MATRICES AND TRANSFORMATIONS

1. 1989 Q12 P2
The point (5, 2) undergoes the transformation [−13 20 ] followed by a

translation [−611] . Determine the coordinates of the image. (3 marks)

2. 1990 Q16 P1
The vertices of a triangle ABC are A(-1, -4), B (3,3) and C (2, 5). Find the

image of the triangle under the transformation whose matrix is [ 46 −2

−3 ]
Draw the triangle and its image on the same axis. (Grid was provided) (3 marks)

3. 1990 Q21 P2
A parallelogram whose vertices are A (1, 0), B (3, 0), C (4, 2) and D (2, 2) is
mapped onto a parallelogram A′ B′ C′ D′ by a transformation whose matrix is M.
Under the transformation the images of A and C are A′ (0, 1) and C′ (-2, 4) respectively.

i) Find matrix M (3 marks)

ii) Plot the parallelogram ABCD on its image on the given grid (3 marks)
iii) A′ B′ C′ D′ is mapped onto ABCD by a transformation T. Obtain the
matrix for T. (2 marks)
4. 1991 Q4 P1
The image of a point A, under the transformation represented by the matrix
T= [ 10 −12 ] is A′ (-2, 4). Find the coordinates of a. (3 marks)

5. 1991 Q21 P2
A rectangle OABC has the vertices O (0, 0), A (2,0), B (2, 3) and C (0, 3).
O′ A′ B′ C′ is the Image of OABC under the translation given by [ 04 ] . O′′A′′ B′′ C′′

is the image of O′ A′ B′ C′ under the transformation given by the matrix [ 01 −10 ] .

Draw the rectangles OABC, O′ A′ B′ C′ and O′′ A′′ B′′ C′′ on the grid provided.
(grid was provided). (6 marks)
Use your diagram to find the centre of rotation which maps OABC onto
O′′ A′′ B′′ C′′ (2 marks)

6. 1992 Q18 P1
A point pʹ (x′, y′) is the image of a point p(x, y) under a transformation, T,
given by the Matrix [ 34 −34 ]
a) Express x′ and y′ in terms of x and y (3 marks)
b) If appoint Q and its image Q′ under the transformation T lie on the same
line y=mx,
Find two possible values of m (5 marks)

7. 1993 Q12 P1

1
 A transformation is represented by the matrix [ 13 22] . This transformation
maps a triangle ABC of the area 3cm2 onto another triangle A′B′C′. Find the
area of triangle A′B′C′. (3 marks)
8. 1993 Q19 P1
A transformation T1 maps the triangle ABC whose coordinates are A (-2, 0),
B (1, -2)C (0, 1) onto triangle ABC whose coordinates A′(2, 4), B′(4,1), and C′(1, 2).
Another transformation T2 maps the same triangle ABC onto triangle A′′B′′C′′
whose coordinates are A′′ (4, 2), B′′ (1,4), C′′ (2,1).

a) On the same axes plot triangle ABC, A′B′C′ and A′′B′′C′′ (2 marks)
b) Determine
i) T1
ii) T2
iii) the matrix of T such that TT1 = T2 (6 marks)

9. 1994 Q23 P1
A rectangle ABCD with vertices A (2, 0), B (4, 0), C (4, 4) and D (2,4) is given a
Stretch transformation with line x = 2 as the invariant and point (4, 0) being
mapped onto point (6,0). The image A′B′C′D′ of the rectangle is enlarged with
a scale factor of -2, centre origin, followed by a reflection in the line y = 0.

On the grid below plot the images of the rectangle ABCD after the successive
transformation. (Grid was provided)
a) Describe the transformation which map the third image onto the first image
(2 marks)
b) Describe the single matrix that will map the matrix on the third image
onto the first image
(2 marks)

10. 1994 Q7 P2
Determine the matrix of transformation that represents the following
transformation:
Reflection in x + y = 0, followed by a positive quarter turnabout (0, 0) (2 marks)

11. 1995 Q 23 P1
On the grid provided ABCE is a trapezium
(NB: Coordinates of the figure are A(2,0), B(6, 0), C(6, 5) and D(2,2)).

(a) ABCD is mapped onto A ′B ′C ′D′ by a positive quarter turn. Draw the
image A′B′C ′D′ on the grid. (1 mark)

(b) A transformation maps [

−2 −1
1 −1 ] A ′B′C ′D′ onto A ″B ″C ″D″

(i) Obtain the coordinates of A ″B″ C″ D ″ on the grid (2 marks)

(ii) Plot the image A″ B ″C ″D″ on the grid (1 mark)

(c) Determine a single matrix that maps A ″B ″C ″D″ and ABCD (4 marks)

2
12. 1997 Q 23 P1
The figure on the grid shows a triangular shaped object ABC and its image A′ B ′C′
(NB: Coordinates of the figure are A(2,0), B(4, 1), C(4,3)).
(NB: Coordinates of the figure are Aʹ(-2,0), Bʹ(-4, -1), Cʹ(-4,-3)).

(a) (i) Describe fully the transformation that maps ABC and A ′B ′C′ (1 mark)
(ii) Find a 2 x 2 matrix that transforms triangle ABC onto triangle A ′B′C′
(1 mark)
(b) The matrix P = [ 21 11] transforms triangle ABC onto A″ B″ C″

(i) Find the coordinates of A ″B ″C″ (2 marks)

(ii) Draw the image A″ B″ C″ (1 mark)

(c) Find the area of triangle ABC (1 mark)

(d) Hence find the area of the image A″B″C″ (2 marks)

13. 1998 Q 19 P1
A quadrilateral ABCD has vertices A (4, -4), B (2, -4), C (6, -6) and D (4, -2)
a) On the grid provided draw the quadrilateral ABCD. (1 mark)
b) A′ B ′C ′D′ is the image of ABCD under positive quarter turn about the
origin. On the same grid draw the image A’B’C’D (2 marks)
c) A′ B ′C′ D′ is the image of A′ B ′C′ D′ under the transformation given

by the matrix [ 10 −21 ] (1 mark)

3
 i) Determine the coordinators of A″B ″C ″D″ (2 marks)
ii) On the same grid draw the quadrilateral A″B ″C ″D″ (1 mark)

d) Determine a single matrix that maps ABCD onto A″B ″C ″D″ (2 marks)

14. 1999 Q 23 P2
The transformation R given by the matrix
A= [ ac bd ] maps [ 170] [ 158] [ 170 ] [−815]
to and to

(a) Determine the matrix A giving a, b, c and d as fractions (3 marks)

(b) Given that A represents a rotation through the origin determine

the angle of rotation
(2 marks)
(c) S is a rotation though 1800 about the point (2, 3). Determine the
image of (1,0) under S followed by R. (3 marks)

15. 2000 Q 23 P2
The diagram on the grid provided below shows a trapezium ABCD On the same grid
(NB: Coordinates of the figure are A (1, 2), B (7, 2), C (5,4)D(3, 4)).

(a) (i) Draw the image A ′B ′C ′D of ABCD under a rotation of 900

clockwise about the origin . (1 mark)

(ii) Draw the image A″B ″C″ D″ of A′ B′ C′ D′ under a reflection in line

y = x. State coordinates of A″B ″C″ D″ (3 marks)

4
 (b) A″′B ″′C″′ D″′ is the image of A″B ″C″ D″ under the reflection in the
line x=0.
Draw the image A′″B ″′C ″′D″′ and state its coordinates. (2 marks)

(c) Describe a single transformation that maps A″′B′″ C″′ D′′′ onto ABCD.
(2 marks)
16. 2001 Q 18 P1
The coordinates of the vertices of rectangle PQRS are P (1, 1), Q (6, 1), R (6, 4)
and S (1, 4)
(a) (i) Find the coordinates of the vertices of its image, P ′Q′ R′S′ under the
transformation defined by the matrix [ 10 −21 ] (2 marks)
(ii) Draw the object and its image on the grade provided (2 marks)

(iii) On the same grid draw the image, P″Q″R″S″ of P ′Q ′R ′S′ under
the transformation given by [−10 10 ] (2 marks)

(b) Find a single matrix which will map P″Q″R″S″ onto PQRS (2 marks)

17. 2002 Q 22 P1
A triangle T whose vertices are A (2, 3) B(5,3) and C (4,1) is mapped onto
triangle T1 whose vertices are A1 (-4,3) B1 (-1,3) and C1 (x,y) by a Transformation

M= [ ac bd ]
Find the: (i)Matrix M of the transformation (4 marks)
(ii)Coordinates of C1 (2 marks)

b) Triangle T2 is the image of triangle T1 under a reflection in the line y = x. (2 marks)

Find a single matrix that maps T and T2

18. 2004 Q 21 P2
Triangle ABC is the image of triangle PQR under the transformation M = [ 20 42]
Where P, Q and P map onto A, B, and C respectively.
(a) Given the points P(5, -1), Q(6,-1) and R (4, -0.5), draw the triangle ABC
on the grid provided below. (3 marks)

5
 (b) Triangle ABC in part (a) above is to be enlarged scale factor 2 with centre
at (11, -6) to map onto A ′B′C′. Construct and label triangle A ′B ′C′ on the
grid above. (2 marks)

(c) By construction find the coordinates of the centre and the angle of
rotation which can be used to rotate triangle A ′B ′C′ onto triangle
A ″B ″C″, shown on the grid above.
(3 marks)
19. 2005 Q 18 P2
Triangles ABC and A ″B ″C″ are drawn on the Cartesian plane provided.
Triangle ABC is mapped onto A ″B″C″ by two successive transformations
(NB: Coordinates of the figure are A (2, 2), B(2, 4), C (4,4)).
(NB: Coordinates of the figure are Aʹʹ (0,- 2), B (-4, -10), C (-4,-12)).

6
 R= [ ac bd ] Followed by P = [−10 −10 ]
(a) Find R (4 marks)

(b) Using the same scale and axes, draw triangle A ″B ″C″, the image of
triangle ABC under transformation R (2 marks)

(c) Describe fully, the transformation represented by matrix R (2 marks)

20. 2006 Q 19 P2
Triangle ABC is shown on the coordinates plane below

7
 (a) Given that A (-6, 5) is mapped onto A′ (6,-4) by a shear with y- axis invariant
(i) draw triangle A′ B′ C′, the image of triangle ABC under the shear
(3 marks)
(ii) Determine the matrix representing this shear (2 marks)

(b) Triangle A′ B′ C′ is mapped on to A″ B″ C″ by a transformation defined by the

[ ]
−1 0
matrix 1
1 −1
2

(i) Draw triangle A″ B″ C″ (3 marks)

(ii) Describe fully a single transformation that maps ABC onto A″ B″ C″
(2 marks)

21. 2008 Q 10 P2
Points A(2,2)and B(4,3) are mapped onto Aʹ(2,8) and Bʹ (4,15) respectively by
a transformation T. Find the matrix of T. (4 marks)

22. 2009 Q 9 P2
The area of triangle FGH is 21cm2. The triangle FGH is transformed using the

matrix [ 41 52] Calculate the area of the image of triangle FGH (2 marks)

23. 2009 Q 20 P2

8
 Triangle PQR shown on the grid has vertices P(5,5) Q(10,10) and R(10,15)

(a) Find the coordinates of the points P′, Q′ and R′, the images of P, Q and R

respectively under transformation M whose matrix is [−0.6

0.8 0.6 ]
0.8

(2 marks)
(b) Given that M is a reflection
(i) Draw triangle P′ Q′ R′ and the mirror line of the reflection (2 marks)
(ii) Determine the equitation of the mirror line of the reflection (1 mark)

(c) Triangle P″Q″R″ is the image of triangle P′ Q′ R′ under reflection N

where N is a reflection in the y-axis
(i) Draw triangle P″Q″R″ (1 mark)
(ii) Determine a 2 x 2 matrix equivalent to the transformation NM
(2 marks)
(iii) Describe fully a single transformation that maps triangle
PQR onto triangle P″Q″R″ (2 marks)
24. 2010 Q 10 P2
The point O, A and B have the coordinates (0,0), (4,0) and (3,2) respectively.
Under shear represented by the matrix [ 10 k1] ,triangle OAB maps onto
triangle OAB’

a) Determine in terms of k, the x coordinates of point B′ (2 marks)

b) If OAB′ is a right angled triangle in which angle OB′ A is acute, find
two possible values of k. (2 marks)

9
25. 2011 Q 19 P2
The vertices of a rectangle are A(-1,-1) , B(-4,-1) , C( -4,-3) and D( -1,-3).

a) On the grid provided, draw the rectangle and its image A′ B′ C′ D′

Under a transformation whose matrix is [−20 −20 ] (4 marks)

b) A″ B″ C″ D″ is the image of A′ B′C′D′ under a transformation matrix,

[ ]
1
1
2
P=
1
1
2
i) Determine the coordinates of A″ B″ C″D″ (2 marks)
ii) On the same grid draw the quadrilateral A″B″C″ D″ (1mark)

c) Find the area of ABCD. (3 marks)

26. 2012 Q18 P2

OABC is a parallelogram with vertices O(0,0), A(2,0), B(3,2) and C(1,2).
O'A'B'C' is the image of OABC under the transformation matrix [−20 −20 ]
(i) Find the coordinates of O'A'B'C' (2 marks)
ii) On the grid provided draw OABC and O'A'B'C' (2 marks)

(a) (i) Find O''A''B''C'', the image of O'A'B'C' under the transformation
matrix [ 10 −20 ] (2 marks)
(ii) On the same grid draw O''A''B''C''. (1 mark)

(c) Find the single matrix that maps O''A''B''C'' onto OABC. (3 marks)

27. 2013 Q16 P2

The vertices of triangle T are A(1, 2), B(4, 2) and C(3, 4). The vertices of triangle
1 3
Tʹ, the image of T are Aʹ( ,1) , Bʹ(2,1) and Cʹ( , 2).
2 2
Determine the transformation matrix M = (ac bd ) that maps T onto Tʹ.
(3 marks)
28. 2014 Q7 P2
In a transformation, an object with an area of 5 cm2 is mapped onto an image

whose area is 30cm2. Given that the matrix of the transformation is [ ]

x x−1 ,
2 4
find the value of x.
(3 marks)
29. 2015 Q24 P2
A quadrilateral with vertices at K (1,1), L(4,1), M(2, 3) and N (1, 3 ) is

transformed by a matrix T = [ 10 31] to a quadrilateral K’L’M’N’

a) Determine the coordinates of the image (3 marks)

10
 b) On the grid provided draw the object and the image (2 marks)
c i)Describe fully the transformation which maps KLMN onto K’L’M’N (2 marks)
ii)Determine the area of the image. (1 mark)
d)Find a matrix which maps K’L’M;N’ onto KLMN (2 marks)
STATISTICS II
1. 1989 Q12 P1
The table below shows the defective bolts from 40 samples

No. of detective bolts (x) 0 1 2 3 4 5

Frequency (y) 20 8 6 4 1 1

Calculate the standard deviation (5 marks)

2. 1989 Q20 P2
The table below shows the life expectancy, in hours, of 106 bulbs

Expectancy 90-94 95-99 100-104 105-109 110-114 115-119 120-124 125-129 130-134 135-139
(hours)

Frequency 5 14 16 17 24 12 11 4 2 1
(f)

(a) Calculate the mean life expectancy, (4 marks)

(b) On the grid provided draw a cumulative frequency curve and use it to
Determine the median
(4 marks)
3. 1990 Q23 P1
For a sample of 100 bulbs the time taken for each bulb to burn out was recorded.
The table below shows the results of the measurements.

Time (hours) 12-19 20-24 25-29 30-34 35-39 40-44

Frequency (Number of bulbs) 6 10 9 5 7 11

Time (hours) 45-49 50-54 55-59 60-64 65-69 70-74

Frequency (Number of bulbs) 15 13 8 7 5 4

Using an assumed mean or otherwise, calculate

(i) The mean
(ii) The standard deviation of the distribution (8 marks)

4. 1991 Q22 P1
The table below gives marks scored by 80 candidates in a test

Marks 1-10 11-20 21-30 31-40 41-50

11
 No. of candidates 5 13 32 27 3

Using an assumed mean of 25.5 calculate the mean, the variance and the
standard deviation of the marks (8 marks)

5. 1992 Q 23 P1
Lengths of 100 mango leaves from a certain mango tree were measured to the
nearest centimeter and recorded as per the table below
Lengths in cm No. of leaves
10 to 12 3
13 to 15 12
16 to 18 40
19 to 21 37
22 to 24 8
(a) On the grid provided, draw a cumulative frequency graph to represent
this data
(4 marks)
(b) Use the graph to estimate
(i) The median length of the leaves (1 mark)

(ii) The number of leaves whose lengths lie between 13 and 17cm
(3 marks)

6. 1993 Q20 P2
The table below shows the distribution of marks scored in a test by standard
8 pupils in one school.

Marks 30-34 35-39 40-44 45-49 50-54 55-59 60-64 65-69 70-74 75-79

No. of 1 5 10 10 19 20 20 8 4 3
pupils

Using 57 as the assumed mean mark, calculate

(i) The actual mean for the grouped marks (3 marks)
(ii) The standard deviation of the marks (5 marks)

7. 1994 Q10 P1
Determine the interquartile range for the following numbers: (3 marks)
4, 9, 5, 4, 7, 6, 2, 1, 6, 7, 8.

8. 1994 Q21 P2
The table below gives marks obtained in a mathematics test by 47 candidates

Marks 31-35 36-40 41-45 46-50 51-55 56-60

No. of candidates 4 6 12 15 8 2

(a) Calculate the mean score

(b) Draw a cumulative frequency graph and use it to estimate

12
 (i) The median score (2 marks)
(ii) The semi-interquartile range (6 marks)

9. 1995 Q18 P2
The table below shows high altitude wind speeds recorded at a weather
station in a period of 100 days.

Wind speed 0 - 19 20 - 39 40 - 59 60-79 80- 99 100- 119 120-139 140-159 160-179

( knots)

Frequency 9 19 22 18 13 11 5 2 1
(days)

(a) On the grid provided draw a cumulative frequency graph for the data
(4 marks)
(b) Use the graph to estimate
(i) The interquartile range (3 marks)
(ii) The number of days when the wind speed exceeded 125 knots (1 mark)

10. 1996 Q10 P1

Five pupils A, B, C, D and E obtained the marks 53, 41, 60, 80 and 56 respectively.
The table below shows part of the work to find the standard deviation.

Pupil Mark x x-x ( x-x)2

A 53 -5
B 41 -17
C 60 2
D 80 22
E 56 -2

(a) Complete the table (1 mark)

(b) Find the standard deviation (3 marks)

11. 1996 Q19 P2

In an agricultural research centre, the lengths of a sample of 50 maize cobs were
measured and recorded as shown in the frequency distribution table below.

Length in cm. Number of cobs.

8 – 10 4
11 – 13 7
14 – 16 11

13
 17 – 19 15
20 – 22 8
23 – 25 5
Calculate;
a) the mean
b) i) the variance
ii) the standard deviation (8 marks)

12. 1997 Q22 P2

The table below shows the frequency distribution of masses of 50 new-born
Calves in a ranch.
Mass (kg) Frequency
15 – 18 2
19 – 22 3
23 – 26 10
27 – 30 14
31 – 34 13
35 – 38 6
39 – 42 2
a) On the grid provided draw a cumulative frequency graph for the data
(4 marks)
b) Use the graph to estimate
(i) the median mass (1 mark)

(ii) the probability that a calf picked at random has a mass lying between
25kg and 28 kg
(3 marks)
13. 2001 Q18 P2
The marks obtained by 10 pupils in an English test were 15, 14, 13, 12,
P, 16,11,13,12 and 17.The sum of the squares of the marks, ∑x2 is 1,794
a) Calculate the:
i) Value of P (2 marks)
ii) Standard deviation. (4 marks)

b) If each mark is increased by 3, write down the:

i) New mean (1 mark)
ii) New standard deviation (1 mark)

14. 2002 Q19

The following distribution shows the masses to the nearest kilogram of
65 animals in a certain farm.

Mass Kg 26-30 31-35 36-40 41-45 46-50 51-55

frequency 9 13 20 15 6 2

a) On the grid provided draw the cumulative frequency curve for the
given information.
(3 marks)

b) Use the graph to find the:-

i) Median mass
ii) Inter-quartile range

14
 iii) Percentage of animals whose mass is at least 42kg. (5 marks)

15. 2003 Q18 P1

The mass of 40 babies in a certain clinic were recorded as follows:
Mass in Kg No. of babies.
1.0 – 1.9 6
2.0 – 2.9 14
3.0 -3.9 10
4.0 – 4.9 7
5.0 – 5.9 2
6.0 – 6.9 1

Calculate
(a) The inter – quartile range of the data. (4 marks)
(b) The standard deviation of the data using 3.45 as the assumed mean (4 marks)

16. 2004 Q18 P2

The table below shows the ages in years of 60 people who attended a conference.

Age in years 30 – 39 40- 49 50- 59 60- 69 70-79

Number of people 10 12 18 17 3

Calculate
(a) The inter-quartile range of the data (5 marks)

(b) The percentage of the people in the conference whose ages were 54.5
years and below.
(3 marks)
17. 2005 Q22 P1
The data below shows the masses in grams of 50 potatoes

Mass (g) 25- 34 35-44 45 - 54 55- 64 65 - 74 75-84 85-94

No of 3 6 16 12 8 4 1
potatoes

(a) On the grid provided, draw a cumulative frequency curve for the data (4 marks)

(b) Use the graph in (a) above to determine

(i) The 60th percentile mass (1 mark)

(ii) The percentage of potatoes whose masses lie in the range 53 g to

68g (3 marks)

18. 2006 Q5 P2
The data below represents the ages in months at which 6 babies started
walking: 9,11, 12, 13, 11, and 10. Without using a calculator, find the exact
value of the variance (3 marks)

15
19. 2008 Q22 P2
The table below shows the distribution of marks scored by 60 pupils in a test.

Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90

Frequenc 2 5 6 10 14 11 9 3
y

a) On the grid provided, draw an ogive that represents the above information (4 marks)

b) Use the graph to estimate the interquartile range of this information. (3 marks)

20. 2009 Q19 P2

The table below shows the number of goals scored in handball matches
during a tournament

Number of goals 0-10 10-19 20-29 30-39 40-49

Number of matches 2 14 24 12 8
(a) Draw a
cumulative frequency curve on the grid provided (5 marks)

(b)Using the curve drawn in (a) above determine

(i) The median; (1 mark)
(ii) The number of matches in which goals scored were not more
than 37;
(1 mark)
(iii) The inter-quartile range

21. 2010 Q8 P2
The heights, in centimeters, of 100 tree seedlings are shown in the table below.

Height (cm) 10-19 20-29 30-39 40-49 50-59 60-69

Number of 9 16 19 26 20 10
seedlings

Find the quartile deviation of the heights. (4 marks)

22. 2012 Q8 P2
The masses in kilograms of 20 bags of maize were ;

90,94,96,98,99,102,105,91,102,99,105,94,99,90,94,99,98,96,102 and 105.

Using an assumed mean of 96kg, calculate the mean mass, per bag, of the maize
(3 marks)

23. 2013 Q24 P2

The table below shows marks scored by 42 students in a test.
35 49 69 57 58 75 48

16
 40 46 86 47 81 67 63
56 80 36 62 49 46 26
41 58 68 73 65 59 72
64 70 64 54 74 33 51
73 25 41 61 56 57 28

a) Starting with the mark of 25 and using equal class intervals of 10,
make a frequency distribution table. (2 marks)

b) On the grid provided , draw the ogive for the data (4 marks)

c) Using the graph in (b) above , estimate:

(i) The median mark (2 marks)
(ii) The upper quartile mark (2 marks)

24. 2015 Q23 P2

The marks scored by 40 students in a mathematics test were as shown in the
table below.
Marks 48-52 53-57 58-62 63-67 68-72 73-77

Number of 3 4 10 12 8 3
students

a) Find the lower class boundary of the modal class (1 marks)

b) Using an assumed mean of 64, calculate the mean mark (3 marks)

ci)On the grid provided, draw the cumulative frequency curve for the data (3 marks)
ii)Use the graph to estimate the semi-interquartile range (3 marks)

17
 GEOMETRIC CONSTRUCTION AND LOCI
1 1989 Q20 P1
Use the straight lines AB and AC given below for the following construction.
A circle centre, O touches the line AC at C and passes through B.

(a) Use ruler and compasses only to locate the centre O. Draw the circle (3 marks)
(b) The circle cuts AB produced at D. Mark D and measure BD (1 mark)
(c) Locate a point R on the minor arc BD such that BR = rd (2 marks)
(d) Locate a point Q on AC such < COQ = < OQR (2 marks)

A B

BD = cm

2 1989 Q5 P2
Construct triangle PST equal in area to quadrilateral PQRS such that T lies
on PQ produced.
(4 marks)
S

P Q

3 1990 Q8 P1
Draw a line AB of length 9cm. On one side of the line AB construct the locus of
a point P such that the area of triangle APB is 13.5cm2. On this locus locate two
positions of P, P1 and P2 such that <AP1B = <AP2B = 900. (4 marks)

18
4 1990 Q17 P2
Use ruler and compass only for all the constructions in this question
A triangular plot of land ABC is such that AC = 300m, AB =280m and angle BAC = 750.

