Transcript Higher Paper 2 Powerpoint Version (Questions

Harris Academy
Supported Study
Session 1
Paper 2 Questions and Answers
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Question 1
(Unit 1 LO1 Straight Line)
Triangle ABC has as its vertices A(-18,6) , B(2,4) and C(10,-8)
.
(a) Find the equation of the median
from A to BC
A
B
(b) Find the equation of the
perpendicular bisector of side AC.
(c) Find the coordinates of T, the
point of intersection of these lines.
marks 3, 4, 3
C
Solution 1(a)
(a) ans:

2
3


1
y  x
3
mid-point of BC
1
M
2 10 4  ( 8 )
2 ,
2
  (6,2)
62
8
1



 18  6  24 3
gradient of median
2
equation of line
1
3 y  6   ( x 18)
3
m AM
Solution 1(b)
(b) ans:
y  2 x 7
1 N 

68
14
1



 18  10  28
2
1810 6  ( 8 )
2
2
,

mid-point of AC

gradient of AC
2

perpendicular gradient
3 m  2

equation of line
4 y 1  2( x  4)
m AC

 ( 4,1)
Solution 1(c)
ans:
T(-3,1)
solving a system of
equations
1
1
 x  2x  7
3

x-coordinate
2
x  3

y-coordinate
3 y  2(3)  7
y 1

Question 2
(Unit 1 LO3 Differentiation)
The diagram below shows the parabola with equation
y 8x  3x
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.
marks (4)
Solution 2
ans: 63




Know to differentiate
Find gradient of
tangent at x = 1
dy
8 6 x
1
dx
2 mtangent  8  6(1)  2
Use m = tanθ

tan

2
3
Complete calculations
4
  tan

  63
1
2
Question 3
(Unit 2 LO4 Circle )
The circle in the diagram has equation
x  y  4x  8 y  5  0
2
2
The line AB is a chord of the circle
and has equation
.
y
O
B
x
x 7 y
.
(a) Show that the coordinates of
A and B are (-1, -8) and (6, -1)
respectively.
A
(b) Establish the equation of the circle which
has AB as its diameter.
marks (4,3)
Solution to question 3a
ans:



A(-1, -8) and B(6, -1)
substituting into circle
equation
 1 7  y 2  y 2  4(7  y)  8 y  5  0
multiplying brackets and
tidying up
2
2
y
 18 y  16  0
2
factorising and values of y
3
2 y 1 y 80
 y  1or 8

corresponding values of x  4
A(-1, -8) and B(6, -1)
Solution to question 3b
ans:



x  2  5
2
 y  4  5  24  5
2
knowing to find midpoint  1
of AB
finding radius
substituting into
equation
2
3
(25, -45)
r  (6  2  5) 2  (1  4  5) 2  24  5
x  2  5
2
 y  4  5  24  5
2
Question 4
(Unit 1 LO3 Differentiation)
The graph of the cubic function y = f (x) is shown in
the diagram. There are turning points at (1,1) and (3,5).
(3,5)
y
y  f (x )
(1,1)
x
Sketch the graph of y = f '(x)
marks (3)
Solution 4
ans: sketch
Interpret stationary
points
1

Parabola
3

Maximum TP

2
1
3
Harris Academy
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Paper 2 Questions and Answers
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Question 5 (Unit 2 LO3 Trigonometry)
Solve algebraically the equation
sin x  3cos 2 x  2  0

where
o
0  x  360
marks 5
Solution 5
ans:


1950, 16050 , 2100, 3300
double angle formula
re-arrange to zero and
factorise
 1 sin x   31  2 sin 2 x    2  0
2
6sin 2 x   sin x  10
(3sin x   1)( 2sin x   1)0

find roots
3

answers from sin x   13


19
.
5
,
160
.
5
4


answers from sin x   12
5 210 , 330
sin x   13
or sin x    12
Question 6
(Unit 1 LO3 Differentiation)
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
x
x
(a) If the length of the base is x centimetres,
show that the volume V (x) is given by
1 3
V ( x )  300 x  x
4
(b) Find the value of x that maximises the volume
of the box.
marks (3,5)
Solution 6 (a)
ans: proof

Equation for surface area
 1 1200  x 2  4 xh

Rearrange with h = …….
2
1200

x
2
h
4x

Find V
2


1200

x
3 V  x 2 

4x


1 3
 300 x  x
4
Solution 6 (b)
ans:
x  20
knowing to differentiate
 1 V ( x )  ............
differentiate
2
V ( x ) 300 

set derivative to zero
3
300 

solve for x



nature table
4
3 2
x 0
4
x  20
r
5
3 2
x
4
V ( x )

