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8/6/2019 Mathematics AH SQP
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Mathematics T i m e : 3 h o u r s
Advanced HigherSpecimen Question Paper
for use in and after 2004
NATIONAL
QUALIFICATIONS
©
Read carefully
1. Calculators may be used in this paper.
2. Candidates should answer all questions .3. Full credit will be given only where the solution contains appropriate working.
[C100/SQP255] 1
[C100/SQP255]
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Marks
2
4
5
2
3
2
2
5
Answer all the questions.
1. (a) Find partial fractions for
(b) By using (a) obtain
2. Use the Euclidean Algorithm to find integers of x, y such that
195x + 239 y = 1.
3. The performance of a prototype surface-to-air missile was measured on a
horizontal test bed at the firing range and it was found that, until its fuel was
exhausted, its acceleration (measured in m s –2) t seconds after firing was given
by
(a) Obtain a formula for its speed, t seconds after firing.
(b) The missile contained enough fuel for 10 seconds. What horizontal
distance would it have covered on the firing range when its fuel wasexhausted?
4. The n × n matrix A satisfies the equation
A2 = 5 A + 3I
where I is the n × n identity matrix.
Show that A is invertible and express A –1 in the form of pA + qI .
Obtain a similar expression for A4.
5. Use the substitution x = 4 sin t to evaluate the definite integral
Page two[C100/SQP255] 2
2
4
.4x −
2
2.
4
xdx
x −∫
238 10 .4
a t t= + −
2
20
1.
16
xdx
x
+−
∫
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Marks
5
3
2
2
2
1
2
3
5
6. Use Gaussian elimination to solve the system of linear equations
7. Use Maclaurin’s theorem to write down the expansions, as far as the term in
x3, of
(i) where –1 < x < 1, and
(ii) where –1 < x < 1.
8. (a) Find the derivative of y with respect to x, where y is defined as an
implicit function of x by the equation
(b) A curve is defined by the parametric equations
x = 2t + 1, y = 2t(t – 1).
(i) Find in terms of t.
(ii) Eliminate t to find y in terms of x.
9. Let u1, u
2, ..., u
n, ... be an arithmetic sequence and v
1, v
2, ... , v
n, ... be a
geometric sequence. The first terms u1
and v1
are both equal to 45, and the
third terms u3
and v3
are both equal to 5.
(a) Find u11
.
(b) Given that v1, v
2, ... is a sequence of positive numbers, calculate
10. Use induction to prove that
for all positive integers n.
Page three[C100/SQP255] 3
0
2 1 1
3 2 0 9.
x y z
x y z
x y z
+ + =⋅− + = −
⋅+ + =
1 ,x+
2(1 ) ,x −−
2 2 1.x xy y+ + =
dy
dx
1
.n
n
v∞
=
∑
1
1( 1) ( 1)( 2)3
n
r
r r n n n=
+ = + +∑
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Marks
3
3
1
1
1
3
4
1
2
6
5
11. Find the general solution of the differential equation
in each of the cases(i) f (x) = 20cos x
(ii) f (x) = 20sinx
(iii) f (x) = 20cos x + 20sin x.
12. Let the function f be given by
(a) The graph of y = f (x) crosses the y-axis at (0, a). State the value of a.
(b) For the graph of f (x)
(i) write down the equation of the vertical asymptote,
(ii) show algebraically that there is a non-vertical asymptote and state
its equation.
(c) Find the coordinates and nature of the stationary point of f (x).
(d ) Show that f (x) = 0 has a root in the interval –2 < x < 0.
(e) Sketch the graph of y = f (x). (You must include on your sketch theresults obtained in the first four parts of this question.)
13. (a) Show that the lines
intersect, and find the point of intersection.(b) Let A, B, C be the points (2, 1, 0), (3, 3, –1), (5, 0, 2) respectively.
Find
Hence, or otherwise, obtain the equation of the plane containing A, B
and C.
Page four [[C100/SQP255] 4
2
25 6 ( )
d y dy y f x
dx dx− + =
3 2
2
2 7 4 5( ) , 2.
( 2)
x x x f x x
x
− + += ≠−
1
2
13 6:
2 3 1
3 6 11:
1 2 2
yx zL
yx zL
+− −= =
− − −= =−
. AB AC
×
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Marks
5
1
1
4
2
4
1
2
14. Let z = cos θ + i sin θ .
(a) Use the binomial theorem to show that the real part of z4 is
Obtain a similar expression for the imaginary part of z4 in terms of θ .