(a) Construct this plot of land using the scale 1cm : 50m (3 marks)
(b) A borehole P is equidistant from BA and lies on the perpendicular from
C to AB.
Locate the position of P (3 marks)
(c) Find the point on this farm which is furthest from the borehole. What
is its distance from the borehole? (2 marks)

5. 1991 Q8 P1

6. 1991 Q22 P2
Using ruler and compasses only construct an acute angled triangle ABC such
that <ABC = 450, BC = 9cm and AC = 7cm.
(3 marks)

19
 Locate appoint x in triangle ABC such that x is equidistant from A, B and C. (2 marks)
Measure AX, AB and < AXC. (3 marks)

7. 1992 Q8 P1
A point P moves so that its distance from the fixed point Q (2,3) is equal to 5 units.
Draw the locus of P on the grid provided. Hence find the coordinates of the points
were the locus of P cuts the x axis. (grid was provided) (3 marks)

8. 1992 Q13 P2
Using a ruler and a pair of compasses only, construct a circle to touch the three
lines AB, BC and CD given below. (3 marks)
B C

D
9. 1992 Q21 P2
(a) Use the points given below to construct
(i) The locus of a point Q such that AQ = AC (2 marks)

(ii) The locus of a point P such that P lies on the same side of AB
as the point C and <APB = 450. (5 marks)

(b) The loci intersect at M and N. measure the distance MN. (1 mark)
XC

X X

A B

20
10. 1993 Q11 P1
In the figure below triangle AOB is isosceles with AO = 0B and <AOB = 1500.Draw
the locus of a point P such that <APB = 750.
O
B
1500

11. 1993 Q19 P2

Using a ruler and a pair of compasses only, construct a triangle ABC in which
ABC = , BC = 7cm and BA = 6cm. Drop a perpendicular from A TO BC to meet BC
at D. Measure AD. Hence calculate the area of the triangle (8 marks)

12. 1994 Q19 P1

On the line AB below and on the same side of the line, use ruler and compasses only to
construct the following:
(a) Triangle ABC whose area is 20cm2 and ACB = 900. (3 marks)

(b) (i) the locus of a point P such that APB = 450 (2 marks)

(ii)locate the position of P such that triangle APB has maximum area
and calculate this area (3 marks)

A B

13. 1995 Q22 P2

Using ruler and compasses only, construct a parallelogram ABCD such that
AB = 10cm, BC = 7cm and < ABC = 1050. Also construct the loci of P and Q within
the parallel such that AP ≤ 4 cm, and BC ≤ 6 cm. Calculate the area within the
parallelogram and outside the regions bounded by the loci. (8 marks)

14. 1996 Q5 P2
Using the equilateral triangle below, construct the locus of a point P
such that <APC = 300
(3 marks)

21
 C

15. 1996 Q23 P1

Use ruler and compasses only in this question. The diagram below shows three
points A, B and D.
(a) Construct the angle bisector of acute angle BAD. (1 mark)
D

A B

(b) A point p, on the same side of AB as D, moves in such a way that

10
<APB = 22 Construct the locus of P (6 marks)
2
(c) The locus of P meets the angle bisector of < BAD at C. Measure <ABC (1 mark)

16. 1997 Q19 P1

Using ruler and compasses only construct triangle ABC such that AB = 4 cm,
BC = 5cm and < ABC = 1200. Measure AC. (3 marks)
On the diagram, construct a circle which passes through the vertical of the
triangle ABC.
Measure the radius of the circle (4 marks)
Measure the shortest distance from the centre of the circle to line BC. (1 mark)

17. 1997 Q4 P2
On the figure below construct
(i) the perpendicular bisector of BC (1 mark)
(ii) The locus of a point P which moves such a way that  APB =  AVB and
P is on the same side of AB on the same side of AB as C (2 marks)

18. 1998 Q23 P1

22
 Use a ruler and a pair of compasses only for all constructions in this question.
(a) On the line BC given below, construct triangle ABC such that
< ABC = 300 and BA = 12 cm (3 marks)
(b) Construct a perpendicular from A to meet BC produced at D.
Measure CD (2 marks)

(c) Construct triangle A’BC such that the area of triangle A’BC is three
quarters of the area of triangle ABC and on the same side of BC as triangle ABC.
(2 marks)

(d) Describe the locus of A’ (1 mark)

B C

19. 1998 Q8 P2
In the figure below a line XY and three points. A,B and C are given. On
the figure construct
(a) The perpendicular bisector of AB (1 mark)
(b) A point P on line xy such that < APB= < ACB (2 marks)

20. 1999 Q11 P1

Given below is line BC. Without using a protractor construct another through
10
B making an angle of 37 with BC. Using the constructed line subdivide BC into
2
7 equal parts.

B C
(4 marks)
21. 1999 Q21 P2
The diagram below shows a garden drawn to scale of 1: 400. In the garden there
are already tow trees marked A and B. The gardener wises to plant more trees.
There are a number of rules he wishes to apply.

23
 Rule 1: Each new tree must be an equal distance from both trees A and B.
Rule 2: Each new tree must be atleast 4 m from the edges of the garden.
Rule 3: each new tree is atleast 14 m from tree B.

(a) draw the locus given by each of these rules on the diagram (6 marks)
(b) If the new trees are to be planted 4m apart, show on your diagram
the possible planting points for the new trees. (2 marks)

22. 2000 Q22 P2

The line segment BC given below is one side of triangle ABC
(a) Use a ruler and compasses to complete the construction of a triangle ABC in
Which < ABC = 450. AC = 5.6 cm and angle BAC is obtuse
(2 marks)
(b) Draw the locus of point P such that P is equidistant from a point O and
passes though the vertices of triangle. (2 marks)

(c) Locate point D on the locus of P equidistant from lines BC and BO. Q
lies in the region enclosed by lines BD, BO extended and the locus of P.
Shade the locus of Q. (2 marks)

B C
23. 2001 Q8 P1
Use a ruler and compasses in this question. Draw a parallelogram ABCD in which
AB = 8 cm, BC = 6 cm and <BAD = 750. By construction, determine the perpendicular
distance between AB and CD. (4 marks)

24. 2001 Q14 P2

The diagram below represents a field PQR.

a) Draw the locus of points equidistant from sides PQ and PR. (1 mark)

24
 b) Draw the locus of points equidistant from points P and R. (1 mark)

c) a coin is lost within a region which is nearest to point P than to R and

closer to side PR than to side PQ. Shade the region where the coin can
be located. (1 mark)

25. 2002 Q10 P1

The figure below shows a triangle ABC.

a) Using a ruler and a pair of compasses, determine a point D on the line

BC such that BD:DC = 1:2. (2 marks)

b) Find the area of triangle ABD, given that AB = AC. (2 marks)

26. 2002 Q21 P1

In this question use a ruler and a pair of compasses.
a) Line PQ drawn below is part of a triangle PQR. Construct the triangle
PQR in which < QPR = 300 and line PR = 8cm (2 marks)

P Q

b) On the same diagram construct triangle PRS such that points S and Q are
no the opposite sides of PR<PS = PS and QS = 8cm (2 marks)

c) A point T is on the a line passing through R and parallel to QS. If <QTS =900,
locate possible positions of T and label them T1 and T2, Measure the length
of T1T2. (2 marks)

25
27. 2003 Q22 P2
The line PQ below is 8cm long and L is its midpoint

P L Q
a) i) Draw the locus of point R above line PQ such that the area of
triangle PQR is 12cm2. (2 marks)
ii) Given that point R is equidistant from P and Q, show the position
of point R.
(2 marks)
b) Draw all the possible loci of a point T such that < RQL = <RTL.
(4 marks)

28. 2004 Q6 P1
Point C divided the line AB given below externally in the ratio 5:2

B
B

A
By construction, determine the position of point c (3 marks)

29. 2004 Q15 P2

The figure below is a triangle XYZ. Using a pair of compasses and a ruler only,
construct an escribed circle such that the centre of the circle and the point x are
the opposite sides of line YZ. (2 marks)

30. 2005 Q20 P2

(a) BCD is a rectangle in which AB = 7.6 cm and AD = 5.2 cm. draw the
rectangle and construct the locus of a point P within the rectangle such
that P is equidistant from CB and CD (3 marks)

(b) Q is a variable point within the rectangle ABCD drawn in (a) above such
that 600 ≤ <AQB ≤ 900
On the same diagram, construct and show the locus of point Q, by leaving

26
 unshaded, the region in which point Q lies (5 marks)

31. 2006 Q8 P1
In this question use a pair of compasses and a ruler only
(i) Construct triangle ABC such that AB = 6 cm, BC = 8cm and ABC 1350
(2 marks)
(ii) Construct the height of triangle ABC in a) above taking BC as the base
(1 mark)
32. 2006 Q7 P2
The figure below shows a circle centre O and a point Q which is outside the circle

o Q

Using a ruler and a pair of compasses, only locate a point on the circle such that
angle OPQ = 90o (2 marks)
33. 2006 Q13
The figure below is drawn to scale. It represents a field in the shape of an equilateral
triangle of side 80m
C

A
The owner wants to plant some flowers in the field. The flowers must be at most,
60m from A and nearer to B than to C. If no flower is to be more than 40m
from BC, show by shading, the exact region where the flowers may be planted
(4 marks)
34. 2007 Q12 P1
(a) Draw a regular pentagon of side 4 cm (1 mark)
(b) On the diagram drawn, construct a circle which touches all the sides of
the pentagon (2 marks)

35. 2007 Q21 P2

In this question use a ruler and a pair of compasses only
In the figure below, AB and PQ are straight lines

27
 A B

(a) Use the figure to:

(i) Find a point R on AB such that R is equidistant from P and Q (1 mark)

(ii) Complete a polygon PQRST with AB as its line of symmetry and hence
measure the distance of R from TS. (5 marks)

(b) Shade the region within the polygon in which a variable point X must lie given
that X satisfies the following conditions
I: X is nearer to PT than to PQ
II: RX is not more than 4.5 cm
III.  PXT > 900 (4 marks)
36. 2008 Q8 P1

Line BC below is a side of a triangle ABC and also a side of a parallelogram BCDE.

B C

Using a ruler and a pair of compasses only construct:

(i) The triangle ABC given that ABC = 1200 and AB= 6cm (1 mark)

(ii) The parallelogram BCDE whose area is equal to that of the triangle ABC and
point E is on line AB (3 marks)

37. 2008 Q3 P2
Line AB given below is one side of triangle ABC. Using a ruler and a pair of
compasses only;

A B

(i) Complete the triangle ABC such that BC=5cm and ABC=450 (1 mark)
(ii) On the same diagram construct a circle touching sides AC, BA
produced and BC
(3 marks)
38. 2009 Q11 P1
Line AB shown below is a side of a trapezium ABCD in which angle ABC = 1050,
BC= 4 CM, CD=5cm and CD is parallel to AB.

A B

Using a ruler and a pair of compasses only.

28
 (a) Complete the trapezium (3 marks)
(b) Locate point T on line AB such that angle ATD = 90◦ (1 mark)

39. 2009 Q4 P2
In the figure below, O is the centre of the circle and radius ON is perpendicular
to the line TS at N

Using a ruler and a pair of compasses only, construct a triangle ABC to inscribe
the circle, given that angle ABC = 60˚, BC =12cm and points B and C are on the
line TS. (4 marks)

40. 2010 Q10 P1

Using a ruler and a pair of compasses only, construct a rhombus QRST in which
an angle TQR =600 and QS = 10cm. (3 marks)

41. 2010 Q13 P2

a) Using line AB given below, construct the locus of a point P such that APB = 900
(1 mark)

A B

b) On the same diagram locate two possible position of point C such that
point C is on the locus of P and is equidistance from A and B. (2 marks)

42. 2011 Q9 P1
Using a ruler and a pair of compasses only:

a) Construct a parallelogram PQRS in which PQ= 6cm, QR = 4cm and

angle SPQ = 750; (3 marks)

b) Determine the perpendicular distance PQ and SR (1mark)

43. 2011 Q12 P2

The figure below represents a scale drawing of a rectangular piece of land, RSTU.
RS =9cm and ST =7cm
U T

29
 7cm

R 9cm S

An electric post, is to be erected inside the piece of land. On the scale drawing,
shade the possible region in which P would lie such that PU > PT and PS ≤ 7cm.
(3 marks)
44. 2012 Q8 P1
Using a pair of compasses and a ruler only, construct a quadrilateral ABCD in
which AB= 4cm,BC = 6cm,AD=3cm, angle ABC =1350 and angle DAB = 600.
Measure the size of angle BCD.

45. 2012 Q21 P2

(a) On the same diagram construct:
(i) Triangle ABC such that AB=9cm,AC=7cm and angle CAB=600 (2 marks)
(ii) The locus of a point P such that P is equidistant from A and B; (1 mark)

(iii) The locus of a point Q such that CQ≤3.5 cm. (1 mark)

(b) On the diagram in part (a):

(i) Shade the region R, containing all the points enclosed by the locus of P
and the locus of Q, such that AP≥BP; (2 marks)

(ii) Find the area of the region shaded in part (b)(i) above. (4 marks)

46. 2013 Q6 P1
2
A point P on the line AB shown below is such that AP = AB. By construction
7
locate P.
(3 marks)

A B

47. 2013 Q5 P2
(a) Using a pair of compasses and ruler only, construct an escribed circle to
touch side XZ of triangle XYZ drawn below (3 marks)
Z

30
 Y

X
(b) Measure the radius of the circle (1 mark)

48. 2013 Q12 P2

A point P moves inside a sector of a circle, centre O, and chord AB such that 2cm
< OP ≤ 3cm and angle APB = 650 Draw the locus of P (4 marks)

49. 2014 Q22 P1

Using a pair of compasses and a ruler only, construct
(a)
(i) Triangle ABC in which AB =5cm, ˂BAC = 300 and ˂ABC = 1050 (3marks)
(ii) A circle that passes through the vertices of the triangle ABC.
Measure the radius
(3marks)

(iii) The height of triangle ABC WITH AB as the base. Measure the height
(2marks)

(b) Determine the area of the circle that lies outside the triangle correct to
2decimal places (2marks)

50. 2014 Q22 P1

Figure ABCD below is a scale drawing representing a square plot of side 80 metres.
D C

A B

(a) On the drawing, construct:

(i) the locus of a point P, such that it is equidistant from AD and BC.
(2 marks)
(ii) the locus of a point Q such that <AQB = 60°. (3 marks)

(b) (i) Mark on the drawing the point Q , the intersection of the locus of
Q and line AD.
Determine the length of BQ1 in metres. (1 mark)

31
 (c) (ii) Calculate, correct to the nearest m2, the area of the region bounded
by the locus of P, the locus of Q and the line BQ1 (4 marks)

51. 2015 Q19 P1

Line AB drawn below is a side of a triangle ABC.

A B

(a) Using a pair of compasses and ruler only construct:

(i) triangle ABC in which BC = 10cm and ZCAB = 90°; (2 marks)
(ii) a rhombus BCDE such that ZCBE = 120°; (2 marks)
(iii) a perpendicular from F, the point of intersection of the diagonals
of the rhombus, to meet BE at G. Measure FG; (2 marks)
(iv) a circle to touch all the sides of the rhombus. (1 mark)

b) Determine the area of the region in the rhombus that lies outside the circle (3marks)

52. 2015 Q10 P2

Below is a lien AB and a point X. Determine the locus of a point P equidistant
from points A and B and 4 cm from X. (3 marks)
-X

A B

32
 TRIGONOMETRY III
1. 1989 Q7 P2
Find the values of θ between 00 and 3600 that satisfy the equation
Sin 2θ = - 0.5 (4 marks)
2. 1991 Q2 P2
Solve for x
4 sin (x + 20)0 = 3 for 00 ≤ x ≤ 3600 (3 marks)
3. 1992 Q4 P2
Determine the amplitude and the period for the graph of

( )
0
x
y=6 sin −90 (3 marks)
2
4. 1994 Q4 P2
Solve for θ in the equation
sin (2θ – 100) = - 0.5 for 00 ≤ θ ≤ 3600 (4 marks)
5. 1996 Q12 P1
Solve the equation
5 1
sin θ= for 00 ≤ 0 ≤ 1800 (2 marks)
2 2
6. 1997 Q11 P1
Find the value of θ between 00 and 3600 satisfying the equation
5 sin θ = - 4 (2 marks)
7. 1998 Q14 P1
Solve the equation

cos (3θ + 1200) = for 0 ≤ θ ≤ 1800 (4 marks)

8. 1999 Q12 P2
Solve the equation 8s2 + 2 s – 3 = 0

Hence solve the equation

8 sin2Ө + 2sinӨ - 3 = 0 for 00 ≤ θ ≤ 1800 (4 marks)
9. 2000 Q8 P2
Solve the equation
2 sin2(x-300) = cos 600 for – 1800 ≤ x ≤ 1800 (3 marks)

10. 2001 Q12 P1

Given that sin (x + 30)0 = cos 2x0 for 00 ≤ x ≤900 find the value of x.
Hence find the value of cos 23x0. (3 marks)

11. 2001 Q15 P2

Solve the equation 4 sin2 θ+4 cos θ =5
o o
For 0 ≤ θ ≥ 360 give the answer in degrees (3 marks)

12. 2003 Q7 P1
Solve the equation
3 tan2 x – 4 tan x -4 =0 for 00 ≤ x ≥ 1800 (4 marks)

33
13. 2004 Q9 P1
Give that xo is an angle in the first quadrant such that 8 sin2 + 2 cos x -5=0
Find:
a) Cos x (3 marks)
b) Tan x (1 marks)

14. 2005 Q9 P2
Given that Cos 2x0 = 0.8070, find x when 00 < x < 3600 (4 marks)
15. 2007 Q3 P2
Solve the equation 3 cos x = 2 sin2 x, where 00 ≤ x ≤ 3600 (4 marks)

16. 2008 Q16 P1

Solve the equation;
2 cos 2 =1 for 00   3600 (4 marks)
17. 2008 Q16 P2
Find in radians, the values of x in the interval 0c x  2c for which
2 cos 2x – sin x=1.
(Leave the answers in terms of ) (4 marks)

18. 2009 Q13 P1

Solve the equation

sin (3x + 300) = for 0  x  90 (4 marks)

19. 2009 Q14 P2
Solve
4 – 4 cos2 ∞ = 4 sin∞ – 1 for 0˚ ≤ ∞ ≤ 360˚ (4 marks)
20. 2012 Q5 P2
Solve the equation
Sin(2t + 10)0 = 0.5 for 00 ≤ t ≤ 1800 (2 marks)
21. 2013 Q8 P1
Given that sin (x + 20)0 = - 0.7660, find x, to the nearest degree, for 00 ≤ x ≥ 3600. (3 marks)

22. 2013 Q14 P2

Solve the equation
6 cos2 x +7 sin x – 8 = 0 for 00 ≤ x ≤ 900 (4 marks)

23. 2014 Q6 P2
Determine the amplitude and period of the function, y = 2 cos (3x — 45) °. (2marks)

34
 GRAPHS OF TRIGONOMETRIC EQUATIONS
1. 1989 Q23 P2
Complete the table given below by filling in the blank boxes.

x0 00 150 300 450 600 750 900 1050 1200 1350 1500 1650 1800

3 cos x0 3.0 2.60 1.50 0 -0.75 -3.0

4 sin - 1.37 3.94 3.76 0.69 -3.06 - -0.69

(2x – 100) 0.69 3.76

Taking 1 cm to represent 150 on the x-axis and 2cm to represent 1 unit on the y axis,
draw the graph of y = 3 cos x and y = 4 sin(2x - 100 ) using the same axes on the graph provided.
Use your graph to find the value of x for which 3 cos x = 4sin (2x - 100) (8 marks)

2. 1990 Q19 P1
Copy and complete the table given below.

x0 300 00 300 600 900 1200 1500 1800 2100 2400

Sin(x+300) 0 0.5 0.87

Cos (x-150) 0.97

(3 marks)

Using the same axes plot the curves y = sin(x+300) and cos (x – 150)
For -300 x 2400 (3 marks)
Use your graph to find the value of x such that cos (x-150) = sin (x=300) (2 marks)

3. 1991 Q18 P1
a) Complete the table given below by filling in the blank boxes. (2 marks)

x0 00 300 600 900 1200 1500 1800 2100

3 sin x0 -1 -1 0.5

cos x0 1 0.87 0.5 0 -0.87 -1

b) On the same axes draw the graph of y = 3 sin x0 -1 and y = cos x0 on the grid below.
(grid was provided) (4 marks)

35
 c) Use your graph to solve the equation
3 sin x0 - cos x0 1 (2 marks)

4. 1993 Q23 P1
Complete the table below by filling in the blank spaces.

x -1800 -1350 -900 -450 00 450 900 1350 1800

2 sin x 0

Cos 2x 1

Draw the graph of y =2 sinx and y = cos 2x using the axes on the grid provided. (2 marks)

1
a) What is the difference in the values of y = 2sin x and y = cos 2x at x = 67 ? (2 marks)
2
b) State the periods of
i) Y = 2 sinx (1 mark)
ii) Y = cos 2x (1 mark)

5. 1995 Q 23 P2
(a) Complete the table for the function y = 2 sin x (2 marks)

x 00 100 200 300 400 500 600 700 800 900 1000 1100 1200

Sin 3x 0 0.5000 -
0.8666

y 0 1.00 -1.73

(b)(i) Using the values in the completed table, draw the graph of y = 2 sin 3x for

00 ≤ x ≤ 1200 on the grid provided

(ii) Hence solve the equation 2 sin 3x = -1.5 (3 marks)

6. 1996 Q 24 P2

Complete the table given below for the functions:

Y= -3 cos 2x0 and y= 2 sin ( 3/2x0 + 300) for 00 ≤ x ≤ 1800 (2 marks)

X0 00 200 400 600 800 1000 1200 1400 1600 1800

36
 -3 cos 2x0 -3.00 1.50 2.82 2.82 -0.52 -2.30
2 sin( 3/2x0 +300) 1.00 2.00 1.73 0.00 -1.00 -1.73

a) Using the grid provided, draw the graphs of y= -3cos 2x0 and y= 2sin 3/2x0 + 300)
on the same axis. Take 1 cm to represent 200 on the x-axis and 2cm to
represent one unit on the y-axis (4 marks)

b) From your graphs, find the roots of 3cos 2x0 + 2sin ( 3/2 x0 + 300) =0. (2 marks)

7. 1997 Q 18 P2

Complete the table below by filling in the blank spaces

X0 00 300 600 900 1200 1500 1800 2100 2400 2700 3000 3300 3600

Cos x0 1.00 0.50 -0.87 -0.87

2 cos ½ x0 2.00 1.93 0.52 -1.00 -2.00

Using the scale 1 cm to represent 300 on the horizontal axis and 4 cm to represent 1
unit on the vertical axis draw, on the grid provided, the graphs of y = cosx0 and
y = 2 cos ½ x0 on the same axis.

1 0
(a) Find the period and the amplitude of y = 2 cos x (2 marks)
2

(b) Describe the transformation that maps the graph of y = cosx0 on the graph
1 0
of y = 2 cos x (2 marks)
2

8. 1998 Q 18 P2
(a) Complete the table below for the value of y = 2 sin x + cos x.

x 00 300 450 600 900 1200 1350 1500 1800 2250 2700 3150 3600

2 sin x 0 1.4 1.7 2 1.7 1.4 1 0 -2 -1.4 0

Cos x 1 0.7 0.5 0 -0.5 -0.7 -0.9 -1 0 0.7 1

y 1 2.1 2.2 2 1.2 0.7 0.1 -1 -2 -0.7 1

(b) Using the grid provided draw the graph of y= 2 sin x + cos x for 00. Take 1 cm

37
 represent 300 on the x- axis and 2 cm to represent 1 unit on the axis.
(3 marks)
(c) Use the graph to find the range of x that satisfy the inequalities 2 sin x cos x > 0.5
(2 marks)

9. 1999 Q 18 P2
(a) Complete the table below, giving your values correct to 2 decimal places.

x 0 10 20 30 40 50 60 70

Tan x 0

2 x + 300 30 50 70 90 110 130 150 170

Sin ( 2x + 300) 0.50 1

(2 marks)

b) On the grid provided, draw the graphs of y = tan x and y = sin (2x + 300)
for 00 ≤ x 700
Take scale: 2 cm for 100 on the x- axis
4 cm for unit on the y- axis

Use your graph to solve the equation tan x- sin (2x + 300 ) = 0 (6 marks)

10. 2000 Q 24 P2
(a) Complete the table for the equation y = 2sin (3x + 300) (2 marks)

x 00 100 200 300 400 500 600 700 800 900

3x + 300 30 60 90 120 150 180 210 240 270 300

Y = 2sin (3x + 300) 1 1.73 2 0 -2 -1.73

(b) Using the grid provided, draw the graph of y = 2sin (3x + 300) for 00 ≤ x ≤ 900 .
Take 1cm to represent 40 on the x-axis and 2cm to represent 1 unit on the y axis
(3 marks)

(c) Use the graph in (b) to find the range of values of x that satisfy the
inequality y ≤ 1.6
(3 marks)
11. 2001 Q 21 P2
a) Complete the table given below in the blank spaces.

