Download Higher Unit 3 Prelim 2011

Madras
College
Mathematics
Higher March Prelim Examination 2010/2011
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 2010
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:
sin  A  B   sin Acos B  cos Asin B
cos  A  B   cos Acos B  sin Asin B
sin 2 A  2sin Acos A
cos2 A  cos 2 A  sin 2 A
 2 cos 2 A 1
 1  2 sin 2 A
Scalar Product:
.
a b  a b cosθ, where θ is the angle between a and b.
or
.
a b  a1 b1  a 2 b2  a 3 b3
Table of standard derivatives:
Table of standard integrals:
f ( x)
f ( x)
sin ax
a cos ax
cos ax
 a sin ax
 f ( x) dx
f ( x)
sin ax
cos ax
Pegasys 2010
 a1 
 b1 
 
 
where a   a 2  and b   b 2 
a 
b 
 3
 3

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.
If f ( x)  (2 x  1) 4 then f (1) equals
A
4
B
1
C
2
D
8
The maximum value of the function g ( x)  3 sin x  2 cos x is
A
13
B
5
C
0
D
2
The radius of the circle with equation x 2  y 2  4 x  2 y  4 is
A
2
B
3
C
1
D
24
When 5 sin x   3 cos x  is written in the form k sin  x    where 0    360 , tan  is equal to

5
3
A
B

5
3
C

3
5
D
Pegasys 2010
3
5
5.
6.
7.
The value of log
A
2
B
4 2
C
1
4
D
4
2
4 is
1 
 p 
 
 
Given that the vectors  4  and   2  are perpendicular, the value of p is
0
 3 
 
 
A
0
B
8
C
4
D
6
Part of the graph of y  log 10 x is shown in each diagram below as a broken line.
Which diagram also shows, as a full line, part of the graph of y  log 10 1x ?
A
y
B
y
o
o
x
C
y
o
Pegasys 2010
D
x
x
y
o
x
8.
 12 
 
a    12  is a unit vector. Which of the following could be the value of g?
 
 g
A
1
2
B
1
C
1
D
1
2
[ END OF SECTION A ]
SECTION B
ALL questions should be attempted
9.
Triangle ABC has vertices A( 1, 0 ,  3 ), B( 5,  4 ,  1 ) and C( 4,  16 , 4 ) respectively.
A, B and D are collinear such that
A( 1, 0 ,  3 )
AB 2
 .
BD 3
B( 5,  4 ,  1 )
D
C( 4,  16 , 4 )
(a)
Show that the coordinates of D are ( 11, 10, 2 ).
2
(b)
Hence show clearly that angle ADC is a right angle.
4
(c)
Prove that angle ABC is obtuse.
3
Pegasys 2010
10.
A function is defined as f ( x)  6 cos 2 12 x   3 sin x  .
(a)
By using the fact that cos 2 x   12 (cos 2 x   1) show clearly that this
function can be expressed in the form
f ( x)  3 cos x   3 sin x   3 .
(b)
(c)
11.
12.
13.
3
Express 3 cos x   3 sin x   3 in the form k cos( x  )   3 where 0    360
and k  0 .
3
Hence solve the equation f ( x)  0 for 200  x  360 .
4
A function f is given by f ( x)  x 2  3 2 .
1
(a)
Find f (x)
(b)
Find algebraically the values of x for which f ( x) 
3
1
.
2
3
Given that ( x  1) and ( x  3) are both factors of 2 x 3  5 x 2  ax  b , find a and b.
4
Solve log 3 ( x 2  4)  log 3 ( x  2)  3
5
[ END OF SECTION B ]
[ END OF QUESTION PAPER ]
Pegasys 2010
Higher Mini Prelim - Unit 3
Section A - Answers
2010/2011 (Answers + Marking Scheme)
1
5
D
D
2
6
A
B
3
7
B
B
4
8
C
D
2 marks each (16 marks)
Section B - Marking Scheme
Give 1 mark for each
9(a)
(b)
ans:
valid method
●1
●2
answer
●2
ans:
proof
evidence of using stepping out/section
formula
(11, – 10, 2)
(4 marks)
finds DA
●1
  10 


DA =  10 
 5 


●2
finds DC
●2
 7
 
DC =   6 
 2 
 
●3
●4
finds DA.DC
conclusion
●3
●4
DA.DC = 70 – 60 – 10 =0
since DA.DC = 0;  ADC is right angled
finds BA and BC
●
  4
 
BA =  4 
  2
 
finds BA.BC
conclusion
●2 BA.BC = 4 – 48 – 10 = – 54
●3 scalar product < 0 so obtuse angle
ans:
●
1
●2
●3
10(a) ans:
(b)
(2 marks)
●1
●1
(c)
(11, – 10, 2)
Illustration(s) for awarding each mark
proof
proof
(3 marks)
1
 1 


BC =   12 
 5 


(3 marks)
●1
applies given info to new function
●1
cos 2 12 x  12 (cos x  1)
●2
knows to substitute in function
●2
6[ 12 (cos x  1)]  3 sin x
●3
simplifies to required form
●3 3(cos x  1)]  3 sin x;3 cos x  3  3 sin x
ans:
●1
●2
●3
√12cos(x – 30)o + 3
finds k
finds tan α
finds α
Pegasys 2010
(3 marks)
●1
●2
●3
k  9  3  12
tan   33
  30
Give 1 mark for each
(c)
ans:
equates to 0
simplifies
●1
●2
√12cos(x – 30)o + 3 = 0
cos(x – 30)o =  312
●3
●4
finds values
discards
●3
●4
x = 180o; 240o
240o
●1 use of chain rule
●1
●2 differentiates bracket
●2
●3 combines and simplifies
●3
1 2
x 3
2
2x
2x
●2
●3
2

3
1
(3 marks)
2
equates derivative to
1
2
●1


1
2 x2  3
1
2
2
ans: x = 25
x
x
2

3
1
1
2
=
2
1
(4 marks)
uses synthetic division to find one equation
uses synthetic division to find other eq.
knows to use system of equations
solves for a and b
●1 b – a = 7
●2 b + 3a = – 9
●3 evidence
●4 a = – 4; b = 3
(5 marks)
●1
applies difference of logs
●1
●2
exponentiates
●2
●3
difference of two squares
●3
●4
simplifies
●4
●5
solves to answer
●5
Pegasys 2010

●2 2 x  x 2  3 2 : 4 x 2  x 2  3
●3 3 x 2  3 : x 2  1 : x  1
rearranges and squares
solves to answer
Sect. B

(3 marks)
ans: a = – 4; b = 3
●1
●2
●3
●4
13(a)
x
x
ans: x =  1
●1
12
(4 marks)
●1
●2
11(a) ans:
(b)
240o
Illustration(s) for awarding each mark
(34 marks)
log 3
x2  4
3
x2
x2  4
 27
x2
x 2  4  x  2x  2
x  2x  2  x  2
x  2
x  2  27
16 + 34
Total: 50 marks
Pegasys 2010