can one element be in its original position?
    \begin{pmatrix}4\\1\end{pmatrix} \times 3!=24
    Choose one element from the 4 to be in its original position, and then arrange the remaining three elements.

    So far, we have D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!=0

    Clearly we are counting some arrangements multiple times, for example ABDC has A and B in the correct position, so we need to add all of the arrangements with 2 elements in their original position.
  • In how many ways can two elements be in their original position?
    \begin{pmatrix}4\\2\end{pmatrix} \times 2!=12

    So now we have, D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!=12

    But once again we have counted some arrangements multiple times, so we need to subtract all of the arrangements with 3 elements in their original position.
  • In how many ways can three elements be in their original position?
    \begin{pmatrix}4\\3\end{pmatrix} \times 1!=4

    So now we have, D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!=8

    We now need to add all of the arrangements with 4 elements in their original position.
  • In how many ways can four elements be in their original position?
    \begin{pmatrix}4\\4\end{pmatrix} \times 0!=1

    So now we have, D_4=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!+\begin{pmatrix}4\\4\end{pmatrix} \times 0!=9

Hence,

D_n=\begin{pmatrix}n\\0\end{pmatrix}\times n!-\begin{pmatrix}n\\1\end{pmatrix}\times (n-1)!+\begin{pmatrix}n\\2\end{pmatrix}\times (n-2)!-... \mp \begin{pmatrix}n\\n\end{pmatrix}\times 0!

Which we can simplify to

D_n=\Sigma_{i=0}^{n}(-1)^n\begin{pmatrix}n\\i\end{pmatrix}\times(n-i)!

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February 25, 2025 · 9:28 am

Differentiating f(x)=e^x

We are going to differentiate f(x)=e^x from first principals.

Remember the definition of a derivative is

(1)   \begin{equation*}f'(x)=\lim_{\limits{h\to 0}}\frac{f(x+h)-f(x)}{h}\end{equation*}

If f(x)=e^x, then

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^{x+h}-e^x}{h}\end{equation}

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x\times e^h-e^x}{h}\end{equation}

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x(e^h-1)}{h}\end{equation}

    \begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}

Let’s think about \lim_{\limits{h \to 0}}\frac{e^h-1}{h}

Remember e is defined as \lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n

(2)   \begin{equation*}\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation*}

Let y=e^h-1, as h \to 0, e^h-1 \to 1-1=0 hence y \to 0

If y=e^h-1, then h=ln(y+1)

(3)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{y}{ln(y+1)}\end{equation*}

We are going to rewrite the equation 3 as

(4)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{\frac{1}{ln(y+1)}}{y}\end{equation*}

And then we can write equation 4 as

(5)   \begin{equation*}{\lim_{\limits{y \to 0}}{\frac{1}{\frac{1}{y}ln(y+1)}\end{equation*}

Using log laws we can write equation 5 as

(6)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{1}{ln(y+1)^{\frac{1}{y}}}\end{equation*}

Let y=\frac{1}{n}

As y \to 0, n \to \infty

equation 6 becomes

(7)   \begin{equation*}\lim_{\limits{n \to \infty}}\frac{1}{ln(\frac{1}{n}+1)^n}\end{equation*}

We can move the limit to inside the natural log

(8)   \begin{equation*}\frac{1}{ln(\lim_{\limits{n \to \infty}}(\frac{1}{n}+1)^n)}\end{equation*}

And we know from the definition of e that e=\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n

Hence, equation 8 is

(9)   \begin{equation*}\frac{1}{ln(e)}=1\end{equation*}

Back to our derivative

    \begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}

We know that \lim_{\limits{h \to 0}}{\frac{e^h-1}{h}=1 hence

    \begin{equation*}f'(x)=e^x\end{equation}

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February 19, 2025 · 6:15 am

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

How do we go about finding the rule for a first order linear recurrence relation?

Something like

    \begin{equation*}5, 7, 11, 19, ...\end{equation}

There isn’t a common difference (arithmetic sequence) or a common ratio (geometric sequence). Sometimes you can just see the rule, but an algorithm will be handy.