38
 X 0o 150 300 450 600 750 900 1050 1200 1350 1500 1650 1800

3 cos 2x 3 2.598 1.5 0 1.5 --3 -2.598 -1.5 0 2.598 3

2 sin 1 2 2.732 1 0 -1 -1.732 -2 -2.732 -2 1

(2x +30o)

b) On the grid provided draw, on the same axis, the graph of y = 3cos 2x and
y = sin(2x +300) for 0o ≤ x ≥ 1800. Take the scale: 1cm for 150 on the axis and
2cm for 1 unit on the y-axis.
(4 marks)
c) Use your graph to estimate the range of value of x for which 3 cos 2x ≤ 2sin (2x+30o).
Give your answer to the nearest degree. (2 marks)

12. 2002 Q 23 P2
a) Complete the table below, giving your values correct to 2 decimal place.

00 150 300 450 600 750 900 1050 1200 1350 1500 1650 1800

Tan Ө0 0 0.27 0.58 1 1.73 ∞ -3.73 -1.73 -1 -0.27 0

Sin Ө0 0 0.5 1 0.87 0.5 0 -0.5 -1 -0.87 -0.5 0

b) Using the grid provided and the table in part (a) draw the graphs of Y = tan θ and
y = sin 2θ. (5 marks)

c) Using your graphs, determine the range of values for which tan θ>Sin 2θ
for 00 ≤ θ ≤ 900. (1 mark)

13. 2003 Q 23 P2

a) Complete the table below, giving your values correct to 2 decimal places.

X 0 15 30 45 60 75 90 105 120 135 15 165 180

0

Cos x 1 0.7 0.87 0.71 0.1 0.24 0 -0.26 -0.5 -0.17 0.5 0.87 1
7 5

Sin 0.5 0.1 0.87 0.97 0.1 0.97 0.87 0.71 0.5 - 0.26 0 -0.26 -0.5
(x +300) 7 0

39
 b) Using the grid provided draw, on the same axes, the graph of y = cos 2x and
y = sin (x+ 300) for 00< x< 1800 Take the scale: 1cm for 150 on the x axis
4cm for 1 unit on the y- axis. (4 marks)

c) Find the period of the curve y = axis (1 mark)

d) Using the graphs in part (b) above, estimate the solutions to the equation
Sin (x + 300) = Cos 2x (4 marks)

14. 2005 Q 21 P2
(a) Complete the table below, giving your values correct to 2 decimal places (2 marks)

X0 0 30 60 90 120 150 180

2 sin x0 0 1 2 1

1 – cos x0 0.5 1
(b) On the grid
provided,
using the same scale and axes, draw the graphs of
y = sin x0 and y = 1 – cos x0 ≤ x ≤ 1800 (4 marks)
Take the scale: 2 cm for 300 on the x- axis
2 cm for I unit on the y- axis

(c) Use the graph in (b) above to

(i) Solve equation
2 sin xo + cos x0 = 1 (1 mark)
(ii) Determine the range of values x for which 2 sin xo > 1 – cos x0
(1 mark)
15. 2007 Q 19 P2
(a) Given that y = 8 sin 2x – 6 cos x, complete the table below for the missing
values of y, correct to 1 decimal place. (2 marks)

X 00 150 300 450 600 750 900 1050 1200

Y = 8 sin 2x – 6 cos x -6 -1.8 3.8 3.9 2.4 0 -3.9

(b)On the grid provided, below, draw the graph of y = 8 sin 2x – 6 cos for
00 ≤ x ≤ 1200
Take the scale 2 cm for 150 on the x- axis
2 cm for 2 units on the y – axis (4 marks)

(c) Use the graph to estimate

(i) The maximum value of y (1 mark)
(ii) The value of x for which 4 sin 2x – 3cos x =1 (3 marks)
16. 2008 Q 19 P2
a) Complete the table below, giving the values correct to 2 decimal places.

x0 0 30 60 90 120 150 180 210 240 270 300 330 360

40
 Sin 2x 0 0.87 -0.87 0 0.87 0.87 0

3cosx-2 1 0.60 -2 -3.5 -4.60 -0.5 1

b) On the grid provided, draw the graphs of y=sin 2x and y=3cosx-2 for
00  x  3600 on the same axes. Use a scale of 1 cm to represent 300 on the x-axis
and 2cm to represent 1 unit on the y-axis.
c) Use the graph in (b) above to solve the equation 3 Cos x – sin 2x = 2. (2 marks)
d) State the amplitude of y=3cosx-2. (1 mark)

17. 2010 Q 17 P2
(a) Complete the table below, giving the value correct to 2 decimal places. (2 marks)

X0 00 200 400 600 800 1000 1200 1400 1600 1800

Cos X0 1.00 0.94 0.77 0.50 -0.17 -0.77 -1.00

Sin x0 – Cos x0 -1.00 -0.60 0.37 0.81 1.37 1.28 1.00

b) On the grid provided and using the same axes draw the graphs of y = Cos x0 and
y = sin x0 – Cos x0 for 00 ≤ x ≤ 1800 .Using the scale; 1 cm for 200 on the x-axis
and 4cm for 1 unit on the y-axis. (5 marks)

c) Using the graph in part (b);

i)Solve the equation sin x0 – cos x0 = 1.2; (1 mark)
1
ii) Solve the equation cos x0 = sin x0; (1 mark)
2
iii)Determine the value of cos x0 in part (c) (ii) above. (1 mark)

18. 2011 Q 16 P2
The table below shows values x and y for the function y=2sin3x0 in the range 00 ≤ x ≤ 1500

x0 0 15 30 45 60 75 90 105 120 135 150

y 0 1.4 2 1.4 0 -1.4 -2 -1.4 0 1.4 2

a. On the grid provided, draw the graph of y= 2sin 3x (2 marks)

b. From the graph determine the period. (1 mark)

19. 2013 Q21 P2

(a) Complete the table below, giving the values correct to 1 decimal place. (2 marks)

x0 0 40 80 120 160 200 240

2 sin(χ +20)0 0.7 2.0 0.0 -2.0
√ 3 cos x 1.7 1.3 -0.9 -1.6

b) On the grid provided, using the same scale and axes, draw the graphs of
y = 2 sin (x +20)0 and y = √ 3 cos x for 00 < x ≤ 2400. (5 marks)

c) Use the graphs drawn in (b) above to determine:

i) The value of x for which 2sin (x + 20) = √ 3 cos x; (2 marks)

41
 ii) The difference in the amplitudes of y =2sin(x + 20) and y =√ 3 cos x. (1 mark)
AREA APPROXIMATIONS

(a). COUNTING SQUARES TECHNIQUE

1. 1999 Q 5 P1

The figure below is a map of a forest drawn on a grid of 1 cm squares

i. Estimate the area of the map in square centimeters (2 marks)

ii. If the scale of the map is 1: 50,000 estimate the area of the forest in hectares

2. 2000 Q 6 P1
The enclosed region shown in the figure below represents a ranch drawn to scale.
The actual area of the ranch is 1075 hectares.
a) Estimate the area of the enclosed region in square centimeters (1 mark)

b) Calculate the linear scale used (2 marks)

42
(b). TRAPEZIODAL

1. 1991 Q24 P2
The travel graph of a cyclist given below represents a journey from 9am up to 11am. By dividing
the shaded region into six strips of equal width and using trapezium rule, estimate
the total distance travelled

2. 1992 Q24 P2
In order to sketch the cross-section of a ditch 175cm wide, the depth of water was
measured at intervals of 25cm from one of the banks. The reading of the depths
were as follows:

Distance (cm) 0 25 50 75 100 125 150 175

Depth (cm) 100 115 132 156 167 200 163 153

i) Sketch the cross-section of the ditch (3 marks)

ii) Use trapezoidal rule to estimate the area of the cross-section (5 marks)

3. 1993 Q18 P1
a) Use the trapezoidal rule to find the area under the curve y = x2 + 1
From x = 1 to x = 15 using seven strips. (5 marks)

b) The cross-section areas in m2 along the length of an 18m wooden log are:
5.0, 5.4, 7.0, 8.0, 5.5, 5.8, 6.0

The cross sectional areas are equally spaced. The first and the last areas represent the
ends of the log. Estimate its volume using the trapezoidal rule.

43
 (2 marks)

4. 1996 Q 11 P2
Complete the table below for the function. y= 3x2 - 8x + 10

X 0 2 4 6 8 10
y 10 6 70 230

Using the values in the table and the trapezoidal rule, estimate the area bounded by the
curve y=3x2 – 8x + 10 and the lines y=0,x=0 and x=10
(3 marks)
5. 1997 Q 8 P2
Use the trapezoidal rule with intervals of 1 cm to estimate the area of the shaded region below

6. 1999 Q 24 P1
The graph below consists of a non- quadratic part ( 0 ≤ x ≤ 2) and a quadrant part
(2 ≤ x 8) The quadratic part is y = x2 – 3x + 5, 2 ≤ x ≤ 8

44
 (a) Complete the table below (1 mark)

x 2 3 4 5 6 7 8

y 3

(b) Use the trapezoidal rule with six strips to

estimate the area enclosed by the
curve, x = axis and the line x = 2 and x = 8 (3 marks)

(c) Find the exact area of the region given in (b) (3 marks)

(d) If the trapezoidal rule is used to estimate the area under the curve between
x = 0 and x = 2, state whether it would give an under- estimate or an
over- estimate. Give a reason for your answer (1 mark)

6. 2001 Q 11 P1
A particle is projected from the origin. Its speed was recorded as shown in
the table below

Time (sec) 0 5 10 15 20 25 39 35

Speed (m/s) 0 2.1 5.3 5.1 6.8 6.7 4.7 2.6

Use the trapezoidal rule to estimate the distance covered by the particle within
the 35 seconds

7. 2002 Q 21 P2
The table below shows the values of x and corresponding values of y for a given curve.
X 0 π π π π 5π π
12 6 4 3 12 2

y 0 0.26 0.48 0.65 0.76 0.82 0.84

45
 a) Use the trapezium rule with seven ordinates and the values in the table only to estimate
the area enclosed by the curve, x – axis and the line x = Л/2 to four decimal places.
(Take Л = 3.142)

b) The exact value of the area enclosed by the curve is known to be 0.8940. Find the
percentage error made when the trapezium rule is used. Give the answer correct to two
decimal places.

8. 2006 Q 16 P1
A circle centre O, ha the equation x2 + y2 = 4. The area of the circle in the first quadrant
is divided into 5 vertical strips of width 0.4 cm

(a) Use the equation of the circle to complete the table below for values of y
correct to 2 decimal places (1 mark)

X 0 0.4 0.8 1.2 1.6 2.0

Y 2.00 1.60 0

(b) Use the trapezium rule to

estimate the area of the circle
(3 marks)

9. 2007 Q 24 P1
The diagram on the grid below represents as extract of a survey map showing two
Adjacent plots belonging to Kazungu and Ndoe.

46
 The two dispute the common boundary with each claiming boundary along different
smooth curves coordinates ( x, y1) and (x, y2) in the table below, represents points on the
boundaries as claimed by Kazungu Ndoe respectively

x 0 1 2 3 4 5 6 7 8 9

y1 0 4 5.7 6.9 8 9 9.8 10.6 11.3 12

y2 0 0.2 0.6 1.3 2.4 3.7 5.3 7.3 9.5 12

(a) On the grid provided above draw and label the boundaries as claimed by
Kazungu and Ndoe (2 marks)

10. 2011 Q 21 P1
a) Using the trapezium rule with seven ordinates, estimate the area of the region
bounded by the curve y= -x2 + 6x + 1, the lines x=0,y=0 and x=6.
(5 marks)

b) Calculate
i) the area of the region in a) above by integration: (3 marks)
ii) the percentage error of the estimated area to the actual area of the
region, correct to two decimal places. (2 marks)

47
 (C). MID-ORDINATE

1. 1990 Q15 P2
The figure below shows the shape of a piece of land OABC. Using the mid-ordinate rule
with 11 ordinates estimate the area of the land

2. 1995 Q 16 P2
The shaded region below represents a forest. The region has been drawn to scale where
1 cm represents 5 km. Use the mid – ordinate rule with six strips to estimate the area of
forest in hectares. (4 marks)

48
3. 1996 Q 21 P1
The table below shows some values of the function y = x2 + 2x – 3

x
-6 -6.75 -5.5 -5 -4.75 -4.5 4.25 -4.0 -3.75 -3.75 -3.5 -3.25 -3

y 21 18.56 14.06 10.06 8.25 5 2.25 1.06 0

a) Complete the table

b) Using the completed table and the mid- ordinate rule with six ordinates, estimate
the area of the region bounded by the y = x2 + 2 x – 3 and the line y = 0, x = -6 and x = -3
(3 marks)
(i) By integration find the actual area of the region in (b) above (2 marks)
(ii) Calculate the percentage error arising from the estimate in (b) (2 marks)

4. 2003 Q 20 P1
The diagram below is a sketch of the curve y =x2 + 5.

a) i) Use the mid –ordinate rule, with six strips to estimate the area
enclosed by the curve, the x – axis and the y – axis and line x =3.
(4 marks)
ii) Calculate the same area using the integration method. (2 marks)

b) Assuming the area calculated in (a) (ii) is exact, calculate the percentage
error made when the mid – ordinate rule is used. (2 marks)

5. 2004 Q 11 P1

The table below shows some values of the function y = x2 +3

X 0 ½ 1 11/2 2 21/2 3 31/2 4 41/2 5 51/2 6

y 3 4 51/4 7 12 151/4 19 28 39

a) Complete the table (1 mark)

b) Use the mid – ordinate rule with six ordinates to estimate the area
bounded by y = x2+3, the y – axis, the x – axis and the line x = 6 (2 marks)

6. 2005 Q 20 P1
The table below gives some of the values of x for the function y=½ x2 + 2x + 1
in the interval 0≤ x ≤ 6.
49
 x 0 1 2 3 4 5 6

y 1 3.5 7 11.5 17 23.5 31

(a) Use the values in the table to
draw the graph of the function
(2 marks)
(b) (i) Using the graph and the mid – ordinate rule with six (6) strips, estimate
the area bounded by the curve, the x- axis, the y- axis and the line = 6 (4 marks)

(ii) If the exact area of the region described in (b) (i) above is 78cm2,
calculate the percentage error made when the mid – ordinate rule is used.
Give the answer correct to two decimal places (2 marks)

7. 2008 Q 18 P1
The figure below is a sketch of the curve whose equation is y=x2+x+5. It cuts the line
y=11 at points P and Q.

a) Find the area bounded by the curve =x2+x+5 and the line y=11 using the
trapezium rule with 5 strips. (5 marks)

b) Calculate the difference in the area if the mid-ordinate rule with 5 ordinates
was used instead of the trapezium rule. (5 marks)

8. 2009 Q 24 P1
(a) On the grid provided, draw a graph of the function
1 2
y= x – x + 3 for 0  x  6.
2
(b) Calculate the mid – ordinates for 5 strips between x = 1 and x=6, and hence
use the mid- ordinate rule to approximate the area under the curve between
x = 1, x=6 and the x – axis. (3 marks)

(c) Assuming that the area determine by integration to be the actual area,
calculate the percentage error in using the mid-ordinate rule. (4 marks)

50
 INTEGRATION
1. 1992 Q22 P1
a) The gradient of the curve y = ax2 + bx at the origin is equal to 8. Find the values
of a and b if the curve has a maximum point at x = 4 (5 marks)

b) Determine the area bounded by the lines x=0, x=6, y=0 and the curve y=ax2+bx,
for the values of a and b obtained in part (a) (3 marks)

2. 1994 Q6 P2
Determine the area bounded by the curve y=x2 – 4, the x axis and the line x=4
(4 marks)
3. 1995 Q 7 P1

Find the area enclosed by the curve y=4x –x2, the x-axis and the lines x=1 and x=2
(3 marks)

4. 1996 Q 8 P2

Find the area bounded by the curve y= 2x3 -5, the x-axis and the lines x=2 and x=4

5. 1998 Q 20 P2
(a) Find the value of x at which the curve y= x2- 2x – 3 crosses the x- axis (2 marks)

(b) Find
∫ ( x 2−2 x−3 ) dx (1 mark)
c) Find the area bounded by the curve y = x – 2x – 3, the axis and the line x= 2
2

and x = 4
(5 marks)
6. 2000 Q 21 P2
−2
The curve of the equation y = 2x + 3x2, has x = and x = 0 and x intercepts.
3
−2
The area bounded by the axis x = and x = 2 is shown by the sketch below.
3
Find:
(a) ∫ ( 2 x +3 x ) dx
2
(2 marks)
−2
(b) The area bounded by the curve x – axis, x = and x =2 (6 marks)
3

51
7. 2006 Q 24 P2
The diagram below shows a sketch of the line y = 3x and the curve y = 4 – x2
intersecting at points P and Q.

(a) Find the coordinates of P and Q (4 marks)

(b) Given that QN is perpendicular to the x- axis at N, calculate

(i) The area bounded by the curve y = 4 – x2, the x- axis and the line QN (2 marks)
(ii) The area of the shaded region that lies below the x- axis (2marks)
(iii)The area of the region enclosed by the curve y = 4-x2, the line y – 3x
and the y axis
(2 marks)
8. 2007 Q 20 P2
The gradient function of a curve is given by the expression 2x + 1. If the curve passes
through the point ( -4, 6);
(a) Find:
(i) The equation of the curve (3 marks)
(ii) The values of x, at which the curve cuts the x- axis (3 marks)
(b) Determine the area enclosed by the curve and the x- axis (4 marks)

9. 2013 Q23 P2
1 2
The equation of a curve is given by y=5 x− x
2
1 2
(a) On the grid provided, draw the curve of y=5 x− x for 0 ≤ x ≤ 6 (3 marks)
2
(b) By integration, find the area bounded by the curve, the line x =6 and
52
 the x-axis. (3 marks)

(c) (i) On the same grid as in,(a).draw the line y = 2x. (1 mark)
(ii) Determine the area bounded by the curve and the line y = 2 x. (3 marks)

10. 2014 Q21 P1

(a) Complete the table below for the function y = x2 – 3x + 6 in range -2 ≤ x ≤ 8 (2marks)

x -2 -2 0 1 2 3 4 5 6 7 8
y

(b) Use the trapezium rule with strips to estimate the area bounded by the curve,
y = x2 – 3x + 6, the lines x = -2, x = 8, and x - axis (3marks)

(c) Use the mid-ordinate rule with 5 strips to estimate the area bounded by the curve,
y = x2 – 3x + 6, the lines x = -2, x = 8, and x –axis (2marks)

(d) By integration, determine the actual area bounded by the curve y = x2 – 3x + 6, the
lines x = -2, x = 8, and x –axis (3marks)
CALCULUS
(a) DIFFERENTIATION

1. 1990 Q15 P1
A farmer has 1200m of wire to fence three sides of a rectangular paddock.
The fourth side is a wall. Find the dimension that will give the maximum
possible area (4 marks)
2. 1991 Q11 P2
Use differentiation to find the x coordinate of the maximum point for the curve
y = x3 + 2x2 – 4x -8 (5 marks)
3. 1992 Q11 P2
Find the equation of the tangent to the curve y + 2x2 at (2, 8) (4 marks)

4. 1993 Q12 P2
Calculate the gradient of the curve y = x2 – 3x -4 at a point where x = -1 (2 marks)

5. 1993 Q24 P2
A projectile is fired vertically upwards. At anytime t (seconds) its height
h(metres) above the ground is given by: h = 30t -5t2

a) How fast is it moving at (2 marks)

i) t = 1 second?
ii) t= 2 seconds? (1 mark)

b) How far up does it travel (5 marks)

6. 1994 Q11 P1
A rectangular plate has a perimeter of 28cm. Determine the dimensions of the
plate that give the maximum area
(4 marks)

7. 1996 Q 19 P1
The equation of a curve us y = 3x2 - 4 x + 1
53
 (a) Find the gradient function of the curve and its value when x = 2 (2 marks)

(b) Determine
(i) The equation of the tangent to the curve at the point (2, 5) (2 marks)

(ii) The angle which the tangent to the curves at the point (2, 5)makes
with the horizontal ( 1 mark)

(iii) The equation of the line through the point (2, 5) which is perpendicular
to the tangent in (b) (i) (3 marks)

8. 1997 Q 10 P1
The curve y= ax3 – 3x2 – 2x + 1 has the gradient 7 when x=1.
Find the value of a
(3marks)

9. 1999 Q 16 P2
Find the equation of the tangent to the curve y = (x2 + 1) (x- 2) when x= 2 (3 marks)

10. 2000 Q 5 P2
The distance from a fixed point of a particular in motion at any time t
seconds is given by
3 5 2
S=t − t +2 t+5 metr es
2
Find its:
(a) Acceleration after t seconds (1 mark)
(b) Velocity when acceleration is Zero (2 marks)

11. 2001 Q 11 P2
A curve is given by the equation: y = 5x3 – 7x2 + 3x +2
Find the: a) Gradient of the curve at x = 1 (2 marks)
b) Equation of the tangent to the curve at the point( 1,3) (2 marks)

12. 2001 Q 22 P2
The displacement x metres a particle after seconds given by.
x = t3-2t2 +6, t>0.
a) Calculate the velocity of the particle in m/s when t = 2 seconds. (3 marks)
b) When the velocity of the particle is zero, calculate its:-
i) Displacement (3 marks)
ii) Acceleration. (2 marks)

13. 2002 Q 16 P1
Given the curve y = 2x3 + ½ x2 – 4x + 1. Find the:

[
i) Gradient of curve at 1,−
1
2 ]
[
ii) Equation of the tangent to the curve at 1,−
1
2 ] (4 marks)

14. 2002 Q 24 P2
The displacement, s metres, of a particle moving along straight line after
t seconds is given by. S = 3t + 3/2 t2 – 2t3
54
 a) Find its initial acceleration (3 marks)
b) Calculate: i) The time when the particle was momentarily at rest. (2 marks)
ii)Its displacement by the time it comes to rest momentarily (1 mark)
c) Calculate the maximum speed attained.
(2marks)
15. 2003 Q 8 P2
Find the coordinates of the turning point of the curve whose equation is
y =6 +2x – 4x2 (3 marks)
16. 2003 Q 21 P2
a) i)Find the coordinated of the stationary points on the curve
y = x3 – 3x + 2 (2 marks)
ii)For each stationary point determine whether it is minimum or maximum.
(4 marks)
b)In the space provided below, sketch the graph of the function y= x3 – 3x +2
(2 marks)

17. 2004 Q 5 P1
The velocity V ms-1, of a moving body at time t seconds is given by
V = 5t2 – 12t + 7.
Calculate the acceleration when t=2 seconds (3 marks)

18. 2005 Q 16 P2
A stone is thrown vertically upwards from a point O. After t seconds, the
stone is S metres from O. Given that S= 29.4t – 4.9t2, find the maximum height
reached by the stone (3 marks)

19. 2005 Q 17 P2
1 3 2
A curve is represented by the function y = x + x – 3x + 2
3
dy
(a) Find (1 mark)
dx
(b) Determine the values of y at the turning points of the curve
1 3 2
y= x + x – 3x + 2 (4 marks)
3
20. 2006 Q 24 P1
A particle moves along straight line such that its displacement S metres
from a given point is S = t3 – 5t2 + 4 where t is time in seconds
Find
(a) the displacement of particle at t = 5 (2 marks)
(b) the velocity of the particle when t = 5 (3 marks)
(c) the values of t when the particle is momentarily at rest (3 marks)
(d) The acceleration of the particle when t = 2 (2 marks)

21. 2007 Q 5 P1
The gradient of the tangent to the curve y = ax3 + bx at the point ( 1,1) is -5.
Calculate the values of a and b ( 4 marks)
22. 2007 Q 13 P1
The sum of two numbers x and y is 40. Write down an expression, in terms
of x, for the sum of the squares of the two numbers.
Hence determine the minimum value of x2 + y2 ( 4 marks)

55
23. 2008 Q 24 P1
The distance s metres from a fixed point O, covered by a particle after t seconds
is given by the equation; S =t3 -6t2 + 9t + 5.
a) Calculate the gradient to the curve at t=0.5 seconds (3 marks)
b) Determine the values of s at the maximum and minimum turning points
of the curve. (4 marks)
3 2
c) On the space provided, sketch the curve of s= t -6t +9t + 5. (3 marks)

24. 2010 Q 24 P1
A rectangular box open at the top has a square base. The internal side of
the base is x cm long and the total internal surface area of the box is 432 cm2.

