Download Higher Unit 3 Prelim 2010

MADRAS COLLEGE
Mathematics
Higher Mini-Prelim Examination 2009/2010
NATIONAL
QUALIFICATIONS
Assessing Unit 3 + revision from Units 1 & 2
Time allowed - 1 hour 10 minutes
Read carefully
1. Calculators may be used in this paper.
2. Full credit will be given only where the solution contains appropriate working.
3. Answers obtained from readings from scale drawings will not receive any credit.
Pegasys 2009
FORMULAE LIST
Circle:
The equation x 2  y 2  2 gx  2 fy  c  0 represents a circle centre ( g ,  f ) and radius
The equation (x  a )2  ( y  b)2  r 2 represents a circle centre ( a , b ) and radius r.
Trigonometric formulae:
Scalar Product:
sin  A  B 
cos  A  B 
sin 2 A
cos 2 A






sin Acos B  cos Asin B
cos Acos B  sin Asin B
2sin Acos A
cos 2 A  sin 2 A
2 cos 2 A  1
1  2 sin 2 A
.
a b  a b cosθ, where θ is the angle between a and b.
or
 a1 
 b1 
 
 
a b  a1b1  a2 b2  a3 b3 where a   a 2  and b   b 2 
a 
b 
 3
 3
.
Table of standard derivatives:
Table of standard integrals:
f ( x)
f ( x)
sin ax
cos ax
a cos ax
 a sin ax
sin ax
cos ax
Pegasys 2009
 f ( x) dx
f ( x)

1
cos ax  C
a
1
sin ax  C
a
g2  f 2  c .
SECTION A
In this section the correct answer to each question is given by one of the alternatives A, B, C or D.
Indicate the correct answer by writing A, B, C or D opposite the number of the question on your
answer paper.
Rough working may be done on the paper provided. 2 marks will be given for each correct answer.
1.
2.
3.
4.
The function f ( x)  2 sin x   cos x  has a minimum value of
A
2
B
0
C
3
D
 5
Which of the following is a correct assumption from the statement log b a  c ?
A
ac  b
B
ca  b
C
bc  a
D
cb  a
What is the value of
A
2
B
1
C
2
D
0


