Mathematics - Ace the HSC
Caringbah High School 2016 Mathematics Trial HSC
Caringbah High School
2016 Trial HSC Examination
Mathematics General Instructions
Total marks – 100
• Reading time – 5 minutes • Working time – 3 hours • Write using black or blue pen (Black pen is preferred)
Section I Pages 2 – 5 10 marks • Attempt Questions 1–10 • Allow about 15 minutes for this section
• Board-approved calculators may be used • A Board-approved reference sheet is provided for this paper
• In Questions 11–16, show relevant mathematical reasoning and/or calculations
Section II Pages 6 – 14 90 marks • Attempt Questions 11–16 • Allow about 2 hour and 45 minutes for this section
Page | 1
Caringbah High School 2016 Mathematics Trial HSC Question 1 - 10 (1 mark each) Answer on the page provided.
1
2
3
If a 1 2c, which expression has c been correctly made the subject?
(A)
c
1 a 2
(B)
c
1 a 2
(C)
c
a 1 2
(D)
c
a 1 2
If f x 2 x2 3x 4, what is the value of f 1 f 1 ?
(A)
6
(B)
2
(C)
6
(D)
2
The number of worms W in a worm farm, at time t is given by W 3000e kt , where k is a positive constant. Over time, which expression describes the change in the number of worms?
4
(A)
decreasing at a constant rate
(B)
increasing at a constant rate
(C)
decreasing exponentially
(D)
increasing exponentially
The second term and the fifth term of a geometric sequence are 3 and 192 respectively. What is the common ratio of the sequence?
(A)
4
(B)
4
(C)
8
(D)
8
Page | 2
Caringbah High School 2016 Mathematics Trial HSC
5
Which of the following circles has the lines x 1, x 5, y 3, and y 7 as its tangents?
2
(A)
x 3 2
7 y 4 2
(C)
x 5 2
y 3 2
4
2
(B)
5 x 2
(D)
x 32
y 5 2
y 5 2
4
4
6 b
y = f (x)
a
O
1
5
Using Simpson’s rule with 3 function values, which expression best represents the area bounded by the curve y f x , the x-axis and the lines x 1 and x 5?
(A)
2 1 4a b 3
(B)
1 1 4a b 2
(C)
1 4a b 2
(D)
2 4a b 3
Page | 3
Caringbah High School 2016 Mathematics Trial HSC
7
How many points of intersection of the graph of y sin x and y 1 cos x lie between 0 and 2 ?
(A)
1
(B)
2
(C)
3
(D)
4
8
In the figure above, which region represents the solution to the following inequalities?
x 3y 0 x y 2 0 x y 4 0 (A)
I
(B)
II
(C)
III
(D)
IV
Page | 4
Caringbah High School 2016 Mathematics Trial HSC
9
In the figure above, the graph of y x2 px q cuts the x-axis at A and B. Which is the value of OA OB ?
10
(A)
p
(B)
p
(C)
q
(D)
q
If log 10 x, log 10 y, log 10 z form an arithmetic progression, which one of the following relationships hold true?
xz 2
(A)
y 10
(C)
y2 x z
xz 2
(B)
y
(D)
y2 x z
END OF MULTIPLE CHOICE QUESTIONS Page | 5
Caringbah High School 2016 Mathematics Trial HSC
Section II 90 marks Attempt all questions 11−16 Allow about 2 hour and 45 minutes for this section Answer each question in a SEPARATE writing booklet. Extra writing booklets are available. In Questions 11−16, your responses should include relevant mathematical reasoning and/or calculations.
Question 11 (15 marks) Start a NEW booklet.
3 2 ,
(a)
Find the value of e
(b)
Factorise 50 2 x2 .
(c)
Simplify
(d)
Solve 3x 5 4 .
(e)
Find the values of a and b if
(f)
State the domain and range of y 2 x 1 .
(g)
2 2 x dx . Find sec 0 3
correct to 3 significant figures.
Marks
2
2
1 1 , expressing the answer as a single fraction. x 1 x 1 2
2
2
1 a b 2. 3 2
2
2
3
Page | 6
Caringbah High School 2016 Mathematics Trial HSC Question 12 (15 marks) Start a NEW booklet.
d dx
x3 1 .
(a)
Find
(b)
Find the limiting sum of 81 27 9 ......
(c)
Find
(d)
Find the exact value of sin
(e)
The gradient function of a curve is given by 3x 2 5 .
3 x 2
5
2
dx .
3 7 cos . 4 6
2
2
When a balloon is being filled with helium, its volume at time t is given by V
t3 12
cm3 , where t is in seconds.
Find the rate at which the balloon is being filled when t 1.
(g)
2
2
If the curve passes through the point 1,1 , find its equation.