(a) Express in terms x:

(i) The internal height h, of the box. (3 marks)
(ii) The internal volume V, of the box. (1 mark)

(b) Find:
i) The value of x for which the volume V is maximum; (4 marks)
ii) The maximum internal volume of the box. (2 marks)

25. 2011 Q 22 P1
The displacement, s metres, of a moving particle after t seconds is given by
s=2t3 – 5t2 + 4t + 2.
Determine:

a) The velocity of the particle when t=3 seconds: (3 marks)

b) The value of t when the particle is momentarily at rest; (3 marks)
c) The displacement when the particle is momentarily at rest; (2 marks)
d) The acceleration of the particle when t=3 seconds. (2 marks)

26. 2012 Q22 P1

The equation of a curve is y=2 x 3 +3 x2
(a) Find (i) The x-intercept of the curve
(ii) The y-intercept of the curve

(b) (i) Determine the stationary points of the curve

(ii) For each point in (b) (i) above, determine whether it is a maximum or
minimum curve
(c) Sketch the curve

27. 2012 Q24 P2

The acceleration of a body moving along a straight line is (4-t) m/s2 and its velocity
is v m/s after t seconds,
(a) (i) If the initial velocity of the body is 3m/s, express the velocity v in terms of t.
(3 marks)
(ii) Find the velocity of the body after 2 seconds. (2 marks)

(b) Calculate:
(i) The time taken to attain maximum velocity; (2 marks)
(ii) The distance covered by the body to attain the maximum velocity
(3 marks)
28. 2013 Q21 P1
The displacement, s metres, of a moving particle from a point O, after t seconds is
56
 given by,
s = t3 – 5t2 + 3t + 10.
a) Find s when t =2. (2 marks)
b) Determine:
i. The velocity of the particle when t = 5 seconds; (3 marks)
ii. The value of t when the particle is momentarily at rest. (3 marks)

c) Find the time, when the velocity of the particles is maximum. (2 marks)

29. 2014 Q24 P1

The equation of acurve is given by y = x3 – 4x2 – 3x

(a) Find the value of y when x = -1 (1mark)

(b) Determine the stationary points of the curve (5marks)
(c) Find the equation of the normal to the curve at x = 1 (4marks)

30. 2014 Q24 P1

A particle was moving along a straight line. The acceleration of the particle
after t seconds was given by (9 — 3t) ms-2. The initial velocity of the particle
was 7 ms-1.

Find:
a) the velocity (v) of the particle at any given time (t); (4 marks)
b) The maximum velocity of the particle; (3 marks)
c)the distance covered by the particle by the time it attained maximum velocity

(b) INTEGRATION

1. 1989 Q15 P1
A particle moves along a straight line PQ. Its velocity v metres per second
after t seconds is given by v = t2 -3t + 5. Its distance from P at the time t = 1
is 6metres. Determine its distance from p when t = 3.
(4 marks)
2. 1991 Q14 P1

Evaluate (3 marks)
3. 1992 Q12 P2

57
 The velocity v m/s of a particle moving along a straight line at any time t (sec)
is given by v = 3t – 2. Its distance x (m) at the time t = 0 is equal to 2.
Calculate x when t = 4
(4 marks)
4. 1994 Q19 P2
The velocity of a particle moving in a straight line after t seconds given by
v = 6t – t2 + 4 m/s.
Calculate
a) The acceleration of the particle after 2 seconds (2 marks)
b) The distance covered by the particle between t = 2 sec and t =5sec. (3 marks)
c) The time when the particle will be momentarily at rest. (3 marks)

5. 1999 Q 16 P1
A particle moves on a straight line. The velocity after t seconds is given by
V = 3t2 – 6 t – 8.
The distance of the particle from the origin after one second is 10 metres. Calculate
the distance of the particle from the origin after 2 seconds. (4 marks)

6. 2000 Q 14 P1
The acceleration a m/s2 of a particle moving in a straight line is given
bya = 18t - 4, where t is time in seconds. The initial velocity of the particle is 2 m/s
a) Find the expression for velocity in terms of t (2 marks)
b) Determine the time when the velocity is again 2m/s ( 4 marks)

7. 2001 Q 21 P1
dy 2
(a) The gradient function of a curve is given by =2 x −5
dx
Find the equation of the curve, given that y = 3, when x = 2 (4 marks)

b) The velocity, vm/s of a moving particle after seconds is given:

v = 2t3 + t2 – 1. Find the distance covered by the particle in the interval
1≤t≤3 (4 marks)

8. 2002 Q 20 P1
The diagram below shows a straight line intersecting the curve y = (x-1)2 + 4
at the points P and Q. The line also cuts x-axis at (7, 0) and y axis at (0, 7)

a) Find the equation of the straight line in the form y = mx +c. (2 marks)
b) Find the coordinates of p and Q. (2 marks)
c) Calculate the area of the shaded region. (3 marks)

58
9. 2003 Q 16 P1
The velocity Vms-1 of particle in motion is given by V =3t2 – t +4, where t is time in
seconds. Calculate the distance traveled by the particle between the time t=1 second
and t=5 seconds. (3 marks)

10. 2004 Q 13 P2
dy 2
The gradient function of a curve is given =x −8 x+ 2. If the curve passes
dx
through the point, (0, 2), find its equation. (3 marks)

11. 2004 Q 22 P2
A particle moves in a straight line. It passes though point O at t = 0 with velocity
v= 5m/s.
The acceleration a m/s2 of the particle at time t seconds after passing through
O is given by a = 6t + 4
(a) Express the velocity v of the particle at time t seconds in terms of t (3marks)
(b) Calculate
(i) The velocity of the particle when t = 3 (2 marks)
(ii) The distance covered by the particle between t = 2 and t = 4 (3 marks)

12. 2005 Q 16 P1
The acceleration, a ms-2, of a particle is given by a =25 – 9t2, where t in seconds
after the particle passes fixed point O.
If the particle passes O, with velocity of 4 ms-1, find

(a) An expression of velocity V, in terms of t (2 marks)

(b) The velocity of the particle when t = 2 seconds (2 marks)

13. 2005 Q 21 P1
The gradient of a curve at point (x,y) is 4x – 3. The curve has a minimum value
of – 1/8
(a) Find
(i) The value of x at the minimum point (1 mark)
(ii) The equation of the curve (4 marks)

b) P is a point on the curve in part (a) (ii) above. If the gradient of the
curve at P is -7, find the coordinates of P ( 3 marks)

14. 2006 Q 15 P2
A particle moving in a straight line passes through a fixed point O with a velocity of
9m/s. The acceleration of the particle, t seconds after passing through O is given by

59
 a = (10 – 2t) m/s2.
Find the velocity of the particle when t – 3 seconds (3 marks)

15. 2007 Q 5 P2
A particle moves in a straight line through a point P. Its velocity v m/s is given by
v= 2 -t, where t is time in seconds, after passing P. The distance s of the particle
from P when t = 2 is 5 metres. Find the expression for s in terms of t. (3 marks)

16. 2008 Q 15 P2
A particle moves in a straight line from a fixed point. Its velocity Vms-1 after t
seconds is given by V=9t2 – 4t +1
Calculate the distance traveled by the particle during the third second. (3 marks)

17. 2009 Q 16 P2
A particle moves in a straight line with a velocity V ms –l. Its velocity after t
seconds is given by V= 3t2 – 6t -9
The figure below is a sketch of the velocity-time graph f the particle

Calculate the distance the particle moves between t = 1 and t = 4

18. 2010 Q 11 P2
A particle starts from O and moves in a straight line so that its velocityV ms-1 after
time t seconds is given by V = 3t – t2. The distance of the particle from O at time t
seconds is s metres.

a) Express s in terms of t and c where c is a constant. (1 mark)

b) Calculate the time taken before the particle returns to O. (3 marks)

19. 2014 Q15 P2

dy
The gradient of a curve is given by = x2 - 4x 4- 3. The curve passes through the
dx
point (1,0). Find the equation of the curve. (3 marks).

20. 2015 Q24 P1

The gradient of the curvey y = 2x3 – 9x2 + px – 1 at x = 4 is 36.
a)Find :
i) the value of p; (3 marks)
ii)The equation of the tangent to the curve at x = 0.5. (4 marks)

b) Find the coordinates of the training points of the curve (3 marks)

60
21. 2015 Q13 P2

Evaluate ∫ 42x + 2x-15dx

2
(3 marks)

LINEAR PROGRAMMING
1. 1989 Q18 P2

At an agricultural research station, a rectangular plot is to be allocated for a certain

experiment. The length of the plot must be greater than the breadth but not more
than twice the breadth. The area must be more than 400m2 and the perimeter
must be less than 200m.
(a) Form inequalities representing the above information and graph them on
the grid below. (graph was provided) (6 marks)

(b) If the plot must also be exactly divisible into the sub plots of 10m by 10m,
determine the size of the plot which will give maximum area (2 marks)

2. 1990 Q20 P1
A farmer has 15 hectares of land on which he can grow maize n beans only. In
a year he grows maize on more land than beans. It costs him sh 4,400 to grow
maize per hectare and sh. 10,800 to grow beans per hectare. He is prepared to
spend at most sh 90,000 per year to grow the crops. He makes a profit of
sh 2,400 from one hectare of maize and sh3,200 from one hectare of beans.
Find the maximum profit he can make from the crops in a year. (8 marks)

3. 1991 Q24 P1
A chemical firm has 160 litres of solution A, 110 litres of solution B and
150 litres of solution C. To prepare a bottle of syrup X, 200ml of solution A,
100ml of solution B and 100ml of solution Care needed. For a bottle of Y 100ml
of A, 200ml of B and 300ml of C are needed. Syrup X sells at sh 60 per bottle
and syrup Y sells at sh 100 per bottle.
How many bottles of each type of syrup should the firm make inorder to obtain
the maximum amount of money? (8 marks)

4. 1992 Q20 P1
A transporter has two types of trucks to transport sugar. Type A truck carries
2000 bags while type B carries 3000 bags per trip. The transporter has to
transport 120000 bags. He has 2 make more than 50 trips. Type B trucks are
to make at most twice the number of trips made type A (B to be at most twice A).

(a) By taking x to be the number of trips made by type A trucks and y to B

the number of trips made by type b trucks, write down the inequalities
representing this information (3 marks)

(b) If the transporter makes a profit of sh 1000 per trip for the type A truck

61
 and sh 2000 per trip for the type B truck use graphical methods to find
the number of trips he should make with each type of truck in order to
maximize his profit
(5 marks)

5. 1993 Q23 P2
A coffee merchant has 400kg of Robusta coffee and 480kg of Arabica coffee.
The coffee is packed by weight as follows:
Type I : 30% Robusta and 70% Arabica
Type II: 50% Robusta and 50% Arabica

Type I is sold at sh 34 per kg while type II is sold at sh 36 per kg. Use

graphical method to determine the number of kilograms of type I and type II
which should be packed to maximize the profit. (8 marks)

6. 1994 Q 18 P2
A factory manufacturers two products which are produced on the three machines
X, Y and Z. The first product requires 2 hours on machine X, 3hours on machine Y
and 1 hour on machine Z. The second product requires 1 hour on machine X,
4hours on machine Y and 2 hours on machine Z.Machine X can be used for at
most 100 hours, machine Y for at most 240 hours and machine Z for at most 90 hours
The profit per unit is sh 300 for the first product and sh 400 for the second product.

Form inequalities representing the above information and represent them on the
grid provided.. From the graph determine the number of units of each product that
should be produced to give maximum profit. (8 marks)

7. 1995 Q24 P2
A manufacture of jam has 720 kg of strawberry syrup and 800 kg of mango
syrup for making two types of jam, grade A and B. Each types is made by
mixing strawberry and mango syrups as follows:
Grade A: 60% strawberry and 40% mango
Grade B: 30% strawberry and 70% mango

The jam is sold in 400 gram jars. The selling prices are as follows:
Grade A: Kshs. 48 per jar
Grade B: Kshs 30 per jar.

(a) Form inequalities to represent the given information (3 marks)

(b) (i) On the grid provided draw the inequalities ( 3 marks)

(ii) From your, graph, determine the number of jars of each grade the
manufacturer should produce to maximize his profit
(1 mark)
(iii) Calculate the total amount of money realized if all the jars are sold (1 mark)

8. 1997 Q19 P2
62
 An institute offers two types of courses technical and business courses. The
institute has a capacity of 500 students. There must be more business students
than technical students but at least 200 students must take technical courses.
Let x represent the number of technical students and y the number of business students.

(a)Write down three inequalities that describe the given conditions (3 marks)

(b) On the grid provided, draw the three inequalities (3 marks)

(c) If the institute makes a profit of Kshs 2, 500 to train one technical students
and Kshs 1,000 to train one business student, determine
(i)the number of students that must be enrolled in each course to maximize the
profit (1 mark)
(ii) The maximum profit. (1 mark)

9. 1997 Q22 P2
A school has to take 384 people for a tour. There are two types of buses available,
type X and type Y. Type X can carry 64 passengers and type Y can carry 48 passengers.
They have to use at least 7 buses.
(a) Form all the linear inequalities which will represent the above information.
(3 marks)
(b) On the grid provided below,draw the inequalities and shade the unwanted
Region (3 marks)
(c ) The charges for hiring the buses are:
Type X: sh 25,000
Type Y: sh 20,000
Use your graph to determine the number of buses of each type that should
be hired to minimize the cost (2 marks)

10. 1998 Q24 P2

A draper is required to supply two types of shirts A and type B.The total number
of shirts must not be more than 400. He has to supply more type A than of
type B however the number of types A shirts must be more than 300 and the
number of type B shirts not be less than 80.

Let x be the number of type A shirts and y be the number of types B shirts.
(a) Write down in terms of x and y all the linear inequalities representing
the information above. (3 marks)

(b) On the grid provided, draw the inequalities and shade the unwanted regions
(3 marks)
(c) Type A: Kshs 600 per shirt
Type B: Kshs 400 per shirt
(i) Use the graph to determine the number of shirts of each type that
should be made to maximize the profit. (1 mark)

(ii) Calculate the maximum possible profit. (1 mark)

11. 2000 Q24 P2

A theatre has a seating capacity of 250 people. The charges are Kshs. 100 for
an ordinary seat and Kshs 160 for a special seat. It cost Kshs 16,000 to stage a
show and the theater must make a profit. There are never more than 200 ordinary

63
 seats and for a show to take place at least 50 ordinary seats must be occupied.
The number of special seats is always less than twice the number of ordinary seats.

(a) Taking x to be the number of ordinary seats and y the number of special
seats write down all the inequalities representing the information above.
(2 marks)

(b) On the grid provided, draw a graph to show the inequalities in (a) above
(4 marks)
(b) Determine the number of seats of each type that should be booked in
Order to maximize the profit. (2 marks)

12. 2001 Q24 P2

Bot juice Company has two types of machines, A and B, for juice production.
Type A machine can produce 800 litres per day while type B machine produces
1,600 litres per day.
Type A machine needs 4 operators and type B machine needs 7 operators. At least
8,000litres must be produced daily and the total number of Operators should not
exceed 41. There should be 2 more machines of each type.

Let x be the number of machines of type A and Y the number of machines for
type B
a) Form all inequalities in x and y to represent the above information. (3 marks)

b) On the grid provided below, draw the inequalities to shade the unwanted regions.
(3 marks)

13. 2003 Q20 P2

Omondi makes two types of shoes: A and B. He takes 3 hours to make one pair
of type A and 4 hours to make one pair of type B.He works for a maximum of
120 hours to x pairs of type A and y pairs of type B.It costs him sh 400 to make
a pair of type A and sh 150 to make a pair of type B.

His total cost does not exceed sh 9000. He must make 8 pairs of type A and
more than 12 pairs of type B.

64
 (a) Write down four inequalities representing the information above (3 marks)
(b) On the grid provided, draw the inequalities and shade the unwanted regions
(3 marks)
(c) Omondi makes a profit of sh40 on each pair of type A and sh70 on
each pair of type B shoes.
Use the graph provided to determine the maximum possible profit he makes.
(2 marks)
14. 2005 Q24 P2
A diet expert makes up a food product for sale by mixing two ingredients N and S.
One kilogram of N contains 25 units of protein and 30 units of vitamins. One
kilogram of S contains 50 units of protein and 45 units of vitamins.
The food is sold in small bags containing atleast 175 units of protein and atleast
180 units of vitamins. The mass of food product in each bag must not exceed 6kg.
If one bag of the mixture contains x kg of N and y kg of S

(a) Write down all the inequalities, in terms of x and representing the
information above (2 marks)
(b) On the grid provided draw the inequalities by shading the unwanted regions
(2 marks)

(c) If one kilogram of N costs Kshs 20 and one kilogram of S costs Kshs 50,
use the graph to determine the lowest cost of one bag of the mixture
(3 marks)
15. 2006 Q23 P2
Mwanjoki flying company operates a flying service. It has two types of aeroplanes.
The smaller one uses 180 litres of fuel per hour while the bigger one uses 300
litres per hour.The fuel available per week is 18,000 litres. The company is allowed
80 flying hours per week while the smaller aeroplane must be flown for y hours
per week.
(a) Write down all the inequalities representing the above information
(3 marks)
(b) On the grid provided on page 21, draw all the inequalities in a) above by
shading the unwanted regions (3 marks)

(c) The profits on the smaller aeroplane is Kshs 4000 per hour while that on
the bigger one is Kshs 6000 per hour
Use the graph drawn in (b) above to determine the maximum profit
that the company made per week. (3 marks)

16. 2007 Q22 P2

A company is considering installing two types of machines. A and B. The
information about each type of machine is given in the table below.

Machine Number of operators Floor space Daily profit

A 2 5m2 Kshs 1,500

B 5 8m2 Kshs 2,500

The company decided to install x machines of types A and y machines of type B

(a) Write down the inequalities that express the following conditions
65
 I. The number of operators available is 40
II. The floor space available is 80m2
III. The company is to install not less than 3 type of A machine
IV. The number of type B machines must be more than one third
the number of type A machines
(4 marks)
(b) On the grid provided, draw the inequalities in part (a) above and shade
the unwanted region (4 marks)

(c) Draw a search line and use it to determine the number of machines of
each type that should be installed to maximize the daily profit.
(2 marks)
17. 2010 Q20 P2
A carpenter takes 4 hours to make a stool and 6 hours to make chair. It takes
the carpenter and at least 144 hours to make x stools and y chairs. The labour
cost should not exceed Ksh.4800 the carpenter must make a least 16 stools
and more than 10 chairs.

a) Write down inequalities to represent the above information. (3 marks)

b) Draw the inequality in (a) above on a grid. (4 marks)
c) The carpenter makes a profit of Ksh 40 on a stool and Ksh 100 on a chair.
Use the graph to determine the maximum profit the carpenter can make.
(3 marks)

18. 2011 Q24 P2

A building contractor has two Lorries, P and Q used to transport atleast 42 tonnes
of sand to a building site. Lorry P carries 4 tonnes of sand per trip while lorry Q
carries 6 tonnes of sand per trip. Lorry P uses 2litres of fuel per trip while lorry Q
uses 4litres of fuel per trip.
The two Lorries are to use less than 32 litres of fuel. The number of trips made by
lorry Pshould be less than 3 times the number of trips made by lorry Q. Lorry p
should make more than 4 trips.

(a) Taking x to represent the number of trips made by the lorry Pand y to
represent the number of trips made by lorry Q, write the inequalities that
represent the above information. (4 marks)
(b) On the grid provided, draw the inequalities and shade the unwanted regions
(4 marks)
(c) Use the graph drawn in (b) above to determine the number of trips made by
lorry P and by lorry Q to deliver the greatest amount of sand

(2 marks)
19. 2014 Q20 P2
The dimensions of a rectangular floor of a proposed building are such that
o the length is greater than the width but at most twice the width;
o the sum of the width and the length is, more than 8 metres but less than
20metres.
If'x represents the width and y the length.
(a) write inequalities to represent the above information.

66
(b) (i) Represent the inequalities in part (a) above on the grid provided.

(ii) Using the integral values of x and y, find the maximum possible area
of the floor. (2 marks)

20. 2015 Q12 P1

A school decided to buy at least 32 bags of maize and beans. The number of
bags of maize were to be more than 20 and the number of bags of beans were to
be at least 6. A bag of maize costs Ksh 2500 and a bag of beans costs Ksh 3500.
The school had Ksh 100 000 to purchase the maize and beans. Write down all the
inequalities that satisfy the above information. (4 marks)

LATITUDES AND LONGITUDES

1. 1989 Q18 P1
A globe representing the earth has a radius 0.5m. Points A(100W), B (00, 350E)
P(600N, 1100E) and Q (600N, 1200W) are marked on the globe.

(a) Find the length of the arc AB, leaving your answer in terms of Л (3 marks)
(b) If O is the centre of the latitude 600 N, find the area of the minor sector OPQ
(5 marks)
2. 1990 Q18 P1
(a) Calculate the distance round the latitude 600 N. (Take the radius of the earth,
R = 6370 km and Л = 22/7) (4
marks)
(b) An aeroplane flew due south from a point A (600N, 450E), to a point B. The
distance covered by the aeroplane was 8000km. Determine the position of B.
(4 marks)
3. 1991 Q13 P2
The latitude and longitude of two stations A and B are (470N, 250 E) and
(470N, 700 E).
Calculate the distance in nautical miles between A and B along latitude 470 N.
(3 marks)

4. 1992 Q11 P2
A point Q is 2000 nm to the west of P (600N, 00). Find the longitude of Q to the
nearest degree.
(3 marks)
5. 1994 Q18 P1
A and B are two points on the latitude 400 N. The two points lie on the longitudes
200 W and 1000E respectively. Calculate
(i) The distance from A to B along a parallel of latitude (5
marks)
(ii) The shortest distance from A to B along a great circle
(Take Л = 22/ 7 and R = 6370 km) (4 marks)

6. 1995 Q24 P1
67
 An aeroplane flies from a point A(1o 15’S, 37oE) to a point B directly north of A.
The arc AB suspends an angle of 45o at the centre of the earth.From B, the
aeroplane flies due West to a point C on longitude 23o W.
(Take the value of л as 22/7 and radius of the earth as 6370)

a) (i)Find the latitude of B (1 mark)

(ii) Find the distance travelled by the aeroplane between Band C (5 marks)

b) The aeroplane left B at 1.00 am local time. When the aeroplane was leaving
B, what was the local time at C? (2 marks)

7. 1996 Q20 P1
The position of two A and B on the earth’s surface are ( 360 N, 490E) and
(3600N, 1310 W) respectively.
(a) Find the difference in longitude between town A and town B (2 marks)
(b) Given that the radius of the earth is 6370, calculate the distance between
town A and town B. (3 marks)

(d) Another town, C is 840 east of town B and on the same latitude as towns A
and B. Find the longitude of town C. (3 marks)

8. 1997 Q18 P1
A ship leaves an island ( 50N, 450E) and sails due east for 120 hours to another
island.
The average speed of the ship is 27 knots.
(a) Calculate the distance between the two islands
(i) in nautical miles (2 marks)
(ii) in kilometers (1 mark)
(b) Calculate the speed of the ship in kilometers per hour (3 marks)
(c) Find the position of the second island (3marks)
(take 1 nautical mile to be 1.853 Km and the radius of the earth to
be 6370 Km)

9. 1998 Q20 P1
The position of two towns X and Y are given to the nearest degree
as X (450 N, 100W) and Y ( 450 N, 700W)
Find
a) the difference in longitude (1 mark)
b) The distance between the two towns in
(i) Kilometers ( take the radius of the earth as 6371) (3 marks)
(ii) Nautical miles ( take I nautical mile to be 1.85 km) (2 marks)

c) The local time at X when the local time at Y is 2.00 pm. (2

marks)

10. 2000 Q22 P1

A plane leaves an airport A (38.50N, 37.050W) and flies dues North to a point B
on latitude 520N.
(a) Find the distance covered by the plane (4 marks)

(b) The plane then flies due east to a point C, 2400km from B. Determine the
68
 position of C
Take the value л of as 22/7 and radius of the earth as 6370 km (4
marks)

11. 2001 Q24 P1

A plane flying at 200 knots left an airport A (300S, 310E) and flew due North to
an airport B (300 N, 310E)
(a) Calculate the distance covered by the plane, in nautical miles (2 marks)

(b) After a 15 minutes stop over at B, the plane flew west to an airport C
(300N, 130E) at the same speed.

