MATHEMATICS EXTENSION 2
2 August 2017
General
Instructions
•
Reading time – 5 minutes
•
Working time – 3 hours
•
Write using black pen.
•
NESA approved calculators may be used.
•
Commence each new question in a new booklet. Write
on both sides of the paper.
•
A reference sheet is provided.
•
In Question 11–16 show relevant mathematical reasoning
and/or calculations
•
At the conclusion of the examination, bundle the booklets used
in the correct order including your reference sheet within
this paper and hand to examination supervisor.
Total Marks: Section 1 – 10 marks (pages 3 - 6)
100
• Attempt Questions 1 – 10
•
Allow about 15 minutes for this section
Section 2 – 90 marks (pages 7 - 14)
•
Attempt Questions 11 – 16
•
Allow about 2 hours and 45 minutes for this section
NESA NUMBER:
........................
# BOOKLETS USED: . . . . .
Marker’s use only.
QUESTION 1-10
MARKS
10
11
12
13
14
15
16
Total
15
15
15
15
15
15
100
This task constitutes 40% of the HSC Course Assessment
1
2017 Mathematics Extension 2 HSC Trial Examination
3
Section I
10 marks
Attempt Question 1 to 10
Allow approximately 15 minutes for this section
Mark your answers on the answer grid provided (labelled as page 15).
1.
The circle |z − 3 − 2i| = 2 is intersected exactly twice by the line given by:
(A) Im(z) = 0
(B) |z − i| = |z + 1|
(C) Re(z) = 5
(D) |z − 3 − 2i| = |z − 5|
2.
If z1 + z2 + z3 = 0 and |z1 | = |z2 | = |z3 | = 1, then z12 + z22 + z32 equals:
(A) −3
(B) −
1
3
(C) 0
(D) 3
3.
The graph of y =
ax2
1
has asymptotes at x = 5 and x = −3.
+ bx + c
1
Given the graph has only one stationary point with y-value of − , it follows that:
8
1
15
(A) a = , b = 1 and c = −
2
2
1
15
(B) a = , b = −1 and c = −
2
2
(C) a = 1, b = 2 and c = −15
(D) a = 1, b = −2 and c = −15
KILLARA HIGH SCHOOL
2 AUGUST 2017
4
2017 Mathematics Extension 2 HSC Trial Examination
4.
√
1 + sin x dx equals:
√
(A) −2 1 − sin x + C
(B) sin
(C) cos
x
2
x
2
+ cos
− sin
x
2
x
2
+C
+C
√
(D) 2 1 − sin x + C
5.
In how many ways can 30 distinct toys be divided into 10 packets?
(A) 1030
(B) 3010
6.
(C)
30!
(3!)10
(D)
30!
10! × (3!)10
If ex + ey = 2 then
dy
is:
dx
(A) ex−y
(B) −ex−y
(C) ey−x
(D) −ey−x
2 AUGUST 2017
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
7.
5
Consider the following shaded region on the Argand diagram.
Which of the following inequations would represent the region?
8.
(A) |z − 1 − i| ≤ 2 and
π
≤ arg(z + 1) ≤ π
4
(B) |z + 1 + i| ≤ 2 and
π
≤ arg(z + 1) ≤ π
4
(C) |z − 1 − i| ≤ 2 and 0 ≤ arg(z + 1) ≤
π
4
(D) |z + 1 + i| ≤ 2 and 0 ≤ arg(z + 1) ≤
π
4
x2
+ y 2 = 1 are:
The equations of the directrices of the ellipse
9
1
(A) x = ± √
2 2
9
(B) x = ± √
2 2
(C) x = ±3
√
2 2
(D) x = ±
9
KILLARA HIGH SCHOOL
2 AUGUST 2017
6
9.
2017 Mathematics Extension 2 HSC Trial Examination
The region bounded by the curves y = x2 and y = x3 in the first quadrant is rotated
about the y-axis. Which integral could be used to find the volume of the solid of
revolution formed?
1
1
1
3
2
(A) V = π
y − y dy
0
1
1
1
y 2 − y 3 dy
1
2
y 3 − y dy
1
(B) V = π
0
(C) V = π
0
(D) V = π
0
x4 − x6 dx
x2 y 2
+
= 1 and
a2 b2
the chord P Q subtends a right angle at (0,0). Which of the following is the correct
expression?
10. The points P (a cos θ, b sin θ) and Q(a cos φ, b sin φ) lie on the ellipse
(A) tan θ tan φ = −
b2
a2
(B) tan θ tan φ = −
a2
b2
(C) tan θ tan φ =
b2
a2
(D) tan θ tan φ = −
2 AUGUST 2017
a2
b2
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
7
Section II
90 marks
Attempt Questions 11 to 16
Allow approximately 2 hours and 45 minutes for this section.
Write your answers in the writing booklets supplied. Additional writing booklets are
available.
Your responses should include relevant mathematical reasoning and/or calculations.
Question 11 (15 Marks)
sin x
(a)
Find
dx
cos3 x
Use a SEPARATE writing booklet
2
(b)
By splitting the integral find
(c)
Find
x2
3x + 1
dx
+ 2x + 3
x tan−1 (x) dx
2 dx
+ x2 + x + 1
Find
(e)
Use the substitution t = tan x2 to find the exact value of
3
π
2
0
KILLARA HIGH SCHOOL
3
3
(d)
x3
Marks
4
dx
5 + 4 cos x
2 AUGUST 2017
8
2017 Mathematics Extension 2 HSC Trial Examination
Question 12 (15 Marks)
Use a SEPARATE writing booklet
Marks
(a)
If p2 = 24 − 70i, express p in the form a + bi, where a and b are real.
(b)
Given that p and q are real and also that 1 − 4i is a root of the equation:
2
x2 + (p + i)x + (q − 5i) = 0
(c)
i. Find the values of p and q.
2
ii. Find the other root of the equation.
2
The complex number z = x + iy is such that
z − 8i
is purely imaginary.
z−6
i. Find the locus of the point P representing z.
2
ii. Sketch the locus of x on an Argand diagram.
2
y
(d)
B
β
A
α
0
x
The diagram show the equilateral triangle OAB in the complex plane. O
is the origin and the points A and B represent the complex numbers α and
β respectively.
π
π
+ i sin .
3
3
−→
i. Write down the complex number BA.
1
ii. Show that α = μ(α − β).
