Module 5 Higher June 2005 Paper 1

Module 5 Higher

June 2005 Paper 1

1. Which is larger 5/6 or ¾?

You must show your working.

You may use the grids to help you.

(2 marks)

You could shade 5/6 of one grid (10 squares) and ¾ of the other (9 squares). This shows easily which is bigger.

Or, think of 5/6 as 10/12,

and think of 3/4 as 9/12.

Then compare these instead

2 Find, using trial and improvement, an exact solution of

3x2 – 2x = 96

x 3x2 – 2x Comment

1 1 Too small

(3 marks)

It’s a non-calc paper, so the answer must be “nice”!!

x 3x2 – 2x Comment

1 1 Too small

5 3x25 – 10 = 65 Too small

(3 marks)

2 Find, using trial and improvement, an exact solution of

3x2 – 2x = 96

x 3x2 – 2x Comment

1 1 Too small

5 3x25 – 10 = 65 Too small

7 3x49 – 14 = 133 Too big

(3 marks)

2 Find, using trial and improvement, an exact solution of

3x2 – 2x = 96

x 3x2 – 2x Comment

1 1 Too small

5 3x25 – 10 = 65 Too small

7 3x49 – 14 = 133 Too big

6 3x36 – 12 = 96 Exact

(3 marks)So x = 6

2 Find, using trial and improvement, an exact solution of

3x2 – 2x = 96

3 (a) A, B and C are three towns which form an equilateral triangle as shown.

Use the bearings given to complete the sentences.

060o 120o 180o 240o 300o

(i) C is on a bearing of from A.

(ii) B is on a bearing of from C

(1 mark)

(1 mark)

Bearings are just angles measured from north, going

clockwise

120°

240°

(b) D, E, F are three towns.E and F are shown on the diagram.D is on a bearing of 070o from E.D is also on a bearing of 320o from F.

Complete the diagram to show accurately the position of D

( 2 marks)

Protractor cross on the town, 0° pointing north!!

Draw the two lines faintly, then mark the cross.

D

4. The diagram shows a parallelogram ABCD.Angle BAC = 20o

Angle ADC = 70o

(a)Show that angle x is a right angle.

(b) The area of the triangle ADC is 8.4cm2

Work out the area of the parallelogram.

(2 marks)

(2 marks)

If D is 70°, then so is B.

In the triangle ABC, 20° + 70° = 90°.

So 90° is left at C. (2)

The two triangles are congruent (exactly the same), so the area is

2 x 8.4 = 16.8 cm2. (2)

5 (a) Factorise 10p – 4

(b) Factorise q2 + 3q

(c) Factorise r2 – r

(d) Simplify t2 x t3

(1 mark)

(1 mark)

(1 mark)

(1 mark)

“factorise” – see what’s in both bits

In the first one, 2 is.

= 2(5p – 2)

= q(q + 3)

= r(r - 1)With multiplying powers,

ADD

= t5

6 (a) Complete the table of values for y = x3 – 4

x -2 -1 0 1 2

y -5 -4(2 marks)

Do some jotting if it will help you get it

right!!

e.g. 23-4 = 8-4 = 4

-3 4 -12

(b) On the grid, draw the graph of y = x3 – 4 for values of x from -2 to 2.

x-6 -4 -2 2 4 6

y

-10

-5

5

7 (a) Solve the inequality 3x + 5 ≤ 16

(2 marks)

(b) Write down the integer value satisfied by the inequality 5 < 2x < 7

(2 marks)

Treat it like an equation!

3x + 5 ≤ 16

3x ≤ 11

x ≤ 113

“Integer” = whole number

2x must be 6, so x = 3

8a Factorise r6 – 3r4

r4 goes into both parts, so use that as the outside the bracket term.

r6 – 3r4 = r4(r2 – 3)

8b (i) Factorise x2 + 5x – 14

(ii) Hence solve the equation x2 + 5x – 14 = 0

Need two numbers that x to make -14 and + to make 5. The numbers are 7 and -2.

So, (x + 7)(x – 2) (2)

(x + 7)(x – 2) = 0 means:

Either (x + 7) = 0 or (x – 2) = 0

So, either x = -7 or x = 2 (1)

9 Two congruent triangles are shown.Angle B = Angle E

(a) Write down the length of DF.

(b) Explain why angle A = angle D(1 mark)

(1 mark)

“congruent” = exactly the same size and shape.

Always draw them the same way round

first.Not the 5.6 side, not the longest side, so must be the 9.4cm side.

e.g. The triangles are congruent, so the smallest angles must be equal.

