GCSE Maths (Edexcel 1MA1) · Higher · Grade 8/9 mixed revision · Sheet 1 of 15

Grade 8/9 mixed revision 1

48 marks

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Date

Exam questions

Edexcel GCSE Higher style. Show all your working. The hints are at the end of the sheet. Use one at a time.

X1
The point $P (4, - 9)$ lies on the circle with equation $x^{2} + y^{2} = 97$ .
Find an equation of the tangent to the circle at $P$ .
Give your answer in the form $y = m x + c$ .
[3 marks]
Hint 1 · What to try
The tangent at $P$ is perpendicular to the radius $O P$ . Start by finding the gradient of $O P$ .
Hint 2 · The first line
The gradient of $O P$ is $- \frac{9}{4}$ . The gradient of the tangent is the negative reciprocal of this.
Hint 3 · The full method
Use $y - (- 9) = \frac{4}{9} (x - 4)$ and rearrange into the form $y = m x + c$ . Take care with the sign of $- 9$ .
Answer $y = \dots$

X2
$A$ , $B$ , $C$ and $D$ are points on a circle.
Angle $B A D = (4 x + 10)^{\circ}$
Angle $B C D = (3 x - 5)^{\circ}$
Angle $A D C = (2 x + 30)^{\circ}$
Work out the size of angle $A B C$ .
Give a reason for each stage of your working.
Diagram NOT accurately drawn
[3 marks]
Hint 1 · What to try
Which pairs of angles in a cyclic quadrilateral add up to $18 0^{\circ}$ ?
Hint 2 · The first line
Angles $B A D$ and $B C D$ are opposite each other. Use them to form an equation in $x$ .
Hint 3 · The full method
$7 x + 5 = 180$ . Find $x$ , then work out angle $A D C$ . Angle $A B C$ is opposite angle $A D C$ .
AnswerAngle $A B C = \dots^{\circ}$
X3
$f (x) = 2 x^{2} - 12 x + 23$
(a)
Write $f (x)$ in the form $a (x + b)^{2} + c$ , where $a$ , $b$ and $c$ are integers.
[2 marks]
Answer $\dots$
(b)
Hence write down the coordinates of the turning point of the graph of $y = f (x + 4) - 6$ .
[1 mark]
Answer $(\dots, \dots)$
Total for X3: 3 marks
X4
Write $\frac{5 + 3}{2 - 3}$ in the form $a + b 3$ , where $a$ and $b$ are integers.
You must show all your working.
[3 marks]
Hint 1 · What to try
To remove the surd from the denominator, multiply the top and the bottom by the same expression.
Hint 2 · The first line
Use $2 + 3$ , because $(2 - 3) (2 + 3)$ has no surd in it.
Hint 3 · The full method
The denominator becomes $4 - 3 = 1$ . Expand $(5 + 3) (2 + 3)$ and collect the terms; remember $3 \times 3 = 3$ .
Answer $\dots$
X5
$n$ is an integer.
(a)
Prove that the sum of any three consecutive integers is a multiple of $3$ .
[2 marks]
(b)
Prove that the mean of any four consecutive integers is never an integer.
[2 marks]
Total for X5: 4 marks
X6
Solve the inequality $(x - 1) (x + 6) > 3 x + 18$
You must show all your working.
[4 marks]
Hint 1 · What to try
Expand the left-hand side, then move every term to one side so that the right-hand side is $0$ .
Hint 2 · The first line
You should get $x^{2} + 2 x - 24 > 0$ . Factorise it to find the critical values.
Hint 3 · The full method
The critical values are $- 6$ and $4$ . Sketch $y = x^{2} + 2 x - 24$ : it is above the $x$ -axis to the left of $- 6$ and to the right of $4$ .
Answer
X7
The functions $f$ and $g$ are such that
$f (x) = 2 x + a g (x) = x^{2} - 3 x$
where $a$ is a constant.
Given that $g f (1) = 10$ , find the two possible values of $a$ .