2
iii. Prove that α2 + β 2 = αβ.
2
Let μ = cos
2 AUGUST 2017
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
Question 13 (15 Marks)
(a)
9
Use a SEPARATE writing booklet
Marks
Two tangents OA and OB are drawn from a point O to a given circle.
Through A a chord AC is drawn parallel to the other tangent OB. OC
meets the circle at E.
Copy the diagram into your answer booklet
(b)
i. Prove that the triangles AF O and EF O are similar.
3
ii. Hence show that OF 2 = AF × EF .
1
iii. Hence prove that AE extended bisects OB.
2
Consider the function f (x) = (5 − x)(x + 1). On separate axes sketch,
using 13 of a page, showing all important features, the graphs of:
i. y = f (|x|)
1
f (x)
1
iii. y 2 = f (x)
1
ii. y =
(c)
1
If x3 + 3mx + n = 0 has a double root, prove that n2 = −4m3
2
Question 13 continues on page 10
KILLARA HIGH SCHOOL
2 AUGUST 2017
10
2017 Mathematics Extension 2 HSC Trial Examination
(d)
4
Use the method of cylindrical shells to calculate the volume of the solid
formed when the area bounded by y = x and y = x2 is rotated about the
line x = −1.
2 AUGUST 2017
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
Question 14 (15 Marks)
11
Use a SEPARATE writing booklet
Marks
(a)
Factorise P (x) = 2(x + 1)2 (2x + 1) + 1.
2
(b)
Prove by mathematical induction that for all integer values of n,
3
tan−1
(c)
π
1
1
1
1
+ tan−1
+ · · · + tan−1 2 = − tan−1
2
2
2×1
2×2
2n
4
2n + 1
Particles of mass m and 3m kilograms are connected by a light inextensible
string which passes over a smooth fixed pulley. The string hangs vertically
on each side, as shown in the diagram.
The particles are released from rest and move under the influence of gravity.
The air resistance on each particle is kv Newtons, when the speed of the
particles is v ms−1 and the acceleration due to gravity is g ms−2 and is
taken as positive throughout the question and is assumed to be constant.
k is a positive constant.
i. Draw diagrams to show the forces acting on each particle.
2
2
ii. Show that the equation of motion is:
mg − kv
2m
iii. Find the terminal velocity V or maximum speed of the system stating
your answer in terms of m, g and k.
ẍ =
1
iv. Prove that the time elapsed since the beginning of the motion is given
by:
2m mg ln t=
k
mg − kv 3
v. If the bodies attain a velocity equal to half of the terminal speed,
show by using the results in iii. and iv. that the time elapsed is
V
equal to ln 4, where V is the terminal velocity.
g
2
KILLARA HIGH SCHOOL
2 AUGUST 2017
12
2017 Mathematics Extension 2 HSC Trial Examination
Question 15 (15 Marks)
Use a SEPARATE writing booklet
(a)
Given that In = xn e2x dx
Marks
xn e2x n
− In−1 .
2
2
ii. Use the above result to find x2 e2x dx.
i. Show that In =
(b)
2
If α, β and γ are the roots of x3 + 5x2 − 2x − 3 = 0:
i. Find α2 + β 2 + γ 2 .
1
ii. Find α3 + β 3 + γ 3 .
2
iii. Find the equation whose roots are
(c)
3
1 1
1
,
and .
α β
γ
2
As shown in the diagram below, a solid has parallel vertical ends and the
base is horizontal. One end is a rectangle with length a and breadth b and
the other end is a scalene triangle with base c and height d. The parallel
ends are at a distance e apart. A typical slice is taken parallel to the ends
at a distance x from the triangular end.
i. Let the top of the trapezium be f , the base of the trapezium be g
and the height of the trapezium be h. Show that:
a−c
x
g =c+
e
and hence write down similar expressions for both f and h in terms
of x.
ii. Hence find the volume of the solid.
2 AUGUST 2017
3
2
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
Question 16 (15 Marks)
(a)
13
Use a SEPARATE writing booklet
The point P (a sec θ, b tan θ) lies on the hyperbola
S are the foci.
Marks
x2 y 2
−
= 1 and S and
a2 b 2
The equations of the tangent and normal at P are as given below:
x sec θ y tan θ
−
=1
a
b
by
ax
+
= a2 + b2
Normal:
sec θ tan θ
Tangent:
DO NOT PROVE THIS
Copy the diagram into your writing booklet.
The tangent at P intersects the x–axis at X and the y–axis at Y .
PX
i. Show that
= sin2 θ
PY
ii. Deduce that if P is an extremity of a latus rectum, then
PX
e2 − 1
.
=
PY
e2
3
1
Let the normal at P (a sec θ, b tan θ) intersect the y–axis at A.
iii. Prove that ASY is right angled at S
2
iv. Explain why ∠P SA = ∠P S A
2
Question 16 continues on page 14
KILLARA HIGH SCHOOL
2 AUGUST 2017
14
(b)
2017 Mathematics Extension 2 HSC Trial Examination
Newton’s method √
may be used to determine numerical approximations to
3
find the value of 2. This can be done by finding the real roots of the
equation x3 − 2 = 0.
Let x1 , x2 , x3 , . . . , xn , . . . be the series of estimators obtained by iterative
applications of Newton’s method.
i. Taking xn as the first root, use Newton’s method to show that:
2
1
xn+1 =
xn +
3
(xn )2
1
ii. Show algebraically that:
2
xn+1 −
iii. Given that xn >
iv. Show that x12
√
3
√
3
2=
xn −
√
√ 2 3
2 2xn + 3 2
3(xn )2
2 show that:
2
√
√
3
3
xn+1 − 2 < xn − 2
√
and 3 2 agree to at least 267 decimal places.
2
2
End of Examination 2 AUGUST 2017
KILLARA HIGH SCHOOL
.*$)%$?(" (!$#
+"*&"%
>4
?4
@4
/"#$*%$1$)%$"/%$
)%"+*%$$%)$%*,$/4
$1*)%"+*%$)
-‫ݖ‬ଵ ' ‫ݖ‬ଶ ' ‫ݖ‬ଷ .ଶ * ‫ݖ‬ଵଶ ' ‫ݖ‬ଶଶ ' ‫ݖ‬ଷଶ ' -‫ݖ‬ଵ ‫ݖ‬ଶ ' ‫ݖ‬ଵ ‫ݖ‬ଷ ' ‫ݖ‬ଶ ‫ݖ‬ଷ .
