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Cambridge Assessment
For examination from 2025
MATHEMATICS
0580/02
Paper 2 Non-calculator
(Extended) For examination from 2025
SPECIMEN PAPER 2 hours
Solved by shahbaz ahmed
February 2024

List of formulas with proof
Area, A, of triangle, base b, height
h
𝐴 = 12 𝑏ℎ

Proof

Reference to the ▱ ABCD in the figure

1

Area of the ▱ ABCD= 𝑏𝑎𝑠𝑒 × 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒
Area of the ▱ ABCD= (𝐷𝐶) × (𝐴𝐿)
Let 𝐷𝐶 = 𝑏 ,𝐴𝐿 = ℎ ⟹
Area of the ▱ ABCD= (𝑏)(ℎ)
Area of the ▱ ABCD= 𝑏ℎ
Area of the △ADC= 12 𝑏ℎ
Area, A, of circle of radius r
...
𝐴 = 2𝜋𝑟ℎ

proof

Reference to the figure

2

Curved surface is made by the rectangle

Curved surface area, A, of cylinder of radius r, height h
...

= 4𝜋𝑟2

Volume, V, of prism,cross-sectional
area A, length l
...
𝑉 = 13 𝐴ℎ

Volume, V, of cylinder of radius r,
height h
...
V=area of the circular base ×ℎ = 𝜋𝑟2 ℎ

Volume, V, of cone of radius r,
height h
...

𝑉 = 43 𝜋𝑟3

For the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
where 𝑎 ≠ 0:
𝑥=

√
−𝑏± 𝑏2 −4𝑎𝑐
2𝑎

6

For the triangle shown,
𝑎
sin 𝐴

=

𝑏
sin 𝐵

=

𝑐
sin 𝐶

𝑎2 = 𝑏2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
Area= 12 𝑎𝑏 sin sin 𝐶

Proof

7

Reference to the triangle

8

Let altitude= 𝐴𝐷
Also in the right angle triangle ADC
𝐴𝐷
𝐴𝐶

= sin 𝐶 ⟹ 𝐴𝐷 = 𝐴𝐶 sin 𝐶

Since 𝐴𝐶 = 𝑏 ⟹ 𝐴𝐷 = 𝑏 sin 𝐶
Also in the right triangle ABD
𝐴𝐷
𝐴𝐵

= sin 𝐵

Since 𝐴𝐵 = 𝑐 ⟹

𝐴𝐷
𝑐

= sin 𝐵 ⟹ 𝐴𝐷 = 𝑐 sin 𝐵

Since
𝐴𝐷 = 𝑐 sin 𝐵 and 𝐴𝐷 = 𝑏 sin 𝐶
⟹ 𝑐 sin 𝐵 = 𝑏 sin 𝐶 ⟹
𝑐
sin 𝐶

=

𝑏
sin 𝐵

Similarly from the same triangle ABC, drawing BE⟂ AC

9

In the right triangle AEB
𝐵𝐸
𝐴𝐵

=

𝐵𝐸
𝑐

= sin 𝐴 ⟹ 𝐵𝐸 = 𝑐 sin 𝐴

Similarly
In the right triangle BEC
𝐵𝐸
𝐵𝐶

= sin 𝐶

Since 𝐵𝐶 = 𝑎 ⟹

𝐵𝐸
𝑎

= sin 𝐶 ⟹ 𝐵𝐸 = 𝑎 sin 𝐶

Now

10

𝐵𝐸 = 𝑐 sin 𝐴 and 𝐵𝐸 = 𝑎 sin 𝐶 ⟹ 𝑐 sin 𝐴 = 𝑎 sin 𝐶
Or

𝑐
sin 𝐶

=

𝑎
sin 𝐴

Combining the results
𝑎
sin 𝐴

=

𝑏
sin 𝐵

=

𝑏
sin 𝐵

=

𝑐
sin 𝐶

and

𝑐
sin 𝐶

=

𝑎
sin 𝐴

𝑐
sin 𝐶

Drawing BE⟂ to AC in the same triangle ABC

In the right angled triangle AEB
cos 𝐴 =

𝐴𝐸
𝐴𝐵

=

𝐴𝐸
𝑐

⟹ 𝐴𝐸 = 𝑐 cos 𝐴

Also

11

we have:

(𝐵𝐸)2 + (𝐴𝐸)2 = (𝐴𝐵)2 Or
(𝐵𝐸)2 = (𝐴𝐵)2 − (𝐴𝐸)2 = 𝑐 2 − (𝑐 cos 𝐴)2 = 𝑐 2 − 𝑐 2 cos2 𝐴
In the right triangle BEC
𝐸𝐶 = 𝐴𝐶 − 𝐴𝐸 = 𝑏 − (𝑐 cos 𝐴)
(𝐵𝐸)2 + (𝐸𝐶)2 = (𝐵𝐶)2
𝑐 2 − 𝑐 2 cos2 𝐴 + (𝑏 − 𝑐 cos 𝐴)2 = 𝑎2
𝑐 2 − 𝑐 2 cos2 𝐴 + 𝑏2 + 𝑐 2 𝑐𝑜𝑠2 𝐴 − 2𝑏𝑐 cos 𝐴 = 𝑎2
𝑐 2 + 𝑏2 − 2𝑏𝑐 cos 𝐴 = 𝑎2
𝑎2 = 𝑏2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
Area of the triangle ABC= 21 (𝐴𝐷)(𝐵𝐶) = 21 (𝑏 sin 𝐶)(𝑎) = 12 𝑎𝑏 sin 𝐶
Q1
Work out (0
...
01)2 = 0
...
01 =
...
3997 correct to 4 significant figures
...
40
Q3
Aimee changes 250 euros into dollars
...
10 dollars
Calculate the number of dollars Aimee receives
...
10 dollars ⟹ 1𝑒𝑢𝑟𝑜 × 250 = 1
...


BDC is a straight line,𝐴𝐵 = 𝐴𝐶 , ∠ ABD= 61◦ and ∠ADC = 81◦
Work out angle DAC
...
17𝑚2 into 𝑐𝑚2
Solution
1𝑚2 = 1𝑚 × 1𝑚 = 100𝑐𝑚 × 100𝑐𝑚 = 10000𝑐𝑚2 Or
1𝑚2 = 10000𝑐𝑚2
⟹ 1𝑚2 × 0
...
17
⟹ 0
...
The volume of the cuboid is 600𝑐𝑚3 Calculate the density of
14

the metal, giving your answer in

𝑔𝑚
𝑐𝑚3

[Density = mass ÷ volume

] Solution
𝑚 = 4𝑘𝑔 = 4𝑡𝑖𝑚𝑒𝑠1000𝑔𝑚 = 4000𝑔𝑚
𝑉 = 600𝑐𝑚3
𝑑=

𝑚
𝑉

=

4000𝑔𝑚
600𝑐𝑚3

=

20 𝑔𝑚
3 𝑐𝑚3

Q7
⎛ ⎞
⎜3⎟
u=⎜ ⎟
⎜−2⎟
⎝ ⎠
⎞
⎛
⎜−12⎟
v=⎜
⎟
⎜ 5 ⎟
⎠
⎝

(a) Find u-2v
(b) Find |v|
Solution
⎞
⎛
⎜−12⎟
2 v = 2⎜
⎟
⎜ 5 ⎟
⎠
⎝

⎛ ⎞ ⎛
⎞
⎜ 3 ⎟ ⎜−12⎟
u-2v = ⎜ ⎟- 2 ⎜
⎟
⎜−2⎟ ⎜ 5 ⎟
⎝ ⎠ ⎝
⎠

⎛ ⎞ ⎛
⎞
⎜ 3 ⎟ ⎜−24⎟
u-2v = ⎜ ⎟- ⎜
⎟
⎜−2⎟ ⎜ 10 ⎟
⎠
⎝ ⎠ ⎝

15

⎛
⎞
⎜ 3 + 24 ⎟
u-2v = ⎜
⎟
⎜−2 − 10⎟
⎝
⎠

⎛
⎞
⎜ 27 ⎟
u-2v = ⎜
⎟
⎜−12⎟
⎝
⎠
(b) ||v||
⎛
⎞
⎜−12⎟ √
=⎜
⎟ = (−12)2 + 52 = ±13
⎜ 5 ⎟
⎝
⎠
Q8

The diagram shows a semicircle with diameter 9cm
...

Give your answer in exact form
...

Here Diameter= 9cm
...
5cm

16

Hence
The total perimeter of this semicircle = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 + 𝜋 × 𝑟𝑎𝑑𝑖𝑢𝑠 = 9 + 4
...
5cm
...

