Calculators are allowed for all questions unless indicated by the question. There are 5 questions for this topic. The difficulty increases with the question number.

Question 1:

A circle has equation  $x^2 +y^2 = z$

A. Find the centre of the circle (1 mark)

B. Find the equation of the tangent to the circle at the point $(-4, 2)$  (4 marks)

C. Calculate the value of $z$ (1 mark)

Centre $(0,0)$

Gradient of the line from the centre of the circle to the point $(-4, 2)$ is$\frac{2}{-4} = -\frac{1}{2}$ (1 mark)

The gradient of the tangent is the negative reciprocal of  $-\frac{1}{2}$ which is +2. (1 mark)

Using $y = mx + c$

$2 = 2(-4) + c$

$c = 10$

$y = 2x + 10$ (2 marks)

$z$ is equal to the radius squared.

$z = 4^2 + 2^2 = 20$ (1 mark)

Question 2:

Solve the simultaneous equations:

$y^2 = x$

$y = 4-3x$ (4 marks)

Solve for y first so you can eliminate the need for negative square roots.

$x=y^2$

$y = 4-3y^2$

$-3y^2 -y + 4 = 0$ (1 mark)

Therefore $y = -\frac{4}{3}$or $1$ (1 mark)

Subbing values of y into the original equation gives$x = (\frac{4}{3})^2 = \frac{16}{9}$ and 1 (2 marks)

Award full marks of any alternative method

Question 3:

You are given 2 odd integers with a difference of 4.

Prove that the difference between the squares of the integers is 8 times the mean of the integers. (5 marks)

Odd integers are $(2n+1)$ and $(2n+5)$. (1 mark for selecting 2 odd integers)

Doubling any number and adding 1 will always be odd, and adding 4 to that number will always be odd.

The difference between the squares of the integers will be $(2n+5)^2 – (2n+1)^2$(1 mark)

$4n^2 + 20n + 25 -4n^2 -4n -1 = 16n + 24$ (1 mark)

The mean of the original integers is the integers added together divided by the number of integers:

$\frac{2n + 1 + 2n + 5}{2} = \frac{4n+6}{2} = 2n+3$ (1 mark)

$16n + 24 = 8(2n+3)$(1 mark)

Therefore, the difference between the squares of the integers is 8 times the mean.

Award full marks for any alternative method including using different odd integers in terms of n.

Question 4:

Simplify the fraction $\frac{(x^2 +2x -15)(2x +4)}{(x^2-4)(x^2 -7x + 12)}$ in the form $\frac{a(x+b)}{(x+c)(x+d)}$ where a,b,c and d are integers to be found (4 marks)

$\frac{(x^2 +2x -15)(2x +4)}{(x^2-4)(x^2 -7x + 12)} = \frac{(x+5)(x-3)(2x+4)}{(x+2)(x-2)(x-3)(x-4)}$ (1 mark)

$\frac{(x+5)(x-3)(x+2)(2)}{(x+2)(x-2)(x-3)(x-4)}$ (1 mark)

$(2x+4) = (x+2)(2), x^2 -4 = (x+2)(x-2)$

$=\frac{(x+5)(x-3)(2)}{(x-2)(x-3)(x-4)}$ (1 mark)

$\frac{2(x+5)}{(x-2)(x-4)}$ (1 mark)

Question 5:

$(x+1)(x+2y+1)=(2n+2)^2$

Where n is a positive integer and y is a positive integer.

Use this statement to prove that $x$ is odd. (4 marks)

As n is a positive integer. $(2n+2)^2$ is also positive. The right side can simply be referred to as m, where m is a positive even integer. (1 mark for recognising that$(2n+2)^2$ is even)

$(x+1)(x+2y+1)=m$

since $(x+1)$ and $(x+2y+1)$ must multiply to make an even number, this can only be true when either both terms are even, or 1 term is odd, and the other term is even.

Scenario 1: Both terms are even.

$x + 1$ is even, therefore x is odd. Then $x + 2y + 1$ is even, so this scenario is possible. (1 mark)

Scenario 2: First term is odd and the second term is even.

$x + 1$ is odd, therefore $x$ is even. Then $x + 2y + 1$ is odd. This is impossible as both $x$ and $y$ are even, thus $x + 2y + 1$ must be odd. This scenario is not possible as both terms are odd. (1 mark)

Scenario 3: the second term is odd and the first term is even.

If $x +1$ is even, $x$ is odd. This will give the same answer as scenario 1 where $x + 2y + 1$ is even, so this scenario doesn’t exist. (1 mark)

Therefore $x$ must be odd.

Award full marks for any alternative method that concludes $x$ is odd.