Calculate the total time to complete the journey from airport C, though airport B.
(6 marks)
12. 2003 Q24 P1
Two towns A and B lie on the same latitude in the northern hemisphere. When
its 8am at A, the time at B is 11.00am.

a) Given that the longitude of A is 150 E find the longitude of B. (2 marks)

b) A plane leaves A for B and takes 31/2 hours to arrive at B traveling along
a parallel of latitude at 850km/h. Find:
(i) The radius of the circle of latitude on which towns A and B lie. (3 marks)

(ii) The latitude of the two towns (take radius of the earth to be 6371km)
(3 marks)
13. 2006 Q16 P2
Two places P and Q are at ( 360N, 1250W) and 360N, 1250W) and 360 N, 1250W)
and 360 N, 550E) respectively. Calculate the distance in nautical miles between
P and Q measured along the great circle through the North pole. (3 marks)

14. 2007 Q13 P2

Two places A and B are on the same circle of latitude north of the equator.
The longitude of A is 1180W and the longitude of B is 1330E. The shorter distance
between A and B measured along the circle of latitude is 5422 nautical miles.
Find, to the nearest degree, the latitude on which A and B lie (3 marks)

15. 2008 Q7 P2
An aero plane flies at an average speed of 500 knots due East from a point
P(53.40E,400E) to another point Q. It takes 2 ¼ hours to reach point Q.
Calculate:
(i) The distance in nautical miles it traveled; (1 mark)
(ii) The longitude of point Q to 2 decimal places (2 marks)

16. 2009 Q13 P2

Point (40˚S, 45˚E) and point Q (40˚S, 60˚W) are on the surface of the Earth.
Calculate the shortest distance along circle of latitude between the two points
(3 marks)
17. 2010 Q19 P2
The position of three points A, B and C are (340N, 160W) , (340N, 240E) and

69
 (260S, 160W) respectively.

a) Find the distance in nautical miles between:

i) Port A and B to the nearest nautical miles; (3 marks)
ii) Ports A and C. (2 marks)
b) A ship left port A on Monday at 1330h and sailed to Port B at 40 knots.
Calculate:

i) The local time at port B when the ship left port A; (2 marks)
ii) The day and the time the ship arrived at port B (3 marks)

18. 2011 Q14 P2

A point M (600N, 180E) is on the surface of the earth. Another point N is situated
at a distance of 630 nautical miles east of M.
Find;
a) the longitude difference between M and N; (2 marks)
b) the position of N (1 mark)

19. 2012 Q22 P2

A tourist took 1h 20 minutes to travel by an aircraft from T(30S, 350E) to town
22
U(90N,350E). (Take the radius of the earth to be 6370km and л = ,
7
a) Find the average speed of the aircraft. (3 marks)

b) After staying at town U for 30 minutes, the tourist took a second aircraft
to town V (90N,50E). The average speed of the second aircraft was 90% that
of the first aircraft. Determine the time, to the nearest minute, the aircraft
took to travel from U to V.
(3 marks)
c) When the journey started at town T, the local time was 0700h.Find the
local time at V when the tourist arrived. (4 marks)

20. 2013 Q15 P2

The position of two towns are (20 S,300 E) and 20S, 37.4 0E) calculate , to the
nearest km, the shortest distance between the two towns.(take the radius of the
earth to be 6370 km)
(2
marks)

21. 2014 Q13 P2

The shortest distance between two points A (40°N, 20°W) and B (0°S, 20°W) on
the surface of y the earth is 8008km. Given that the radius of the earth is 6370km,
22
determine the position of B. (Take n = ). (3 marks)
7

22. 2015 Q14 P2

The positions of two points P and Q, on the surface of the earth are
P(45 °N, 36°E) and Q(45 °N, 71°E). Calculate the distance, in nautical miles,

70
 between P and Q, correct to 1 decimal place. (3 marks)

THREE DIMENSIONAL GEOMETRY

1. 1990 Q23 P2
The figure below shows a right pyramid VABCD whose rectangular base is 18cm
by 12 cm. The altitude is 24cm. The plane PQRS and ABCD are parallel and 6cm apart

Calculate

(a) The angle between the planes ABCD and VAB ( 4 marks)
(b) The area of the rectangle PQRS (4 marks)

2. 1991 Q23 P2
The figure below shows a shape of a roof with a horizontal rectangular base ABCD.
The ridge
EF is also horizontal. The measurements of the roof are AB =8m, BC = 5m,
EF = 4.5m and EA = ED = FC = 3.5m

71
 Calculate:

i) the height of the ridge EF above the base ABCD (4marks)

ii) the angle between the face and the base ABCD (4marks)

3. 1992 Q19 P2
A pyramid VABD has a square base ABCD of side 4m. The slant edges
VA, VB, VC and VD are 2 m long

Calculate

i) The height of the pyramid (3 marks)

ii) The angle between the plane VAB and the base ABCD (2
marks)

a) C′ and D′ are midpoints of VC and VD respectively.

Calculate the angle between the planes ABCD and ABC′D′ (3 marks)

4. 1993 Q22 P2

The figure below shows a triangular prism with dimensions as shown

72
 Calculate

a) The angle between the faces FBCE and ABCD (2marks)

b) the volume of the prism (3 marks)
c) the angle between the planes DFC and ABCD (3 marks)

5. 1994 Q22 P1
In the figure below ABCDEFGH is a frustum of a right pyramid. The altitude of the
frustum is 2cm.

Calculate
a) The altitude of the pyramid (2 marks)
b) The volume of the frustum (3 marks)
c) The angle between the base of the frustum and the face ABGF (3 marks)

6. 1996 Q 13 P2
The base of a right pyramid is a square ABCD of side 2a cm. The slant egdes
VA, VB, VC and VD are each of length 3a cm.
a) Sketch and label the pyramid (1 mark)
b) Find the angle between a slanting edge and the base (3 marks)

7. 1997 Q 6 P2
A pyramid of height 10cm stands on a square base ABCD of side 6 cm
(a) Draw a sketch of the pyramid (1mark)
(b) Calculate the perpendicular distance from the vertex to the side AB
(2marks)
8. 1998 Q 16 P2

73
 The triangular prism shown below has sides AB= DC = EF = 12 cm.
The ends are equilateral triangle of sides 10cm. The point N is the midpoint FC.

(a) Find the length of

(i) BN (1 mark)
(ii) EN (1 mark)
(b) Find the angle between the line EB and the plane CDEF (2 marks)

9. 1999 Q 14 P1
An equilateral triangle ABC lies in a horizontal plane, A vertical flag AH stand
at A. If AB = 2 AH find the angle between the places ABC and HBC (3 marks)

10. 1999 Q 24 P2
The diagram below shows a right pyramid VABCD with V as the vertex. The
base of the pyramid is rectangle ABCD, with ab = 4 cm and BC= 3 cm. The
height of the pyramid is 6cm

(a) Calculate the

(i) length of the projection of VA on the base (2 marks)
(ii) Angle between the face VAB and the base (2 marks)

(b) P is the mid- point of VC and Q is the mid – point of VD.

Find the angle between the planes VAB and the plane ABPQ (4 marks)

11. 2000 Q 11 P1
A pyramid VABCD has a rectangular horizontal base ABCD with AB= 12 cm
and BC = 9cm.

74
 The vertex V is vertically above A and VA = 6cm. calculate the volume of the pyramid.
(2 marks)
12. 2002 Q 18 P1
The figure below represents a right prism whose triangular faces are isosceles.
The base and height of each triangular face are 12cm and 8cm respectively.
The length of the prism is 20cm
A 20cm E

F
D
B
12cm C
Calculate the:

a) Angle CE (2 marks)
b) Angle between
i) The line CE and the plane BCDF (2 marks)
ii) The plane EBC and the base BCDF (2 marks)

13. 2002 Q 20 P2
The figure VPQR below represents a model of a top of a tower. The horizontal base
PQR is an equilateral triangle of side 9cm. The lengths of edges are
VP = VQ = VR = 20.5cm. Point M is the midpoint of PQ and VM = 20cm
Point N is on the base and vertically below V.

Calculate:

a) i) Length of RM (2 marks)
ii) Height of model (2 marks)
iii) Volume of the model (2
marks)

b) The model is made of material whose density is 2,700 kg/m3. Find the
mass of the model. (2 marks)

75
14. 2003 Q 15 P1
Three points O, A and B are on the same horizontal ground. Point A is 80 metres
to the north of O. Point B is located 70 metres on a bearing of 0600 from A.
A vertical mast stands at point B.

The angle of elevation of the top of the mast from o is 200.

Calculate: a) The distance of B from O. (2 marks)
b) The height of the mast in metres (2 marks)

15. 2003 Q 24 P2
The figure below represents a right pyramid with vertex V and a rectangular
base PQRS. VP = VQ = VR= VS = 18cm and QR =16cm and QR = 12cm. M and
O are the midpoints of QR and PR respectively.

Find:
a) The length of the projection of line VP on the plane PQES (2 marks)
b) The size of the angle between line VP and the plane PQRS. (2 marks)
c) The size of the angle between the planes VQR and PQRS. (2 marks)

16. 2004 Q 24 P2
The figure below shows a model of a roof with a rectangular base PQRS .
PQ = 32 cm and QR = 14 cm. The ridge XY = 12 cm and is centrally placed.
The faces PSX and QRY are equilateral triangles M is the midpoint of QR.

76
 Calculate
(a) (i) the length of YM (1 mark)
(ii) The height of Y above the base PQRS (2 marks)
(b) The angle between the planes RSXY and PQRS (3 marks)
(c) The acute angle between the lines XY and QS (2 marks)

17. 2005 Q 23 P2
The diagram below represents a cuboid ABCDEFGH in which FG= 4.5 cm,
GH = 8cm and HC = 6 cm

Calculate:
(a) The length of FC (2 marks)
(b) (i) the size of the angle between the lines FC and FH (2 marks)
(ii) The size of the angle between the lines AB and FH (2
marks)

(c) The size of the angle between the planes ABHE and the plane FGHE (2marks)

18. 2008 Q 14 P2
The figure below represents a triangular prism. The faces ABCD, ADEF and CBFE
are rectangles. AB=8cm, BC=14cm, BF=7cm and AF=7cm.

Calculate the angle between faces BCEF and ABCD. (3 marks)

77
 19. 2009 Q 22 P2
The figure below shows a right pyramid mounted onto a cuboid
AB = BC = cm, CG=8 and VG = cm

Calculate
(a) The length of AC; (1 mark)
(b) The angle between the line AG and the plane ABCD; (3 marks)
(c) The vertical height of point V from plane ABCD; (3 marks)
(d) The angle between the planes EFV and ABCD (3 marks)

20. 2011 Q 22 P2
The figure below represents a rectangular based pyramid VABCD.
AB=12cm and AD=16 cm.
Point O is vertically below V and VA=26cm.
V

26cm

B ………………………………………………… C

12cm O

A D
16 cm

Calculate
a) The height ,VO, of the pyramid; (4 marks)
b) The angle between the edge VA and the plane ABCD; (3 marks)
c) The angle between the planes VAB and ABCD. (3 marks)

21. 2012 Q16 P2

In the figure below, VABCD is a right pyramid on a rectangular base. Point
78
 O is vertically below the vertex V. AB=24cm, BC=10cm and CV=26cm.

Calculate the angle between the edge CV and the base ABCD. (3
marks)

22. 2013 Q20 P2

The figure ABCDEF below represents a roof of a house. AB=DC=12 m,
BC = AD = 6m, AE = BF = CF= DE = 5m and EF = 8m

E
F

D
C
A

B
(a) Calculate, correct to 2 decimal places, the perpendicular distance of EF
from the plane ABCD. (3 marks)

(b) calculate the angle between :

(I) the planes ADE and ABCD (2 marks)
(II) The line AE and the plane ABCD, correct to 1 decimal place; (2 marks)
(III) The planes ABFE and DEFE, correct to 1 decimal place. (3 marks)

23. 2014 Q20 P1

The figure below shows a right pyramid VABCDE. The base ABCDE is
regular pentagon. AO = 15cm and VO = 36 cm.

79
 Calculate:

(a) The area of the base correct to 2decimal places (3marks)

(b) The length AV (1mark)

(c) The surface area of the correct to 2decimal places (4marks)
(d) The volume of the pyramid correct to 4 significant figures (2marks)

24. 2014 Q10 P2

The figure below represents a cuboid PQRSTUVW. PQ = 60cm, QR = 11 cm and
RW - 10cm.

Calculate the angle between line PW and plane PQRS, correct to 2 decimal places.

(3marks)

25. 2015 Q20 P2

The figure below represents a cuboid EFGHJKLM in which EF = 40cm, FG=9cm
and GM=30 cm. N is the midpoint of LM.

80
Calculate correct to 4 significant figures

a)The length of GL: (1 mark)

b)The length of FJ (2 marks)
c)The angle between EM and the plane EFGH; (3 marks)
d)The angle between eh planes EFGH and ENH; (2 marks)
e)the angle between the lines EH and GL (2 marks)

81
 MATRICES AND TRANSFORMATIONS-MARKING SCHEME
NO SOLUTION mark 2d = 0
s d =0
1.
(−13 20) (52) = (−5+ 0)
15+ 4
=
3M

(−5
19
)
(−5
19
)+¿ (−611) = (136 )
Image = (13, 6)
1989Q12
2.
( 46 )(
−2 −1 3 2
−3 −4 3 5
= ) 3M

B D B' D'
( 46 6 −2
9 −3 ) (ii) (
0 −1 3 2
1 0 0 2
= )( )
Image points are : A' (4, 6),
B' (6, 9), C' (-2, -3)
(0 −2
3 2 )
A'(0, 1), B(0, 3) C (-2, 4)
D' (-2, 2)

(iii) T = ( ac db)
( ac db) (01 −2
4
= ) ( )
1 4
0 2

0 + b =1
-2a + 4b =4

0 + d =0
-2c +4d =2
1990Q16 b=1
3. (i) 8M -2a+4 = 4
( ac db) (10 42) = (01 −2
4 ) -2a=0
d=0
a+0=0 -2c=2
4a + 2b = -2 c=1

c+0=1
4c + 2d = 4
T= (−10 10) 1990Q21
a=0
2b = -2
4.
(10 −12 ) (ba) = (−24) 3M

b =-1
a-b =-2
c=1 0+2b=4
2b=4 b= -2
4 + 2d = 4 M= (01 −1
0 ) a-2 = -2 a= -2
82
 a=0 12cm2
a(0, 2) 1991Q4
1993Q12
5.
O' = (00) + ( 04) = ( 04)
A' = ( ) + ( ) = ( )
2 0 2
8. (a) 8M
0 4 4
8M
B' =
2
3 () () ()
+
0
4
=
2
7

C' = (03) + ( 04) = (07)

(b) (i)T1= reflection in the line

x
+y=2
(ii)t2 = Half turn about point
(1, 1)
(iii) T is a reflection in the
line y = x. Thus T = (01 10)
O' A' B' C' O'' A'' B'' C''

(01 −10 ) ( 04 24 27 07) =

1993Q19
9. 4M

(−40 −42 −72 −70)

O' (0, 4), A'(2, 4), B'(2, 7),
C'(0, 7); O'' (-4, 0), A"(-4, 2),
B"(-7, 2), C"(-7, 0)

Centre of rotation = (-2, -2)

1991Q21

6.
(a)( 34 4
−3 ) ( xy ) = ( 34 xx−3 y)
+4 y 8M

x = 3x + 4y
y = 4x – 3y
Aʹ (2,0) Bʹ (6,0) Cʹ (6,4)Dʹ(2,4)
Aʹʹ(-4,0). Bʹʹ (-12,0)
(b) m = 0.5 or -2
Cʹʹ (-12,-8), Dʹʹ (-4,-8)
1992Q18
Aʹʹʹ(-4,0)Bʹʹʹ (-12,0)
Cʹʹʹ (-12,8) Dʹʹʹ (-4,8)
7. A.S.F = (2 -6) 8M
= -4
3x4 = 12
83
 ( ac db)(−40 −12
8
= ) ( )
2 6
0 4

-4a = 2
-12a + 8b = 6

-4c =0
-12c + 8d = 4

-4a = 2

-12 ( −12 )+8b = 6

8b = 6 -6
8b = 0
b=0
-4c =0
c= 0
0 + 8d = 4
d = 0.5

( )
−1
0

( )
2 −1 −2
1994Q23
1 3 3
0 (c)
2 1 −1
3 3
10.
( 1 0
0 −1 ) 2M
1995Q23
1994Q7 12. a). ± 1800 rotation centre
origin(0,0) B1
11. (a)A'(0,2), B'(0,6), C'(-5, 6), 8M matrix M = B1
D'(-2, 2)
(b)(i)

(−21 −1 (−10 −10 )( ac bd)( 20 4 4

)
−1) ( 2 )
0 0 −5 −2 1 3
= B1
6 6 2

(−2 −6 4
−2 −6 −11 −4
2
) ¿ ( 20
4
−1 −3
4
)
A"(-2,-2) B"(-6,-6) C(4, -11)
D"(2, -4)
( )( )(
2 a+ 0 4 a+b 4 a+ 8 b
2c +0 4 c+ d 4 c +d ) B1

= (
−2 4
0 −1 −3
4
)
2a = -2 4c + d = -1
a = -1 d = -1
4a + h = 4 m (10 0
−1 )
b=4–4 b=0
2c – 0 = 0

84
 M1
A1
2m

A(4,4) B(4,2) C (6,6) D (2,4)

(b)

( )( ) (
2 1 2 4
1 1 0 1
=
4 4 9 11
3 25 7 ) C(i)[ 10 −21 ] [ 44 22 2
26 4
= ]
( )(2 ×2+0 2× 4+ 1
2 ×1+0 1× 4+ 1 ) [−44 02−106 −64 ]
( )(2 × 4+3 4 9 11
4+3 2 5 7 )
Aʹʹ (4.2) Bʹʹ(9.5) Cʹʹ(11.7)
c). Area of ∆ ABC = ½ x 2 x 2
d) ) [ 10 −21 ] [ 01 −10 ] =

=2cm2
determinant of 2 1 = 2 -1 = 1
[−21 −10 ]
(21 11) 1998Q19
14. a = 15, 17c = 8, 17b = -8, M1
Area of ∆ A" B" C" 17d = 15
= 1 x 2 = 2cm2 A = 15/17 8/17 A1
8
1997Q23 /17 15/17
b) cos = 15/17 = 0.8824 M1
(or sin = 8/17 or tan = 8/15
13.
(b) [ 01 −10 ] B1
= 280 4 (28.070)

[ 4 2 6 4
−4 −4 −6 12
= ] M1
c). S(O) = ( 36 ) M1

A1

( ) ( )
A1 15 −8 83
[ 4 22 2
4 26 4 ] 17
8
17 3
15 6
=
13
114
17 17 12

Image of (1, 0) – (3/17 114/12)

1999Q23

15. (a)(i) Aʹ (2,1) Bʹ (2,-7) B1

Cʹ (4,-5) Dʹ(4,-3) B1
(ii) Aʹʹ(1,-2). Bʹʹ (7,-2) B1
Cʹʹ (5,-4), Dʹʹ (3,-4) B1

(b) Aʹʹʹ(-1,-2)Bʹʹʹ (-7,-2) B1

85
 Cʹʹʹ (-5,-4) Dʹʹʹ (3,-4)
Half term centre (0,0) B1
2000Q23 B1
B1 B1

b). Single matrix in the

inverse of

16 a)P Q R S
(121 −10 )(−10 10)( 10 −21 )=(−10 12)
(10 −2
¿ ) Which (
1 0)
1 2 −1
Pʹ
Q ʹ
R Sʹ
ʹ

(−11 4 −2 −7
1 4 4 ) 17.
2001Q18
M1
Pʹ(-1, 1) Qʹ(4 ,1) Rʹ(-2, 4)
Sʹ(-7,4)
( ac db) (23 53)
a). i). M1
M1
=(
3 3)
4 −1
A1
ii) object drawn
image P1 Q1 R1 S1 drawn
M1
2a+3b=-4 2c+3d=3
5a+3b=-1 5c+3d=3 A1
iii ¿ ( )(
1 1 1 4 −2 7
) A=1. b=-2 c=0, d=1 M1
0 1 1 1 4 4
Therefore m (10 −21 ) A1

¿ ( 11 1 4
−4 2
4
7 ) ii)(
0 1 )( 1 )( y ) ( 1)
1 −2 4 x
=
2 8
mark
s

(−10 10) Negative quarter C1 = (2,1)

turn about the origin. b).(01 10) (10 1

=)(
−2 0 1
1 −2 )
( )( )
18. P Q R B1-
2 4
5 6 4
0 2 B1
−1 −1 −0.5

( )
A B C B1
= 6 8 6
−2 −2 −1 B1

86
 c). Shear x-axis invariant and
B(2,4) Bʹ(10,4)or
C(4,4) Cʹ(12,4)

A(6,-2) ,B (8,-2) C (6,-1)

Aʹ(1,2). Bʹ(5,2)Cʹ (1,4)
c). Centre (-3,2) angle of
rotation 2700 or -900 2005Q18
2004Q21
19. 20. i).

( )( )
a b 2 2 4 B1
c d 0 4 4

¿ ( 22 ac 22ac +4+4 db)(44 a+4

c +4 d )
d M1

(−1 0 )(2 c 2 c+ 4 d 4 c + 4 d )
0−1 2 a 2 a+ 4 b 4 a+ 4 d M1

=¿

(−2−10−12
0−4−4
)
(−2 a−2 a−4 d )(−4 a−4 d )
−2 c−2 c−4 d −4 c −4 d
B1
B1
(−2c
−2 a=2=a=1 )( d =1 )
=0=c=0 −4 d=−4 M1 Shear maps (1,0) -1(1,11/2)
ii). Shear maps 1(1,0) 1(1,1 ½
B1
M1
M1

hence R
ab
¿ ( )( )
12 (1.51 01)
matrix = A1
M1
cd 01

1.5 −1 )( )
A 1 B1 C1
b). (
1 0
A1 −6 −4 −2 A1
b( )( )
12 224
01 0 44
−4 −1 −2 B1

( )
A1 B1 C 1 B1
6 4 −3 B1
(0+2+00 2+8
0+4 0+4 )
4 +8
−5 −1 −2
ii). Half turn, about (0,0)
10
mark
s
= A' B' C' 2006Q19

(20104 124 ) 21.

87
 Let T = ( ac db) M1
Rotation about (0,0)
Through angle 530 or -3070
2009Q20
( ac db)(22 43) (82154 )=¿ M1

2a + 2b = 2 2c + 2d = 8
A1
24.
(a) = (10 k1) (32) = 3+22 k
4a + 3b = 4 4c + 3d = 15 x – coordinates = 3+ 2k
4a + 4b = 4 4c + 4d = 16
4a + 3b = 4 4c + 3d = 15 (b) Δ at A
A1 3 + 2K = 4 K=½
b=o d=1 4
a=1 c= 3 Δ at O
mark 3 + 2K = 0 K = -1.5
T= (13 01) s 2010Q10

2008Q10

( )(
25. 1 B1
22. M1 1 M1
Image area = [ ( 4 ×2 )− (5 × 1 ) ]
= 63cm2
A1
2
bi)
1
2 2 88 2
1 2 66 6
= ) A1
2009Q9 mark 2
s
23. M1 ( 3 6 10 7
3 9 11 5 ) B1

P Q R M1

( )(
−0.6 0.8 5 1010
0.8 0.6 5 1015 ) A1 Aʹʹ(3,3) , Bʹʹ(6,9) , Cʹʹ(10,11),
Dʹʹ (7,5)
A1

Rectangle AʹʹBʹʹCʹʹ Dʹʹ drawn

Pʹ Qʹ Rʹ

(71214 617) B1
M1
Pʹ(1,7) Qʹ(-2,14) Rʹ(-6,17) B1
A1 B1
c). i). PʹʹQʹʹRʹdrawn
B1

( )(
a b 5 10 10 −1
c d 5 10 15 7 ) B1
M1
A1
10 10
mark

( ac db)(0.6−0.8
0.8 0.6 )
s

3
(c) Det of p=
4
Area AʹʹBʹʹCʹʹ Dʹʹ = ¾ x 6 x4
18sq units

26.