(f)
Marks
2
For the parabola 6 y x 2 2 x 13 , use completing the square or otherwise to find:
(i)
the vertex.
2
(ii)
the focal length.
1
Page | 7
Caringbah High School 2016 Mathematics Trial HSC Question 13 (15 marks) Start a NEW booklet.
Marks
d cos x e . dx
(a)
Find
1
(b)
Find the shortest distance of the point 1, 3 from the line 2 x 5 y 4 .
(c)
In the figure below, ABCD is a square, CEF is an equilateral triangle,
2
DCE 130 and DC = CE.
B
F A
C 130 D
E
Find the size of CBF , giving clear reasons.
(d)
3
Find the value of k for which the equation 3x2 10 x k 0 has:
(i)
one root which is the reciprocal of the other.
1
(ii)
equal roots.
2
Question 13 continues on page 9
Page | 8
Caringbah High School 2016 Mathematics Trial HSC Question 13 (continued)
Marks
(e)
Maeve walks from her house for 6 km, on a bearing of 310 to point B. She then walks on a bearing of 215 to a point C which is due west of her house.
(f)
(i)
Copy the diagram and clearly indicate the information mentioned above.
1
(ii)
Calculate to 1 decimal place the distance of C to her house.
2
Part of the graph of the function y x2 ax 12, is shown below.
If the shaded area is 45 square units, find the values of a.
3
End of Question 13 Page | 9
Caringbah High School 2016 Mathematics Trial HSC Question 14 (15 marks) Start a NEW booklet. (a)
Marks
In the diagram below, the straight line L1 has equation 2 x y 4 0 and cuts the x-axis and y-axis at A and B respectively. The straight line L2 , passing through B and perpendicular to L1 cuts the x-axis at C. From the origin O, a straight line perpendicular to L2 is drawn to meet L2 at D.
(i)
Write down the coordinates of A and B.
1
(ii)
Show that the equation of L2 is x 2 y 8 0 .
2
(iii)
Find the equation of the line OD.
1
(iv)
8 16 Show that the coordinates of D are , . 5 5
1
(v)
Hence or otherwise, find the area of quadrilateral OABD.
2
Question 14 continues on page 11 Page | 10
Caringbah High School 2016 Mathematics Trial HSC Question 14 (continued)
(b)
Marks
(i)
x State the period for y 3cos . 2
(ii)
Neatly sketch the graph of y 3cos
1
x for 0 x 2 . 2
tan cos tan sin . sin
(c)
Prove that
(d)
Michael is attempting to solve the equation 2ln x ln 3x 10 .
2
3
He sets his work out using the following correct steps:
ln x2 ln 3x 10
x2 3x 10
x2 3x 10 0
He is now unsure of what to do next.
Complete the solution for Michael.
2
End of Question 14
Page | 11
Caringbah High School 2016 Mathematics Trial HSC Question 15 (15 marks) Start a NEW booklet.
(a)
Marks
A solid is formed by rotating the curve y x3 about the y-axis for 1 y 8 .
Find the volume of the solid in exact form.
3
(b)
A particle is observed as it moves in a straight line between t 0 and t 11 . Its velocity V m/s at time t is shown on the graph above.
(i)
What is the velocity of the particle after 4 seconds?
1
(ii)
What is the particle’s acceleration after 4 seconds?
1
(iii)
At what time after t 0 is the particle at rest?
1
(iv)
At what time does the particle change direction?
1
(v)
Explain what the shaded area represents.
1
Question 15 continues on page 13 Page | 12
Caringbah High School 2016 Mathematics Trial HSC Question 15 (continued)
(c)
Marks
In Daniel’s first year of fulltime employment his annual salary is $75 000. At the beginning of the second year his salary increases to $79 000. His salary continues to increase by $4000 at the beginning of each successive year.
(i)
What will Daniel’s salary be at the beginning of his 25th year?.
1
(ii)
Calculate Daniel’s total earnings after working for 25 years.
1
(iii)
During which year of employment will his total earnings
2
first exceed $2 000 000?
(d)
Determine the range of values of x when the curve y x 2 ln x is concave up.
3
End of Question 15
Page | 13
Caringbah High School 2016 Mathematics Trial HSC Question 16 (15 marks) Start a NEW booklet.
Marks
2
(a)
ex dx expressing the answer in simplified form. Evaluate x 1 e 1
3
X (b)
In ∆ XYZ, XY = XZ = 13 cm and YZ = 10 cm. A rectangle PQRS is inscribed in the triangle with PQ
Find the perpendicular height of ∆ XYZ.
(ii)
Using similar triangles, show that y 12
(iii)
(c)
1
y cm
(i)
x cm Q
P
parallel to YZ. Let PQ = x cm and QR = y cm.
Y
S
R
Z
2
6x . 5
If the area of the rectangle PQRS is A cm², find the maximum value of A.