20 

0

shape
max TP at x = 20
Question 7
(Unit 1 LO3 Differentiation
Unit 2 LO1 Polynomials)
Part of the curve y  x 3 10 x 2  24 x is shown in the
diagram
Also shown is the tangent to the curve at the point P
where x 1
(a) Find the equation of
the tangent.
y
Q
y  x 3 10 x 2 24 x
P
O
x=1
x
(b) The tangent meets the curve
again at Q.
Find the coordinates of Q.
marks (4,4)
Solution 7(a)
(a) ans:
y 7 x  8

differentiate
1
dy
3 x 2  20 x  24
dx

gradient
2
at x  1 , m  3(1) 2  20(1)  24
2
3
m  3  20 24  7

y-coordinate
3
y  (1)3 10(1)2  24(1) 15

equation
4
y 15  7( x 1)
Solution 7(b)
(a) ans:
Q (8,64)

form equation

rearrange to zero
2
3

factorise
 1 x 3 10 x 2  24 x  7 x  8
2
3
x 3 10 x 2 17 x  8  0
1
1
1
-10
17
-8
1
-9
8
8
0
-9
( x  1)( x 2  9 x  8)
( x  1)( x 1)( x  8)

coordinates of Q
4
x  8 y  7(8)  8  64
Question 8
(Unit 2 LO2 Integration )
The diagram shows the parabolas
y  x 2 x  4
2
y  4 4 x  x 2
and
y  x 2 2 x  4
y
A
4
y  4 4 x  x 2
O
x
(a) Find the coordinates of the point A
(b) Calculate the area enclosed between the two curves.
marks (4, 4)
Solution 8a
(a) ans:
A(3,7)
x 2  2 x  4  4 4 x  x 2

Form equation
1

Rearrange to = 0
 2 2 x 2 6 x 0

Factorise and solve
3 2 x ( x 3) 0
x  0 or x  3

Coordinates of A
4
A (3,7)
Solution 8b
y  x 2 2 x  4
y
A
4
(b) ans: 9 square units
y  4 4 x  x 2
x
O

1
 (2 x
Integrate
2

2x
3

Substitute limits
3

2
3

Answer
4
 23 (27)  27
 (top  bottom)dx

3
0
3
2
6 x)dx
3x
2

3
0

(3)3  3(3) 2  0
 9 square units
Harris Academy
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Session 3
Paper 2 Questions and Answers
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Question 9
(Unit 2 LO4 Circle )
A circle, centre C, has equation
x  y 4 x 20 y  84  0
2
Show that the line with equation
2y = x + 8 is a tangent to the circle
and find the coordinates of the point
of contact P
.
x 2  y 2 4 x 20 y  84  0
2
C
2 y  x 8
P
marks (5)
Solution 9
ans:


P(4, 6)
substitute into circle
equation
 1 2 y 82  y 2 4(2 y 8)20 y 840
multiply out brackets and
simplify
2
5
y
60 y 1800
2
factorise and solve for y
3 5 y6 y6 0  y6

complete proof
4

point of contact
5 y  6 x  2(6)  8  4

One point of intersection so
line is a tangent
Question 10
(Unit 2 LO2 Integration)
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)
(-2,0)
O
x
P
Q
(a) Write down the coordinates of P and Q
(b) Find the total shaded area
marks (2,6)
Solution 10a
ans:
P (1,0) Q (2,0)

Coordinates of P
 1 (1,0)

Coordinates of Q
2 ( 2,0)
Solution 10b
2
1
2
units
ans: 2
1

( x 2)( x 1)( x 2)dx
0
two integrals
1 
multiply out brackets
2
y  x3  x 2  4 x  4

integrate
3
1
4

integral from 0 to 1
11
4 112

integral from 1 to 2
7

5 12

total area
6


2
1
( x  2)( x 1)( x 2)dx
x 4  13 x 3  2 x 2  4 x
11
112
 127  2 12 units2
Question 11
(Unit 2 LO3 Trigonometry)
The diagram shows a sketch of part of the graph of
a trigonometric function whose equation is of the form
y  a sin bx  c
Find the values of a, b and c
y
5
y  a sin bx  c
0
π
x
-3
marks (3)
Solution 11
ans: a  4, b  2, c 1

Interpret amplitude
1
a 4

Interpret period
2
b 2
Interpret vertical
displacement
3
c 1

Question 12
(Unit 1 LO1 Straight Line)
.(a) The diagram shows line OA with equations x 2 y  0

.The angle between OA and the x-axis is a
y
B
Find the value of a.
A
O
30 
a
x
(b) The second diagram shows lines OA and OB.
The
angle between these two lines is 300
.
Calculate the gradient of line OB correct to 1 decimal place
marks 3,1
Solution 12a
ans: 26.6



gradient of line
 1 gradient  12
gradient = tan (angle)
and apply
2 tan a  gradient
process
0
3
1
tan    26.6 
2
1
Solution 12b
ans: 1.5

angle = tan-1(angle)
 1 m  tan( 3026.6)  1.5