* ‫ݖ‬ଵଶ ' ‫ݖ‬ଶଶ ' ‫ݖ‬ଷଶ ' ‫ݖ‬ଵ ‫ݖ‬ଶ ‫ݖ‬ଷ ൬ ' ' ൰
‫ݖ‬ଷ ‫ݖ‬ଶ ‫ݖ‬ଵ
* ‫ݖ‬ଵଶ ' ‫ݖ‬ଶଶ ' ‫ݖ‬ଷଶ ' ‫ݖ‬ଵ ‫ݖ‬ଶ ‫ݖ‬ଷ -‫ݖ‬ଵ ' ‫ݖ‬ଶ ' ‫ݖ‬ଷ .
* ‫ݖ‬ଵଶ ' ‫ݖ‬ଶଶ ' ‫ݖ‬ଷଶ ' ‫ݖ‬ଵ ‫ݖ‬ଶ ‫ݖ‬ଷ ൫‫ݖ‬ଵ '‫ݖ‬ଶ ' ‫ݖ‬ଷ ൯
* ‫ݖ‬ଵଶ ' ‫ݖ‬ଶଶ ' ‫ݖ‬ଷଶ ' ‫ݖ‬ଵ ‫ݖ‬ଶ ‫ݖ‬ଷ -. * ‫ݖ‬ଵଶ ' ‫ݖ‬ଶଶ ' ‫ݖ‬ଷଶ * ‫ *ݕ‬ଶ
ܽ‫ܿ ' ݔܾ ' ݔ‬
)/#&*%*)*‫! * ݔ‬$‫ * ݔ‬െ
* ‫ ׵‬ଶ
- െ !.-‫ݔ‬
.- ' .
݇-‫ݔ‬
ܽ‫ܿ ' ݔܾ ' ݔ‬
ଵ
$1-$‫ * ݕ * ݔ‬െ ଼
*െ $
݇- െ !.- ' .
‫ ׵‬െ"݇ * െ$
ଵ
‫ * ݇ ׵‬
ଶ
ଵ
ଵ
Hభ మ
భ
-௫ିହ.-௫ାଷ.
మ
-௫ ିଶ௫ିଵହ.
మ
ܽ * ܾ * െ
!
ܿ * െ ‫ ׬‬ξ ' ‫ݔ݀ݔ݊݅ݏ‬
A4
* ‫׬‬
* ‫׬‬
B4
>#(!3**#&*)*%.&())
‫ ݔ‬ଶ െ ‫ ݕ‬ଶ * ܽ݊݀‫! * ݕݔ‬
>#(!3,)*$)-($*
%(#(-ܽ ' ܾ݅.+)$*( $ 4
$/)*+$*)+)*#%(
"%(%+)#*%*%)%",*)
&(%"#
%+$*%"($)(
#*%)
>#(!3)+)**+*) െ ݅$*%
*'+*%$$)&(*)
("$#$(/&(*)
>#(!3%",)%($ %#)*+$*)+"*/$
)&(*$("$#$(/
&(*)
?#(!)3%((*$)-((%#
%((*-%(!$
>#(!3#$%(((%($
"+"*%$)
*+$*)-%+))+#)$
&(%+*%(%%*)#*%-(
#%()+))+"4
%$*)($%*("4
$*%$ +*$6*
(%%*
ξଵା௦௜௡௫ξଵି௦௜௡௫
݀‫ݔ‬
ξଵି௦௜௡௫
௖௢௦௫
ξଵି௦௜௡௫
݀‫ݔ‬
* െξ െ ‫ݔ݊݅ݏ‬G
30 distinct toys need to be equally
divided into 10 packets.
Number of toys in each packet =
ଷ଴
* ଵ଴
Since packets do not have distinct
identity, we can consider that all
groups are identical (not distinct).
i.e., we need to divide 30 distinct
toys into 10 identical groups
containing 3 toys each.
‫ ׵‬Total no. of ways =
ଷ଴
ଵ଴)-ଷ.భబ
+)*%$>?
ଶ
:9: ߩ * ଶ െ #݅
* # െ )#)!݅ ' -!݅.ଶ
* -# െ !݅.ଶ ‫( * ߩ ׵‬-# െ !݅.
9:
‫ ݔ‬ଶ ' -‫݅ ' ݌‬.‫ ' ݔ‬-‫ ݍ‬െ !݅.H=
െ ݅)(%%*4
+)**+* െ ݅
‫ ׵‬- െ ݅.ଶ ' -‫݅ ' ݌‬.- െ ݅. ' ‫ ݍ‬െ !݅ * െ! െ $݅ ' ‫ ݌‬െ ‫ ݍ ' ' ݅ ' ݅݌‬െ !݅ * '+*$("$#$(/&(*)1
െ! ' ‫ * ݍ ' ' ݌‬
‫ * ݍ ' ݌ ׵‬9>:
െ$ െ ‫ ' ݌‬െ ! * ‫ * ݌ ׵‬െ
‫ * ݌ ׵‬1'H>A
9:
‫ ݔ‬ଶ ' -െ ' ݅.‫ ' ݔ‬- െ !݅.H=
%%*)+# െ ݅ ' ߚ * െ ݅
‫݅ ' * ߚ ׵‬
:9:
9:
௭ି଼௜
)&+("/#$(/
௭ି଺
‫ ݖ‬െ $݅ ‫ ݖ‬െ "
)
‫ݖ‬െ" ‫ݖ‬െ"
‫ ݖ‬െ $݅ ‫ ݖ‬െ "
)
‫ݖ‬െ" ‫ݖ‬െ"
‫ ݖݖ‬െ "‫ ݖ‬െ $݅‫ ' ݖ‬$݅
‫ ݖ‬െ "ଶ ଶ
ଶ
‫ ݕ ' ݔ‬െ "-‫ݕ݅ ' ݔ‬. െ $݅-‫ ݔ‬െ ݅‫ݕ‬. ' $݅
-‫ ݔ‬െ ".ଶ ' ‫ ݕ‬ଶ
)&+("/#$(/4
‫ ݔ ׵‬ଶ ' ‫ ݕ‬ଶ െ "‫ ݔ‬െ $‫ * ݕ‬
‫ ݔ‬ଶ െ "‫ ' ݔ‬% ' ‫ ݕ‬ଶ െ $‫! * " ' ݕ‬
-‫ ݔ‬െ .ଶ -‫ ݕ‬െ .ଶ * !