Consider the function f x
2
4 . x 1 2
4 is an even function. x 1
(i)
Show that f x
(ii)
Explain why there are no x-intercepts for f x
(iii)
Find any stationary points and determine their nature.
3
(iv)
Neatly sketch the graph of y f x .
2
1
2
4 . x 1 2
1
End of paper Page | 14
CHS
YEAR 12 MATHEMATICS 2U 2016
Multiple Choice Section:
TRIAL HSC SOLUTIONS
Question 7.
1.B
2.A
3.C
4.B
5.D
6.D
7.B
8.C
9.A
10.D
Question 1. 2c= 1 − a 1− a ∴c = 2
−−−−−−−−− B
Question 2.
f (1) = 2 − 3 + 4 = 3
∴ 2 points of intersection
−−−−−−− B
f ( −1) = 2 + 3 + 4 = 9 f (1) − f ( −1) =− 6
−−−−−−−−− A
Question 3. decreasing exponentially
−−−−−−−−− C
Question 8. Using the test point (3,3) it can be determined that each inequality is true – hence III − − − − − C Question 9.
Question 4. OA + OB represents the sum of the roots.
T2 = − 3 → ar = − 3 − − − − 1 4 T5 192 → ar= 192 − − − − 2 =
∴ OA + OB = −
b = −b a
−−−−−−−−−−− A
− 64 2 ÷ 1 → r3 =
∴r= −4
−−−−−−−−−−− B
∴ log 10 y − log 10 x = log 10 z − log 10 y
Question 5.
( x − 3)2
+ ( y − 5) = 4 −−−−−−−−−−− D 2
=
5 −1 ( f (1) + 4 f ( 3) + f ( 5) ) 6 2 ( 4a + b ) 3
∴ 2log 10 y = log 10 x + log 10 z ∴ log 10 y 2 = log 10 ( xz )
Question 6. = A
Question 10.
−−−−−−−−−−− D
∴ y2 = xz − − − − − − − − − − − D
Question 11
a) e
π +3 2
−
81, r =− b) 81 − 27 + 9 −...... → a =
= 0.04638420
≈ 0.0464 ( 3Sig fig )
(
b) 50 − 2 x 2 = 2 25 − x 2
)
c)
= 2 ( 5 − x )( 5 + x ) c)
1 1 −= 2 x −1 x −1 =
d) 3 x − 5 =4
or
dy e) = 3 x 2 − 5 dx
g) ⌠ ⌡0
∴ y = x3 − 5 x + 5
(
π
)
=
3 3
1 − 1 3 2 × 3x 2 x 1 + ( ) 2 3x 2
2 x3 + 1
→
6y =
(x
2
π 3 cm / s 4
+ 2 x + 1) + 12 2
hence the vertex is ( −1, 2)
x x2 sec2 dx = 3 tan 3 3 0
x3 + 1 =
cm
dV π t 2 = dt 4
∴ 6 ( y − 2) = ( x + 1)
Question 12.
d dx
12
3
dV when = t 1,= dt
3 2 ____________________________________________
ii) for the focal length 4a = 6
→
Question 13 a)
a)
π t3
f) = V
3 1 → a= , b= 7 7
π = 3 tan − tan 0 = 6
+C
18
when x =1, y =1 → C =5
1 f) Domain: all real x ≥ 2 Range: all real y ≥ 0
2
6
∴ y = x3 − 5 x + C
3 x − 5 =− 4
g) i)
π
243 4
3π 7π 1 3 + cos = − 4 6 2 2
1− x −1 = ( x − 1)( x + 1)
−x ( x − 1)( x + 1)
(3 x − 2)
dx=
d) sin
1 3+ 2 × 3− 2 3− 2
3+ 2 = 7
∫ (3 x − 2)
5
1 x +1 − ( x − 1)( x + 1) ( x − 1)( x + 1)
= 3 x 9= or 3 x 1 1 ∴ = x 3 or = x 3 e)= LHS
81 = 1 1− − 3
= ∴ S∞
1 3
(
)
d cos x e = − sin x e cos x dx
b) d =
2 ×1 − 5 × − 3 − 4 = 2 22 + ( − 5 )
13 29
a=
c) ∠ FCE = 60 [∆ FCE is equilateral]
Question 14.
