RBHS
Grade 12
SC
A P Mathematics
Algebra and Calculus
17 September 2015
2 hours
200 marks
Instructions:
 Answer all questions.
 Start each question on a new page (side).
 All necessary working must be shown in its proper place with the answer.
 A calculator may be used unless specified otherwise. In fact, the use of the silver
calculator is encouraged for all tedious calculations.
 Give answers to two decimal places, where applicable.
 Blue or black pen must be used in answers although pencil may be used on diagrams.
 The use of correcting fluid is not allowed.
 This examination paper consists of 13 pages including a diagram sheet and a formula
sheet.
___________________________________________________________________________
Timing Guideline
1
After 2 an hour
After 1 hour
1
After 1 2 hours
After 2 hours
Starting 3.2
End of question 5
On 9.1
Finished
Page 1 of 13
MODULE 1
CALCULUS AND ALGEBRA
Question 1
Given: 1.2 + 2.5 + 3.8 + ⋯ + 𝑛(3𝑛 − 1) = 𝑛3 + 𝑛2
1.1
Prove the above statement true for all values of 𝑛 with 𝑛 ≥ 1.
1.2
Hence find the value of 2 + 10 + 24 + ⋯ + 660
(12)
(6)
[18]
Question 2
2.1
Given: 𝑓(𝑥) = 𝑥 4 − 𝑥 3 − 9𝑥 2 + 29𝑥 − 60
If 𝑥 = 1 + 2𝑖 is a root of the equation 𝑓(𝑥) = 0, use algebra to find the other three
roots of the equation 𝑓(𝑥) = 0.
(7)
2.2
2.3
Solve for 𝑥:
2 log 4 (2𝑥 + 3) − log 4 𝑥 − log 4 (2𝑥 − 1) = 1
(9)
Given 𝑓(𝑥) = ln(2𝑥 − 3) and 𝑔(𝑥) = 𝑒 2𝑥 − 4
2.3.1
Find 𝑔(𝑓(𝑥)), giving your answer in the form (𝑎𝑥 − 𝑏)2 − 𝑐 where 𝑎, 𝑏
and 𝑐 are integers.
(4)
2.3.2
Write down an expression for 𝑓(𝑔(𝑥)) and hence find the solution of the
equation 𝑓(𝑔(𝑥)) = ln 5, correct to 2 decimal places.
(5)
[25]
Page 2 of 13
Question 3
3.1
The diagram below shows the graphs of 𝑓(𝑥) = |2𝑥 − 3| and 𝑔(𝑥) = |𝑥|
3.1.1
Find the 𝑥-co-ordinates of the points of intersection of 𝑓 and 𝑔.
(5)
3.1.2
Hence, or otherwise, solve the inequality
𝑓(𝑥) ≥ 𝑔(𝑥)
(4)
3.2
2
3.2.1
Express (𝑥+1)(𝑥+3) in partial fractions.
3.2.2
Using your answer in 3.2.1, show that
2
2
1
1
(5)
1
1
((𝑥+1)(𝑥+3)) = (𝑥+1)2 − 𝑥+1 + 𝑥+3 + (𝑥+3)2
Page 3 of 13
(5)
3.3
The diagram shows 𝑓(𝑥) = {
3.3.1
3.3.2
11 − 𝑥 2
5−𝑥
𝑖𝑓 0 ≤ 𝑥 ≤ 3
𝑖𝑓 3 < 𝑥 ≤ 8
On the diagram sheet, sketch the graph of 𝑓 −1 (𝑥) showing clearly all
important information.
Write the equation of 𝑓 −1 (𝑥) in the form
. . . . . . . . 𝑖𝑓 . . . . . . . .
𝑓 −1 (𝑥) = {
. . . . . . . . 𝑖𝑓 . . . . . . . .
(5)
(5)
[29]
Question 4
Given 𝑔(𝑥) = {
2𝑥 + 1
1
3
1
2
𝑖𝑓 𝑥 ≤ 𝑝
7
−4𝑥 + 2𝑥 + 4𝑥 + 1
𝑖𝑓 𝑥 > 𝑝
4.1
For which value(s) of 𝑝 is 𝑔(𝑥) continuous at 𝑝?
(7)
4.2
For which value(s) of 𝑝 is 𝑔(𝑥) differentiable at 𝑝?
(7)
[14]
Page 4 of 13
Question 5
The diagram shows part of the curve
𝑦 =2−
18
2𝑥 + 3
which crosses the 𝑥-axis at A and the 𝑦-axis at B. The normal (a line perpendicular to a
tangent) to the curve at A crosses the 𝑦-axis at C.
5.1
Find the co-ordinates of A.
(2)
5.2
Find 𝑑𝑥
(4)
5.3