))("-*$*(- .$(+)!4
?#(!)3%((*$)-((%#
%((*-%(!$
>#(!3#$%(((%(
$1+))(#*%)
%#&(*%*"(7
,/&&(%4
?#(!)3%((*(#
*+$*)$*%."+9=1E:
$9C1=:)0$$%*%$*
.*(#*)%*#*(4
>#(!39=1E:$9C1=:($%*
."+
:9:
ሬሬሬሬሬԦ
‫ ߙ * ܣܤ‬െ ߚ
>#(!3%((*,*%(
(&()$**%$
9:
>#(!3%((*(#
>#(!3.&()))
ሬሬሬሬሬሬሬԦ
ሬሬሬሬሬԦ$$
‫ܣܤ‬Ԣ ‫ܣܤ‬
*()+"*
(-$*(#)'+*
))$*"$*))4*+$*)
)%+"(-6$.&"$
ሬሬሬሬሬሬሬԦ * ሬሬሬሬሬԦ
ܱ‫ܣ‬
)%-$‫ܣܤ‬Ԣ
9:
ሬሬሬሬሬԦ * ሬሬሬሬሬሬሬԦ
ߤ‫ܣܤ‬
‫ܣܤ‬Ԣ * ߙ
ߤ-ߙ െ ߚ. * ߙ(%#9:
%-1ߚ * ߤߙ
ߙെߚ
ߙ
- .
‫* ׵‬
ߙ
ߚ
ߙ ଶ * ߙߚ െ ߚ ଶ ߙ ଶ ' ߚ ଶ * ߙߚ
>#(!3.&()))
ሬሬሬሬሬԦ
ሬሬሬሬሬԦ
ܱ‫ܣܱ ܤ‬
>#(!3-(*)*-%
.&())%$)%(ߤ$
'+*)*%**
()+"*
$*)'+)*%$1/%+$*%
&(%,*()+"*2)+)**+*$
$*%*)$%*)+$*4
)*%"#$*ߤ 4*)
#+)*+$()*%%/*
)*+$*)()*4
ߚ * ߤߙ
ߤ-ߙ െ ߚ. * ߙ +)*%$>A
:9:
ܲ-‫ݔ‬. * -‫ ' ݔ‬.ଶ -‫ ' ݔ‬. ' * ‫ ݔ‬ଷ ' ‫ ݔ‬ଶ ' $‫ ' ݔ‬
ܲ ൬െ ൰ * $‫ ' ݔ‬
>#(!3%((*"$(
*%(
"/%$4 $/)*+$*)
.&$*&%"/$%#"1+*
$%*%*(*%*%()*4
>#(!3,)/*
"$(*%($$
**%(%(#
(+)%#&()%$%
%$*)
‫ ݔ‬ଷ ' ‫ ݔ‬ଶ ' $‫* ' ݔ‬
-‫ ' ݔ‬.-‫ ݔ‬ଶ ' ‫ ' ݔ‬.
9:
*&>
%(H>1$)*%&(%,
ߨ
ିଵ
* െ ିଵ
ଶ
)
) ' ଵ
ଵ
%$)(ିଵ ଶ ' ିଵ ଷ
ଵ
ଵ
‫ ݊ܽݐ‬ቀିଵ ଶ ' ିଵ ଷቁH
భ భ
ା
మ య
భ భ H>
ଵି )
మ య
ߨ
ିଵ ' ିଵ * $1*()+"*)*(+ %($H>4
*&?
))+#*()+"*)*(+%($H!4
$1
ଵ
ଵ
‫ݏ‬௞ * ିଵ ଶ)ଵమ ' ିଵ ଶ)ଶమ '
ଵ
గ
ଵ
‫ି ' ڮ‬ଵ ଶ௞ * ସ െ ିଵ ଶ௞ାଵ
To prove the result is true for n =
k+1, we need to prove that
Step 3
*
‫ݏ‬௞ ' ିଵ
-݇ ' .ଶ
ߨ
' ିଵ
݇ ' -݇ ' .ଶ
4-$*%&(%,
*
െିଵ
@#(!)3%((*&(%%
?#(!)3
*&>&(%,)+(*"/4
#+)*#%$)*(**+)%
*$9G:%(#+"1-(*)*
()+"*)%($H!$$H!G>$
#!))$$*&(%())4
>#(!3&(%,)*()+"*%($
H>9$%*$))("/+)$
*$9G:$-(*)*()+"*
%($H!$$H!G>
ଵ
ଵ
ଵ
ିଵ ଶ-௞ାଵ.మ െ ିଵ ଶ௞ାଵHିଵ ଶ௞ାଷ
%$)(1
െ ିଵ
൰
ଶ
-݇ ' .
݇ ' భ
భ
మି
൬ିଵ
మ-ೖశభ. మೖశభ
భ
భ
)
మ-ೖశభ.మ మೖశభ
H
ଵି
ଶ௞ మ ାଶ௞ାଵ
Hെ ଶ-௞ାଵ.మ -ଶ௞ାଵ.ାଵ
ଶ௞ మ ାଶ௞ାଵ
ିଵ
Hെ -ଶ௞ାଷ.-ଶ௞ మ ାଶ௞ାଵ. * ଶ௞ାଷ
$1ܵ௞ ܵ௞ାଵ $1+)$&($&"%
#*#*"$+*%$1*()+"*
%")%%4
:9:
>#(!3%((*
(%/(#)
9:
$*%/%#))!3
݉‫ݔ‬ሷ * ܶ െ ݇‫ ݒ‬െ ݉݃9>:
$*%/%#)) !3
݉‫ݔ‬ሷ * ݉݃ െ ݇‫ ݒ‬െ ܶ9?:>#(!