You must show all your working.
[4 marks]
Hint 1 · What to try
Work out $f (1)$ first. It is an expression in $a$ .
Hint 2 · The first line
$g f (1)$ means put $f (1)$ into $g$ : $g f (1) = (2 + a)^{2} - 3 (2 + a)$ .
Hint 3 · The full method
Set $(2 + a)^{2} - 3 (2 + a)$ equal to $10$ , expand and collect to get $a^{2} + a - 12 = 0$ , then factorise.
Answer $a = \dots$ or $a = \dots$
X8
$O A B$ is a triangle.
$O A = 4 a$ and $O B = 4 b$ .
$P$ is the point on $O A$ such that $O P : P A = 1 : 3$ .
$Q$ is the point on $A B$ such that $A Q : QB = 3 : 1$ .
Diagram NOT accurately drawn
(a)
Find $A B$ in terms of $a$ and $b$ .
[1 mark]
Answer $A B = \dots$
(b)
Prove that $P Q$ is parallel to $O B$ , and find the ratio $P Q : O B$ .
[3 marks]
Total for X8: 4 marks
X9
Solve the simultaneous equations
$x^{2} + y^{2} = 41$
$y = 2 x - 3$
You must show all your working.
[5 marks]
Hint 1 · What to try
Replace $y$ in the first equation by $2 x - 3$ .
Hint 2 · The first line
Expand $(2 x - 3)^{2}$ carefully, collect terms and rearrange to a quadratic equal to $0$ .
Hint 3 · The full method
You should get $5 x^{2} - 12 x - 32 = 0$ , which factorises as $(5 x + 8) (x - 4) = 0$ . Find $y$ for each value of $x$ using $y = 2 x - 3$ .
Answer $x = \dots, y = \dots$ or $x = \dots, y = \dots$
X10
This question is about the expression $\frac{2}{x - 3} + \frac{5}{x + 2}$ .
(a)
Write $\frac{2}{x - 3} + \frac{5}{x + 2}$ as a single fraction in its simplest form.
[2 marks]
Answer
(b)
Hence solve $\frac{2}{x - 3} + \frac{5}{x + 2} = 1$
Give your answers in the form $a \pm b$ , where $a$ and $b$ are integers.
[3 marks]
Answer $x = \dots$
Total for X10: 5 marks
X11
Priya cycles $1250$ m, correct to the nearest $10$ m.
Her time is $96.4$ seconds, correct to $1$ decimal place.
Work out the upper bound for her average speed in kilometres per hour.
Give your answer correct to $3$ significant figures.
You must show all your working.
[5 marks]
Hint 1 · What to try
Write down the bounds of the distance and of the time. The distance is to the nearest $10$ m, so the half-unit is $5$ m.
Hint 2 · The first line
Speed is distance divided by time. For the largest speed, divide the largest distance by the smallest time.
Hint 3 · The full method
Work out $1255 \div 96.35$ to get m/s, then multiply by $3600$ and divide by $1000$ to change to km/h.
Answer $\dots$ km/h
X12
The equation $x^{3} - 5 x - 7 = 0$ has one real solution.
(a)
Show that $x^{3} - 5 x - 7 = 0$ can be rearranged to give $x = 3 5 x + 7$
[2 marks]
(b)
Starting with $x_{0} = 2$ , use the iteration formula $x_{n + 1} = 3 5 x_{n} + 7$ three times to find an estimate for the solution of $x^{3} - 5 x - 7 = 0$
Give your answer correct to 4 decimal places.
[3 marks]
Answer $x_{3} = \dots$
Total for X12: 5 marks

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can prove results with vectors, algebra and circle theorems, giving a reason for each step.
I can find tangents to circles and solve non-linear simultaneous equations and quadratic inequalities.
I can work with composite and inverse functions, algebraic fractions, surds and transformations of graphs.
I can use bounds with division and subtraction, and find roots by iteration.

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