Higher Still - 2011 / 2012
MATHEMATICS
Higher Grade Extended Unit Test - UNIT 3
Time allowed - 50 minutes
Read Carefully
1.
2.
3.
4.
Full credit will be given only where the solution contains appropriate working.
Calculators may be used.
Answers obtained by readings from scale drawings will not receive any credit.
This Unit Test contains questions graded at all levels.
FORMULAE LIST
Scalar Product:
.
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
Trigonometric formulae:
Table of standard derivatives:
Table of standard integrals:
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
f ( x)
f ( x)
sin ax
cos ax
a cos ax
 a sin ax
 f ( x) dx
f ( x)
sin ax

cos ax
1
cos ax  C
a
1
sin ax  C
a
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.
Rough working may be done on the paper provided. 2 marks will be given for each correct answer.
1.
In the diagram ABCD, GHJK represents a cuboid.
AB represents vector p, BC represents vector q
and HB represents vector r.
K
DH is represented by:
G
J
A.
p+q+r
B.
–p–q+r
C.
p–q+r
D.
p–q–r
H
r
D
A
C
p
B
2.
P(−2, 1, 3) , Q(2, y, 9) , and B(4, – 5, 12) are three collinear points.
The value of y is
3.
4.
A.
3
B.
–3
C.
5
D.
–1
 
f ( x)  sin 3 x , the value of f ' ( x)    is
3
Find
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
A.
9
4
B.
3 3
4
C.
9
8
D.
3 3
8
6 sin( 3x  4) dx
q
5.
A.
2 cos(3x  4)  c
B.
6 cos(3x  4)  c
C.
18 cos(3x  4)  c
D.
 2 cos(3 x  4)  c
If 2 cos x  3 sin x  k sin( x   ), 0  x  360,
The value of α is:
A.
34o
B.
56o
C.
214o
D.
236o
End of Section A
Section B
ALL QUESTIONS SHOULD BE ATTEMPTED
In this section credit will be given for all correct working.
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6.
The diagram shows a cuboid with edges
parallel to the axes.
D(5, 3, 9)
C(5, 7, 4)
A(2, 3, 4)
B(5, 3, 4)
A, B, C and D are the points (2, 3, 4), (5, 3, 4), (5, 7, 4) and (5, 3, 9) respectively.
7.
(a)
State the lengths of AB, BC and BD.
2
(b)
Establish the components of DA and DC and then calculate the size
of angle ADC.
5
(a)
If log 2 k  3  log 2 7 , find the value of k.
3
(b)
Given that log 9 x 
1
log 3 x , solve the equation
2
log 2 x  log 4 x  9
8.
.
3
(a)
Express 5 cos x  3sin x in the form kcos(x – α)º where k > 0.
4
(b)
Solve the equation 5 cos x  3sin x = 2, 0 ≤ x ≤ 360.
4
(3 cos x  2 sin x) dx
9.
Find the exact value of
10.
Given that F (t )  ke18t and F(1) = 30, find t when F(t) = 12
END OF QUESTION PAPER
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5
4
Higher Grade Unit Tests 2011/2012
Marking Scheme - UNIT 3
Give 1 mark for each
1
D
2
B
Illustration(s) for awarding each mark
3
C
Award 2 marks for each
correct answer
4
D
10 marks
5
C
6(a)
ans:
3; 4; 5
(2 mark)
1&2
1 gives any two lengths correctly
●2 third length given correctly
6(b)
  3
 
ans:  0  ;
  5
 
(5 marks)
finds components
●
●2
finds magnitudes of both vectors
●2
  3
 0 
 
 
DA=  0  ; DC =  4 
  5
  5
 
 
DA= 34 ; DC = 41
●3
finds scalar product
●3
(–3 × 0) + (0 × 4) + (–5 × –5) = 25
●4
applies formula
●4
cos ADC 
●5
finds angle
●5
48o
●
7(a)
 0 
 
o
 4  , 48
  5
 
AB = 3; BC = 4; BD = 5
1
ans: 56
1
25
34 41
(3 marks)
●1
changes constant to log form
●1
....... = log28 +.....
●2
uses appropriate law of logs
●2
log2k = log2(8 × 7) [or log 2
●3
states value of T
●3
t = 56
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t
 log 2 8 ]
7
Illustration(s) for awarding each mark
Give 1 mark for each
7(b)
8(a)
8(b)
9
10
ans:
64
(3 marks)
1
log 2 x  9
2
●1 applies given information
●1
log 2 x 
●2
simplifies
●2
●3
solves for x
●3
3
log 2 x  9; log 2 x  6
2
x  2 6  64
34cos( x  31)
ans:
(4 marks)
●1
●2
finds value of k
equates coefficients
●1
●2
●3
finds tan α
●3
●4
finds α
●4
101o, 321o
ans:
k  52  32  34
k sin   3; k cos   5
3
tan  
5
o
α = 31
(4 marks)
●1
substitutes for original equation
●1
●2
rearranges
●2
●3
●4
finds 1st value of x
finds 2nd value of x
●3
●4
34cos( x  31) = 2
2
cos( x  31) 
34
x – 31 = 70o; x = 101o
x – 31 = 290o; x = 321o
ans: 2 
3 3
2
(5 marks)
●1
●2
integrates 1st term
integrates 2nd term
●1
●2
[3 sin x..........
..........  2 cos x]
●3
substitutes limits
●3
●4
substitutes exact values
●4
●5
evaluates to answer
●5
 2 cos )  (3 sin  2 cos )
2
2
3
3
3
1
(3  1  2  0)  (3 
 2 )
2
2
3 3
2
2
ans:
0∙486
(3 sin




(4 marks)
1 substitutes for t and finds k
 2 substitutes given values
3 takes logs
1
2
3
 4 finds t
4
30  ke181 ; k  5
12  5e18t or e18t  2  4
1 8t ln e  ln 2  4
ln 2  4
t
 0  486
1 8
Total: 40 marks
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