(%#9>:1
ܶ * ݉‫ݔ‬ሷ ' ݇‫݃݉ ' ݒ‬
+4 $9?:1
݉‫ݔ‬ሷ * ݉݃ െ ݇‫ ݒ‬െ ݉‫ݔ‬ሷ െ ݇‫ ݒ‬െ ݉݃
A݉‫ݔ‬ሷ * ݉݃ െ ݇‫ݒ‬
ଶ௠௚ିଶ௞௩
‫ݔ‬ሷ *
‫ݔ‬ሷ *
ସ௠
௠௚ି௞௩
ଶ௠
*+$*)$*%(-
)&(*%((#)%(
&(*"
>#(!3-(*)*
%((*%(
'+*%$)%(
&(*"4
>#(!3"#$*)
$&(%,)*()+"*
9*!/&(*%*
'+)*%$)*%(")**
**$)%$$*)*($)
*)#*(%+%+**
)*($:
$ *#*)1%(
'+*%$)%(%/#+)*
-(**$%((,$*
()+"*)
7""()+"*)#+)*&(%,1
$%*+)%(#+"
$/)*+$*)()%(**%
+$*($)-(
9:
9,:
%(*(#$","%*/1‫ݔ‬ሷ * ݉݃ െ ݇‫ * ݒ‬
݉݃
‫*ݒ‬
݇
௠௚
$1ܸ * ௞
(%#9:1
ௗ௩
௠௚ି௞௩
* ଶ௠ ௗ௧
݉
݀‫* ݐ‬
݀‫ݒ‬
݉݃ െ ݇‫ݒ‬
௧
௩
݉
න ݀‫ * ݐ‬න
݀‫ݒ‬
଴
଴ ݉݃ െ ݇‫ݒ‬
݉ ௩ )െ݇
න
݀‫ ݒ‬
‫* ݐ‬െ
݇ ଴ ݉݃ െ ݇‫ݒ‬
* െ ଶ௠
/݈݊-݉݃ െ ݇‫ݒ‬.0௩଴ ௞
*െ
݉݃ െ ݇‫ݒ‬
݉
൬
൰
݉݃
݇
*
݉
݉݃
ฬ
ฬ
݇
݉݃ െ ݇‫ݒ‬
>#(!3)*)‫ݔ‬ሷ * $
#!))+ *4
9#+)*)%--%(!$:
""%$
@#(!)3%((*&(%%
?#(!)3%((*"/
$*(*)1+*#$%(((%(
$,"+*$1*
%$)*$*%$*(*%$$
$$%((*()+"*4
>#(!3.&()))
௠௚ି௞௩
‫ݔ‬ሷ *
)
ௗ௩
ௗ௧
$/)*+$*)$%*%*(
*%+)*()+"*,$$9:*%
+)$*)'+)*%$4
ଶ௠
௠௚ି௞௩
$
ଶ௠
*
**#&*)*%$*(*4
"*($*-%(!$3
ௗ௩
௠௚ି௞௩
*
ௗ௧
ଶ௠
݉
݀‫* ݐ‬
݀‫ݒ‬
݉݃ െ ݇‫ݒ‬
݉
݀‫ݒ‬
‫׬ * ݐ݀ ׬‬
݉݃ െ ݇‫ݒ‬
݉
)െ݇
‫* ݐ‬െ
‫׬‬
݀‫ݒ‬
݇
݉݃ െ ݇‫ݒ‬
ଶ௠
* െ /݈݊-݉݃ െ ݇‫ݒ‬.0 ' ‫ܥ‬
௞
$*H=1‫ * ݒ‬
݉
/݈݊-݉݃.0 ' ‫ܥ‬
* െ
݇
݉
/݈݊-݉݃.0
‫* ܥ‬
݇
ଶ௠
‫ * ݐ׵‬െ /݈݊-݉݃ െ ݇‫ݒ‬.0 '
௞
ଶ௠
/݈݊-݉݃.0
௞
4
݉
݉݃
‫ * ݐ‬
ฬ
ฬ
݇
݉݃ െ ݇‫ݒ‬
݉݃ െ ݇‫ݒ‬
݉
൬
൰
*െ
݉݃
݇
9,:
‫ * ݐ‬
݉
݉݃
ฬ
ฬ
݇
݉݃ െ ݇‫ݒ‬
௠௚
௏
* $‫* ݒ‬
ଶ௞
‫ * ݐ‬
ଶ
݉݃
݉
ቮ
݉݃ቮ ݇
݉݃ െ ݇)
݇
""%$
>#(!3+)**+*)
௠௚
‫ * ݒ‬ଶ௞ $*%*.&())%$
%(*#*$9,:$
)#&")
* ݉
݉݃
ቮ ݉݃ ቮ
݇
݉
݇
௠
௏
+* * ௞
௚
ܸ
‫ * ݐ‬
݃
ܸ
* ݃
ܸ
* ݃
*
>#(!3.&()))*
()+"*$*(#)%*
)&,"%*/$
&(%,)*()+"*
+)*%$>C
:9: ௫௦௘௖ఏ
௬௧௔௡ఏ
$$*)
െ
* ௔
௕
8$*(&*3 H=
ܾ
‫*ݕ‬െ
‫ߠ݊ܽݐ‬
௕
ቁ$ܳ- ܾ‫ߠ݊ܽݐ‬.
ܻ ቀ െ
௧௔௡ఏ
$οܱܻܺοܱܲܳ
ܻܺ ܱܻ
*
*
ܻܲ ܱܳ
ܾ
‫ߠ݊ܽݐ‬
ܾ
' ‫ߠ݊ܽݐ‬
‫ߠ݊ܽݐ‬
ܾ
ܾ ' ܾ‫ ݊ܽݐ‬ଶ ߠ
ଶ
* ߠ༃
ܻܲ ܲܺ ' ܻܺ
*
* ଶ ߠ
ܻܺ
ܻܺ
ܲܺ
*
' * ଶ ߠ
ܻܺ
௉௑
‫ * ׵‬ଶ ߠ༄
௑௒
௉௑
༃)ᬆ,) * ଶ ߠ
௉௒
)*"*+)(*+#1*$ܽ‫݁ܽ * ߠܿ݁ݏ‬
‫݁ * ߠܿ݁ݏ ׵‬
ܲܺ
݁ଶ െ ‫ ׵‬
* ܻܲ
݁ଶ
/7$*(&*%$%(#"3
௔௫
௕௬
௔௫
' ௧௔௡ఏ * ܽଶ ' ܾ ଶ ௦௘௖ఏ '
௦௘௖ఏ
* 9:
9:
௕௬
* ܽଶ
௧௔௡ఏ
' ܾ ଶ $1+))(
#*%)*$()%(*$
*%"%$*(%$%#*(
(,*%$)
>#(!3"+"*$*
%%($*)%$
$-(*$*)#"(*/
(*%
>#(!3%(,"%&$
༃
ϭŵĂƌŬ͗ĨŽƌĚĞǀĞůŽƉŝŶŐ༄
$$*$"()+"*4
>#(!3%(%((*&(%%
-*-%(!$
""%$4
ܾ‫ݕ‬
* ܽଶ ' ܾ ଶ ‫ߠ݊ܽݐ‬
ܾ‫ * ݕ‬-ܽଶ ' ܾ ଶ .‫ߠ݊ܽݐ‬
‫ * ݔ‬
* ܽଶ ݁ ଶ ‫ߠ݊ܽݐ‬
ܽଶ ݁ ଶ ‫ߠ݊ܽݐ‬
‫* ݕ ׵‬
ܾ
ଶ ଶ
ܽ ݁ ‫ߠ݊ܽݐ‬
‫ ܣ‬ቆ
ቇ ܵ-ܽ݁ .