Hence
𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 = 17 + (𝑛 − 1)(−5) = 17 − 5𝑛 + 5 = −5𝑛 + 22
Q10
2
1
Work out 2 + 3 Give your answer as a mixed number in its simplest form
3
2
Solution
11
3

+

7
2

11×2
3×2

+

7×3
2×3

22
6

+

21
6

=

=
=

22
6

+

21
6

22+21
6

=

43
6

17

Hence
2
1
2 +3 =
3
2

43
6

Q11
2

Find the value of 64 3
2

2

64 3 = (43 ) 3 = 42 = 16
Q 12 Work out, giving your answer in standard form,
(a) (7
...
1 × 2) × (10−15 × ×103 )
= 14
...
2 × 10−12
= 1
...
2 × 107 ) + (5
...

= (5
...
2 × 106 )
...
2 × 106 )
...
2) × 106
= 57
...
2 × 10−1 × 107
= 5
...

Since interior angle 162◦
180(𝑛−2)
𝑛

= 162◦

18

180(𝑛−2)
𝑛

180(𝑛 − 2) = 162◦ 𝑛
180𝑛 − 360 = 162𝑛
180𝑛 − 162𝑛 = 360
18𝑛 = 360
𝑛=

360
18

= 20

Q 14 The range, mode, median and mean of five positive integers are all equal to 10
...
Solution
Let numbers are in ascending values are such that 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 , 𝑥5
Let highest=𝑥5 , lowest= 𝑥1
Range= 𝑥5 − 𝑥1 = 10 ⟹ 𝑥5 = 𝑥1 + 10
Median term 𝑥3 = 10
Mean=

𝑥1 +𝑥2 +𝑥3 +𝑥4 +𝑥5
5

⟹

𝑥1 +𝑥2 +𝑥3 +𝑥4 +𝑥5
5

= 10

⟹ 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 = 50
Putting 𝑥5 = 𝑥1 + 10, 𝑥3 = 10
𝑥1 + 𝑥2 + 10 + 𝑥4 + 𝑥1 + 10 = 50
2𝑥1 + 𝑥2 + 𝑥4 = 50 − 20 = 30
Let 𝑥2 , 𝑥4 = 10 as mode value ⟶
2𝑥1 + 10 + 10 = 30
2𝑥1 = 30 − 20
2𝑥1 = 10
𝑥1 = 5
𝑥5 = 𝑥1 + 10 = 5 + 10 = 15
Hence numbers are 5, 10, 10, 10, 15

19

Range= 𝑥5 − 𝑥1 = 10
Q15

Describe fully the single transformation that maps triangle T onto triangle A
...

(a) The cumulative frequency diagram shows the results
...

Number of plants up to height 60 cm=364
Number of plants with a height greater than 60cm=400-364=36
(b) The heights are also shown in the frequency table
...


24

Solution

25

Q17

The diagram shows a cyclic quadrilateral ABCD
...


26

(a) Angle 𝐵𝐴𝐷 = 74◦ and angle BCA = 34◦
Find
(i) angle BDA
(ii) angle BCD
(iii) angle ABD
...

BC = 4
...
3cm
...

Solution
(i)Name angles as in figure
∠𝐵𝐶𝐴 = 34
...

Reference to the figure
∠𝐴𝐶𝐵 = ∠𝐴𝐷𝐵 = 34 = 𝛾

27

In triangle △ ADB
∠𝐵𝐴𝐷 + ∠𝐴𝐵𝐷 + ∠𝐴𝐷𝐵 = 74 + 𝛿 + 𝛾 = 180 Or
74 + 𝛿 + 34 = 180 Or
𝛿 + 108 = 180 Or
𝛿 = 180 − 108 = 72◦ Or
𝑎𝑛𝑔𝑙𝑒𝐴𝐵𝐷 = 72◦
(b) In the diagram, triangle ADX is similar to triangle BCX
...
5 cm, AD = 9cm and CX = 3
...

To Work out XD
...
5
3
...
5
)(9) = ( 35
)(9) = 7𝑐𝑚
4
...

(ii) Find gf(–3)
...

(d) Find x when ℎ−1 (𝑥) = 5
...

g(x)=2x+3
g(g(x)=g(2x+3)=2(2x+3)+3=4x+6+3=4x+9=7
4𝑥 = 7 − 9 = −2
𝑥=

−2
4

= − 21

(d) To find x when ℎ−1 (𝑥) = 5
...

1
√
2+1

Solution
√
√
(a) 32 + 98
√
√
= 42 × 2 + 72 × 2
√
√
√
√
= 42 × 2 + 72 × 2
√
√
=4× 2+7× 2
√
= (4 + 7) × 2
√
= 11 2
(b) Rationalise the denominator
...