Let’s say our relationship is

(1)   \begin{equation*}T_{n+1}=bT_n+c, T_1=a\end{equation*}

Referring back to our sequence 5, 7, 11, 19, ..., we know

(2)   \begin{equation*}7=b\times5+c\end{equation*}

and

(3)   \begin{equation*}11=b\times7+c\end{equation*}

We can solve equation 2 and 3 simultaneously

equation 2- equation 3

    \begin{equation*}-4=-2b\end{equation}

Hence b=2

Substitute b=2 into equation 2

    \begin{equation*}7=2\times 5+c\end{equation}

    \begin{equation*}7=10+c\end{equation}

Hence c=-3

    \begin{equation*}T_{n+1}=2T_n-3, T_1=5\end{equation}

Let’s try to generalise

If T_{n+1}=bT_n+c, T_1=a, then

(4)   \begin{equation*}T_2=bT_1+c\end{equation*}

and

(5)   \begin{equation*}T_3=bT_2+c\end{equation*}

Equation 5 - equation 4

    \begin{equation*}T_3-T_2=(T_2-T_1)b\end{equation}

    \begin{equation*} b=\frac{T_3-T_2}{T_2-T_1}\end{equation}

Hence, b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}

Once you know b, substitute into either equation to find C.

Example

Find the recursive rule for the following

-8, -12, -20, -36, ...

    \begin{equation*}b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}=\frac{-20-(-12)}{-12-(-8)}=\frac{-8}{-4}=2\end{equation}

    \begin{equation*}-12=2\times-8+c\end{equation}

    \begin{equation*}-12=-16+c\end{equation}

Hence c=4 and T_{n+1}=2T_n+4, T_1=-8

It is also possible to find the rule using a Classpad (if it’s in the calculator section} by using an e-activity.

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February 15, 2025 · 8:35 am

Counting Techniques

Four teachers decide to swap desks at work. How many ways can this be done if no teacher sits at their previous desk?
Mathematics Specialist Units 1&2 Cambridge

I like this question as it seems easy until you start thinking about it. I think the best approach is a tree diagram.

If we think of the four teachers as A, B, C and D. Then A can no longer sit in A, so the options are B, C and D for the first desk.

For the second desk, If B is in the first desk, then A, C or D could be in the second. If C is in the first desk, then A or D could be in the second (B can’t be in the same desk). If D is in the first desk, then A or C can be in the second desk.

And so on, leaving 9 possibilities

BADC
BCDA
BDAC
CADB
CDAB
CDBA
DABC
DCAB
DCBA

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February 4, 2025 · 9:17 am

Differentiating the Tangent Function

Remember tan(x)=\frac{sin(x)}{cos(x)}.

I use the quotient rule to differentiate f(x)=tan(x).

(1)   \begin{equation*}\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\end{equation*}

If h(x)=tan(x)=\frac{sin(x)}{cos(x)} then from equation 1

(2)   \begin{equation*}h'(x)=\frac{cos(x)cos(x)-(-sin(x)sin(x))}{[cos(x)]^2}\end{equation*}

(3)   \begin{equation*}h'(x)=\frac{cos^2(x)+sin^2(x)}{cos^2(x)}\end{equation*}

Remember the Pythagorean identity

(4)   \begin{equation*}sin^2(x)+cos^2(x)=1\end{equation*}

Hence

    \begin{equation*}h'(x)=\frac{1}{cos^2(x)}=sec^2(x)\end{equation}

(5)   \begin{equation*}\frac{d}{dx}tan(x)=sec^2(x)\end{equation*}

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February 3, 2025 · 9:11 am

Differentiating Trigonometric Functions

In the last post we looked at two trig limits:

(1)   \begin{equation*}\lim_{x \to 0}\frac{sin(x)}{x}=1\end{equation*}

(2)   \begin{equation*}\lim_{x \to 0}\frac{1-cos(x)}{x}=0\end{equation*}

We are going to use these two limits to differentiate sine and cosine functions from first principals.

    \begin{equation*}f(x)=sin(x)\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x+h)-sin(x)}{h}\end{equation}

Use the trig identity

    \begin{equation*}sin(A+B)=sinAcosB+sinBcosA\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}(\frac{sin(x)(cos(h)-1)}{h}+\frac{sin(h)cos(x)}{h})\end{equation}

    \begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{(cos(h)-1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

    \begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{-(-cos(h)+1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

Evaluate the limits

    \begin{equation*}f'(x)=sin(x)\times 0+cos(x)\times (1)=cos(x)\end{equation}

Hence, \frac{d}{dx}sin(x)=cos(x).

Now we are going to do the same for f(x)=cos(x).

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x+h)-cos(x)}{h}\end{equation}

Use the trigonometric identity

    \begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)(cos(h)-1)-sin(x)sin(h)}{h}\end{equation}

    \begin{equation*}f'(x)=cos(x)\lim\limits_{h \to 0}\frac{-(1-cos(h))}{h}-sin(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

Evaluate the limits

    \begin{equation*}f'(x)=cos(x)\times(0)-sin(x)\times (1)=-sin(x)\end{equation}

Hence \frac{d}{dx} cos(x)=-sin(x)

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February 2, 2025 · 7:46 am

Trigonometric Limits

\lim\limits_{x \to 0}\frac{sin(x)}{x}=?