ܾ
ܽଶ ݁ ଶ ‫ߠ݊ܽݐ‬
െ
ܾ
݉ௌ஺ *
ܽ݁ െ െ
௔௘௧௔௡ఏ
௕
ܾ
' ‫ߠ݊ܽݐ‬
>#(!3"+"*)
*(%*
($*)
>#(!3&(%,)*
()+"*
ܾ
ܽ݁ െ ܽ݁‫ߠ݊ܽݐ‬
ܽ݁‫ߠ݊ܽݐ‬
ܾ
)
݉ௌ஺ )݉ௌ௒ * ܾ
ܽ݁‫ߠ݊ܽݐ‬
݉ௌ௒ *
9,:
:9:
* * െ
ܵ‫ ܣ‬٣ ܻܵο‫ܵ ܻܵܣ‬
‫ * ܻܵܣס * ܻܲܣס‬%&
- . %&
$)/"'+("*("$/
)/##*(/ܵԢ")%)%+""%$*4
‫ܵܣס * ܲܵܣס‬Ԣܲ9$")$*)#
)#$*%(('+"4>#(!
݂-‫ݔ‬. * ‫ ݔ‬ଷ െ ݂ ᇱ -‫ݔ‬. * ‫ ݔ‬ଶ ݂-‫ݔ‬௡ .
‫ݔ‬௡ାଵ * ‫ݔ‬௡ െ
݂Ԣ-‫ݔ‬௡ .
‫ݔ‬௡ଷ െ ‫ݔ‬௡ାଵ * ‫ݔ‬௡ െ
‫ݔ‬௡ଶ
‫ݔ‬௡ଷ െ ‫ݔ‬௡ଷ ' * ‫ݔ‬௡ଶ
‫ݔ‬௡ଷ ' * ‫ݔ‬௡ଶ
ଶ
ଵ
* ቀ‫ݔ‬௡ ' మ ቁ
ଷ
௫
>#(!3(%,)**
(%$7/"
>#(!3&(%,)*()+"*
-*()%$$
>#(!3%((*
&&"*%$% -*%$6)
#*%*%‫ ݔ‬ଷ െ * $&(%,)*()+"*4
""%$
೙
9:
య
ଶ
య
൫‫ݔ‬௡ െ ξ൯ ൫‫ݔ‬௡ ' ξ൯
‫ݔ‬௡ଶ
ଶ
య
య
൬‫ݔ‬௡ଷ െ ξ‫ݔ‬௡ ' ଷ ൰൫‫ݔ‬௡ ' ξ൯
*
‫ݔ‬௡ଶ
య
ହ
‫ ݔ‬ଷ െ ξ‫ݔ‬௡ଶ ' ଷ ‫ݔ‬௡ '
ቍ
ቌ ௡
ହ
య
ξ‫ݔ‬௡ଷ െ ଷ ‫ݔ‬௡ ' *
‫ݔ‬௡ଶ
?#(!)3%((*&(%%
>#(!3)$$*
&(%())*%*()+"*
య
* ‫ݔ‬௡ െ ξ ' ଶ ‫ݔ‬௡
9:
య
* ቆ‫ݔ‬௡ െ ଶ ቇ െ ξ
‫ݔ‬௡
య
* ‫ݔ‬௡ାଵ െ ξ
య
య
‫ݔ‬௡ , ξ֜ ‫ݔ‬௡ െ ξ , య
‫ݔ‬௡ାଵ െ ξ *
ଶ
య
య
൫‫ݔ‬௡ െ ξ൯ ൫‫ݔ‬௡ ' ξ൯
‫ݔ‬௡ଶ
ଶ
య
൫‫ݔ‬௡ െ ξ൯ -‫ݔ‬௡ ' ‫ݔ‬௡ .