∠ BCD = 90 [D ABCD is a square] ∴ ∠ BCF = 360 − 130 − 90 − 60 = 80 and = since BC CF = as ( DC CE )
180 − 80° then θ = sides] ∠CBF = [∠ 's opp = 2 50 [ in isosceles ∆ ]
a) i)
A ( −2,0); B ( 0, 4) ;
1 ii) mL1 =→ 2 mL2 = − 2 1 ∴ L2 : y − 4 = − ( x − 0) using B(0, 4) 2 ∴ 2y − 8 = −x
d) i) Product of roots = 1 ∴
k = 1 3
→
∴ x + 2y − 8 = 0
k= 3
iv) Solve y = 2 x and x + 2 y − 8 = 0
ii) ∆ = 0 for equal roots
0 ∴ x + 2 ( 2x) − 8 =
∆= b 2 − 4ac= 100 − 12k 100= 12k
→
y = 2x
iii)
5x = 8
25 3
k=
→ x=
8 5
8 8 16 when x = , y = 2 × = 5 5 5
e)
v) C has coordinates (8, 0). 85°
•
6 km
55°
Area of quadrilateral OABD = Area of ∆ ABC − Area of D ODC
40° x km
=
1 1 16 36 2 × 10 × 4 − × 8 × = u 2 2 5 5
b) i) = P x = sin85
6 sin55
→ = x
6sin85 sin55
∴x = 7.3 km
f)
3
∫0
− x 2 + ax + 12 dx = 45 3
x3 ax 2 ∴ − + + 12 x =45 2 3 0 −9 +
9a += 36 45 2
→ = a 4
ii)
2π = 4π 1 2
sin θ 1 × − cos θ cos θ sin θ
c) LHS =
iv) t = 9s v) The distance travelled in the first 9 seconds.
1 − cos θ cos θ
=
c) i) $ 75000 + 24 × 4000 = $171000
1 − cos 2 θ = cos θ
ii) Total earnings for 25 years =
sin 2 θ = cos θ
$ 75000 + $79 000 + $83000 + ...... + $171000
sin θ sin θ × cos θ 1
=
= sin = θ tan θ
d)
=
25 ( $ 75000 + $171000 ) 2 = $3075000
RHS
0 ( x − 5)( x + 2 ) =
4 000 000 = 150000n + 4 000n 2 − 4 000n
∴ x= −2 5 or x = but since x > 0 then x = 5 only ___________________________________________
∴ 2n 2 + 73n − 2 000 = 0
Question 15. n=
y =x3 → x 2 =y 2 3
8
a) V = π ∫ x 2 dy 1
8
n ( 2 × 75000 + ( n − 1) × 4 000 ) 2
iii) 2 000 000=
2
∴ V= π ∫ y 3 dy
−73 ±
732 − 4 × 2 × − 2 000 4
and since n > 0 , n ≈ 18.26 hence during the 19th year.
1
8
3 5 = π y 3 5 1 =
=
b) i) 12 m/s ii) 0 m/s 2 iii) t = 9s
3π 5
5 53 3 8 −1
3π = ( 31) 5
93π 3 u 5
d) Concave up when y′′ > 0 ∴ y′ = x 2 ×
1 + 2 x × ln x x
= x + 2 x ln x
∴ y′′ = 1 + 2 x ×
1 + 2 × ln x x
= 3 + 2ln x ∴ 3 + 2ln x > 0
∴ ln x > −
3 2
→
x > e
−3
2
Question 16. 2
⌠ e a) x = dx ⌡1 e − 1 x
∴ 12 −
ln ( e − 1) 1 x
2
→
x = 5cm
6×5 Hence maximum area = 5 × 12 − 5 = 30cm 2
= ln ( e 2 − 1) − ln ( e − 1)
(e − 1)(e + 1) (e − 1) ln ( e + 1)
c) i) A function is even if f ( x= ) f (− x) .
= ln =
12 x = 0 5
4 f (− x) = 2 (− x) + 1
b) i) Perpendicular height is 12 cm (Pythagoras)
4 = 2 x +1
=
ii) Let D be the midpoint of PQ.
x ∴ DQ = and XD = 12 − y . 2 X
f ( x ) , hence even.
ii) x-intercepts occur when f ( x ) = 0 , and since the numerator is constant then f ( x ) ≠ 0. X iii)
f= ( x ) 4( x 2 + 1)
−1
∴ f ′( x ) = − 8 x ( x 2 + 1)
−2
−8 x = 2 ( x 2 + 1)
12
D C
Q
For stationary points f ′( x ) = 0 When = x 0, y 4. = Test for maximum or minimum:
12 − y 12 = x 5 2
∴ 6 x = 60 − 5 y ∴ y = 12 − iii) A = x y
x = 0.