10 (a) Write down the equation of a line that is parallel to the line y = 5x

(1mark)

(b) Work out the gradient of the line y + 2x = 6

(2 marks)

If it’s parallel, then the gradient is the same.So, “y = 5x + anything” would work.e.g. y = 5x + 1, y = 5x – 7, etc etc

Change the equation round so that it looks like “y = ▒x + ▓”. Then the number with the x is the gradient.

y + 2x = 6

y = -2x + 6, so the gradient is -2

11 The diagram shows a cylinder.The diameter of the cylinder is 10cm.The height of the cylinder is 10cm.

(a)Work out the volume of the cylinder.

Give your answer in terms of π.

(3 marks)

Use V = πr2 x h

= π x 52 x 10

= 250 π cm3.

Base area x height works for any prism

π250 loses a mark – always put the number

first!

(b) Twenty of the cylinders are packed in a box of height 10cm.The diagram shows how the cylinders are arranged inside the box.The shaded area is the space between the cylinders.

Work out the volume inside of the box that is not filled by the cylinders.

Give your answer in terms of π.

(4 marks)

The box must be 50 by 40 cm, and 10 cm high.

Volume of the box is

50 x 40 x 10 = 20 000 cm3.

There are 20 cylinders inside, each with a volume of 250π.

Volume of cylinders is 5000π cm3.

Volume of empty space is 20 000 - 5000π cm3. (4 marks)

Write down the value of a. (1)

12a) In the diagram, O is the centre of a circle.

Remember, “write down” is different from

“work out”

a is 45° (because angle at the middleis double the angle at the edge)

Write down the value of b. (1)

12b)

b = 53° (angles made from the same points are equal).

Write down the value of c. (1)

12c) In the diagram, O is the centre of a circle.

c = 90° (because the angle in a semicircle is always

90°)

Write down the value of d. (1)

12d)

d = 80° (because in a “cyclic quadrilateral”, opposite angles add up to 180°)

13 Shape B is an enlargement of shape A

a)Write down the centre of enlargement

b)Write down the scale factor of the enlargement

(-1,0) (1)

sf = -0.5 (1)

(not -2!!)

14 In triangle ABC, M is the midpoint of BC.

and sAB tACa) Find in terms of s and t.

AM

BCABAM2

1

)(2

1sts AM

Why??

sts2

1

2

1AM ts

2

1

2

1AM or (3)

b)

AB is not equal to AC.

(i) Write down the name of the shape ABDC.

(ii) Write down one fact about the points A M and D

tsAD

Parallelogram

Either AMD are all on the same straight line, because

and

so they’re parallel and from the same point

)(2

1tsAM

)( tsAD

Or M is halfway between A and D, because

and)(2

1tsAM

)( tsAD

15 When y = 2x + 3 and y = -2x -1 are drawn on a grid, there are two lines of symmetry.

Find their equations.

x-4 -2 2 4

y

-4

-2

2

4

The equations are x = -1 and y = 1

16a Find the values of a and b so that

x2 + 10x + 40 = (x + a)2 + b

Do ½ of the x term, and multiply

out (x + 5)2 = x2 + 5x + 5x + 25

So, x2 + 10x + 40 = (x + 5)2 + 15

a = 5, b = 15

(x + 5)2 = x2 +10x + 25

16b Hence write the minimum value of x2 + 10x + 25.

Minimum value is 15 (which happens when (x + 5)2 = 0)

17 The diagram shows the graph of y = cos x° for 0° ≤ x ≤ 360°

For each equation, write down the number of solutions in the range 0° ≤ x ≤ 360°

a) cos x = -0.5

2 solutions

x90 180 270 360

y

-2

-1

1

2

For each equation, write down the number of solutions in the range 0° ≤ x ≤ 360°

b) 2cos x = -0.5

2 solutions

x90 180 270 360

y

-2

-1

1

2

For each equation, write down the number of solutions in the range 0° ≤ x ≤ 360°

c) ½ cos x = -0.5

1 solution

x90 180 270 360

y

-2

-1

1

2

For each equation, write down the number of solutions in the range 0° ≤ x ≤ 360°

d) cos 2x = -0.5

4 solutions

x90 180 270 360

y

-2

-1

1

2

18 Prove that

“Prove” means start from the question, and

manipulate until you end up with the “answer”!

)1(

)12(2

1

12

xx

x

x

x

x

x

1

12

x

x

x

x)1(

)1()1)(2(

xx

xxxx

)1(

)()22( 22

xx

xxxxx

)1(

23 22

xx

xxxx

)1(

23

xx

xx

)1(

24

xx

x

)1(

)12(2

xx

xas required

19 The graph shows four curves A B C and D.Match each curve to its equation.

xy

1 xy 2

xy

4xy 3

A D

C B