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Scholarship Calculus (93202) 2015 – page 1 of 9 Assessment Schedule – 2015
Scholarship Calculus (93202)
Evidence QUESTION ONE (8 marks)
(a)
2
3⎛
1 ⎞
1 ⎞
⎛ 2
S = 2π ∫ ⎜ x 3 +
⎟⎠ 1+ ⎜⎝ 3x −
⎟ dx
1 ⎝
12x
12x 2 ⎠
3⎛
1 ⎞
1
1
4
= 2π ∫ ⎜ x 3 +
dx
⎟ 1+ 9x − +
1 ⎝
12x ⎠
2 144x 4
2
3⎛
1 ⎞ ⎛ 2
1 ⎞
= 2π ∫ ⎜ x 3 +
⎟⎠ ⎜⎝ 3x +
⎟ dx
1 ⎝
12x
12x 2 ⎠
3⎛
1 ⎞⎛ 2
1 ⎞
= 2π ∫ ⎜ x 3 +
⎟ ⎜ 3x +
⎟ dx
1 ⎝
12x ⎠ ⎝
12x 2 ⎠
3⎛
3⎛
x x
1 ⎞
x
1 ⎞
= 2π ∫ ⎜ 3x 5 + + +
dx = 2π ∫ ⎜ 3x 5 + +
⎟ dx
3⎟
1 ⎝
1
⎠
⎝
12 4 144x
3 144x 3 ⎠
3
⎡ x6 x2
1 ⎤
S = 2π ⎢ + −
= 2295.5
2 ⎥
⎣ 2 6 288x ⎦1
(b)
f ′( x)
( f ( x ))
3
=1⇒ ∫
f (0) = 2 ⇒
Hence,
f ′( x)
( f ( x ))
3
dx = ∫ 1dx ⇒
−1
=c
8
−4
4
=
8x − 1 1− 8x
4
f ( x) = ±
1− 8x
1
x≠
8
( f ( x ))
2
=
−1
2 ( f ( x ))
2
= x+c
Scholarship Calculus (93202) 2015 – page 2 of 9 (c)
ds
6s
= 0.8 × 6 −
dt
200
ds 960 − 6s
=
dt
200
ds
200 ∫
= dt
960 − 6s ∫
960 − 6s
ln
A
200
=t
−6
6
t
960 − 6s = Ae 200 = Ae −0.03t
960 − Ae −0.03t
= 160 − Ce −0.03t
6
t = 0, s = 0.5 × 200 = 100
s=
⇒ 100 = 160 − Ce0
s = 160 − 60e −0.03t
t = ?, s = 130
⇒ 130 = 160 − 60e −0.03t
ln0.5
= 23.1 min
−0.03
QUESTION TWO (8 marks)
(a)
(3 )
2 x+y 2
Scholarship Calculus (93202) 2015 – page 3 of 9 − 32 x+y − 6 = 0
32 x+y = 3, 32 x+y ≠ −2
2x + y = 1 ⇒ y = 1− 2x
Substituting:
log x+1 ( 4 − 2x ) ( 5 − x ) = 3
20 + 2x 2 − 14x = ( x + 1)
3
x 3 + x 2 + 17x − 19 = 0
x =1
y = −1
(b)
The form of the parabola is y = kx 2 .
At (100,100),
100 = 100 2 k ⇒ k =
1
100
1 2
x
100
Gradient of the tangent at ( x0 , y0 ) :
y=
dy x0
=
dx 50
y0 =
1 2
x0
100
1 2
x0 50 − 100 x0
gradient of the line
Matching gradients: Gradient of tangent: =
50
100 − x0
100x0 − x02 = 2500 −
1 2
x0
2
1 2
x0 − 100x0 + 2500
2
x0 = 29.3
Coordinates of the point (29.3,8.6); 29.3 m east, 8.6 m north.
0=
(c)
dS
= k S(N − S)
dt
dS
∫ S ( N − S ) = ∫ k dt
Using partial fractions:
1
A
B
1
1
= +
⇒A= , B=
S(N − S) S N − S
N
N
1 ⎛1
1 ⎞
⎜⎝ +
⎟ dS = ∫ k dt
∫
N
S N − S⎠
Scholarship Calculus (93202) 2015 – page 4 of 9 S
1 N −S
ln
= kt
N
A
S
= Ae kNt
N −S
t = 0, S = 2
2
⇒A=
N −2
S
2
=
e kNt ⇒ S ( N − 2 ) e − kNt = 2 ( N − S )
N −S N −2
S 2 + ( N − 2 ) e − kNt = 2N
(
S=
)
N
1
1+ ( N − 2 ) e − kNt
2
OR Using differentiation:
N
2N
2Ne kNt
S (t ) =
=
=
− kNt
kNt
1
1+ e − kNt ( N − 2 ) 2 + e ( N − 2 ) 2e + ( N − 2 )
2
2 kNt
kNt
kNt
kNt
dS 2kN e 2e + ( N − 2 ) − 2Ne 2kNe
=
2
dt
2e kNt + ( N − 2 )
(
)
(
)
( )
(
2
⎡ 2N 2 e kNt
4N 2 e kNt
= k ⎢ kNt
−
⎢ 2e + ( N − 2 ) 2e kNt + ( N − 2 )
⎣
(
)
(
= k NS − S 2 = kS ( N − S )
)
)
2
⎤
⎥
⎥
⎦
Scholarship Calculus (93202) 2015 – page 5 of 9 QUESTION THREE (8 marks)
(a)
z = cosθ + isin θ , z n = cos nθ + sin nθ , z − n = cos ( −nθ ) + isin ( −nθ ) = cos nθ − isin nθ
.