Z
5
The triangles are similar (equiangular), hence corresponding sides are in the same ratio. ∴
→
→
∴ A = 12 x −
6x A = x 12 − 5
6 x2 5
dA d2A For a maximum= area 0 and < 0. dx dx 2 dA 12 x d 2 A 12 =− 12 ; = − < 0 hence maximum. 2 dx 5 dx 5
−1
0
𝑓𝑓′(𝑥𝑥 )
2
0
/
1 −2
\
Hence (0, 4) is a maximum stationary point. iv)
6x 5
x
Caringbah High School
2016 Trial HSC Examination
Mathematics General Instructions
Total marks – 100
• Reading time – 5 minutes • Working time – 3 hours • Write using black or blue pen (Black pen is preferred)
Section I Pages 2 – 5 10 marks • Attempt Questions 1–10 • Allow about 15 minutes for this section
• Board-approved calculators may be used • A Board-approved reference sheet is provided for this paper
• In Questions 11–16, show relevant mathematical reasoning and/or calculations
Section II Pages 6 – 14 90 marks • Attempt Questions 11–16 • Allow about 2 hour and 45 minutes for this section
Page | 1
Caringbah High School 2016 Mathematics Trial HSC Question 1 - 10 (1 mark each) Answer on the page provided.
1
2
3
If a 1 2c, which expression has c been correctly made the subject?
(A)
c
1 a 2
(B)
c
1 a 2
(C)
c
a 1 2
(D)
c
a 1 2
If f x 2 x2 3x 4, what is the value of f 1 f 1 ?
(A)
6
(B)
2
(C)
6
(D)
2
The number of worms W in a worm farm, at time t is given by W 3000e kt , where k is a positive constant. Over time, which expression describes the change in the number of worms?
4
(A)
decreasing at a constant rate
(B)
increasing at a constant rate
(C)
decreasing exponentially
(D)
increasing exponentially
The second term and the fifth term of a geometric sequence are 3 and 192 respectively. What is the common ratio of the sequence?
(A)
4
(B)
4
(C)
8
(D)
8
Page | 2
Caringbah High School 2016 Mathematics Trial HSC
5
Which of the following circles has the lines x 1, x 5, y 3, and y 7 as its tangents?
2
(A)
x 3 2
7 y 4 2
(C)
x 5 2
y 3 2
4
2
(B)
5 x 2
(D)
x 32
y 5 2
y 5 2
4
4
6 b
y = f (x)
a
O
1
5
Using Simpson’s rule with 3 function values, which expression best represents the area bounded by the curve y f x , the x-axis and the lines x 1 and x 5?
(A)
2 1 4a b 3
(B)
1 1 4a b 2
(C)
1 4a b 2
(D)
2 4a b 3
Page | 3
Caringbah High School 2016 Mathematics Trial HSC
7
How many points of intersection of the graph of y sin x and y 1 cos x lie between 0 and 2 ?
(A)
1
(B)
2
(C)
3
(D)
4
8
In the figure above, which region represents the solution to the following inequalities?
x 3y 0 x y 2 0 x y 4 0 (A)
I
(B)
II
(C)
III
(D)
IV
Page | 4
Caringbah High School 2016 Mathematics Trial HSC
9
In the figure above, the graph of y x2 px q cuts the x-axis at A and B. Which is the value of OA OB ?
10
(A)
p
(B)
p
(C)
q
(D)
q
If log 10 x, log 10 y, log 10 z form an arithmetic progression, which one of the following relationships hold true?
xz 2
(A)
y 10
(C)
y2 x z
xz 2
(B)
y
(D)
y2 x z
END OF MULTIPLE CHOICE QUESTIONS Page | 5
Caringbah High School 2016 Mathematics Trial HSC
Section II 90 marks Attempt all questions 11−16 Allow about 2 hour and 45 minutes for this section Answer each question in a SEPARATE writing booklet. Extra writing booklets are available. In Questions 11−16, your responses should include relevant mathematical reasoning and/or calculations.
Question 11 (15 marks) Start a NEW booklet.
3 2 ,
(a)
Find the value of e
(b)
Factorise 50 2 x2 .
(c)
Simplify
(d)
Solve 3x 5 4 .
(e)
Find the values of a and b if
(f)
State the domain and range of y 2 x 1 .
(g)
2 2 x dx . Find sec 0 3
correct to 3 significant figures.
Marks
2
2
1 1 , expressing the answer as a single fraction. x 1 x 1 2
2
2
1 a b 2. 3 2
2
2
3
Page | 6
Caringbah High School 2016 Mathematics Trial HSC Question 12 (15 marks) Start a NEW booklet.
d dx
x3 1 .
(a)
Find
(b)
Find the limiting sum of 81 27 9 ......
(c)
Find
(d)
Find the exact value of sin
(e)
The gradient function of a curve is given by 3x 2 5 .
3 x 2
5
2
dx .
3 7 cos . 4 6
2
2
When a balloon is being filled with helium, its volume at time t is given by V
t3 12
cm3 , where t is in seconds.
Find the rate at which the balloon is being filled when t 1.
(g)
2
2
If the curve passes through the point 1,1 , find its equation.
(f)
Marks
2
For the parabola 6 y x 2 2 x 13 , use completing the square or otherwise to find:
(i)
the vertex.