૚࢓ࢇ࢘࢑
+ ‫ݔ‬௡ଶ
య
*
య
>#(!3)+)**+*) ξ
-*‫ݔ‬௡ $,)*
$'+"*/
>#(!3&(%,)
ଵ
+ $,)*
ଶ
൫‫ݔ‬௡ െ ξ൯
‫ݔ‬௡
య
ଶ
య
+ ൫‫ݔ‬௡ െ ξ൯ )‫ݔ‬௡ , ξ , ଵ
$ + ௫೙
య
య
ଶ
‫ݔ ׵‬௡ାଵ െ ξ + ൫‫ݔ‬௡ାଵ െ ξ൯ 9,:
௫೙
('+(()+"*
)$9:
య
ଶ
య
‫ݔ‬ଶ െ ξ + ൫‫ݔ‬ଵ െ ξ൯ &&"/$‫ݔ‬ଵ * య
ଶ
య
‫ݔ‬ଶ െ ξ + ൫ െ ξ൯ య
?#(!3%((*&(%%
ଶ
య
‫ݔ‬ଷ െ ξ + ൫‫ݔ‬ଶ െ ξ൯ ସ
య
య
ଶమ
+ ൫ െ ξ൯ H൫ െ ξ൯ య
ଶ
య
‫ݔ‬ସ െ ξ + ൫‫ݔ‬ଷ െ ξ൯ ଼
య
య
ଶయ
+ ൫ െ ξ൯ H൫ െ ξ൯ 3
య
ଶ
య
‫ݔ‬ଵଶ െ ξ + ൫‫ݔ‬ଷ െ ξ൯ య
଼
య
ଶభభ
+ ൫ െ ξ൯ H൫ െ ξ൯ H%$)ିଶ଺଼ H=4======54=>F>>E5
9?CD0(%):
య
‫ݔ‬ଵଶ ξ()*%?CD#"&")
>#(!3%(#!$
)$$*&(%())$
,"%&$*)'+$4
16
2017 Mathematics Extension 2 HSC Trial Examination
Solutions
Question 11 (15 Marks)
(a)
1 mark
1 mark
Makes a substitution or equivalent working
Correct result
sin x
dx
cos3 x
Let u = cos x =⇒ du = − sin x dx
sin x
du
∴
dx = −
3
cos x
u3
1
= u−2 + C
2
1
+C
=
2 cos2 x
tan2 x
+C
alternatively =
2
(b)
1 mark
1 mark
1 mark
Splits the integral correctly
Evaluates the log part correctly for their split
Correctly evaluates the tan−1 part for their integral
(c)
1 mark
1 mark
1 mark
3(2x + 2)
2
dx
− 2
2
2(x + 2x + 3) x + 2x + 3
1
3 log(x2 + 2x + 3)
−2
dx
=
2
(x + 12 ) + 2
3 log(x2 + 2x + 3) √
x+1
−1
√
=
− 2 tan
+C
2
2
3x + 1
dx =
2
x + 2x + 3
Substituting into parts formula correctly
Evaluating tan−1 component correctly
2
Correctly evaluating 2(xx2 +1) dx
x tan−1 (x) dx
u = tan−1 x
u =
∴
−1
x tan (x) dx =
=
=
=
LAST UPDATED AUGUST 23, 2017
x2
x2
−1
tan x −
dx
2
2(x2 + 1)
1
x2
1
−1
tan x −
1− 2
dx
2
2
(x + 1)
x2
1
tan−1 x −
x − tan−1 (x)
2
2
1 2
( x + 1 tan−1 (x) − x + C
2
x2
1
+1
v=
x2
2
v = x
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
(d)
1 mark
1 mark
1 mark
17
Correctly factorising and expressing in partial fractions
Splitting into three distinct parts
Correctly evaluating all parts
2
2
= 2
+ +x+1
x (x + 1) + 1(x + 1)
2
= 2
(x + 1)(x + 1)
c
ax + b
+
≡ 2
x +1 x+1
∴ (ax + b)(x + 1) + c(x2 + 1) = 2
Put x = −1
∴ 2c = 2 =⇒ c = 1
2
2
Now ax + ax + bs + b + x + 1 = 2
∴ (a + 1)x2 + x(a + b) + b + 1 = 2
a+1=0
=⇒ a = −1
a+b=0
=⇒ b = 1
2
1−x
1
∴
dx =
+
dx
3
2
2
x +x +x+1
x +1 x+1
1
x−1
dx
+
dx
=−
x2 + 1
x+1
1
x
dx +
dx + log(x + 1)
=−
2
2
x +1
x +1
1
= − log(x2 + 1) = tan−1 x = log(x + 1) + C
2
x3
x2
Markers Comment: Most did this very well - a few errors when people skipped
steps or failed to recognise standard integrals.
KILLARA HIGH SCHOOL
LAST UPDATED AUGUST 23, 2017
18
(e)
2017 Mathematics Extension 2 HSC Trial Examination
Solutions
2 marks Changing the limits and finding dx in terms of t
1 mark
1 mark
1
dt
Correct substitution leading to 0 t22 +9
Correctly evaluating their integral
2π
0
dx
dx
5 + 4 cos x
t = tan x2 =⇒ cos x =
∴
2π
0
1 − t2
1 + t2
∴ x = tan−1 (2t)
2 dt
∴ dx =
1 + t2
x = 0 =⇒ t = 0
π
x=
=⇒ t = 1
2
2 dt
1
dx
1 + t2 dx =
1 − t2
5 + 4 cos x
0
5+4
1 + t2
2 dt
1
1 + t2
=
2
2
0 5(1 + t ) + 4(1 − t )
1 + t2
2 dt
= 01 2
t +9
1
t
1
−1
tan
=2
3
3 0
2
1
= tan−1
3
3
Markers Comment: Very well done.
LAST UPDATED AUGUST 23, 2017
KILLARA HIGH SCHOOL
20
2017 Mathematics Extension 2 HSC Trial Examination
Solutions
Question 13 (15 Marks)
(a)
i. 3 marks Correct proof with all correct reasons using correct terminology
2 marks Correct proof with 2 correct reasons
1 mark
Significant progress towards proof with correct reasons
∠OAF = ∠ACO
∠OAF = ∠ACO = ∠COB
In AF Oand EF O
∠AF O = ∠EF O
∠OAF = ∠EOF
∴ ∠AOF = ∠OEF
∴ F AO ||| F OE
(the angle between a tangent and a chord
through the point of contact is equal to
the angle in the alternate segment)
(alternate angles AC OB)
(common)
(proven above)
(angle sum of a triangle)
(equiangluar)
Markers Comment: Generally quite well done but many still not being
precise enough with proof - don’t take the risk of losing marks
ii. 1 mark
Correct ratios with correct reasoning
FO
FA
=
(corresponding sides in congruent triangles)
FE
OF
∴ OF 2 = EF × AF
Markers Comment: Well done - but reason often not given
LAST UPDATED AUGUST 23, 2017
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
21
iii. 2 marks Correct solution with correct reasoning
1 mark
Significant progress but incorrect reasoning
But AF × F E = F B 2
(The square of the length of the tangent fro
external point is equal to the product o
intercepts of the secant passing through this po
∴ OF 2 = F B 2
∴ OF = F B
∴AE extended bisects OB
Markers Comment: Very poorly done - this is the forgotten circle geometry
theorem - Learn it!
(b)
For each of the graphs
1 mark
Correct accurate shape with relevant intercepts and asymptotes labeled
i.
ii.
KILLARA HIGH SCHOOL
LAST UPDATED AUGUST 23, 2017
22
2017 Mathematics Extension 2 HSC Trial Examination
Solutions
iii.
Markers Comment:Most were fairly well done although some students still
not accurate enough with their graphs
(c)
1 mark
1 mark
Uses the double root property to establish a solution
Logical reasoning leading to the desired result
x3 + 3x + n = 0
If has a double root then :
3x2 + 3m = 0
√
3
√
∴
−m + 3m −m + n = 0
√
√
∴ − −m + 3m −m = −n
√
∴ 2m −m = −n
∴ −4m3 = n2
√
=⇒ x = ± −m
Markers Comment:Generally well done - no need to use complex numbers!