88
 10
= O'' A'' B'' C''

(00 −4 −6 −2
0 8 8 )
M1 c) (10
0 −2 0
−2 0 −2
= )( )
(
−2 0
0 4 )
A1
Inverse =
1 4 0
−8 0 −2 ( )

( )
−1
0
2
O A B C M1 =
1
0
(a) (
−2 0 0 2 3 1
0 −2 0 0 2 2 )( ) A1
4
2012 Q18 P2
O' A' B' C'
27.
( 0.5 0
0 0.5 )
= (
0 −4 −6 −2
0 0 −4 −4 ) M1 28.
2013 Q16 P2

30 B1
∴ O' ( 0 , 0 ) Area scale factor = =6 M1
5
A' (−4 , 0 ) M1 A1
4x – 2x + 2 = 6
B' (−6 ,−4 ) 2x = 4 3
C' (−2 ,−4 ) x=2
2014 Q7 P2
(b) (i) A1

(10 )(
0 0 −4 −6 −2
−2 0 0 −4 −4 )

89
 STATISTICS II MARKING SCHEME
1.
Mean =
∑ fx = 41 = 1.025 5M

∑ f 40
x f fx fx2
0 20 0 0
1 8 8 8
2 6 12 24
3 4 12 36
4 1 4 16
5 1 5 25
Σ 40 41 109

√ ( )
2

S.D= ∑ fx2 − ∑ fx
∑f ∑f

√
= 109 − (1.025 )2
40

= 1.293976

= 1.294(4s.f) 1989Q12

2.
a) =
∑ fx 8M

∑f
= 11612

106

= 109.5471

= 109.5hours(4s.f)

hours x(mid pt) f fx cf u.c.i

90-94 92 5 460 5 94.5
95-99 97 14 1358 19 99.5
100-104 102 16 1632 35 104.5
105-109 107 17 1819 52 109.5
110-114 112 24 2688 76 114.5
115-119 117 12 1404 88 119.5
120-124 122 11 1342 99 124.5
125-129 127 4 508 103 129.5
130-134 132 2 264 105 134.5
135-139 137 1 137 106 139.5

Σ 106 11612

(b)(i)cumulative frequency curve

90
 (ii) median at 106 = 53rd position, median = 109.7 ± 0.1
2
1989Q20
3.
(i) mean =
∑ fx = 4340 = 43.4 8M

∑f 100

time x(mid pt) f fx fx2

15-19 17 6 102 1734
20-24 22 10 220 4840
25-29 27 9 243 6561
30-34 32 5 160 5120
35-39 37 7 259 9583
40-44 42 11 462 19404
45-49 47 15 705 33135
50-54 52 13 676 35152
55-59 57 8 456 25992
60-64 62 7 434 26908
65-69 67 5 335 22445
70-74 72 4 288 20736
Σ 100 4340 211610

√ ( )
2

(ii) S.D= ∑ fx2 − ∑ fx

∑f ∑f

√
= 211610 −( 43.4 )2
100

= 15.24926.....
= 15.25(4s.f) 1990Q23

4.
Mean = a +
∑ fy 8M

∑f
100
= 25.5 +
80

91
 = 26.75

Marks x(midpoint) f y = x-a fy Fy2

1-10 5.5 5 -20 -100 2000
11-20 15.5 13 -10 -130 1300
21-30 25.5 32 0 0 0
31-40 35.5 27 10 270 2700
41-50 45.5 3 20 60 1200

Σ 80 100 7200
∑ fy
( )
∑ fy
2

Variance = =
∑f ∑f
7200 2
= −( 1.25 )
80
= 88.4375 = 88.44(4s.f)
Standard deviation = √ variance
= √ 88.44
= 9.4042543
= 9.404(4s.f)
1991Q22
5. (a)

lengths u.c.l f c.f

10-12 12.5 3 3
13-15 15.5 12 15
16-18 18.5 40 55
19-21 21.5 37 92
22-24 24.5 8 100

Σ 100

(b)(i)median = 50.5 -15 x3

40
35.5 x 3
40
= 2.67
15.5 +2.67

92
 = 18.17± 0.1 1992Q23

6.
(i)Mean =
∑ fy 8M

∑f
−170
= 57 + 100

= 55.3marks
marks x(mid pt) f y=x-a fx fx2

30-34 32 1 -25 -25 625

35-39 37 5 -20 -100 2000
40-44 42 10 -15 -150 2250
45-49 47 10 -10 -100 1000
50-54 52 19 -5 -95 475
55-59 57 20 0 0 0
60-64 62 20 5 100 500
65-69 67 8 10 80 800
70-74 72 4 15 60 900
75-79 77 3 20 60 1200
Σ 100 4340 211610

∑ fy
( )
∑ fy
2

(ii) S.D = =
∑f ∑f
√
= 9750 −(−1.7 )2
100
= 9.726767192
= 9.727(4s.f) 1993Q20

7. Inter quartile range = 3 1994Q10

8.
(a)mean =
∑ fx 8M

∑f
= 2136
47

= 45.44680851

= 45.45(4s.f)

Marks Midpoint(x) f fx

93
 31-35 33 4 132
36-40 38 6 228
41-45 43 12 516
46-50 48 15 720
51-55 53 8 424
56-60 58 2 116

Σ 47 2136

(b) (i)

median = Q2 = 46.5 ± 0.2

(ii)Q3 = 49.75± 0.1, Q1 = 41.5 ± 0.1
Semi-interquatile range = Q3 –Q2
2
= 4.125 ± 0.2
1994Q21
9 (a) 8M

Wind speed u.c.i f c.f

0-19 19.5 9 9
20-39 39.5 19 28
40-59 59.5 22 50
60-79 79.5 18 68
80-99 99.5 13 81
100-119 119.5 11 92
120-139 139.5 5 97
140-159 159.5 2 99
160-179 179.5 1 100

Σ 100

94
 (b) (i)Q3 = 90 ± 1, Q1 = 36 ± 1
(ii) 8days ± 1
Interquartile range = Q3 – Q1
= 90-36
= 54 ± 2 1995Q18
10. 25, 289, 4,484, 4806 B1
O=
√
806
5
= √ 161.2
M1

= 12.7 1996Q10 A1
11. Md f fx Fx 2

9 4 36 324
12 7 84 1008 M1
15 11 165 2475
18 15 270 4860 M1
21 8 168 3528
24 5 120 2880 M1

∑ fx=843 843 15075

M1
fx : 36,84,165,270,168,120
A1
843
(a) Mean = = 16.862
50
(b) (i) fx2 : 324, 1008, 2475, 4860, 3528,2880
M1
15075 2
Variance = −16.86
50 A1
= 301.5 – 284.2 =17.3(17.24)
(ii) S.D = √ 17.3 = 4.159 4marks
OR 4.152 1996Q19
12. (a)

Class frequency u.c.l c.f

95
 14.5-18.5 2 18.5 2
18.5-22.5 3 22.5 5
22.5-26.5 10 26.5 15
26.5-30.5 14 30.5 29
34-5-38.5 13 34.5 42
38.5-42-5 6 38.5 48
2 50
Σ 50

b) (i) median = 29.5

(ii) reading at mass 25.28
= 11 and 20
20 11
Probability = - = 0.8 1997Q22
5 0

96
13. a) i) 22.5 +1196+144=169 + P2 + 256 +121+169+144 + 1289 M1
= 1794
P2 + 1713 = 1794 A1
P2 = 81
P=9 M1
ii). Standard deviation A1
Mean = (915+14+12=13+9+16+11+13+12+17)+10
= 13.2

√ √
51.6 or 1794 2 M1
s.d = −( 13.2 )
10 10
A1
=2.272 OR 2.28

b (i) New mean = 16.2

ii) new S.d = 2.272 2001Q18 B1
B1
8
marks

14.

97
15.
B1

M1

M1

A1

M1

M1

M1

A1

2003Q18 8 marks
16.

M1

2002 Q19
A1

M1

M1

B1

M1

A1

2004Q18

17.

98
99
22.

23.

100
 GEOMETRIC CONSTRUCTION AND LOCI
MARKING SCHEME

1. (a) draw a perpendicular through C, 8M intersection of the semi-circle and

join B and bisect. the locus of P.
O is the intersection of the two
perpendiculars.Draw the circle 1990Q8
using the radius CO.

(b) BD = 7.4cm ± 0.1 4. (a) construct angle BAC

complete ΔABC 8M
(c) Construct the perpendicular (b)drop a perpendicular from C to
bisector of chord BD and locate R AB bisect angle ABC
on the circumference of circle. point P is the bisector of the two
above.
(d)construct a line QR which is (c) B is the furthest point.
parallel to CO, then locate Q PB=225m ± 5

1990Q17
1989Q20 5. Bisect line AB to get locus of P
Draw a circle centre A,radius 5cm
2. Measure PQ and PS, 4M to show the locus of Q.
Area PQRS = area of ΔPQR + area Points of intersection of two loci
of Δ PSR are
Drop perpendiculars from R to PS, (3,-3) and (4,4)
and from R to PQ, measure the 1991 Q8
heights of the two triangles. P1
Equate area of ΔPST = area of 6. Construct 900 and bisect to get 8M
PQRS, to help you get the height of angle ABC
ΔPST, h= 10.4cm. Complete the Δ using the given
Using the above lines draw the dimensions
locus of T Bisect any two sides of Δ to locate
X
AX = 5cm ± 0.1, AB = 9.5cm ±
1989Q5 0.1,
3. Construct a semi circle with mid- 4M <AXC = 900± 0.10
point of AB as the centre. AB is the
diameter of the circle. 1991Q22
Compute h = 3cm, and draw the
locus of P. 7. Draw a circle centre Q (2, 3), 3M
Points P1 and P2 are the points of radius 5cm.

101
 Read off intersection of the 12 (a) Draw a semi-circle with AB as 8M
circumference of circle with x axis. . diameter.
Points are (-2, 0) and (6,0). Height of Δ ABC = 4cm. Draw a
parallel line to AB, 4cm above AB.
1992Q8 C1 and C2 are the points of
intersection of the line and the
8. Bisect <ABC and <BCD 3M semicircle.
the bisectors intersect at the centre
of the required circle, draw the (b)(i) Construct base <OAB = <
circle.. OBA = 450.
radius of circle is 4cm ± 0.1 Locus of p is a major arc of a
circle, radius OA or OB with AB as
1992Q13 an internal chord.

9. (a) (i) draw the locus of Q which is 8M (ii) Position of P with greatest area
the circumference of a circle centre is that which gives the greatest
A, radius AC. distance from AB. Area ABP =
(ii) draw the locus of P which is a 60cm2.
major
arc of a circle with AB as an 1994Q19
internal
chord < OAB = < OBA = 450 13 Construct 900, 600, bisect 300,
are the base angles. . combine 900 + 150 = 1050

(b) M and N are the points of Locus of p is a continuous arc of a

intersection of the two loci circle radius AP = 4cm drawn
above, within the parallelogram, shaded on
MN = 2cm ± 0.1 the outside.
(Note : in this case a protractor
may be Locus of Q is a continuous arc of
used to measure the angle) the circle radius AQ = 6cm drawn
within the parallelogram, shaded on
1992Q21 the outside
Area = area of //gram – area of the
10 This is constant angle loci two sectors = 24.036cm2
. Draw two major arcs, one with
centre O and radius OB or OA.
For the other arc, identify O1 by 1995Q22
drawing ΔAO1B or O1BA =
O1AB=150, then draw the arc with 14 Construct angle loci – base angles
centre O1 and radius O1B or O1A. . = 600. Point B is centre of one
Note that the locus of P is above circle, identify, the other centre by
and below AB constructing < OAC = <OCA = 600
on the opposite side of AC as B.
1993Q11 two major arcs on either side of AC
11 Construct 900, then 600, then 750 8M from the required loci
. then 371/2 1996Q5
(other possibilities are : 600, 300, 15 (a) bisect angle BAD
1 . (b) constant angle loci – base
150, 7 0r 600,1500,750, 371/2 angles: <OBA = <OAB
2
AD = 3.6cm ± 0.1 = 67.50
Area of ΔABC = 12.6cm2 ± 0.35 The locus of p is a major arc of a
circle, radius OA or OB with AB as
1993Q19 an internal chord.
(c) locate c, the point of

102
 intersection between the locus of P 19 (a) Bisect AB
and the bisector of < BAD. . (b) For angles APB to be equal to
<ABC = 1300 ± 10 ACB,it means that P and C are on
1996Q23 the circumference of a circle.
Thus bisect line AC or CB to get
16 Construct 600 so as to form 1200 on B1 centre of the circle.Locate P where
. the other side of it. B1 the arc ACB meets with line XY.
complete Δ ABC B1
AB = 4cm, BC=5cm B1 1998Q8
Length of AC=7.7± 0.1cm B1
Bisect any two sides to get the B1 20 Construction marks for 37 ½ B1
centre (O) of the circle,radius OA B1 . < ABC = 37 ½ 0 + 10 B1
or OB or OC. B1 Subdivision of AB B1
Radius 4.5 ±0.1cm B1 Subdivision f BC(ruler and set B1
Shortest distance 3.8± 0.1 cm square) A 4
ma
rks

B C

1999Q11

21 Bisector of AB drawn
. Interpretation of the scale path B1
1 – 0.1 cm wide all round
There are 3.5 0.1cm from B d B1
1997Q19 Five points 1cm apart on bisector
17 √const of bisector of BC B1 of AB B1
. √const. of bisector of AC or AB to B1 A
get the centre of the circle B1 B2
Locus of P drawn 3
ma 8cm
6.5cm
1997Q4 rks
18 (a) Construct 600 then bisect to get
. 300
(b) CD = 5.4cm ± 0.1 B
(c ) Find h for which the triangle
AʹBC
1999Q21
3
is of the area of triangle 22 B1
4 . Angles 450 constructed B1
ABC. Completion of ABC with <BAC
h=4.5cm obtuse B1
3 x 6 = 4.5 Two 1 bisector √ constructed or B1
4 drawn (may be implied centre)
Construct a line 4.5cm above and Identified and circle drawn bisector B1
parallel to BC to give the locus of of <OBC drawn B1
Aʹ
(d) The locus of Aʹ is a line 4.5cm 2000Q222
above and parallel to BC such that
the area of triangle AʹBC is 23 D
constant.

1998Q23
5.8 cm

103
 B1
B1 28 C
.
B1 B
A B
B1
4m
2001Q8
3m
24 B1 A
D E
B1
29 y>x
. y<-x+4
B1
7<3x+3

2004Q15
Q R 3m 30
2001Q14 . M1

25 a) making 3 equal length from B B1 M1

. along BA and joining the last point B1
Z to C Construction of angle at x M1
equal to angle Z and M1
identify D A1 M2
Area of DABD = ½ x 2.7 x 8 4
= 7.83cm2 M A1

2002Q10 a). P lies on any

point along cp
26 a). Construction of 300 B1
AQB<600<900
. completion of DPQR B1
B1
b). Q lies on the unshaded
b). bisector of PR (must be seen) B1
drawing region
location of S1 QS=8cm and drawing B1
DPRS B1
B1 2005Q20
c). Construction of semicircle with B1 31 Construction
diameter SQ construction of 8 .
parallel line to QS through R ma
location of T1 and T2. rks 2006Q8
T1 T2 =4.7+0.1cm 32
. B1
2002Q21
B1
27 M

33
. Construction
Perpendicular to PQ drawn at L B1
R identified Arc centre A radius 6cm drawn B1
Locating centre 01 bisector of BC drawn & dotted B1
parallel 4cm form BC drawn region B1
2003Q22 shaded

104
 2006Q13 37
. B1
34 (i) construct <1080, sides 4cm
. (ii) bisect two angles to produce B1
centre O A1
(iii) draw a circle touching the M1 B1
vertices A1 C

3 B1
m
3m
A

2007Q12 2008Q3
38
35 B1 . D
. construction (i) C

I bisector of PQ constructed and B1

point R marked B1
T 1050
1 dropped from Q to AB or <PRB B1 A B
B1
Transferred to < BRS RS 1 from P
to AB constructed B1
2009Q11
PT=2 length of 1 and polygon B1
39
completed R from TS=4.6+0.1
. B1
B1
Bisect of < QPT drawn dotted B1
B1
Arc centre R with radium 4.5cm B1
drawn semi circle with PT as
B1
diameter drawn dotted correct
region shaded. B N S C
B1
2007Q21
Check 600 at 0
120 at 0
36 CONSTRUCTION B1
. B1
2009Q4P2
< 1200 completion
Drop (From A to CB produced B1
40 <TQR = 600 ; QS = 10cm and 3M
bisect height to determine E) B1
. bisects < TQR mediator ( ┴ or
Determination of point D and 4
bisector) of QS drawn or < RSQ =
completion of parallelogram ma
QST =300
rks
Rhombus completed
2008Q8

2010Q10
41 Locus of P drawn
. ┴ bisector of AB constructed.
Positions of C indicated

2010Q13

105
42 CONSTRUCTION B1 43 CONSTRUCTION B1
. B1 .
Construction of 750 R= 2.5 ± 0.1cm B1
construction of 2 adjacent sides B1 B1
Completion of //gm B1 3m
HEIGHT = 3.9 ± 0.1cm 4
2011Q9

106
44 Construct AB= 4cm,construct at B
<ABC=1350(900+450),BC=6cm,
construct at <DAB=600 and
AD=3cm. < BCD =310

45. B1

B1

B1

B1

(b)
(i)
(ii)
B2

B1

M1

M1

A1

2012 Q21 P2 10

107
46

47

48

49.
(a) (i) B1

B1

B1

B1

B1

B1
(ii) Radius = 3.5 ± 0.1
B1
(iii) Height construction
Height = 3.5 ± 0.1 B1

(b) Area of circle outside triangle

M1
21
= π ×3.5 − × 3.4 ×5
2 A1

= 30 2014 Q22 P1

108
50. a)

B1
B1
B1
B1
(i) B1
(ii)
b) (i) 9.2 x 10 = 92 m B1

(ii) Area of region bounded y locus of P

Locus of Q and line BQ1 B1
Angle = 600 radius = 46 m ∓ 0.1
B1
2 60
= π ×46 × M1
360
= 1107.94
~ 1108 m2 A1
−¿ ¿ 10
2014Q23 P2

109
 TRIGONOMETRY III
MARKING SCHEME

1. Sin 2θ = -0.5 4 θ = 30 x 2/5 = 12

θ = -300 M 5
/2 θ = 150
2 θ = 150 x2/5 = 600
360 – 15 = 3450 θ = 120 OR 600
270 + 15 =2850 1996Q12
180 -15 = 1650
90 + 15 = 1050 6. 5Sin= θ = - 4 B
θ = 1050, 1650,2850, 3450 Sin θ = -0.8 1
1989Q7 Sin θ = sin 306.87 = sin 233.13 B
2. 4sin(x +20)0 = 3 3 θ = 306.870 or 233.130 1
Sin (x + 20) 0 = sin 48.60 M 1997Q11 B
x + 20 = 48.6 1
x = 48.6 – 20 2
x =28.6 m
7. 1.731
sin(x+20)0 = sin(131.4)0 Cos (30θ + 1200) – = 0.8660
2
x + 20 = 131.4 3θ + 1200 = 3900
x = 131.4 -20 3θ + 120 = 330
x= 111.40 3θ=270 3θ = 210
θ = 90 θ = 700
x = 28.590 OR 111.40 1998Q14
1991Q2 8. 82 + 2s – 3 = (4s + 3) (2s – 1) = 0 M
S = ¾ or s = ½ 2
3. 360 ÷ ½ = 360 x 2 = 720 3 S = ½ = θ = 300 or 1500
Amplitude = 6, period = 7200 M
1992Q4 For all θ +ve no -1 A
2
4. Sin(2θ – 10)0 = sin 3300 = sin 210 4 4
= sin 690 = sin 570 M m
2θ – 10 = 330 9 Sin2 (x – 300) = ½ x ½ = ¼ M
2θ = 340 Sin (x – 30)0 = + 0.5 1
θ=1700 X – 300 = 300, 1500, -30 , -1500 - M
2100 1
2θ – 10 = 210 X=600, 1800, 00, -1200, -1800 A
2θ = 220 2000Q8 1
θ=1100 3
m
2θ – 10 = 690
2θ = 700 1 Cos 2x = sin (900 -2x) M
θ=3500 0. Sin(x+300)=Sin (900 2x) 1
x+300=900-2x
2θ – 10 = 570 x =600 A
2θ = 570 + 10 cos 3x=cos2 600 1
2θ=580 =(½) 2
θ = 2900 =1/4 or 0.25 B
2001Q12 1
θ = 1100, 1700 2900 and 3500 3
1994Q4 m
5. Sin 5/2 θ = sin30 = sin150 2 1 4(1-Cos2θ)+ 4 Cos θ= 5 M
5
/2θ = 30 M

110
1. 4 Cos 2θ – 4Cos θ + 1 =0 1 1
(2Cos θ -1) (2Cosθ-1)=0 M 4
Cosθ= ½ 1 m
θ = 600, 3000 1 2 Cos 2θ = 1 B
2001Q15 A 6. Cos2θ = ½ 1
1 2θ = 600, 3000, 4200, 8600 B
3 θ = 300, 1500,2100, 3300 1
m 2008Q16 B
1 3 tan 2 x – 4 tan x – 4 = 0 1
2 Let tan x = y B
3y2 – 4y – 4 = 0 1
(3y + 2) ( y – 2) = 0 4
Y = -2 or y = 2 m
3 1 2(1-sin2x) –sinx =1 M
7. 2sin2x + sin x – 1= 0 1
Tan x = -2/3 x = 146.300, 146.310, B (2sinx -1) (sinx +1) = 0 M
146.320 1 Sin x = ½ or sin x = -1 1
Tan x = 2 x = 63.430 B x = 1/6 π c, 5/6 π c, 3/2 π c A
2003Q7 1 2008Q16 1
4 B
m 1
1 8(1-cos2x)+2=cos x-5 =0 M 4
3. 8 cos x-2cos x-3 =0 1 m
92cosx+1)(4cosx-3)=0 1 x=100 or 300
Cos x=3/4 M 8
1 1 Sin (3x+30) = Sin 600 B
4 A
√7 9. Sin (3x+30) =Sin 1200 1
1 3x+30 = 600 B
3x+30 =1200 1
3 x=100, x=300 B
Tan x = 7/3 B 2009Q14 1
(tan 41.41)=0.8519 1 B
(tan 41.4)=0.9316 1
2004Q9 4 4
m m
1 Cos 2x0 = 0.870 B 2 4 -4 Cos2α = 4 sin x -1 M
4. 2x0 = 36.2, 143.8, 216;2 1 0. 4 -4 (1-sin2α) = 4 sin x -1 1
323;8, 396;2, 503.8, 576.2, 683.8 M 4 sin2α -4sinα+1 = 0 M
Hence x0 = 18;1, 71;9, 108;1, 1 (2sinα-10 (2sin α-1) =0 1
161;9, 198;1, 251;9, 288;1, 341;9 M Sinα= ½ α = (300, 1500) B
2005Q9 1 1
A
A 1
1 2 x + 20 = 2300 or a + 20 = 3100
1 3 Cos x = 2 (1-cos2x) 1 x = 2100
5. 3 cos x = 2 -2 cos2x M Or
2Cos2x+3Cosx-2 = 0 1 x = 2900
2y2+3yz-2=0 x = 2100,2900
(2y-1) (y+2)=0 M
y= ½ or y=-2 1 2013 Q8 P1
Cos x = 0.5 x=600, 3000 2 6(1 – sin2 x) + 7 sin x - 8 = 0
2007Q3 A 2
1 6 - 6 sin2 x + 7 sin x - 8 = 0
B

111
 6 sin2 x + 7 sin x + 2 = 0 2 3x – 45 = 0
(3 sin x - 2) (2 sin x - 1) = 0 3 x = 15
2 1 3x – 45 = 360
sin x = or Sin x x = 135
3 2
135 – 15 = 1200
x=300,41.810 2014 Q6 P2

2013 Q14 P2

GRAPHS OF TRIGONOMETRIC EQUATIONS

MARKING SCHEME
1. a)
x0 150 300 4500 7500 1050 1200 1350 1500 1650
3cosx0 2.9 2.12 0.78 -1.50 -2.12 -2.60 -2.90
4sin 3.06 2.57 -1.37 -3.94 -2.57
(2x0-10)

(b) graphs of y = 3cosx and y = 4sin(2x – 10) on the axes

(c)to find x, read off the values of x at the points of intersection between
the two curves x = 25.50 ± 10 , 97.50 ± 10 , 1620± 10
8marks
1989Q23

112
2.

x0 -300 00 300 600 900 1200 1500 1800 2100 2400

Sin(x+30)0 1 0.87 0.5 0 -0.5 -0.87 -1
Cos(x-15)0 0.71 0.97 0.71 0.26 -0.26 -0.71 -0.97 -0.97 -0.71

(b)scale: x-axis;1cm rep 300

y-axis;1cm rep 0.25 units
c) to find x read off the values of x at at the points of intersection between 8marks
the two curves x = 39 ± 1 , 219 ± 1
0 0 0 0
1990Q19
3. (a) 6M
x0 600 900 1200 1500 1800 2100
0
3sinx -1 1.60 2 1.60 0.5 -1 -2.5
Cos x0 -0.5 -0.87

(b) scale: x-axis;1cm rep 150

y-axis;1cm rep 0.25 units (an alternative scale may be
used)

(c) x= 370 ± 10 , 1800 ± 10 1991Q18

113
4.
x0 -1800 -1350 -900 -450 00 450 900 1350 1800
2sinx0 0 -141 -2 -141 1.41 2 1.41 0
cos2x0 1 0 -1 0 0 -1 0 1

scale: x-axis;1cm rep 22.5 y-axis;1cm rep 1 unit

0
1,
a) at x = 67 2sinx = 1.85 ± 0.05 and cos 2x = -0.7 ± 0.05.
2
subtract to get difference = 2.55 ± 0.1

b) (i)3600 (ii)1800 (6marks) 1993Q23

114
5. (a)
x 200 300 400 500 600 700 900 1000 1100 1200
Sin3 0.866 1.00 0.866 0.5 0.00 -0.5 -1.00 -0.866 -0.5 0.00
x
y 1.73 2.00 1.73 1.00 0.00 -1.00 -2.00 -1.73 -1.00 0.00

(b)(i) scale: x-axis;1cm rep 100 y-axis;1cm rep 0.5 units

(ii)to solve the equation draw the line y = -1.5 and read off the values of x at the
points of intersection between the curves and the line x = 760 ± 10 , 1030 ± 10
1995Q23
6. 6M

(a)
x0 200 400 80 1200 1400 1600 1800
-3cos 2x0 -2.30 -0.52 1.50 -3.00
2sin (3/2x0 + 300) 1.73 1.00 -1.73 -2.00

scale: x-axis;1cm rep 200

y-axis;1cm rep 1 unit (this scale must be used)

115
 (b) to find x read off the values of x at at the points of intersection of

the two curves x = 620, 1560 ( ± 10 allowed for each)

1996Q24
7.
x 300 600 900 1200 1500 1800 2100 2400 270 3000 3300 3600
0

cosx 0.87 0.5 0 -0.5 -0.87 -1.0 -0.87 -0.5 0 0.5 0.87 1.0
2cos1/2x 1.93 1.73 1.41 1.0 0.52 0 0.52 -1.00 1.4 1.73 1.93 -2.00
1

Cos x √ row B1 Allow d.p apply PA once

2 cos ½ x row √ B1 Allow B1 for any 12 √
Graph of cos x√
Graph of 2 cox ½ x √ B1 (√) all points must be
For any error in fitting B1 correctly
Table the graph drawn should have < that 2 points out (B) √ B1 B1 ( √ ) Plotted using
Period = 7200 B1 given scale
Amplitude = 2 units Apply Ow-1 If scale not
Enlargement of 2 about centre (0.0) 8 marks used.
1997Q18
8.
x 0 30 45 60 90 120 135 150 180 225 270 315 360
2sinx 0 1 1.4 1.7 2 1.7 1.4 1 0 -1.4 -2 -1.4 0
Cosx 1 0.9 0.7 0.5 0 -0.5 -0.7 -0.9 -1 -0.7 -1 0.7 1
y 1 1.9 2.1 2.2 2 2.3 0.7 0.1 -1 -2.1 -2 -0.7 1

116
 b)scale used S1
all points by plotted P1
smoth curve C1
B1
c)1400 ± 30 <1400 ±30 B1
range 0 ≤ x ≤1400 ±30 8marks
3480 ±30 ≤ x ≤ 3600 1998Q18
9. M1
M1 T
A1
100m
P
M1
A1 F

1999Q18
10. B2
Mark for scale used
B1 Mark for alt points
Marks for all points
B1 Mark for smooth
curves
B1
Mark for 250 1160
B1 stated accept (240-27)
From his graph read

117
 B1 within 10 mark for 25<
B1 x <116

2000Q24

11. B1 Allow B1 for any 2√

B1 Tan curve
S1 All points plots maybe
Implied
P1 Both parts of curve
P1 Asymptote drawn or
P1 implied
C1 Since curve
B1 All points plotted
B1 Smooth sine curve
8 marks Apply PA-1 once on
plotting (if 1 dp used)

118
 12. B2

P1

C1
B1

B1
C1

B1
8m

2002Q23
13. B2
S1

P1
P1
C1
B1
B1
8 marks

2003Q23

119
14. √ for x =300
√ for x = 1050

2005Q2
15. B1

B1

B1
B1
B1
10
marks

2007Q19

16.