(b) Use de Moivre’s theorem to write down an expression for z4 in terms
of 4θ .
(c) Use your answers to (a) and (b) to express cos4θ in terms of cos θ and
sinθ .
(d ) Hence show that cos 4θ can be written in the form k(cosmθ – cosn
θ ) + p
where k, m, n, p are integers. State the values of k, m, n, p.
15. In a chemical reaction, two substances X and Y combine to form a third
substance Z . Let Q(t) denote the number of grams of Z formed t minutesafter the reaction begins. The rate at which Q(t) varies is governed by the
differential equation
(a) Express in partial fractions.
(b) Use your answer to (a) to show that the general solution of the
differential equation can be written in the form
where A and C are constants.
State the value of A and, given that Q(0) = 0, find the value of C.
Find, correct to two decimal places,
(i) the time taken to form 5 grams of Z ,
(ii) the number of grams of Z formed 45 minutes after the reaction
begins.
[END OF SPECIMEN QUESTION PAPER]
Page five[C100/SQP255] 5
4 2 2 4cos 6cos sin sin .θ θ θ θ − +
(30 )(15 ).
900
dQ Q Q
dt
− −=
900
(30 )(15 )Q Q− −
30ln ,
15
Q A t C
Q
− = + −
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[C100/SQP255] 6 Page six
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Mathematics
Advanced HigherSpecimen Solutions
for use in and after 2004
NATIONAL
QUALIFICATIONS
[C100/SQP255] 7
[C100/SQP255]
©
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1. (a)
(b)
2. 239 = 1 × 195 + 44
195 = 4 × 44 + 19
44 = 2 × 19 + 6
19 = 3 × 6 + 1
So 1 = 19 − 3 × 6
= 19 − 3(44 − 2 × 19)
= 7 × (195 − 4 × 44) − 3 × 44
= 7 × 195 − 31(239 − 195)
= 38 × 195 − 31 × 239
ie 195x + 239 y = 1 when x = 38 and y = −31
3. (a)
(b)
Marks
[2]
[4]
[5]
[2]
[3]
Page two[C100/SQP255] 8
2
4 4
4 ( 2)( 2) 2 2
1 1
2 2
A B
x x x x x
x x
= = +− − + − −
= −− +
2
2 2
414 4
1112 2
ln( 2) ln( 2)
x dx dxx x
dxx x
x x x c
= +− −
= + −− +
= + − − + +
∫ ∫
∫
2
2
2 3
2 3
38 104
38 104
18 5 4
0; 0 0
18 54
a t t
v t t dt
t t t c
t v c
v t t t
= + −
= + −
= + − += = ⇒ =
= + −
∫
2 3 45 143 16
0; 0 0
5000 2when 10, 400 625 14413 3
s v dt t t t c
t s c
t s
′= = + − +
′= = ⇒ =
∴ = = + − =
∫
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Page three
4.
5.
6.
Marks
[2, 2]
[5]
[5]
[C100/SQP255] 9
2
2
51
3 3
1 13
5 3
5 3
( )
is invertible and ( 5 )
A A I
A A I
A A I I
A A A I −
= +
∴ − =
− =
∴ = −
4 2
2
(5 3 )
25 30 9
155 84
A A I
A A I
A I
= +
= + +
= +
4sinx t=
4 cos
0 0;
26
dx tdt
x t
x t π
⇒ =
= ⇒ =
= ⇒ =
1 1 1 0
2 1 1 1 1
1 3 2 0 9
1 1 1 0
0 3 1 1 1
0 2 1 0 9
1 1 1 0
0 3 1 1 1
0 0 1 0 5
− − ⋅
⋅
− − − ⋅
⋅
− − − ⋅
⋅
2 2 1
3 3 1
( 2 )
( )
r r r
r r r
′ = −
′ = −
3 3 2( 3 2 )r r r ′′ = +
Hence z = 0.5; y = (1.1 − 0.5)/3 = 0.2;
x = −0.2 − 0.5 = −0.7
2
20
/ 6
2
0
/ 6
0
/ 6
0
/ 6
0
1
16