Hence show that the equation of the line AC is 9𝑥 + 4𝑦 = 27
(6)
5.4
Find the length of BC.
(3)
[15]
𝑑𝑦
Question 6
1
6.1
Differentiate 𝑦 = cos 𝑥 by using the quotient rule, hence showing that
𝑑
𝑑𝑥
(sec 𝑥) = sec 𝑥 . tan 𝑥
(4)
1
6.2
Show that sec 𝑥−tan 𝑥 = sec 𝑥 + tan 𝑥
6.3
Deduce that (sec 𝑥−tan 𝑥)2 = 2 sec 2 𝑥 − 1 + 2 sec 𝑥 tan 𝑥
6.4
Hence find the value of
1
𝜋
4
∫0
(2)
(3)
1
(sec 𝑥−tan 𝑥)2
You may only use a calculator for the calculations at the end.
Page 5 of 13
(8)
[17]
Question 7
C
D
B
10 cm
12 cm
O
A
E
F
The diagram shows a metal plate ABCDEF which has been made by removing the two
shaded regions from a circle of radius 10 cm and centre O. The parallel edges AB and ED are
both of length 12 cm.
7.1
Show that angle DOE is 1,287 radians, correct to 3 decimal places.
(2)
7.2
Find the perimeter of the metal plate.
(4)
7.3
Find the area of the metal plate.
(6)
[12]
Page 6 of 13
Question 8
The diagram shows the curve with equation 𝑦 = 𝑥(𝑥 − 2)2 . The minimum point on the curve
has co-ordinates (𝑎; 0) and the 𝑥-co-ordinate of the maximum point is 𝑏, where 𝑎 and 𝑏 are
constants.
8.1
State the value of 𝑎.
(2)
8.2
Find the value of 𝑏.
(5)
8.3
Find the area of the shaded region.
(4)
8.4
The gradient, 𝑑𝑥 , of the curve has a minimum value 𝑚. Find the value of 𝑚.
𝑑𝑦
Page 7 of 13
(5)
[16]
Question 9
9.1
A pupil is asked to find the maximum distance between 𝑓(𝑥) = cos 2𝑥 + 1 and
𝑔(𝑥) = 𝑥 2 − 2𝑥 − 3 in the interval −1 ≤ 𝑥 ≤ 3
Quite correctly, he started off as follows:
𝑙 = (cos 2𝑥 + 1) − (𝑥 2 − 2𝑥 − 3 )
∴ 𝑙 = cos 2𝑥 − 𝑥 2 + 2𝑥 + 4
𝑑𝑙
∴ 𝑑𝑥 = −2 sin 2𝑥 − 2𝑥 + 2 = 0
∴ sin 2𝑥 = −𝑥 + 1
He then found that he couldn’t solve that equation.
Use Newton’s method, with a starting value of 𝑥0 = 0,5 to find the value of 𝑥 that
gives the maximum distance between 𝑓 and 𝑔. Give your answer correct to five
decimal places.
(7)
Page 8 of 13
9.2
A solid building is in the shape of an ellipse with equation 𝑥 2 + 4𝑦 2 = 5.
9.2.1
Find, using implicit differentiation, the equation of the tangent to the ellipse
at the point (−1; 1)
(8)
9.2.2
A person stands on this tangent at the point (3; 2). Can he see his friend
who is standing at the point (−8; −1)? Explain your answer carefully,
using appropriate calculations.
(4)
[19]
Question 10
Determine:
10.1
∫ sin 3𝑥 sin 2𝑥 𝑑𝑥
(5)
10.2
∫ 𝑥 sin 3𝑥 𝑑𝑥
(8)
10.3
Find the value of 𝑝 if
𝑝 2
∫0 √𝑥+4 𝑑𝑥 = 4
You may not simply give an answer, i.e. trial and error on a calculator is not
permissible.
Page 9 of 13
(8)
[21]
Question 11
The diagram shows part of the curve
𝑎
𝑥
where 𝑎 is a positive constant. Given that the volume obtained when the shaded region is
rotated about the 𝑥-axis is 24𝜋, find the value of 𝑎.
𝑓(𝑥) =
[8]
Question 12
Consider the following integral:
1
1
𝑥
∫ 𝑎2 +𝑥 2 𝑑𝑥 = 𝑎 tan−1 (𝑎) + 𝑐
1
Find ∫ 9+4𝑥 2 𝑑𝑥
[6]
TOTAL MARKS: 200
Page 10 of 13
INFORMATION SHEET
General Formulae
– b ± b 2 – 4ac
x=
2a
n(n  1) n 2 n
i
 

