Session 2
1.
Paper 2 Questions
Triangle ABC has as its vertices A(-18,6) , B(2,4) and C(10,-8)
A(-18,6)
B(2,4)
T
C(10,-8)
(a) Find the equation of the median from A to BC
(b) Find the equation of the perpendicular bisector of side AC
(c) Find the coordinates of T, the point of intersection of these lines.
(a) ans:
1
2
3
(b) ans:
1
2
3
4
y   13 x (or equiv.)
2.
(a)
For mid-point of BC
For gradient of altitude
For equation of line
y  2 x  7 (or equiv.)
4 marks
For mid-point of AC
For gradient of AC
For perpendicular gradient
For equation of line
(c) ans: T(-3,1)
1
2
3
3 marks
(b)
3
4
3
 1 M (6,2)
 2 mAM =  13
3
y - 6   13 ( x 18)
1
N(-4,-1)
2
m AC 
68
14
1


1810 28
2
 3 m  2
4 y 1 2( x  4)
3 marks
Solving system of equations
For x-coordinate
For y-coordinate
(c)
1
2
3
 13 x  2 x 7 (or equiv.)
x = -3
y  2(3)7  y 1
The diagram below shows the parabola with equation y  8 x  3 x 2
and a line which is a tangent to the curve at the point T(1,5).
Find the size of the angle marked θ, to the nearest degree.
4
Question 2 continued
ans: 63
4 marks
●1
know to differentiate.
●1
●2
find gradient of tangent
●2
dy
 8 3 x
dx
m  8  6(1)  2
●3
uses m = tanθ
●3
tan   2
●4
complete calculations
●4
  63
3.
The circle in the diagram has equation
y
x 2  y 2  4x  8 y  5  0 .
O
B
The line AB is a chord of the circle and
has equation x  7  y .
A
(a)
(b)
Show that the coordinates of A and B are (-1, -8) and (6, -1)
respectively.
4
Establish the equation of the circle which has AB as its diameter.
3
x
Question 3 continued
(a)
Ans: proof
marks
●
1
4
7  y 2  y 2  4(7  y)  8 y  5  0
●1
substituting into circle eq.
●2
2 y 2  18 y  16  0
factorising and values of ‘y’
●3
2 y  1 y  8  0  y  1 or  8
●4
corresponding values of ‘x’
●4
A(-1, -8) and B(6, -1)
(b)
Ans: x  2  5  y  4  5  24  5
3 marks
●1
(25, -45)
●2
multiplying brackets and tidying up
●3
2
2
●1
knowing to find midpoint of AB
●2
finding radius
●3
substituting into equation
4.
The graph of the cubic function y = f (x) is shown in the diagram.
There are turning points at (1,1) and (3,5). Sketch the graph of y = f '(x)
y
●2 r  (6  2  5) 2  (1  4  5) 2  24  5
●3
x  2  52  y  4  52  24  5
(3,5)
y  f ( x)
(1,1)
x
3
ans: a  4, b  2, c 1
3 marks
●1
●2
●3
interpret stationary
points.
parabola
maximum turning
point
●1
●2
●3
1
3
5.
Solve algebraically the equation
sin x o  3cos2 x o  2  0,
ans: 195o, 1605o, 210o, 330o
where 0  x  360 .
5 marks
 1 sin x 0  3(12sin 2 x  )  2  0
 1 Substitute double angle formula
 2 Re-arrange to zero and factorise
 3 Find roots
 4 Answers from one root
 5 Answers from other root
2
6sin 2 x  sin x 1 0
(3sin x  1)( 2sin x  1)  0
 3 sin x   13 or sin x    12
 4 19.5o , 1605o
 5 210o , 330o
5
6.
An open box is designed in the shape of a
cuboid with a square base.
h
The total surface area of the base and
four sides is 1200cm2
(a)
x
x
If the length of the base is x centimetres,
show that the volume V(x) is given by
1
V ( x )  300 x  x 3
4
(b)
3
Find the value of x that maximises the volume
of the box.
(a) ans: Proof
3 marks
Equation for surface area
1
2
Rearrange with h = …….
2
3
Find V
3
1
5
(b) ans: x = 20
1200  x 2  4 xh
1200  x 2
4x
 1200 x 2 
1
 = 300 x  x 3
V  x 2 
4
 4x 
h
5 marks
●1
knowing to differentiate
●1
●2
differentiate
●2
●3
set derivative to zero
●3
●4
solve for x
●4
●5
nature table
●5
V ( x )  ...........
3
V ( x )  300  x 2
4
3
300  x 2  0
4
x = 20