2
(ii)
the focal length.
1
Page | 7
Caringbah High School 2016 Mathematics Trial HSC Question 13 (15 marks) Start a NEW booklet.
Marks
d cos x e . dx
(a)
Find
1
(b)
Find the shortest distance of the point 1, 3 from the line 2 x 5 y 4 .
(c)
In the figure below, ABCD is a square, CEF is an equilateral triangle,
2
DCE 130 and DC = CE.
B
F A
C 130 D
E
Find the size of CBF , giving clear reasons.
(d)
3
Find the value of k for which the equation 3x2 10 x k 0 has:
(i)
one root which is the reciprocal of the other.
1
(ii)
equal roots.
2
Question 13 continues on page 9
Page | 8
Caringbah High School 2016 Mathematics Trial HSC Question 13 (continued)
Marks
(e)
Maeve walks from her house for 6 km, on a bearing of 310 to point B. She then walks on a bearing of 215 to a point C which is due west of her house.
(f)
(i)
Copy the diagram and clearly indicate the information mentioned above.
1
(ii)
Calculate to 1 decimal place the distance of C to her house.
2
Part of the graph of the function y x2 ax 12, is shown below.
If the shaded area is 45 square units, find the values of a.
3
End of Question 13 Page | 9
Caringbah High School 2016 Mathematics Trial HSC Question 14 (15 marks) Start a NEW booklet. (a)
Marks
In the diagram below, the straight line L1 has equation 2 x y 4 0 and cuts the x-axis and y-axis at A and B respectively. The straight line L2 , passing through B and perpendicular to L1 cuts the x-axis at C. From the origin O, a straight line perpendicular to L2 is drawn to meet L2 at D.
(i)
Write down the coordinates of A and B.
1
(ii)
Show that the equation of L2 is x 2 y 8 0 .
2
(iii)
Find the equation of the line OD.
1
(iv)
8 16 Show that the coordinates of D are , . 5 5
1
(v)
Hence or otherwise, find the area of quadrilateral OABD.
2
Question 14 continues on page 11 Page | 10
Caringbah High School 2016 Mathematics Trial HSC Question 14 (continued)
(b)
Marks
(i)
x State the period for y 3cos . 2
(ii)
Neatly sketch the graph of y 3cos
1
x for 0 x 2 . 2
tan cos tan sin . sin
(c)
Prove that
(d)
Michael is attempting to solve the equation 2ln x ln 3x 10 .
2
3
He sets his work out using the following correct steps:
ln x2 ln 3x 10
x2 3x 10
x2 3x 10 0
He is now unsure of what to do next.
Complete the solution for Michael.
2
End of Question 14
Page | 11
Caringbah High School 2016 Mathematics Trial HSC Question 15 (15 marks) Start a NEW booklet.
(a)
Marks
A solid is formed by rotating the curve y x3 about the y-axis for 1 y 8 .
Find the volume of the solid in exact form.
3
(b)
A particle is observed as it moves in a straight line between t 0 and t 11 . Its velocity V m/s at time t is shown on the graph above.
(i)
What is the velocity of the particle after 4 seconds?
1
(ii)
What is the particle’s acceleration after 4 seconds?
1
(iii)
At what time after t 0 is the particle at rest?
1
(iv)
At what time does the particle change direction?
1
(v)
Explain what the shaded area represents.
1
Question 15 continues on page 13 Page | 12
Caringbah High School 2016 Mathematics Trial HSC Question 15 (continued)
(c)
Marks
In Daniel’s first year of fulltime employment his annual salary is $75 000. At the beginning of the second year his salary increases to $79 000. His salary continues to increase by $4000 at the beginning of each successive year.
(i)
What will Daniel’s salary be at the beginning of his 25th year?.
1
(ii)
Calculate Daniel’s total earnings after working for 25 years.
1
(iii)
During which year of employment will his total earnings
2
first exceed $2 000 000?
(d)
Determine the range of values of x when the curve y x 2 ln x is concave up.
3
End of Question 15
Page | 13
Caringbah High School 2016 Mathematics Trial HSC Question 16 (15 marks) Start a NEW booklet.
Marks
2
(a)
ex dx expressing the answer in simplified form. Evaluate x 1 e 1
3
X (b)
In ∆ XYZ, XY = XZ = 13 cm and YZ = 10 cm. A rectangle PQRS is inscribed in the triangle with PQ
Find the perpendicular height of ∆ XYZ.
(ii)
Using similar triangles, show that y 12
(iii)
(c)
1
y cm
(i)
x cm Q
P
parallel to YZ. Let PQ = x cm and QR = y cm.
Y
S
R
Z
2
6x . 5
If the area of the rectangle PQRS is A cm², find the maximum value of A.