2
2 2
i 1
n
n
1 n
i 1
n
i 2 
i 1
x0
x0
 x if
x 
 x if
nn  12n  1 n 3 n 2 n



6
3
2 6
n 2 n  1
n 4 n3 n 2
i 




4
4
2
4
i 1
2
n
3
z  a  bi
z*  a  bi
n A  n B  n  AB
 A
n A  n B  n  
B
n An  n n A
log a x 
log b x
log b a
Calculus
f '( x)  lim
h 0
n
f ( x  h) – f ( x )
h
 f ' g ( x).g ' ( x) dx 
b
 x n1 
x
dx




 n  1 a
a
b
ba n
Area  lim 
  f  xi 
n  n 
i 1
dy dy dt


dx dt dx
f ( g ( x))  c
 f ( x).g ' ( x)dx  f ( x).g ( x)   g ( x). f ' ( x) dx  c
xr 1  xr 
f ( xr )
f ' ( xr )
b
V    y 2 dx
a
Page 11 of 13
Function
xn
Derivative
nx n 1
sin x
cos x
cos x
 sin x
sec 2 x
 cosec 2 x
tan x
cot x
sec x
sec x. tan x
 cosec x. cot x
cosec x
f ( g ( x))
f ( x). g ( x)
f ( x)
g ( x)
f ' ( g ( x)). g ' ( x)
g ( x). f ' ( x)  f ( x).g ' ( x)
g ( x). f ' ( x)  f ( x). g ' ( x)
g ( x)2
Trigonometry
A
1 2
r 
2
In ABC:
s  r
a
b
c
=
=
sin A sin B sin C
a 2  b 2  c 2 – 2bc. cos A
Area 
sin 2 A  cos 2 A  1
1
ab.sin C
2
1  tan 2 A  sec 2 A
1  cot 2 A  cosec 2 A
sin  A  B  sin A. cos B  cos Asin B
cos A  B  cos A cos B  sin Asin B
sin 2 A  2 sin A cos A
cos 2 A  cos 2 A  sin 2 A
1
sin( A  B)  sin( A  B)
2
1
sin A. sin B  cos( A  B)  cos( A  B )
2
1
cos A. cos B  cos( A  B)  cos( A  B)
2
sin A. cos B 
Matrix Transformations
 cos   sin  


 sin  cos  
 cos 2

 sin 2
sin 2 

 cos 2 
Page 12 of 13
NAME: __________________________________________________________________
3.3
Page 13 of 13