120
 2008Q19
17.

B1
B1
10
B1

2010Q17

121
18. P1

C1

B1
3

a)
for plotting
for smooth sine curve

b)period = 1200
2011Q16
19.

2013 Q21

122
 AREA APPROXIMATIONS
MARKING SCHEME

a). COUNTING SQUARES TECHNIQUE

1 a). Area = 6 + 22/2 6 + 22/7 or 22/7 M1 2 a) 29 + 28 = 43cm2

. = 20cm2 . 2 B1
A1 b) 43 : 1075 x 104 x 104 M1
b). Area in hectares M1 1 : 25 x 104 x 108 A1
= 17 x 50000 x 50000 4 1 : 5 x 104 = 1: 50000 3
10 x 100 x 1000 mark 2000Q6 mark
=425ha s s

1999Q5

b).TRAPEZIODAL

1 (a) 28km =½x2

[2+226[10+26+50+82+122+17
2. Sketch the cross section 8M 0]
distance (x-axis) depth (y –axis) = [ 228 + 2x460]
A= 1/2h [a1 + an + 2(a2 + a3 + = 1148sq units
……an-1] b) 1/2h [a1 + an + 2(a2 + a3 +
= ½ x 25 [ 100 + 153 + ……an-1]
2(45+132+1156+167+200+163 V = ½ x3
)] [ 5+6+2(5.4+7+8.0+5.5+5.8)
= ½ x 3 [ 11 + 11 + 2 x31.7]
= ½ x 3 x 74.4
= 111.6m3 1993Q18

Missing values of y : 26, 138 B1

Area = ½ x 2(10 + 230) + 2(6 + M1
26 +70138) M1
= 240 + 480 A1
= 720 1996Q11 4m
= 25/2[253 + 2x933]
5. 3.55 + 0.05 , 4.85 + 0.05, 5.7, B1
= 25/2 [ 253+1866]
= 25/2 x 2119 6.3, 6.7 & 6.9
M1
= 26487.5cm2 Area = ½ x M1
1992Q24 A1
3 a) y =x2 + 1 8M 1(0+7+2(3.6=4.9+5.7+6.3+6.7+6.9)
A1
= ½ x 1 ( 7 + 68.20) 4m
x 1 3 5 7 9 11 13 15
= 37.6
y 2 10 26 50 82 122 170
226 1997Q8
6. a).
A= 1/2h [a1 + an + 2(a2 + a3 + B1
……an-1] M1

123
 X 2 3 4 5 6 7 8 M1 10 a). M1
Y 3 5 9 15 23 33 45 .
b).A = 1 x 1 x (3 + 45) + 2(5 A1 B1
+9+15+23+33) M1
= ½ (48 + 170) = 109 sq. units M1
M1 M1
2
c). (x – 3x + 5)dx
= x2 – 3x2 – 5x2 A1 M1
3 2 B1 A1
= 8 – 3 x 8 + 5 x 5) 2 – 3 x 2
3 2 2 3 2

3 2 3 2 8m M1
+5x2 = 108 A1
d). It would given A1
underestimate because the line
for the trapezium run below the 10
curve in the region. b). i) Area below upper curve;
1999Q24 = ½ x 1 x (12 +2 (12 + 2(4 +
7. Distance =5/2 (2.6 + M1 5.7
2(2.1+5.3+5.1+6.8+6.7+4.7) M1 + 6.9 + 8 + 9 +
= 5/2 (2.6 + 61.4) A1 9.8+10.6+11.3)
= 160 m 3m = ½ (12+130.6) = 71.3
2001Q11

8. a) π used as constant width B1 Area below lower curve;

12 (or 0.26) M1 = ½ x 1x (12 + 2(0.2+1.3 + 2.4
π M1 + 3.7 + 5.3+7.3+9.5)
½ x 12 (0 + 0.84) + M1 = ½ 912+60.6) = 36.3
2(0.26+0.48+0.65+0.76+0.82 A1 Area in dispute = 713 – 36.3 =
π M1 35
(0.84 + 2 ((2.97)
24 M1
A1 ii). Area in hectares = 35 x 400
3.142 x 6.78 = 0.8875 (4) (6)
10000
24
8 = 1.4
Absolute error: 2007Q24
0.8940 - 0.8875 = 0.0065 ma
11 (a) ordinates
0.0065 x 100% rks
x =0 y1 = 1
0.894 % x=1 y2 =6
= 0.73% x=2 y3=9
2002Q21 x=3 y4=10
9. x 0 0.4 0.8 1.2 x=4 y5=9
x=5 y6=1 B3
y 2.0 1.96 1.83 1.60 M1
x =6 y7=1
A1 area = ½ x 1 x{ 1 + 1 + 2 M1
b). Area of ¼ circle (6+9+10+90)}
½ (0.4)x (2+0) + (1.96 + 1.83+ M1 = ½{2 +, 2(40)} A1
1.60+1.20) = ½ (82) = 41 sq. units
= 3.036cm2 A1 M1
8
M1
= area of a circle (b)∫ −x +6 x +1 =
2

= 4 x 3.036 4 0
A1
= 12.144cm2
2006Q16
m
[ −1 3 6 2
3 2 ]
x + x +x 6
0
M1
A1

124
 = -72 + 108 + 6 10 42
= 114 -72 = 42 sq. units = 2.38% 2011Q21
(ii) 42-41 x 100%
c). MID-ORDINATE

1. 540m2
1990 Q15
P2
2. (8 + 6.5 + 5.6 + 6.1 + 6.4 + 4.7) 4
x5 M
= 374 x 5 = 187km
187 x 2 x5 = 1870km2
1870 = 0.187ha
10000
1995Q16

3. (a) x -5.5 -5 4.25 -3.75 6

y 16.25 12 6.25 3.56 M

(b)0.5(18.56 + 14.06 +
10.06+6.56 + 3.56 +1.06)
A =0.5 x 53.86
= 26.93sq units
−3

(c)∫ ( x +2 x −3 ) dx =
2

−6

[ ]
3
x 2
+ x −3 x
3
= 9 + 18 = 27sq units

(ii) E = 27 – 26.93 x 100 =

0.259%
27
2003Q20

125
4.

y= x2+5

B2
M1
x 0.25 0.75 1.25 1.75 2.25 2.75
y 5.0625 5.5625 6.5625 8.0625 10.0625 12.5625 M1
A1
5 1/16 5 1/16 691/16 8 1/16 10 1/16 129/16
M1
Area = ½ (5.0625+5.5625+6.5625+8.0625+10.0625+12.5625)
= ½ (47.875) A1
= 23.9375 M1
= 23.94
a). ii) exact area A1
3 3 8m
∫ (x+ 5) dx3 = x3 +3 x 30
0
= 9+15
=24

b) % error = 24-23.94x100
24
=0.06x100
24
=0.25% 2003Q20
5. M1
x ½ 2½ 4½ 5½
y 3¼ 9¼ 23 ¼ 33 ¼ M1

M coordinates 3 1/4 , 5 1/4 , 9 1/4 , 15¼ 23 1/4 , 33 ¼ A1

1 1 4 marks
Area = 1 (3 ¼ : 5 ¼ - 9 ¼ + 15 ¼ 23¼ -33 = 89 sq. units
4 4
2004Q11

6. y1 y2 y3 y4 y5 y6
2 5 9 14 20 27

126
Mid point
Area = h(y1+y2+y3+y4+y5+y6)
=1(2+5+9+14+20+27)
=77cm2
b). error = 78cm2-77cm2
=1cm
% error = 1/78 x 100
1232/39% 0r 12.82 2005Q20

127
7. a). Trapezium X -2.5 -1.5 -0.5 0.5 1.5 rule
Y 8.75 5.75 4.75 5.75 8.75 B1
x -1 -2 0 1 M1
y 7 5 5 7
are a= ½ x 1 ( 11+11) + 2(7+5+5+7) A1
= ½ ( 22 + 48) = 35
Area = 11 x 5 = 55 M1
= 55-35
= 20 square units A1
Mid-ordinates
B2

M1
AC = (8.75+5.75+4.75+5.75+8.75x1) = 33.75
A = 55-33.75 M1
= 21.25
Differnce = 21.25 -20 A1
= 1.25 square units 2008Q18 10 marks

8. x 0 1 2 3 4 5 6
y=1/2 x2- x+3 3 2½ 3 4½ 7 10 ½ 15

B1

M1

A1

M1

A1

y1 = ½ x 1.52 -1.5 +3 = 2.625 M1

y2 = ½ x 2.52 – 2.5+3 = 3.625
y3 = ½ x 3.52 -3.5 +3 = 5.625 A1
y4 = ½ x 4.52 -4.5+ 3 = 8.625
y5 = ½ x 5.52 -5.5+3 =12.625

Approximate area = 1(2.625+3.625+5.625+8.625+12.625)

= 33.125 square units

[ ]
6 3 2
X x 6
Area = ∫ (½ x – x+3) dx =
2
− +3 x
1 6 2 1

( )( )
3 2 3 2
6 6 1 1
= − + 3× 6 − − +3 = 33.3 % error
3 2 6 2
= 33.3 -33.125 x 100
33.3 = 0.6151% 10 marks

128
 INTEGRATION
1. a)y = ax2 + bx 8 =4
dy M 1998Q20
= 2ax + b
dx
6.
2a (o) + b = 8 a). (2x + 3x2)dx = x2+ x8+c B2
B=8 = 250
2ax + b b). Area below x-axis M
2a( 4)+b = 0 2 1
(x2 + x3)- =(-2/3)2 + (-2/3)2=12
8a + 8 =0 a = -1 3

b)y = -x2 + 8 4
= (4/9 – 8/27) = −¿

[ ] 27 M
6 3
∫ (−x 2+ 8 ) dx −x3 +8 x
6
0 Area above x-axis 1
0
2
(x2+x3) = (4 + 8) = -0 = 12
0 A1
= 72 sq. units Total area = 4/27 + 12
1992Q22 = 12 4/27
2000Q21
2 102/3 sq.units
1994Q6 7. a). 3x =4 -x2 M
(x+4) 9x-1) =0 1
3. 2 3
3 sq.units x = -4 or x=1 A1
3 M The coordinator of P(1,3) B1
1995Q7 The coordinator of Q(-4,-12) B1
4. Area = s4(2x3 -5)dx
[
b). (14-x2) dx = 4 x− x ]
1 3 −3
3 −4
M
1
[ ]
2
4x 4 = 4 x 2 - /3x (-2) -(4x-4-1/3 (-
1 3 A1
= −5 x
2 2 4)3
= 108 + 2 = 10 2/3 M
= 110squnits 1996Q8 The shaded area 1
= ½ x 4 x 12 – 102/3 A1
5. (a)x2 – 2x – 3 = 0 Below x axis
(x-3) (x +3) = 0 = 13 1/3 M
x=3 or -1 Shaded area 1
x
3
= 131/3 + (4x – 1/3x3)0 A1
(b)x + 2x –3)dx= - x2 –3x+ c
2
3 = 13 1/3 + 0 = (4x-2-1/3 (8) 10
= 12/3 = 131/3 + 5 1/3 m
(c)x3 – x2 + 3x 4/3 = 182/3 2006Q24 ar
27 ks
= (64/3 – 16 -12) – – 9 -9
3 8. a). i). y= 2/2x2 + x + c M
= 21/3 at x = -4, y= 6 1
Sum of areas = 12/3 + 21/3 6= (-4)2 -4 + c

129
 6 = 16 -4 + c M P2
c = -6 1 10
y = x2+x-6 A1
(a) B2
ii). x2+x-6 = 0 M
x -2 - 0 1 2 3 4 5 6 7
(x-2) (x+3) = 0 1 1
x =2, x=-3 M y 1 1 6 4 4 6 1 1 2 3
6 0 0 6 4 4
b). (x2 + x - 6)dx 1

( ) A1
3 2
x x
+ −6 x (b) Area using trapezium rule
3 2
( /3 + 4/2 -12 ) – (-27/3+9/2+18)
8 M M
= 7 1/3 -13.5 1 =
1
= -20 5/6 M 1
1 ×1 [ 16+ 46+2 ( 10+6+ 4+ 4+6 +10+16+24 +34 ) ]
Area = 20 5/6 square units 2
A1 M
B1 1 1
= [ 62+2 ( 114 ) ]
2007Q20 10 2
m
ar = 145
A1
ks (c) Area using mid-ordinate rule
= 2 × ( 10+ 4+6 +16+34 ) M
9 a) 1
= 140
A1
(d) Area using integration method
8 3 2
∫ x 2−3 x+ 6= x3 − 32x +6 x
M
−2
8 1
−2

= [ 512 192
3
−
2
+ 48 -
] M
1

[ −8 3 x 4
3
−
2
−12
]
A1
2 2
= 122 +20
3 3
2
b) area = 54sq units = 143
3
c) (i) as on the graph
(ii) area = 18sq units 2014 Q21
2013 Q23

130
 CALCULUS
MARKING SCHEME
DIFFERENTIATION
1. Length be x 4M
Then width = ( 1200−
2
X
) 3. y = 8x -8
1992Q11
4M

Area = x ( )
1200− X 4. y = x2 – 3x -4 2M
2 dy
= 600x – ½ x2 = 2x – 3
dx
da = 600 – x dt x = -1
dx dy
for maximum da = 0 = 2(-1) -3 = -2 -3 =-5
dx dx
600-x = 0 gradient = -5
x = 600 1993Q12
Width = 1200 – 600 = 300m
2 5. (a)(i)20m/s
1990Q15 (ii) 10m/s

2. y = x3 + 2x2 – 4x -8 5M (b) max. height = 45m

dy 1993Q24
= 3x2 + 4x – 4
dx 6. width = 7cm, length = 7cm
1994Q11
dy
maximum =0
dx 7 y = 3x2 – 4x + 1
dy
3x2 + 4x – 4 = 0 (a) = 6x -4
3x2 + 6x – 2x -4 = 0 dx
3x (x + -2) (x +2) = 0 dy
where x = 2 =8
x = -2 0r x = 2/3 dx
x -3 -2 0 2/3 b)let m(xy) be a point on curve
dx / \ min / (i) y–5 =8
dy x–2
x =-2 (ii) Tanθ = 8 θ
1991Q11 = 82.80

131
 (iii) Gdt of perpendicular = -1/8 ii). D2x = 6t – 4 8
(iv) y–5 =1 dt2 ma
x–2 8 a = 6 x 4/ 5 – 2 rks
8y – 40 = -x + 2 = 4m/s
8y + x = 42 2001Q22
1996Q19 13. dy
8. dy M1 i) = 6 x 2 + x + -4
= 3ax2 -6x-2 dx
dx when x = 1
3ax2 – 6x – 2=7 at x=1 M1
3a = 15 A1 dy
a=5 =6+1–4
dx
1997Q10 Grad=3
9. (x2 ÷ 1)(x – 2) = x2 –2x2 + x – 2 M1 ii) y + ½ = 3(x – 1)
dy y = 3x – 3 – ½
= 3x2 – 4x + 1 M1
dx y = 3x – 3½
dy y=3x - 3 ½ 2002Q16
when x = 2 =5 A1
dx
y=0 14. ds
a) = 3 + 3t – 6t2
y–0=5 dt
x–2 2
d s
y = 5x – 10 3 = 3 – 12t
1999Q16
dt
a = 3 – 12 (0)
= 3ms-2
10 ds B1
. a) V = = 3t – 5t + 2
dt b) i) 3 + 3t – 6t2 = 0
B1
(t – 1) (2t + 1) = 0
dv t = 1 or -1/2 t = 1sec
a = = 6t – 5 B1
dt
b) 6t - 5 = 0 = t = 5/6 ii) at t = 1 s = 3(1) + 3/2 (1)2 – 2(1)3
v = 3 (5/6)2 - 5 (5/6) + 2 = 3 + 3/2 -2 = 21/2m
3m
= 25/12 – 25/6 + 2
= -1 (0.0833) c) 3 – 12t = 0
12 v = 3 + 3 ( ¼ ) - 6(1/4 )2
2000Q5 = 3 + ¾ - 6/16
11 dy M1 = 33/8 m/s 2002Q24
. a). = 15x2 – 14x + 3 A1 15. dy
dx = 2 - 8x = 0
gradient = 4 M1 dx
A1 x =¼
b). y – 3 y = 6 + 2(¼) - 4(1/4 )2
x–1= 4 = 6 1/ 4
y = 4x -1 Turning point ( 1/4, 6 ¼ )
2001Q11 2003Q8
12 dx M1 16. dy
. a). = 3t2-4t M1 = 3x2 - 3 = 0
dt dx
velocity = 3 x 22- 4x2 A1 3(x2 – 3) = 1=0
= 4m/s M1 (x –1) (x +1) =0
b). i). 3t2-4t=0=t(3t-4)=0 Coordinates are (1,0) & (-1 4)
t=¾ M1
x = (4/3) -2(4/3)2 + 6
A1
4 32
= 6 /27 – /7 + 6 = 64-96 + 6
27
= 4 22/27 ( 4.815) M1
A1

132
 dy 2005Q17
For (1, 0) x <1, is
dx 20. a). S = 53 -5 x 52 + 3 x 5 + 4
dx = 125 – 125 + 15 + 4
dy = 19m
x > 1, is+
dx
dx ds
b). V = = 3t2 -10t +3
dy dt
x >-1, is – 3t2 -10t + 3
dx
= (-1, 4) is a maximum. = 3(5)2 – 10(5) + 3
2003Q21 = 75 – 50 + 3
17. A=10t -12 = 28ms-1
=10 x 2 -12
8 ms-2 2004Q5 c). Momentarily at rest v =O
ds 3t2 -10t + 3 = 0
18. = O at maximum (3t-1) (t-3) = 0
dt
t = 1/3 or 3
= 29.4 – 9.8t
d). Acceleration when t = 2
9.8t =29.4
dv
t=3 a=
dt
Hence S = 29.4 x 3 – 4.9x32 = 6t -10
= 44.1m 2005Q16 6x 2-10
= 2m/s2 2006Q24
19. ds 21. dy
= x2+2x-3 =3ax2 + b
dt dx
ds 3a + b = -5
At turning points ; =0 a+b=1
dt
a = -3 b = 4
x2+2x-3 =0
2007Q5
2 22. x+y = 40 y = 40 –x
x -x+3x-3=0
sum of the squares in terms of x
x(x-1)+3(x-1)=0
s = x2 + (40-x)
(x-1)(x+3)=0
2x2-80x+1600
x=11 or -3
dy
= 4x- 80 = 0
Subtract y = 1/3 or 11 dx
The turning points are (1, 1/3 ) and
(-3, 11) 4x = 8- 80
x=20

Sum of the squares

=202+(40-20)2
202+202
400+400
=800 2007Q13

133
23. ds = 26m/s2
a). = 3t2-12+9 2011 Q22
dt
26 (a) (i) x-intercept = -1.5
ds (ii) y-intercept = 0
(0.5) = 3(0.5)2- 12(0.5)+9 (b) (i) (0,0) and (-1,1)
dt
(ii) (0,0) is minimum and
= 3.75
(-1,1) is maximum
ds
b). = 0 = 3t2 - 12t + 9 = 0 2012 Q22 P1
dt 27.
t2-4t + 3 = 0
dv
4(t-3) (t-1) = 0 (a) (i) = 4-t
t=3 t=1 dt
When t =3 s=33 -6 x 32+9x3+5 = 5 v = ∫ ( 4−t )dt
When S= 13-6 x 1+9 x 1+5 - 9 1 2
= 4 t− t +c
2
When t=0, v=3m/s
1 2
∴ 3 = 4 ×0− ×0 +c
2
C=3
1 2
V = 4 t− t +3
2

(ii) when t= 2s
2008Q24 1 2
24. 2 V= 4 ×2− × 2 + 3
432−x 2
(a) (i) h=
4x = 8−¿ 2+ 3
= 9m/s
3
432−x dv
(ii)V= (b) (i) At maximum velocity =0
4 dt
(b) (i) x=12cm i.e 4-t =0
(ii)h=6,V=864cm3 t= 4s
2010 Q24 P1 4
1
25. ds (ii) ∫ 4 t− t +3 =
2
(a) V = = 6t2 – 10t + 4 2
dt 0

[ ]
When t = 3 4 2 1 1
3
v = 6(9) – 10(3) + 4 t − × t +3 t
= 28m/s
2 2 6

[ ]
3
1
2
(b) v = 0 => 6t – 10t +4 =0 = 2t 2− t +3 t
6
3t2 – 5t + 2 = 0
(3t-2) (t-1) = 0 1
= 2 ×16− ×64 +12−0
t = 2/3 or t = 1 6
2
2
=32−10 3 +12 = 331 /2
(c) t = /3;
s = 2(2/3)3 – 5(2/3)2 + 4(2/3)
= 3.037m 2012 Q24 P2
t = 1; s = 2(1)3 – 5(1)3 + 4(1) + 2 28. (a) s=4m
= 3m (b) (i) 28m/s
dv 1
(ii) t= sec or 3 sec
(d) a= = 12t – 10 3
dt
t = 3; a = 12(3) – 10

134
 5 Stationary points
(c) t =
3

29.
2013 Q21
( −13 , 1427 ) and ( 3 ,−18 )
(a) Value of y when x=−1 (c) Equation of normal to curve:
y=−1 -4 +3= -2 Gradient of tangent at x=1

(b) Stationary points dy

=3−8−3=−8
dy 2 dx
=3 x −8 x−3
dx
1
Gradient of normal =
For stationary points 8
2
3 x −8 x−3 = 0 Equation of normal at x=1