V / ( x) 
x
20
shape

0
max TP at x = 20


7.
Part of the curve y  x3  10 x 2  24 x
y
Q
is shown in the diagram.
Also shown is the tangent to the curve at
the point P where x  1 .
y = x3  10x2 + 24x
P
(a) Find the equation of the tangent.
x
O
4
(b) The tangent meets the curve again at Q. Find the coordinates of Q.
Question 7 continued
(a) ans: y  7 x  8
1
2
3
4
equate
equate to zero
factorise
coordinates of Q
(a)
dy
 3 x 2 20 x  24
dx
 2 at x 1, m  7
1
differentiate
calculate gradient
y-coordinate
equation of tangent
(b) ans: P(8,64)
1
2
3
4
4 marks
3
4
4 marks
y  (1) 3 10(1) 2 24(1) = 15
y 15  7( x 1)
(b)
 1 x 3 10 x 2 24 x  7 x  8
 2 x 3 10 x 2 17 x  8 0
 3 ( x 1) 2 ( x  8)  0
 4 Q (8,64)
4
8. The diagram shows the parabolas y  44 x x 2 and
y  x 2 2 x 4
y  x 2 2 x  4
y
A
4
y  4 4 x  x 2
x
O
(a) Calculate the coordinates of the point A
4
(b) Find the area enclosed between the two curves.
4
(a)ans : A(3,7)
4 marks
●1
Equate
●1
x 2  2x  4  4 4x  x 2
●2
Rearrange to = 0
●2
2x2 6x 0
●3
factorising and solving
●3
2 x( x  3)  0, x  0, x  3
●4
coordinates of A
●4
A(3,7)
●1
 (2 x
(b)ans: 9 units2
●1
●
2
●3
●
4
top bottomdx
4 marks
●
2
integrate
●3
substitute limits
●4
answer
3

0
2
 6 x)dx
2
2 3
x

3
x
3

3
0
( 23 (3) 3  27)  0
9 square units
9.
x 2  y 2 4 x 20 y  84  0
A circle, centre C, has equation
x 2  y 2  4 x  20 y  84  0 .
C
Show that the line with equation
2y = x+ 8 is a tangent to the circle.
Find the coordinates of the point of contact P
2 y  x 8
P
5
Ans: P(4,6)
5 marks
2 y  82  y 2  4(2 y  8)  20 y  84  0
●1
substitute into circle eq.
●1
●2
multiply out brackets and simplify
●2
5 y 2 60 y  180  0
●3
factorise and solve for y
●
4
complete proof
●3
●4
●5
point of contact
●5
5 y6 y60 y 6
One point of intersection so line
is a tangent
P(4,6)
10.
The diagram shows a sketch of the graph of y = (x + 2)(x – 1)(x – 2)
and the points P and Q.
y
y  ( x  2)( x 1)( x  2)
(0,4)
x
(-2,0)
O
2
(b) Find the total shaded area
6
2marks
2 12 units 2
two integrals
(a)
1
2
coordinates of P
coordinates of Q
(b) ans:
1
Q
(a) Write down the coordinates of P and Q
(a) ans: P (1,0,) Q(,2,0)
1
2
P
6 marks
P(1,0,)
Q(2,0)
(b)
1


1
( x  2)( x  1)( x  2)dx
0
2
1
2
multiply out brackets
2
3
integrate
3
4
integral from 0 to 1
11
 4 112
5
integral from 1 to 2
5
6
integral from 1 to 2
6
( x  2)( x  1)( x  2)dx
y  x3  x2 4x  4
2
1 4
1 3
4 x  3 x 2x  4x
 127
11
112
 127  2 12 units 2
11.
The diagram shows a sketch of part of the graph of a trigonometric
function whose equation is of the form y  asin bx  c
y
5
y  a sin bx  c
0
π
x
Find the values of a, b and c
ans: a  4, b  2, c 1
●1
●2
●3
3
3 marks
interpret amplitude.
interpret period
interpret vertical displacement
●1
●2
●3
a4
b2
c 1
12.(a) The diagram shows line OA with equation x 2 y  0 .
The angle between OA and the x-axis is a  .
Find the value of a.
(b) The second diagram shows lines OA and OB. The angle between these
y
two lines is 30  .
B
Calculate the gradient of line OB correct to 1 decimal place.
A
O
30 
a
x
(a) ans: 26.6 
●1
gradient of line.
●
2
●
3
3 marks
gradient = tan(angle) and apply
●1
●2
process
(b) ans: 1.5
●1
angle = tan-1(angle)
●3
1 mark
●1
1
2
0
tan a = gradient
1
tan 1    26.6 
2
m  tan( 3026.6)  1.5
gradient =