1
= 2 cos nθ
zn
Using this
zn +
6
1⎞
15 6 1
⎛
(2 cosθ )6 = ⎜ z + ⎟ = z 6 + 6z 4 + 15z 2 + 20 + 2 + 4 + 6
⎝
z⎠
z
z
z
1⎞
1⎞
1⎞
⎛
⎛
⎛
64 cos 6 θ = ⎜ z 6 + 6 ⎟ + 6 ⎜ z 4 + 4 ⎟ + 15 ⎜ z 2 + 2 ⎟ + 20
⎝
⎝
⎝
z ⎠
z ⎠
z ⎠
= 2 cos 6θ + 12 cos 4θ + 30 cos 2θ + 20
32 cos θ = cos 6θ + 6 cos 4θ + 15 cos 2θ + 10
1
cos 6 θ = ( cos 6θ + 6 cos 4θ + 15 cos 2θ + 10 )
32
6
(b)
1
Restrictions: sin x > , y < 1
2
log ( 2sin x − 1) − log 2 = log(1− y) − log ( 2sin x − 1)
2 log ( 2sin x − 1) = log 2 + log(1− y)
log ( 2sin x − 1) = log 2 (1− y )
2
( 2sin x − 1)2 = 2 (1− y )
( 2sin x − 1)2
y = 1−
2
0 < ( 2sin x − 1) ≤ 1
The range of possible values for y is
1
∴ ≤ y <1
2
(c)
LHS =
=
=
(
cos 2x − 1
8 cos 2 2x − 7 cos 2x − 1
2
=
8sin 4 x
4 sin 2 x 1− cos 2 x
4 cos 2 2x − 4 cos 2x +
(
)
( 8 cos 2x + 1)( cos 2x − 1) = ( 8 cos 2x + 1)( cos 2x − 1) = 8 cos 2x + 1 = RHS
2
2 ( cos 2x − 1)
2 ( cos 2x − 1)
(1− cos 2x )2
8
4
)
4 cos 2 2x − 4 cos 2 x − sin 2 x − sin 2 x
4 cos 2 2x − 4 cos 2 x + 3sin 2 x
=
⎛ 5π
⎞
4 sin 2 x − 4 sin 2 x cos 2 x
4 cos 2 ⎜
− x ⎟ − sin 2 2 ( x − π )
⎝ 2
⎠
QUESTION FOUR (8 marks)
(a)
Scholarship Calculus (93202) 2015 – page 6 of 9 Assume the real root exists and let it be p.
3p 3 + ( 2 − 3ai ) p 2 + ( 6 + 2bi ) p + 4 = 0
Equating real and imaginary:
Imaginary:
−3ap 2 + 2bp = 0
p ( −3ap + 2b ) = 0
p ≠ 0,
p=
2b
3a
Real:
3p 3 + 2 p 2 + 6 p + 4 = 0
p 2 ( 3p + 2 ) + 2 ( 3p + 2 ) = 0
( 3p + 2 )( p 2 + 2 ) = 0
p 2 ≠ −2,
Solution exists if
(b)
p=
−2
3
b
= −1
a
Option 1
AX + BY
= OD ⇒ AX + BY = 2OD Trapezium
2
AX = AF and BY = BF distance to directrix = distance to focal point
AF + BF = 2OD
AB = 2OD
2AO = 2OD
AO = OD
Scholarship Calculus (93202) 2015 – page 7 of 9 (b)
Option 2
Every hour, the maximum number of each type of decoration available is 500 type A, 400 type B and 300 type C.
6x + 2y + 2z ≤ 500 ⇒ 3x + y + z ≤ 250 (1)
3x + 4y + 5z ≤ 400
(2)
2x + 4y + 2z ≤ 300 ⇒ x + 2y + z ≤ 150
( 3)
Every hour, the factory must pack at least 1000 decorations.
11x + 10y + 9z ≥ 1000 ( 4 )
Every hour, the factory must pack more type B decorations than type A decorations.