( 3 x+ 1 )( x−3 )=0 y +6 1
=
x−1 8
−1
𝑦+6 =
x= ∨x=3 1 1
3 x−
8 8
−1 14
𝑦=
When x= , y= 1 1
3 27 x−6
8 8
When x=3 , y=−18
2014 Q24

2. INTEGRATION
1. s=1¼ 4 = 18m
1989Q15 M 1992Q12

2. 3
3 3 4. (a)2m/s2 8
∫ ( 2 X +3 ) dx = x + 3x −1
2
M (b)36m M
1
= (9 + 9) – (1-3) (c)t=6.606seconds
= 18 + 2 1994Q19
= 20
1991Q14 5. V=3t2-6t-8
S=(3t2-6t-8)dt M
3. V = 3t -2 4 S=t3 -3t2-8t+c 1
∫ dt = 3t2 – 2t + c M 10=1-3-8+c M
2 C=20 1
T=0 x=2 S=t3-3t2-8t+c A1
S = 3t2 – 2t + 2 0m
1999Q16 3
2
dt = 4 ma
S = 3 (16) -2 (4) + 2 rks
2 6 a) V = 9t2 - 4t + c
= 24-8 + 2 initial velocity : t = 0 when V M

135
 = 2ms – 1 c = 2 1 1
V = 9t2 – 4t + 2 Co-ordinates of P,Q are A1
A1 P(-1,8) q(2,5)
b) 9t – 4t + 2 ½ (8+5)3 -2(x2- 2x + 5)
9t2 – 4t = 0 B1 M
( )
3
39 x 2
t(9t – 4) = 0 allow transfer of = - −x +5 x 2 1
2 3
measures here 3 M
4 2 1
t = 0 or t = /9 can be given ma =19.5 – 8 + 6 1
rks 3 3
early
A1
= 19.5 – 15 8
t= 4 sec = 4.5 or 4 ½
9 ma
2000Q14 rks
2002Q20

7. 2
a)∫ (2 x −5) dx= x3 –x + c
2
3 M 9. V=3t2+4 M
2 3 1 1
S= ∫ (3 t −t + 4)dt
2
y= x -5x + c
3
2 M 5
3 - × 2x3 - 5x2 + c t3 - t2 + 4t 1
3 1 M
2 23 2
C = 7 0r /3 1
3 A1 =(25- 25+20) – (1 -1 + 4)
2 3 2 2 2
y = x -5x + 7 = 256 – 9
3 3 A1
M 2 2
256 3
b) ∫ (2 t +t −1) dt
2 1 1 = /2 ma
4 3 S = 128m rks
2t t
= + −¿t +c
4 3 M 2003Q16
1 10 dy
= 3x2-8x+2
∫ (2 t2 +t 2 – 1) dt = . dx M
M 3 2
y = x -4x +2x+c 1

( 24 ×3 + 33 −3)−( 24 + 13 −1)
4
3 1 At x =0 y=2 2 = 0-0+0+c
C=2 M
A1 3 2 1
8 y = x - 4x = 2x+2 A1
¿ ( 812 +9−3) - ( 12 + 13 −1) ma
rks
2004Q13

2004 Q13 P2
11 a) V= ∫ (6 t + 4) dt=3t2+4t +c
= 46 ()
1 1
2 6
. At t = 0.5 = 3x02 + 4 x 0+c M
1
2 V=3t2+4t+5
= 46
3
2001Q21 b). at t=3, v=3x32+4x3+5
M
8. a). Gradient = -1 B1 = 44m/s
1
y = x+7 B1 ii). Distance from t = 2 tn =
b). 7-x = (x-1)2 +4 4
4 A1
x2-x-2=0 M
1 ∫ (3 t + 4 t+5)dt
2

(x-2) (x+1) = 0 2

x= 2, y = 5 x= -1, y = 8
M

136
 4 . S = 2t –t2/2 + c 1
t3+2t2+5t 2
When s =5, t=2
= 43 +2 (42)+5x4) – 5 = 2x2 -22/2 + c =3
M
(23+2x22+5x2) 2 1
S=2t – ½ t + 3 A1
= 116-26
2007Q5 P2 3
= 90m
ma
2004 Q22 P2 rks
2004Q22 16 ∫ (3 t2−6 t −9) dt = t3-3t2-9t+c B1
.
V = ∫ Adt ∫ (25−9 t )dt
2
12 B1 3 M
.
= 25t - 3t3+ c t3-3t2-9t 1 = 1
M
4 = 25t-3t+c when t = 0
1 (33-3(32)- 9(3) - (13-3(1)2- (1)2
4=c M
-9(1)
1
Hence V = 25t -3t2 + 4 = -16
V = 25 x 2 -3 x22+ 4 A1 4
t3-3t2-9t 3
= 50+4-12 A1
= 42ms-1 A1 = (43-3(42)-9(4) - 4
2005Q16 (33-3(32)- 9(3) ma
=7 rks
13 dy Distance travelled
a). = O at turning points
. dx = 16+7 =23m 2008 Q15 P2
hence = 4x – 3 = 0 2009Q16
x=¾ 17 a) i) Area of base x2
min. value = y at min. Point . or area of sides = 4xh
hence at minimum point x = ¾ x2 + 4xh = 432
1 h = 432 - x2
and y = /8 4x
= (4x-3)dx ii) Volume = x2h
y= 2x2-3x+c subs x= ¾
= x2(432 - x2)
c=1 y = 1/8 4x
hence = 2x2-3x+ 4 b) i) Volume (v)
= 108x - 1/4 x3
dy dy dv/ = 108 - 3/ x
b). =4x – 3 and /dx = 7 dx 4
dx
therefore 4x -3=7 3 2
108 - /4 x = 0
5 x = 12
x = /2
subsitute for x y=6 ii) Vol = 108x - 1/4 x3
hence the point is (2.5,6) (108x 12) - 1/ x 123
2005Q21 4
2005 Q21 P1 = 864cm3
14 V = ∫ adt = 10t – /2 t2 +c
2 M 2010Q24
. At t = 0, v = 9 c = 9 1
18 3 13
10t – t2 +9
M (a)s = t2 – t + c
. 2 3
1
At t = 3, =10(3) -32+9
A1
= 30m/s (b) when t = 0, s = 0 c = 0
2006Q15 P2 3
32 13
ma s = t – t =0
rks 2 3
15 S = Σ(2-t) dt M

137
 t
2
( 32 − 13 ) = 0 0=
1
3
+2+3+ c

t = 4.5s
2010Q11 −4
∴ c= A1
3
19 Y=∫ ( x −4 x +3 ) dx M1
2
3
. 1 3 2 4
∴ y= x −2 x +3 x−
1 3 3 3
2
= x −2 x +3 x +c
3 M1 2014 Q 15 P2

LINEAR PROGRAMMING
MARKING SCHEME
1. Let the length of the plot be xcm and 8M Objective function. Max .point at
the width be y cm (12,3) max profit = sh 38400
(a) x > y, xy > 400, x < 2y, x + y < 1990Q20
100
3. let no. of syrup x made be x, and syrup 8M
(b) size of plot = 50m by 48m, 24sub y made be y.
plots (a)2x + y ≤ 1600, x + 2y ≤ 1100, x + 3y
1989Q18 ≤ 1500, x ≤ 0
y≤0
2. Let maize land be x hectares and beans 8M
be y cm. (b)graph
(a) x > y, x + y ≤ 15 11x + 27y ≤ 225;
x > 0, y > 0 (c)60x + 100y…………objective
function. x = 700 bottles, y
(b) graph = 200 bottles
1991Q24
(c) 2400x + 3200y 4. (a)x + y ≤ 50, y ≤2x, 2x + 3y ≥ 120 8M

138
 (c) 25000x +20000y
(b) (i) graph x = type X
(ii)1000x + 2000y…….objective y = type Y
function. Point for max profit = (17,33) 1997 Q22 P2
thus x = 17 trips and y = trips 10 (a) x + y ≤ 400, x.y; x ≤ 300, y ≥ 80 B3
. (if A and B are used throughout)
1992Q20 (b)All 4 inequalities by drawn and
5. Let the number of kg of type I be x 8M shaded
and type II be y. (c)(i)x = 300 and y 100
(a) 3x + 5y ≤ 4000, 7x + 5y ≤ 4800, x (ii)max profit
≤ 0, y ≤ 0 = 600 x 300 + 400 x 100
= 220, 000 1998Q24
(b) Graph 11 B2
GRAPH
(c) 34x + 36y…………objective x x y =250
function. Type I = 200kg, type II 100x + 460y > 16000 B1
= 680kg x < 200
x < 50 B1
1993Q23 y = 2x
6. let no. of units of the first product be x 8M x =50 and x =200 drawn and shading B1
and of second product be y. y = 250 drawn and shading
(a)2x + y ≤ 100, 3x + 4y ≤ 240, x + 2y y=2x drawn and shading indentification B1
≤ 90, x ≥ 0, y≥0 of best point
(b) graph Ordinary seats = 84 B1
(c)300x + 400y………. Special seats = 160 B1
Objective function. x = 36, 2000Q24
y= 27 12 a). 800x+1600y>8000(x+2y>10) B1
. 4x+7y<41 B1
1994Q18 x>2 and y>2 B1
Mark for each line draw and shaded B1
7. let x be no. of jars of grade a and those 8M Mark for type A=3 & type B=4 B1
of grade B Mark for numbers of operators B1
(a)2x + y ≤ 6000, 4x + 7y ≤ 20000, x ≥ 3x4+4x7=40 2001Q24 B1
0, y ≤ 0. B1
8
(b)(i) graph 13 (a)3x + 4y < 120 B1
(ii) 48x + 400x + 150y < 9000
30y………………….objective x>8 M1
function, y > 12 B1
x = type x = 3, y = type y = 4 2003Q20 A1
1995Q24 8
8. a) x+y< 500 B1 m
y>x 14. a). i). = 75x + 75y>6 M1
x> 200 B1 = 25x + 25y>2
b). x+y <500 drawn and shaded ii). 75x <175
y> x B1 3x < 7 M1
c). (i) no. enrolled in technical iii). 75y<180
= 249 L1 5y < 12
no . enrolled in boys = 251 L1 iv). y > 0
v). x> 0 M1
(ii). max. profit B1
249 x 2500 + 251 x 1000 – 873.500 B1 b). graph 4
1997Q19 8m m
9. Let the number of buses of type X be x c). i). Lowest cost =20x + 50y
and of type Y be y. at ( 0.1,0) c= 20 x 0.1 + 50x 0 A1
(a) 4x + 3y ≥ 24 c = 2/=
x + y ≥7 ii). Max cost= 20 x 2.3 +50 x 2.4
x≥0 c = 46 + 120
c=166/= 2005Q24
y≥0
15 a). 300x + 180y<18000 B1
(b) Graph

139
 5x+3y < 300 (c)when x = 5, y = 5 B1
x + y < 80 S1 5 x 4 + 5x6 = 50tons B1
x > o, y> 0 B1 when x = 8, y = 3 B1
B1 3 x 4 + 3x6 = 50tons
B1 when x = 7, y = 4, M1
x = 30, y=50 7x4 + 4 x 6 = 52tons A1
Max profit = 50 x 4000 + 30 x 6000 B1 7trips by P and 4 trips by Q 10
380000 B1 2011Q24
M1
2006Q23 A1 19
10 . (a) (i) 38392 + 2108 M1
16 = 41000 A1

[ ]
( 0 , 8 ) ( 10 , 4 ) 10
a) m (ii) 10164 x 0.1 + 9576 x 0.15 + B1
( 0 , 10 ) ( 8 ,5 ) 9576 x 0.2 M1
+ 9576 x 0.25 + 2108 x 0.3 M1
2x + 5y < 40
5x + 8y < 80 = 1016.4 + 1436.4 + 1915.2 +
x>3 y > 1/3x 2394 + 632.4
= 7394.4
2007Q22 M1
17 a) 4x + 6y ≥ 144 Monthly income tax A1
100x + 200y ≤ 4800 = 7394.4 – 1162
b) = 6232.4 M1
x ≥ 16
y > 10
GRAPH (b) amount saved in coop society
(c)z = 40x + 100y drawn OR 2 feasible 5
pts inspected = × 41000−15000 M1
Profit = 40 x 16 + 100 x 16 100
A1
= sh. 2240 10
2010Q20 = 1300
18 4x +6y ≥ 42 B1 Net pay
2x + 4y < 32 B1 41000 – (6232.4 + 1300)
x < 3y B1 = 33467.6
x <4 B1 2014 Q20
graph B1

LATITUDES AND LONGITUDES

MARKING SCHEME

1. (a) θ = 10 + 35= 450 8M 2 x 22 x 6370 x 0.866

Length of AB = 45 x π x 1 7
360 = 20020km
= /8 π meters
1

(b) 0.07092m2 (b) 2 x 22 x 6370 x θ = 8000

1989Q18 7 360
θ = 8000 x 9
2. (a) Rcos = r 8M 91 x 11
r = 6370 cos 60 = 720
Circum = 600 + x = 720

140
 x = 11.93
B (11.930S, 450E) b) speed in km/h M1/
1990Q18 6003.72 = 50.031 km/h A1
3. (70 – 25 ) x 60cos 470
0 0
3M 120
45 x 60cos 470 c). /360 x 2 x 22/7x 6370 cos 5
θ

= 1841.395572 = 6003.72
= 1841.4nm θ= 6003.72 x 360 x 7
1991Q13 2 x 22 x 6370 cos 5 A1
4. 2000 = θ x 60cos600 3M = 54.190 8m
200 = 30θ
θ= 66.6 Position (50N.99.190E)
= 670W (nearest degree) 1997Q18
1992Q11 9. Longitudinal difference 70 – 10 = B1
5. (i) 100 + 20 =120 9M 600
2 x 22 x 6370 x 120xcos400 (i)distance between x and y M1
7 360 60 x 22 x 2 x 6371 cos 450
= 10224.13983km 360 7 A1
= 10224km
1 x 22 x 2 6371 x 0.7071 B1
(ii)9245km 6 7 B1
1994Q18 = 4719km
6. (a) (i) 450 -1075' = 430 15'N 7M (ii)distance between x and y 8mar
(ii) 370 + 23 = 60 4919.45 = 2551.0.5mm ks
60 x2 x 22 x6370 xcos43045' 1.85
360 7 (c) time diff = 60 x 4 = 240min
= 4820.57550 = 4hrs
= 4821km (nearest km) Local time at x = 10.00a.m
1998Q20
(b) 230 + 370 = 600
60x4 = 240min 10 a) Angle change 52 – 38.
240 = 4hrs . S = 2 x 22 x 6370 x 13.5 M1
60 7 360 M1
1.00a.m – 4hrs = 9.00p.m = 1501.5km M1
1995Q24 θ x 2 x 22 x 6370 cos 520 = 2400 A1
7. (a) 131 + 49 = 1800 360 7 M1
M1
(b) 180 x 22 x 6370 cos 36 θ= 2400 x 7 x 360 A1
360 7 2 x 22 x 6370 cos 520 B1
= 16,196.18km -35.050
16196.18km(along a latitude) or C= (520N 210W) 8m
along a longitude)
2000Q22
(c) x x 22 x 2 x 6370 cos 36
360 7 11 (a) (300 + 300) = 600 x 600
= 840 . = 3600n.m
x = 840 x 9 = 9.340
11x91x0.8090 (b)18 x 60 cos300 = 935nm +
3600nm = 4535nm
Town C longitude = 1310 – 9.340 4535 hrs = 22.765hr
= 121.70W 20
1996Q20 2001Q24
8. a). i). 120 x 27 M1
= 3240 A1 12 a). different time =3h
ii). 120 x 27 x 1.853 longitude difference
= 6003.72km M1 =3 x 150 or 450 B1

141
 = 600E (b) (i) 16 10 h
B1 (ii) Wednesday at 1755h.
b). i). distance travelled 2010 Q19 P2
850 x 31/2 M1
=2974km M1 18. Angle subtended (longitude) B1
Arc = AB=297 M1 16 + 24 = 400
45 x 3.142= 2975 i) Arc AB = 40 x (60 x cos 340)
360 A1 = 1989.69 = 1990nm M1
r = 2975 x 360
45x3.14x2 M1 ii) Arc AC: latitude difference
ii). 6371 cos θ=788 A1 M1
cos θ = 3788 B1 = 26 + 34 = 600
= 6371-0.594 8m .: Arc AC = 60 x 60 nm A1
θ=53.510 = 3600
Latitude of two towns is 53.510N
2003Q24 b) i) real time at B M1
13 < POG = 0 – (36 x 2) B1 1330 + 40h
. = 1080 M1 15 B1
Dist PQ 08 x 60 A1 1330 + 2h 40min
6480 mm 3m = 1610h B1
2006Q16
14 Longitude difference M1 ii) Time taken to travel from A to B M1
. = 3600 – (1330+1180) M1 = 1990/
40
= 109 A1
= 49h 45min
109 x 60 cos = 5422
Time of arrival
Cos x = 0.8291 A1
Wednesday at 1610 + 1h 45 min
x = 22.990 A1
= Wed at 1755h
Longitude of A or B =340N 3m 10
2011Q14
2007Q13
15 i). Distance = 500 x 9/4 B1
. = 1125nm 19 (a) Distance from T to U
ii). θ x 60 x Cos 53.4 = 1125 M1 22 12 M1
=2 ×6370 × ×
θ = Cos 53.40 A1 7 360
= 31.450 3m M1
Longitude of θ = 71.450E Speed =
2008Q7 22 12 M1
16. B1 2× 6370 × ×
7 360
M1
1 M1
A1 1
3m 3

= 1001 km/h
A1

Longitude difference = 45+60 b) Time =

=1050 22 30 0 M1
2× 6370 × × cos 9
Distance in km 7 360
= 105 x 2 x 3.142 x 6370Cos 400 90 M1
360 1001×
100
= 8946.12 km when 22/7 is used = 3.658104965 h
for π = 3h 39 min
2009Q13
17. (a) (i) 1990nm c) Arrival time at U = 0700 +
(ii) 3600nm 1h 20min M1

142
 = 0820h 2013 Q15 P2
Departure time at U
0820 + 30min 21.
0850h
M1
Y=∫ ( x 2−4 x +3 ) dx
35−5 A1 M1
Time difference = ×24 1 3 2
360 10 = x −2 x +3 x +c
3
¿2h
1 M1
Arrival time at V (Location) 0 = +2+3+ c
0850 + 3h 39 min – 2h
3
= 1029h A1
2012 Q22 P2 −4 3
∴ c=
20 Distance between towns K and S 3
. 374−30 M1 1 3 2 4
0
¿ 2 π × 6370 cos 2 × ∴ y= x −2 x +3 x−
360 A1 3 3
= 822.2121281 2 2014 Q13
= 823km

THREE DIMENSIONAL GEOMETRY

MARKING SCHEME
0
1. (a) 75.97 8M 1991Q23
(b) 121.5cm2
1990Q23 3. (a) (i) 2√ 3 or 3.464m 8M
(ii)600
2. (i)1.714m
(ii) 44.40 (b)300

143
 1992Q19 10 . a) OA = ½ √ 3 2+ 4 2
. V
M1
V
4. (a)41.410 8M
(b)119.05cm3
(c)15.410 M1
A1
1993Q22 A
O O
M1.5
5. (a) 4cm
2 3 A1
(b) 46 cm
3 ii) tan α = 6/12 = 4
(c ) 580 1994 α = 750 58 (75.960)
Q22 Identification of <√VMR
6. (a) Tan β -3 B1
2¼ = 1.333
(b)AC3 – 2(a)2 + (2a) – 8a2 Β = 530
AC = 2a√ 2 = 1/2 AC = a√ 2 θ = 750 58 - 530 7
= 220 51 or 22.830
Cos θ = a
√ 2 = √2 = 1.414 = 1999Q24
3a 3 3
0.4713
11 Volume = 1/3 x 12 x 9 x 6
. = 216cm3
θ = 610 530 (61.880)
2000Q11
1996Q13
12 a). AC=√(82+62)=√100=10 B1
. EC=√(102+202) M1
7.
=√s x100 ,
Labelled sketch of the pyramid
= 10√5 =23.36cm A1
B1
M1
(dimesnsions may be implied) M1
A1,
VN=102 + 32 = 109 A1
b). i).SinQ= 8 M1
=10.44cm 3m
√ 5= 20.960 M1
1997Q6 ,
ii). Tan x = 8 A1
20 8m
8 (a)BN2 = 102 – 52 = 75
BN = 8.65 x =21.80
EN2 = 52 + 123 = 169
2002Q18
EN = 13

8.66
(b)tan α = = 0.6662
13
α = 330 40(i) (33.670)

1998Q16
9. For tangent height of ∆ ABC
= x √3 M1

=Tan -1 x M1
X√ 3
= tan -1 1 A1
√3
0
= 30
1999Q14

144
13 a) M1 14
. .
a). 70m B
A1
A
1200
RM = √ 9 – 4.5 =
2 2
√ 60.75 M1

= 7.794

M1 80m A1
ii) Cos M =
202 + 7.7942 – 20.52
2 x 20 x 7.794 O
= 0.1299 A1
< M – 820 330 Ob2=702+802-2x70x80 Cos 1200
<R = 750 190 M1 =11300 +500
= 16900 M1
= 20 sin 820 330 A1
20 x 0.9915 A1
= 19.83 4m
M1
Volume
= 1/3 x ½ x 9 x 7.794 x 19.84 A1
= 231.8 or 232
8m
h) Mass
= 231.8 x 2.7
= 625.9g
Or = 626.1

2002Q20
OB = √ 16900 =130m

b).
O
B

Tan 200 = BT
130
BT=130 tan 200
= 130 x 0.3640
= 47.32m

2003Q15
15 (a) 10 cm
. (b) 56.250
(c) 61.870 2003
Q24
16 a). i). YM = √ 142 2−7 2 2
. = √ 147=12.12 B1
ii). YL = √147-102
= 6.856 A1

145
 b). Identifying angle θ 4.5cm
tan θ = 6.856 B1 Q
7
= (0.9804) M1 c). Cosine rule
θ= 440 241 62 =102+82-2 x 8 x 10 cos θ
36 = 100 +64 -160 Cosθ
c). tan x = 7/16 A1 36 = 164 -160 cos θ
=0.4375 Cos θ = 128/16
x=230 381 (23.630) Cos θ =0.8
= 36.91
2004Q24 2005Q23
18 Cos θ = 4/7 = 0.5714286 B1
17 . θ = 55.15009540 M1
. PH2 =√ 4.5 2+8 2 θ = 55.15 A1
A1 3m
= √ 20.25+ 64 2008Q14
19 B1
= 9.2 .
√ 2
a). AC = (15 √ 2) +(15 √ 2) =
2 B1
M1
30cm
A1 A1
FC = √ FH 2+ HC 2
b). Identification of A M1
= 9.22 + 62 = 10.97cm
F Tan θ=8/30 or equivalent
θ = 14.930 M1
A1 A1
c). Pyramid height = B1
M1
√(17 √2) −15
2 2
A1
= 18.79cm 10

d). Identification of α
B1 Tan α = 18.79

6 7.5√2 or equivalent
b). i).tanθ =
9.2
tan θ = 0.6522 A1 Α = 60.550
θ = 330 2009Q22
20 AC2 = 162 + 122 = 400
10.97cm
. AC = √ 400 = 20
AD = 10cm
M1 VO2 = 262 - 102
C VO = √ 576 = 24cm
9.2cm H A1
(b)the angle btn VA and ABCD IS
ii). Tan θ = 8/4.5 10 a
Al tan α = 24/10
Tan θ 1.7750 α = 67.380 (9)
θ = 60.600 (c)the angle btwn the planes is β
Tan β = 24/8
β = 71.570

2011Q22
8cm
M1
21

146
.
OC = √24 2 +102 = 17.63 M1
M1
2
A1
= 13
3 Area =
13

√{ }
¿ VCO = Cos -1 1 M1
26 ( 39+39+ 17.63 ) (30.185 ) ( 8.815A1
2
)
2
0
= 60
2012 Q16 P2 = 334.89 M1

22 (a) 3.46m A1
. 0 0 0 Total area = 334.89 × 5 + 534.97
(b) (i) 60 (ii) 43.9 (iii) 81.8
= 2209.42
2013 Q20
P2 (d) Volume of pyramid
23. 1
= ×534.97 × 36
(a) Base area B1 3
1
= ×15 ×15 sin 72 ×5 M1
2 =6419.63 cm2 ≈ 6420(4 s . f )
= 534.97
2014 Q20
(b) Length AV B1
= √ 362 +152 = 39

(c) Area of triangular faces

24. PR = √ 602 +112=61 B1
AB 15 10 M1
= = M1 Tan θ ¿ A1
sin 72 sin 54 61
3
θ = 9.31
15 sin72 2014 Q10
= AB=
sin 54

147