sin x dx ?
0
1 
 3
 
 
P and Q have position vectors  2  and   1 respectively.
0
 2
 
 
The length of PQ is
A
1
B
17
C
21
D
13
Pegasys 2009
5.
6.
Given that cos P  1 , where 0  p  2 , the value of cos 2P is
6
A
1
3
B
2
6
C
5
6
D
 23
An equation is such that log x  log( x  1)  log 6 , where x  0 .
The value of x is
7.
A
2
B
1
C
3
D
6
The gradient of the tangent to the curve y  sin x  at the point where x  60  is
3
2
A
8.
B
1
2
C
 12
D
0
Vectors a and b are such that a  b  2 with P being the angle between the vectors.
If a . b  0 8 , the value of cos P is
A
3 2
B
0 4
C
0 2
D
0 05
a
P
b
[ END OF SECTION A ]
Pegasys 2009
SECTION B
ALL questions should be attempted
9.
A function is defined on a suitable domain as f ( x) 
(a)
Show clearly that the derivative of this function can be written in the form
f ( x) 
(b)
10.
 16
.
(2 x  1) 2
k
(2 x  1) n
and write down the values of k and n.
4
Hence find x when f ( x)  1 and x  0 .
3
In the diagram below A, B and C have coordinates (4 , 0 ,13) , (6,  3 , 4) and
(6 ,1,12) respectively.
P lies on BC and has coordinates (3 , 0 , k )
C (6 ,1,12)
P (3 , 0 , k )
B (6,  3 , 4)
A (4 , 0 ,13)
11.
(a)
Find the value of k.
3
(b)
Hence calculate the size of angle APB.
5
A formulae for mass decay is given as M t  M 0 e 002t , where t is time elapsed in hours,
M 0 is the initial mass in grams and M t is the mass remaining after t hours.
How long will it take for an initial mass of 40 grams to decay down to 28 grams?
Give your answer correct to the nearest minute.
Pegasys 2009
5
dy
 4 x  1 , find an expression for y in terms of x given that y  9  5 when x  2 .
dx
12.
If
13.
Part of the graph of y  2 cos x   2 sin x  is shown below.
5
y
1 7
C
B
O
A
x
y  2 cos x   2 sin x 
(a)
Express y  2 cos x   2 sin x  in the form y  k cos( x  a)  , where k  0 .
3
(b)
Hence state the coordinates of A and B rounding the coordinates to 3 significant
figures where necessary.
2
(c)
By solving the equation
2 cos x   2 sin x   1 7 , find the coordinates of point C.
[ END OF SECTION B ]
[ END OF QUESTION PAPER ]
Pegasys 2009
4
Higher Mini Prelim - Unit 3
Section A - Answers
2009/2010 (Answers + Marking Scheme)
1
5
D
D
2
6
C
A
3
7
C
B
4
8
B
C
2 marks each (16 marks)
Section B - Marking Scheme
Give 1 mark for each
9(a)
(b)
ans: k = 64; n = 3
(4 marks)
●1
●2
●3
prepares to differentiate
starts to differentiate
completes differentiation
●1
●2
●3
●4
simplifies and states values of k and n
●4
ans:
x
5
2
f ( x)  16(2 x  1) 2
f ' ( x)  32(2 x  1) 3 .....
.....×2
64
f ' ( x) 
; k  64; n  3
(2 x  1) 3
(3 marks)
●1
equates derivative to 1
●1
64
1
(2 x  1) 3
●2
starts to simplify
●2
(2 x  1) 3  64
●3
completes simplification
●3
(2 x  1)  4; x 
10(a) ans:
(b)
Illustration(s) for awarding each mark
k = 10
(3 marks)
●1
finds change in x – coords/y – coords
●2
●3
realises CP = ¼ CB
establishes z – coordinate of P
ans:
●1
139∙5o
5
2
●1 CB – 6 to 6 = 12; CP – 6 to – 3 = 3 [x]
CB 1 to – 3 = 4; CP 1 to 0 = 1
2
● evidence
●3 12 – 4 = 8 so change in z – coord is 2;
k = 10
(5 marks)
evidence of cos  
a.b
a b
know how to find angles
●1
finds PA and PB
●
●3
finds scalar product
●3
●4
finds magnitudes of vectors
●5
substitutes in formula and finds angle
●4 |PA| = √10; |PB| = √126
 27
●5 cos  
;   139  5
10 126
●
2
Pegasys 2009
2
  1
 9 
 
 
PA   0  ; PB    3 
 3
  6
 
 
PA . PB = – 9 + 0 – 18 = – 27
Give 1 mark for each
11
12
ans:
17 hours 50 minutes
Illustration(s) for awarding each mark
(5 marks)
28  40e 002t ; e 002t  0  7
log e e 002t  log e 0  7
●1
●2
substitutes values in formula and simplifies
takes natural logs of both sides
●1
●2
●3
releases power and removes log e e
●3  0  02t log e e  log e 0  7; 0  02t  log e 0  7
●4
evaluates for t
●4
●5
changes hours to hours and minutes
●5
log e 0  7
 17  8337....
 0  02
17 hours 50 minutes
knows to integrate and prepares
●1
y   (4 x  1) 2 dx
starts to integrate
●
●3
completes integration and adds C
●3
●4
knows to substitute to find C
●4
●5
evaluates for C
●5
ans:
●1
3
1
y  (4 x  1) 2  5
6
t
(5 marks)
1
3
●
13(a)
2
●1
●2
●3
(b)
ans:
●1
●2
(c)
y  2cos(x  45)
ans:
●1
●2
●3
●4
Pegasys 2009
........
3
1
1
 C ; y  (4 x  1) 2  C
4
6
3
1
9  5  (4(2)  1) 2  C ;
6
27
C  95 
5
6
....... 
(2 marks)
●1
●2
2cos(x – 45)o = 0; x = 135o
y = 2 cos (– 45)o; y = 1·4
(4 marks)
replaces LHS
finds value for cos(x – 45)o
finds values of x
chooses appropriate value and states C
Sect. B
3
2
●1 k cos x cos  k sin x sin 
●2 k = 2
●3 tan   1;   45 Quadrant I
makes y = 0 and finds x
makes x = 0 and finds y
ans: C(77o, 1·7)
y
(4 x  1) 2
(3 marks)
uses correct expansion
finds k
finds 𝛼
A(135o, 0); B(0, 1∙4)
2
(34 marks)
●1 2 cos( x  45)  1  7
●2 cos( x  45)  0  85
●3 (x  45)  32 or 328 ; x = 77o or 373o[13o]
●4 C(77o, 1·7)
16 + 34
Total: 50 marks