Calculators are allowed for all questions unless indicated by the question. There are 5 questions for this topic. Alternatively, you can download the question paper and mark scheme using the links below.

Question 1:

A circle has equation $x^2&space;+y^2&space;=&space;z$

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

B. Find the equation of the tangent to the circle at the point $(-4,&space;2)$  (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  $\frac{2}{-4}&space;=&space;-\frac{1}{2}$  (1 mark)

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

Using $y=&space;mx&space;+c$

$2&space;=&space;2(-4)&space;+&space;c$

$c&space;=&space;10$

$y&space;=&space;2x&space;+&space;10$ (2 marks)

$z$ is equal to the radius squared.

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

Question 2:

Solve the simultaneous equations:

$y^2&space;=&space;x$

$y&space;=&space;4-3x$ (4 marks)

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

$x&space;=&space;y^2$
$y&space;=&space;4-3y^2$
$-3y^2&space;-&space;y&space;+&space;4&space;=&space;0$ (1 mark)

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

Subbing values of y into the original equation gives $x&space;=&space;(-\frac{4}{3})^2&space;=&space;\frac{16}{9}$ 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 $(2n+5)^2&space;-&space;(2n+1)^2$ (1 mark)

$=&space;4n^{2}&space;+&space;20n&space;+&space;25&space;-&space;4n^{2}&space;-4n&space;-1$ $=&space;16n&space;+&space;24$ (1 mark)

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

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

$16n&space;+&space;24=8(2n+3)$ (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 $\\\frac{(x^2&space;+&space;2x-15)(2x&space;+&space;4)}{(x^2-4)(x^2-7x+12)}$ 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)      $(2x+4)&space;=&space;(x+2)(2),&space;(x^2&space;-4)&space;=&space;(x+2)(x-2)$

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

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

Question 5:

$(x&space;+&space;1)(x&space;+2y&space;+&space;1)&space;=&space;(2n&space;+&space;2)^2$

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)

$(x&space;+&space;1)(x&space;+2y&space;+&space;1)&space;=&space;m$

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.