Unit Circle

Remember cos(x)=\frac{OA}{OB}=\frac{OA}{1}, hence OA=cos(x) and the co-ordinate of A is (cos(x), 0).

sin(x)=\frac{AB}{OB}=\frac{AB}{1}, hence AB=sin(x) and the co-ordinate of B is (cos(x), sin(x))

And from the definition of tan(x) we know D is the point (1, tan(x))

Consider the areas of triangle OAB, sector OBC, and triangle OCD.

We know from inspection of the above diagram that

Area OAB< Area OCB<Area OCD

Which means,

\frac{1}{2}b_1 h_1<\frac{1}{2}r^2x<\frac{1}{2}b_2 h_2

We can ignore all of the halves.

cos(x)sin(x)<x<(1)tan(x)

Remember tan(x)=\frac{sin(x)}{cos(x)}

cos(x)sin(x)<x<\frac{sin(x)}{cos(x)}

Divide everything by sin(x) (as we are in the first quadrant we know sin(x)>0, so we don’t need to worry about the inequality)

cos(x)<\frac{x}{sin(x)}<\frac{1}{cos(x)}

Invert everything and change the direction of the inequalities)

\frac{1}{cos(x)}>\frac{sin(x)}{x}>cos(x)

I am going to rewrite it as follows

cos(x)<\frac{sin(x)}{x}<\frac{1}{cos(x)}

because I like to use less thans rather than greater thans.

Now what happens as x tends to 0?

cos(0)=1

1<\frac{sin(x)}{x}<\frac{1}{1}

Hence by the squeeze theorem \lim\limits_{x \to 0}\frac{sin(x)}{x}=1

Now we know this limit, we are going to use it to find \lim\limits_{x \to 0}\frac{1-cos(x)}{x}

Multiply by \frac{1+cos(x)}{1+cos(x)}

\lim\limits_{x \to 0}\frac{1-cos(x)}{x}\times \frac{1+cos(x)}{1+cos(x)}

\lim\limits_{x \to 0}\frac{1-cos^2(x)}{x(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin^2(x)}{x(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin(x)}{x}\times sin(x)(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin(x)}{x}\times \lim\limits_{x \to 0}sin(x)(cos(x)+1)

If we evaluate the limits,

(1)(sin(0)(cos(0)+1)=1\times 0 \times 2=0

Hence, \lim\limits_{x \to 0}\frac{1-cos(x)}{x}=0

In the next post we are going to use these limits to differentiate sine and cosine functions.

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January 16, 2025 · 8:08 am

HSC Advanced 2024 Question 30

HSC Advanced 2024

Two circles have the same centre O. The smaller circle has a radius of 1 cm, while the larger has a radius of (x+1) cm. The circles enclose a region QRST, which is subtended by angle of {\theta} at O, as shaded.

The area of QRST is A cm2, where A is a constant and A>0

Let P cm be the perimeter of QRST

(a) By finding expressions for the area and perimeter of QRST show that P(x)=2x+\frac{2A}{x}

(b) Show that if the perimeter is minimised, then {\theta} must be less than 2.

(a) A=\frac{1}{2}\theta((x+1)^2-1^2)
A=\frac{1}{2}\theta(x^2+2x)
2A=\theta x^2 +2x \theta
\frac{2A}{x}=\theta x +2\theta

P=\theta(1)+\theta(1+x)+2x
P=2\theta +\theta x +2x
P=\frac{2A}{x}+2x

I like it when the first part requires the student to show something and the second part has them use it (that way they can still do the second part even if they couldn’t do the first part).

(b) \frac{dP}{dx}=2-\frac{2A}{x^2}
0=2-\frac{2A}{x^2}
x=\sqrt{A}

\frac{d^2P}{dx^2}=\frac{4A}{x^3}
Both x and A are greater than zero, therefore \frac{d^2P}{dx^2}>0 and x=\sqrt{A} is a minimum.

Substitute x=\sqrt{A} into the Area formula
2A=A\theta+2\sqrt{A}\theta
\theta=\frac{2A}{A+2\sqrt{A}}
\theta=\frac{2A}{\sqrt{A}(\sqrt{A}+2)}
\theta=\frac{2\sqrt{A}}{\sqrt{A}+2}

Now \frac{2\sqrt{A}}{\sqrt{A}+2}<\frac{2\sqrt{A}}{\sqrt{A}}
Hence \theta<\frac{2\sqrt{A}}{\sqrt{A}}
and \theta<2

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