4sin 1 4cos16 16sin
(4sin 1) 4cos
4cos
(4sin 1)
[ 4 cos ] 2 3 4 1 0596
x dxx
t t dt
t tdt
t
t dt
t t
π
π
π
π π
+−
+=−
+ ×=
= +
= − + = + + ≈ ⋅
∫
∫
∫
∫
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7. (i)
(ii)
8. (a)
(b) (i)
(ii)
Page four
Marks
[3]
[2]
[2]
[2]
[1]
[C100/SQP255] 10
1/ 2
1/ 2
3/ 2
5/ 2
2 3
( ) 1 (0) 1
(1 )
1 1( ) (1 ) (0)2 2
1 1( ) (1 ) (0)4 4
3 3( ) (1 ) (0)8 8
1 1 11 12 8 16
f x x f
x
f x x f
f x x f
f x x f
x x x x
−
−
−
= + =
= +
= + =′ ′
= − + = −′′ ′′
= + =′′′ ′′′
+ ≈ + − +
2 21
2 2 0
(2 )
2
x xy y
dy dyx x y y
dx dx
dy x y
dx x y
+ + =
+ + + =
− +
= +
2 1; 2 ( 1)
2; 4 2 2 1
dy
dt
dxdt
x t y t t
dx dy dyt t
dt dt dx
= + = −
= = − ⇒ = = −
1 1( 1) ( 1) ( 1) 12 2
1 ( 1)( 3)2
t x y x x
x x
= − = − − −
= − −
2
3
4
5
2 2 3
( ) (1 ) (0) 1
( ) 2(1 ) (0) 2
( ) 6(1 ) (0) 6
( ) 24(1 ) (0) 24
(1 ) 1 2 3 4
f x x f
f x x f
f x x f
f x x f
x x x x
−
−
−
−
−
= − =
= − =′ ′
= − =′′ ′′= − =′′′ ′′′
− ≈ + + +
∴
∴
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9. (a)
(b)
10.
Marks
[2]
[3]
[5]
Page five[C100/SQP255] 11
3 1
11
2 5
2 5 45
20
45 10( 20)
155
u d u
d
d
u
= + =
= −
= −
= + −
= −
2
113
11 23
45 5
since , ... are positive
45 671
r
r v
S
=
=
= =−
13
1
1 1
13
13
1 LHS 1 2 2
RHS 1 2 3 2
True for 1.
Assume true for and consider
( 1) ( 1) ( 1)( 2)
( 1)( 2) ( 1)( 2)
( 1)( 2)( 3)
Thus if true for then true for 1.
Therefore since
k k
r r
n
n
k
r r r r k k
k k k k k
k k k
k k
+
= =
= = × =
= × × × =
=
+ = + + + +
= + + + + +
= + + +
+
∑ ∑
true for 1, true for all 1.n n= ≥
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11.
(i)
(ii)
(iii)
Marks
[3]
[3]
[1]
Page six[C100/SQP255] 12
2
2
2
2 3
5 6 ( )
A.E. 5 6 0
2 or 3
C.F.x x
d y dy y f x
dx dx
m m
m m
y Ae Be
− + =
− + =
= =
= +
2 3
( ) 20 cos ; P.I. cos sin
cos sin 5 sin 5 cos 6 cos 6 sin 20cos
5 5 20
5 5 0
10 20 2; 2
Solution 2 cos 2 sinx x
f x x a x b x
a x b x a x b x a x b x x
a b
a b a b
b b a
y Ae Be x x
= = +
⇒ − − + − + + =
− =
+ = ⇒ = −
− = ⇒ = − =
= + + −
2 3
( ) 20sin ; P.I. cos sin
5 5 0
5 5 20 2
Solution 2 cos 2 sinx x
f x x c x d x
c d c d
c d c d
y Ae Be x x
= = +
− = ⇒ =
+ = ⇒ = =
= + + +
2 3
( ) 20 cos 20 sin
Solution 4 cosx x
f x x x
y Ae Be x
= +
= + +
∴
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12.
(a)
(b) (i) x = 2
(ii) After division, the function can be expressed in quotient/remainder
form:
Thus the line y = 2x + 1 is a slant asymptote.
(c) From (b), Turning point when
The stationary point at (3, 8) is a minimum turning point.
(d )
Hence a root between –2 and 0.
(e)
Marks
[1]
[1]
[3]
[4]
[1]
[2]
Page seven[C100/SQP255] 13
3 2
2
2 7 4 5( )( 2)
x x x f xx
− + +=−
5 5
4 40x y a= ⇒ = ⇒ =
2
1( ) 2 1( 2)
f x xx
= + +−
32( ) 2 .( 2)
f xx
= −′−
3
3
22 0( 2)
( 2) 1
2 1 3
x
x
x x
− =−
− =
− = ⇒ =
4
6( ) 0 for all .