3x + 4y + 5z > 6x + 2y + 2z ⇒ 2y + 3z > 3x ( 5 )
(ii)
3x + y + z ≤ 250 ⇒ 3x + y + 100 − x − y ≤ 250 ⇒ x ≤ 75
(1)
3x + 4y + 5z ≤ 400 ⇒ 3x + 4y + 500 − 5x − 5y ≤ 400 ⇒ 2x + y ≥ 100
x + 2y + z ≤ 150 ⇒ x + 2y + 100 − x − y ≤ 150 ⇒ y ≤ 50
11x + 10y + 9z ≥ 1000 ⇒ 2x + y ≥ 100
2y + 3z > 3x ⇒ y + 6x < 300
(5)
(4)
( 3)
(2)
QUESTION FIVE (8 marks)
(a)
Scholarship Calculus (93202) 2015 – page 8 of 9 Equation of the Normal
The gradient of the tangent line at the point (x0,y0):
y = kx 2
dy
= 2kx = 2kx0
dx
Gradient of the Normal line:
−1
m=
2kx0
Equation of the normal:
Using y = mx + c at the point (x0,y0) to find c.
−1
1
1
y0 =
x0 + c ⇒ c = y0 +
⇒ c = kx0 2 +
2k x0
2k
2k
y=
(b)
−1
1⎞
⎛
x + ⎜ kx02 + ⎟
⎝
2kx0
2k ⎠
Intersection of Normal and Parabola:
Using the general form of the normal y = mx + c and y = kx2.
kx 2 = mx + c
kx 2 − mx − c = 0
m ± m 2 + 4kc
2k
−1
1
Substitute m =
and c = kx0 2 +
2kx0
2k
x=
x=
m ± m 2 + 4c
2
2
⎛ 1
⎞
−1
⎞
−1
1
−1 ⎛ 1
± ⎜
+ 2kx0 ⎟
±
+ 4k 2 x02 + 2
±
+ 2kx0 ⎟
2 2
2kx0
⎝ 2kx0
⎠
⎛
2kx0
4k x0
2kx0 ⎜⎝ 2kx0
⎠
1 ⎞
x=
=
=
= x0 , − ⎜ x0 + 2 ⎟
2k
2k
2k
2k x0 ⎠
⎝
To find the value of
⎛
⎝
x0 that makes x = − ⎜ x0 +
1 ⎞
a maximum (least negative):
2k 2 x0 ⎟⎠
⎛
dx
1 ⎞
1
1
= − ⎜ 1− 2 2 ⎟ = 0 ⇒ x02 = 2 ⇒ x0 =
d ( x0 )
2k
x
2k
⎝
2k
0 ⎠
Test for maximum
d2x
−1
1
2 = 2 3 < 0 at x0 =
k
x
2k
d ( x0 )
0
Equation of Extreme Normal
Substituting into y =
−1
1
− 2
1
x + kx0 2 +
gives y =
x+
2kx0
2k
2
k
Scholarship Calculus (93202) 2015 – page 9 of 9 (c)
Area
The area is an integral given as:
x0
⎛ −1
⎞
1
A=∫
x + kx02 +
− kx 2 ⎟ dx
1
− x0 − 2 ⎜
2k
⎠
2 k x0 ⎝ 2kx0
x0
⎡ −1 2 ⎛ 1
kx 3 ⎤
⎞
A=⎢
x +⎜
+ kx02 ⎟ x −
⎝ 2k
⎠
3 ⎥⎦ − x −
⎣ 4kx0
0
1
2 k 2 x0
⎡ −1 2 ⎛ 1
kx 3 ⎤
⎞
=⎢
x0 + ⎜
+ kx02 ⎟ x0 − 0 ⎥
⎝ 2k
⎠
3 ⎦
⎣ 4kx0
3
⎡
⎛
1 ⎞ ⎤
k ⎜ −x0 − 2 ⎟ ⎥
⎢
2
2k x0 ⎠ ⎥
⎝
−1 ⎛
1 ⎞ ⎛ 1
1 ⎞
⎞⎛
−⎢
−x −
−
+ kx02 ⎟ ⎜ −x0 − 2 ⎟ +
⎢ 4kx0 ⎜⎝ 0 2k 2 x0 ⎟⎠ ⎜⎝ 2k
⎥
⎠⎝
2k x0 ⎠
3
⎢
⎥
⎢⎣
⎥⎦
3
x 4kx0
1
1
+
+ 3
Simplifying: A = 0 +
5 3
k
3
48k x0 4k x0
Differentiate to find the minimum
dA 1
1
1
= + 4kx02 −
− 3 2 =0
5 4
dx0 k
16k x0 4k x0
=
(
)
1
16k 4 x04 + 64k 6 x06 − 1− 4k 2 x02 = 0
16k 5 x04
16k 5 x04 ≠ 0; 16k 4 x04 + 64k 6 x06 − 1− 4k 2 x02 = 0
(
) (
)
16k 4 x04 1+ 4k 2 x02 − 1+ 4k 2 x02 = 0
(1+ 4k x )(16k x
2
2
0
4
4
0
)
−1 = 0
1+ 4k x ≠ 0; 16k x − 1 = 0
1
x0 = ±
( x0 > 0 )
2k
2
2
0
4
4
0
1
2k
Test minimum using the first derivative test.
Equation of the normal:
3
y = −x +
4k
The real positive root is x0 =
Minimum Area:
4
Amin = 2
3k