Consider the function f x
2
4 . x 1 2
4 is an even function. x 1
(i)
Show that f x
(ii)
Explain why there are no x-intercepts for f x
(iii)
Find any stationary points and determine their nature.
3
(iv)
Neatly sketch the graph of y f x .
2
1
2
4 . x 1 2
1
End of paper Page | 14
CHS
YEAR 12 MATHEMATICS 2U 2016
Multiple Choice Section:
TRIAL HSC SOLUTIONS
Question 7.
1.B
2.A
3.C
4.B
5.D
6.D
7.B
8.C
9.A
10.D
Question 1. 2c= 1 − a 1− a ∴c = 2
−−−−−−−−− B
Question 2.
f (1) = 2 − 3 + 4 = 3
∴ 2 points of intersection
−−−−−−− B
f ( −1) = 2 + 3 + 4 = 9 f (1) − f ( −1) =− 6
−−−−−−−−− A
Question 3. decreasing exponentially
−−−−−−−−− C
Question 8. Using the test point (3,3) it can be determined that each inequality is true – hence III − − − − − C Question 9.
Question 4. OA + OB represents the sum of the roots.
T2 = − 3 → ar = − 3 − − − − 1 4 T5 192 → ar= 192 − − − − 2 =
∴ OA + OB = −
b = −b a
−−−−−−−−−−− A
− 64 2 ÷ 1 → r3 =
∴r= −4
−−−−−−−−−−− B
∴ log 10 y − log 10 x = log 10 z − log 10 y
Question 5.
( x − 3)2
+ ( y − 5) = 4 −−−−−−−−−−− D 2
=
5 −1 ( f (1) + 4 f ( 3) + f ( 5) ) 6 2 ( 4a + b ) 3
∴ 2log 10 y = log 10 x + log 10 z ∴ log 10 y 2 = log 10 ( xz )
Question 6. = A
Question 10.
−−−−−−−−−−− D
∴ y2 = xz − − − − − − − − − − − D
Question 11
a) e
π +3 2
−
81, r =− b) 81 − 27 + 9 −...... → a =
= 0.04638420
≈ 0.0464 ( 3Sig fig )
(
b) 50 − 2 x 2 = 2 25 − x 2
)
c)
= 2 ( 5 − x )( 5 + x ) c)
1 1 −= 2 x −1 x −1 =
d) 3 x − 5 =4
or
dy e) = 3 x 2 − 5 dx
g) ⌠ ⌡0
∴ y = x3 − 5 x + 5
(
π
)
=
3 3
1 − 1 3 2 × 3x 2 x 1 + ( ) 2 3x 2
2 x3 + 1
→
6y =
(x
2
π 3 cm / s 4
+ 2 x + 1) + 12 2
hence the vertex is ( −1, 2)
x x2 sec2 dx = 3 tan 3 3 0
x3 + 1 =
cm
dV π t 2 = dt 4
∴ 6 ( y − 2) = ( x + 1)
Question 12.
d dx
12
3
dV when = t 1,= dt
3 2 ____________________________________________
ii) for the focal length 4a = 6
→
Question 13 a)
a)
π t3
f) = V
3 1 → a= , b= 7 7
π = 3 tan − tan 0 = 6
+C
18
when x =1, y =1 → C =5
1 f) Domain: all real x ≥ 2 Range: all real y ≥ 0
2
6
∴ y = x3 − 5 x + C
3 x − 5 =− 4
g) i)
π
243 4
3π 7π 1 3 + cos = − 4 6 2 2
1− x −1 = ( x − 1)( x + 1)
−x ( x − 1)( x + 1)
(3 x − 2)
dx=
d) sin
1 3+ 2 × 3− 2 3− 2
3+ 2 = 7
∫ (3 x − 2)
5
1 x +1 − ( x − 1)( x + 1) ( x − 1)( x + 1)
= 3 x 9= or 3 x 1 1 ∴ = x 3 or = x 3 e)= LHS
81 = 1 1− − 3
= ∴ S∞
1 3
(
)
d cos x e = − sin x e cos x dx
b) d =
2 ×1 − 5 × − 3 − 4 = 2 22 + ( − 5 )
13 29
a=
c) ∠ FCE = 60 [∆ FCE is equilateral]
Question 14.