( 2) f x x
x= >′′
−
2
16 28 8 5( 2) 0; (0) 0.( 4) 4
f f − − − +− = < = >−
y = 2x + 1
x = 2
(3, 8)
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13. (a) L1: x = 3 + 2s; y = –1 + 3s; z = 6 + s
L2: x = 3 – t; y = 6 + 2t; z = 11 + 2t
∴ for x: 3 + 2s = 3 – t ⇒ t = –2s
∴ for y: 3s – 1 = 6 + 2t
7s = 7 ⇒ s = 1; t = –2
∴ L1: x = 5; y = 2; z = 6 + s = 7
∴ L2: x = 5; y = 2; z = 11 + 2t = 11 – 4 = 7ie L1 and L2 intersect at (5, 2, 7)
(b)
Marks
[6]
[5]
Page eight[C100/SQP255] 14
(2,1,0); (3,3, 1); (5,0,2)
2 ; 3 2
1 2 1 3 5 7
3 1 2
Equation of plane has form 3 5 7
(2,1,0) 1
Equation is 3 5 7 1.
A B C
AB AC
AB AC
x y z k
k
x y z
−
= + − = − +
× = − = − −
−
− − =
⇒ =
− − =
i j k i j k
i j ki j k
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14. (a) z4 = (cos θ + i sin θ )4
= cos4θ + 4 cos3
θ (i sin θ ) + 6 cos2θ (i sin θ )2 + 4 cosθ (i sin θ )3 + (i sin θ )4
= cos4θ + 4i cos3
θ sin θ – 6 cos2θ sin2
θ – 4i cosθ sin3θ + sin4
θ
= cos4θ – 6 cos2
θ sin2θ + sin4
θ + i (4cos3θ sinθ – 4 cosθ sin3
θ )
Hence the real part is cos4θ – 6 cos2
θ sin2θ + sin4
θ .
The imaginary part is (4 cos3θ sin θ – 4 cos θ sin3
θ )
= 4 cosθ sin θ (cos2θ – sin2
θ )
(b) (cos θ + i sinθ )4 = cos 4θ + i sin 4θ
(c) cos 4θ = cos4θ – 6 cos2
θ sin2θ + sin4
θ .
(d ) cos 4θ = cos4θ – 6 cos2θ sin2θ + sin4 θ
= cos4θ – 6 cos2
θ (1 – cos2θ ) + (1 – cos2
θ )2
= cos4θ – 6 cos2
θ + 6 cos4θ + 1 – 2 cos2
θ + cos4θ
= 8 cos4θ – 8 cos2
θ + 1
= 8 (cos4θ – cos2
θ ) + 1
ie k = 8, m = 4, n = 2, p = 1.
Marks
[5]
[1]
[1]
[4]
Page nine[C100/SQP255] 15
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15. (a) 900 = A(15 – Q) + B(30 – Q)
Letting Q = 30 gives A = –60
and Q = 15 gives B = 60
(b)
(i)
(ii)
[END OF SPECIMEN MARKING SOLUTIONS]
Marks
[2]
[4]
[1]
[2]
900 60 60
(30 )(15 ) (30 ) (15 )Q Q Q Q
−= +− − − −
(30 )(15 )900
900 (30 )(15 )
60 60 (30 ) (15 )
60ln (30 ) 60ln (15 )
30ie 60ln 15
60
60ln 2 41 59 to 2 decimal places
dQ Q Qdt
dQ dtQ Q
dQ dtQ Q
Q Q t C
Q t CQ
A
C
− −=
=− −
− + =− −
− − − = +
− = +−
=
= = ⋅
∫ ∫
∫ ∫
30 3060ln 60ln 2 60ln
15 2(15 )
25When 5, 60ln 13 3920
Q Qt
Q Q
Q t
− −= − =
− −
= = = ⋅
/ 60
/ 60 / 60
/ 60
/ 60
30ln
602(15 )
30 2(15 )
(2 1) 30( 1)
30( 1)
2 1
When 45, 10 36 grams to 2 decimal places.
t
t t
t
t
Q tQ
Q Q e
Q e e
eQ
e
t Q
−=
−
− = −
− = −
−=
−
= = ⋅
∴
∴
( )
( ) ( )
( )
minutes to 2 decimal places