Find y when x = 49
...

√
8 4=𝑘
8(2) = 𝑘
√
𝑘 = 16 𝑦 𝑥 = 16
when x = 49
...


31

The height of the triangle is h and the height of the rectangle is (h + 2)
...

The area of the triangle is 11𝑐𝑚2
and the area of the rectangle is 39𝑐𝑚2
(a) Write down an expression, in terms of x, for the height of the rectangle
...

Solution
(a)The area of the triangle = 12 ℎ𝑥 = 11𝑐𝑚2
The area of the triangle ℎ𝑥 = 22𝑐𝑚2
⟹ℎ=

22
𝑥

The height of the rectangle= h+2= 22
+2=
𝑥

2𝑥+22
𝑥

(b) The area of rectangle = (𝑥 + 1)(ℎ + 2) = 𝑥ℎ + 2𝑥 + ℎ + 2 = 39𝑐𝑚2
The area of rectangle = 𝑥ℎ + 2𝑥 + ℎ + 2 = 39𝑐𝑚2
Putting ℎ =

22
𝑥

The area of rectangle = 𝑥( 22
) + 2𝑥 +
𝑥
The area of rectangle = 22 + 2𝑥 +

22
𝑥

22
𝑥

+ 2 = 39

+ 2 = 39

The area of rectangle = 22𝑥 + 2𝑥2 + 22 + 2𝑥 = 39𝑥

32

The area of rectangle = 2𝑥2 + 24𝑥 − 39𝑥 + 22 = 0
The area of rectangle = 2𝑥2 − 15𝑥 + 22 = 0
(c) Factorising 2𝑥2 − 15𝑥 + 22 = 0
2𝑥2 − 11𝑥 − 4𝑥 + 22 = 0
𝑥(2𝑥 − 11) − 2(2𝑥 − 11) = 0
(2𝑥 − 11)(𝑥 − 2) = 0
2𝑥 = 11𝑂𝑟𝑥 =

11
2

Or 𝑥 = 2

Putting in :
ℎ=

22
𝑥

ℎ=

22
11
2

𝑎𝑡𝑥 =

11
2

ℎ = 4𝑐𝑚 at 𝑥 =
ℎ=

22
2

11
2

at 𝑥 = 2

ℎ = 11𝑐𝑚
Two possible heights of the triangle are:
ℎ = 4𝑐𝑚 and ℎ = 11𝑐𝑚
Q22

33

Find the exact value of x
...
Therefore
2

8
𝑥

√
3𝑥 = 16
√
√
√
3 × 3𝑥 = 16 3
√
3𝑥 = 16 3
𝑥=

√
16 3
3

Q23 Write as a single fraction in its simplest form
...

The area of the larger shape is 36𝑐𝑚2
and the area of the smaller shape is 25𝑐𝑚2
The height of the larger shape is 9cm and the height of the smaller shape is x cm
...

Solution
Since shape are similar
𝑥
9

=

𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑓 𝑖𝑔
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑙𝑎𝑟𝑔𝑒𝑟 𝑓 𝑖𝑔𝑢𝑟𝑒

𝑥
9

=

25
36

𝑥=

25
36

×9=

25
4

=

25
36

= 6
...

(b) Expand and simplify
...

The tangent to the graph of y = f(x) at A meets the y-axis at B
...

Solution (a)
𝑓 (𝑥) = 𝑥(𝑥 + 2)(𝑥 − 3)
Putting 𝑥 = −3, −2, −1, 0, 1, 2, 3, 4 ⟹ 𝑓 (−3) = (−3)(−3 + 2)(−3 − 3) = −18
𝑓 (−2) = (−2)(−2 + 2)(−2 − 3) = 0
𝑓 (−1) = (−1)(−1 + 2)(−1 − 3) = 4
𝑓 (0) = (0)(0 + 2)(0 − 3) = 0
𝑓 (1) = (1)(1 + 2)(1 − 3) = −6
𝑓 (2) = (2)(2 + 2)(2 − 3) = −8
𝑓 (3) = (3)(3 + 2)(3 − 3) = 0
𝑓 (4) = (4)(4 + 2)(4 − 3) = 24
The values of the intersections with the axes are :
(-2,0),((0,0),(3,0)
(b) Expanding and simplifying
...

The tangent to the graph of y = f(x) at A meets the y-axis at B
...

The tangent intersect the y-axis at B(0,-1) as shown in figure

40



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