LAST UPDATED AUGUST 23, 2017
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
(d)
23
1 mark
Finds correct radius and height
1 mark
1 mark
1 mark
Establishing the integral V = 2π 0 (1 + x)(x − x2 ) dx
Correct primitive function for their integral
Correct solution for their integral
1
ΔV ∼
= 2πrhδx
Radius of a typical shell = x + 1
Height of a typical shell = x − x2
1
∴ V = 2π
(1 + x)(x − x2 ) dx
0 1
= 2π
(x − x3 ) dx
0
1
x2 x4
−
= 2π
2
4
0
1 1
= 2π
−
2 3
π
= units3
2
Markers Comment:Generally very well done - those that made mistakes did
so because they skipped steps
KILLARA HIGH SCHOOL
LAST UPDATED AUGUST 23, 2017
2017 Mathematics Extension 2 HSC Trial Examination
25
Question 15 (15 Marks)
(a)
i. 1 mark
Splitting into parts correctly
1 mark
Substituting into parts formula correctly
1 mark
Resolving the parts formula to get desired reduction formula
In =
xn e2x dx
u = xn
∴ In = (xn )
xn e2x
2
n 2x
x e
=
2
=
n−1 1 2x
e
−
nx
2
n
−
xn−1 e2x dx
2
n
− In−1
2
u = nxn−1
1 2x
e
2
1
v = e2x
2
v = e2x
dx
Markers Comment:Generally well done - some were not explicit enough
with their parts integration and made errors.
ii. 1 mark
1 mark
Getting to 12 x2 e2x −
1
xe2x
2
− 12 I0 or similar place
Correctly using reduction formula to obtain final result
I2 =
x2 e2x dx
1
= x2 e2x − I1
2
1 2 2x
1 2x 1
= x e − xe − I0
2
2
2
1 2 2x 1 2x 1
= x e − xe +
x0 e2x dx
2
2
2
1
1
1
= x2 e2x − xe2x + e2x + C
2
2
4
e2x 2
2x − 2x + 1 + C
=
4
Markers Comment:Skipping steps lead to sign errors - show all working
KILLARA HIGH SCHOOL
LAST UPDATED AUGUST 23, 2017
26
(b)
2017 Mathematics Extension 2 HSC Trial Examination
i. 1 mark
Solutions
Correct result with working
x3 + 5x2 − 2x − 3 = 0
∴
α2 = (α + β + γ)2 − 2(α + β + γ)
= 25 − 2(−2)
= 29
Markers Comment:Well done
3
Generates the expression
α + 15 α2 − 8 α − 9 = 0 or
similar equivalent progress towards solutions
ii. 1 mark
1mark
substitutes in correctly to obtain desired result
Substitute the roots into the polynomial
∴α3 + 5α2 − 2α − 3 = 0
β 3 + 5β 2 − 2β − 3 = 0
γ 3 + 5γ 2 − 2γ − 3 = 0
α2 − 2
α−9=0
∴
α3 + 5
∴
α3 = 9 − 5
α2 + 2
α
= 9 − 5(29) + 2(−5)
= −146
Markers Comment:Some students made very careless errors trying to skip
steps - some were very inefficent
iii. 1 mark
Attempts to substitute in roots and manipulate or equivalent
1
1
1
significant progress producing the expression x − α x − β x − γ =
0
1mark
Obtains desired solution
1 1
1
,
and
then
α β
γ
1
1
1
x−
x−
x−
α
β
γ
1
1
1
∴
−α
−β
−γ
x
x
x
5
2
1
∴ Equation is: 3 + 2 − − 3 = 0
x
x
x
∴1 + 5x − 2x2 − 3x3 = 0
∴ 3x3 + 2x2 − 5x − 1 = 0
If the equation has roots
=0
=0
Markers Comment:Generally well done - but solution must be a polynomial
in x
LAST UPDATED AUGUST 23, 2017
KILLARA HIGH SCHOOL
2017 Mathematics Extension 2 HSC Trial Examination
(c)
27
ı
2 marks
i. Correctly deriving expression for g with appropriate reasoning.
1 mark
Corresponding expression for h and f
The base of the solid is a trapezium and it’s area is the sum of the two
trapezia.
x
(e − x)
e
(a + g)
∴ (a + c) = (g + c) +
2
2
2
∴ e(a + c) = x(g + c) + (e − x)(a + g)
+ ce = + cx + ∴
ae
gx
ae
−gx
∴ ge = ce + ax − cx
=⇒ g = c +
a−c
x
e
Similarly (using the same reason as part i.)
h=d+
b−d
e
x
and
a
x
e
Markers Comment:Not very well done - many tried to use linear expressions
without defining where any axes are - lots of fudging to get desired result.
Most only got 1 mark
f=
KILLARA HIGH SCHOOL
LAST UPDATED AUGUST 23, 2017
28
2017 Mathematics Extension 2 HSC Trial Examination
ii. 1 mark
1 mark
Solutions
Correct expression for V (x)
Significant progress towards the final solution
x
d + (b−d)
(2a − c)
e
x
Now A(x) = c +
e
2
e
cd
bc + 2ad − 2cd
2a − c
b−d
2
∴ V (x) =
.dx
+
x+
x
2
2e
2e
e
0
e
bc + 2ad − 2cd x2
2a − c
b − d x3
cd
x+
+
=
2
2e
2
2e
e
3
0
2
3
b−d e
cd
bc + 2ad − 2cd e
2a − c
=
e+
+
2
2e
2
2e
e
3
e
{6cd + 3bc + 6ad − 4cd + 4ab − 4ad − 2bc + 2cd}
=
12
e
=
{2cd + bc + 2ad + 4ab}
12
b+d 2
Alternatively solve by Simpson’s rule using A1 = 12 cd, A2 = a2 + a+c
2
2
and A3 = ab which gives:
b+d
e 1
a a+c
2
V =
cd + 4
+
+ 2ab
6 2
2
2
2
yielding the same result.
Markers Comment:Quite poorly done - it seems as if many were intimidated
by a simple integral - no one fully simplified and no one thought to use
Simpson’s rule since it is quadratic - much easier!
LAST UPDATED AUGUST 23, 2017
KILLARA HIGH SCHOOL