∠ BCD = 90 [D ABCD is a square] ∴ ∠ BCF = 360 − 130 − 90 − 60 = 80 and = since BC CF = as ( DC CE )
180 − 80° then θ = sides] ∠CBF = [∠ 's opp = 2 50 [ in isosceles ∆ ]
a) i)
A ( −2,0); B ( 0, 4) ;
1 ii) mL1 =→ 2 mL2 = − 2 1 ∴ L2 : y − 4 = − ( x − 0) using B(0, 4) 2 ∴ 2y − 8 = −x
d) i) Product of roots = 1 ∴
k = 1 3
→
∴ x + 2y − 8 = 0
k= 3
iv) Solve y = 2 x and x + 2 y − 8 = 0
ii) ∆ = 0 for equal roots
0 ∴ x + 2 ( 2x) − 8 =
∆= b 2 − 4ac= 100 − 12k 100= 12k
→
y = 2x
iii)
5x = 8
25 3
k=
→ x=
8 5
8 8 16 when x = , y = 2 × = 5 5 5
e)
v) C has coordinates (8, 0). 85°
•
6 km
55°
Area of quadrilateral OABD = Area of ∆ ABC − Area of D ODC
40° x km
=
1 1 16 36 2 × 10 × 4 − × 8 × = u 2 2 5 5
b) i) = P x = sin85
6 sin55
→ = x
6sin85 sin55
∴x = 7.3 km
f)
3
∫0
− x 2 + ax + 12 dx = 45 3
x3 ax 2 ∴ − + + 12 x =45 2 3 0 −9 +
9a += 36 45 2
→ = a 4
ii)
2π = 4π 1 2
sin θ 1 × − cos θ cos θ sin θ
c) LHS =
iv) t = 9s v) The distance travelled in the first 9 seconds.
1 − cos θ cos θ
=
c) i) $ 75000 + 24 × 4000 = $171000
1 − cos 2 θ = cos θ
ii) Total earnings for 25 years =
sin 2 θ = cos θ
$ 75000 + $79 000 + $83000 + ...... + $171000
sin θ sin θ × cos θ 1
=
= sin = θ tan θ
d)
=
25 ( $ 75000 + $171000 ) 2 = $3075000
RHS
0 ( x − 5)( x + 2 ) =
4 000 000 = 150000n + 4 000n 2 − 4 000n
∴ x= −2 5 or x = but since x > 0 then x = 5 only ___________________________________________
∴ 2n 2 + 73n − 2 000 = 0
Question 15. n=
y =x3 → x 2 =y 2 3
8
a) V = π ∫ x 2 dy 1
8
n ( 2 × 75000 + ( n − 1) × 4 000 ) 2
iii) 2 000 000=
2
∴ V= π ∫ y 3 dy
−73 ±
732 − 4 × 2 × − 2 000 4
and since n > 0 , n ≈ 18.26 hence during the 19th year.
1
8
3 5 = π y 3 5 1 =
=
b) i) 12 m/s ii) 0 m/s 2 iii) t = 9s
3π 5
5 53 3 8 −1
3π = ( 31) 5
93π 3 u 5
d) Concave up when y′′ > 0 ∴ y′ = x 2 ×
1 + 2 x × ln x x
= x + 2 x ln x
∴ y′′ = 1 + 2 x ×
1 + 2 × ln x x
= 3 + 2ln x ∴ 3 + 2ln x > 0
∴ ln x > −
3 2
→
x > e
−3
2
Question 16. 2
⌠ e a) x = dx ⌡1 e − 1 x
∴ 12 −
ln ( e − 1) 1 x
2
→
x = 5cm
6×5 Hence maximum area = 5 × 12 − 5 = 30cm 2
= ln ( e 2 − 1) − ln ( e − 1)
(e − 1)(e + 1) (e − 1) ln ( e + 1)
c) i) A function is even if f ( x= ) f (− x) .
= ln =
12 x = 0 5
4 f (− x) = 2 (− x) + 1
b) i) Perpendicular height is 12 cm (Pythagoras)
4 = 2 x +1
=
ii) Let D be the midpoint of PQ.
x ∴ DQ = and XD = 12 − y . 2 X
f ( x ) , hence even.
ii) x-intercepts occur when f ( x ) = 0 , and since the numerator is constant then f ( x ) ≠ 0. X iii)
f= ( x ) 4( x 2 + 1)
−1
∴ f ′( x ) = − 8 x ( x 2 + 1)
−2
−8 x = 2 ( x 2 + 1)
12
D C
Q
For stationary points f ′( x ) = 0 When = x 0, y 4. = Test for maximum or minimum:
12 − y 12 = x 5 2
∴ 6 x = 60 − 5 y ∴ y = 12 − iii) A = x y
x = 0.
Z
5
The triangles are similar (equiangular), hence corresponding sides are in the same ratio. ∴
→
→
∴ A = 12 x −
6x A = x 12 − 5
6 x2 5
dA d2A For a maximum= area 0 and < 0. dx dx 2 dA 12 x d 2 A 12 =− 12 ; = − < 0 hence maximum. 2 dx 5 dx 5
−1
0
𝑓𝑓′(𝑥𝑥 )
2
0
/
1 −2
\
Hence (0, 4) is a maximum stationary point. iv)
6x 5
x
