Specialist Maths

• Problems that achieved <60% for Year 2010-2013

The MEDIAN score is C+. 49% of students get B or greater21% of students get better than 33/40. Losing any marks will have a big impact on your Study Score. This was about the same as previous years.23/40 is the MEDIAN grade.

MEDIAN is C+: 43/80 B: 51/80

A: 65/80 A+: 70/80

Exam 1 Assessors Report

Hand writing and general setting out of work a concern.Use logical connectors between lines.

Assessors Report Exam 2

2013

Short Answer 2012

• CALCULATOR FREE

6 + xx2 + 4

=6

x2 + 4+

xx2 + 4

6x2 + 4

+x

x2 + 4∫ ⋅dx

=3⋅Tan−1 x2

⎛⎝⎜

⎞⎠⎟+12⋅ln x2 + 4 +c

Most students who recognised to split the fraction into two parts performed quite well.

Not very well done. Average 0.9/3. Many student omitted cos(x)=0 as a possible solution. Many students only gave solutions for 0≤x≤2π.

Real coefficients so look for conjugate pairThird solution must have real root.

Many students attempted long division and were unable to find the correct solutions.

Many students incorrectly put the solution z=3 at z=2 confusing the question with a root of a single complex number and not an expression.

Fr

N

mg

T

T sinθ =mgT =100g

Sign errors were common with students omitting du/dx=-1.

Quite well done once students recognized the correct expression for a.

multiple choice 2012

Conjugate is a reflection in the Re axisMultiply by i is a rotation of pi/4 anti-clockwise, in this case 3 timesThis is the same result as C.

Perpendicular bisector of origin and z=-2iLine z=-i. This is BNote: Im(z) must be R, C is incorrectly represented.

Use calculator program or other to show y(0.2)=1.1995In this region cos(x) is concave down OR gradient is decreasingAnswer will be an over estimate: B

Correct option is E

The options B,C, D, E cannot be correct.MUST be A.

dx

dt=x ⇒

dtdx

=1x

t=ln x +cwhent=3, x=−1c=3

t=ln x + 3

t−3=ln x ⇒ x=et−3

Extended answer 2012

Area=π8

r22 −r1

2( )

Area=π8

22 −12( ) =3π8

units2

z=cisπ12

⎛⎝⎜

⎞⎠⎟

Re z( ) =0, when Arg z( ) =π2

or−π2

⇒ zn ∀n=6 or −6

General soln n=6+12k ,k∈J

z=±i

0=−1100

145t−t2( )+ 25

t=20 sec

Question 3

17 tan−1

0

19

∫πt6

⎛⎝⎜

⎞⎠⎟⋅dt

120×25

−1100

145t−t2( )+ 25⎛⎝⎜

⎞⎠⎟⋅dt

0

20

∫

Need to have completed an earlier section to get first part. Most errors were because of incorrect terminals

3637m use CASDon’t round each answer this may lead to errors, add separately and then round.

400 km

When directly above P, i component is zero, which makes sin=0. When sin=0, cos=1. 6800-6400=400 km

a(t)=&&r t( ) =−1.32π 2 ⋅6800sin π 1.3t−0.1( )( )i −1.32π 2 ⋅6800cos π 1.3t−0.1( )( ) j

a t( )⊥v t( ) as

v(t)×a(t) =0where −Acosθ sinθ + Acosθ sinθ =0

speed=1.3×π ×6800=27,772 m/ s

x =6800sin π 1.3t−0.1( )( )

y=6800cos π 1.3t−0.1( )( )−6400

circlemoved← 6400

⇒ x2 + y+6400( )2 =68002

Use Pythagora’s Thm and a substitution of the time function:

x2 + y2 =10002

6800sinθ( )2 + 6800cosθ −6400( )

2 =10002

whereθ =π 1.3t−0.1( )−π ≤θ ≤2πgivesθ =−0.139, 0.139, 6.144t=0.04, 0.11, 1.58,...

t=1.58 is outside the domain, only two answers.

x =12

u+v( )t

10 =12

0 + 6( )t

t=103

sec

3.3 sec scores zero. Do not give approximate answers unless specified.

u =10m/ s x=ut+12

at2

v= −6 =10t−4.9t2

a=−9.8m/ s2 t=2.5 sec

x=−6m(asit falls)t=

u =? s=ut+12

at2

v= −6 =u⋅223

−4.9223

⎛⎝⎜

⎞⎠⎟

2

a=−9.8ms−2 u=35.1sx=−6 m

t=10 3+ 4 =223sec

dv

dt=− 196−v2

10dtdv

=−10

196−v2

⇒ t=−10⋅Sin−1 v14

⎛⎝⎜

⎞⎠⎟+c

whent=0, v=7

0 =−10⋅Sin−1 12

⎛⎝⎜

⎞⎠⎟+c

c=5π3

5π30

−t10

=Sin−1 v14

⎛⎝⎜

⎞⎠⎟

v=14sinπ6−

t10

⎛⎝⎜

⎞⎠⎟

or 14cosπ3+

t10

⎛⎝⎜

⎞⎠⎟

v=0, 0 =14sinπ6−

t10

⎛⎝⎜

⎞⎠⎟

t=5π3

sec

dx

dt=14⋅Sin

π6−

t10

⎛⎝⎜

⎞⎠⎟

x=140⋅Cosπ6−

t10

⎛⎝⎜

⎞⎠⎟+c

whent=0, x=0

⇒ c=−70 3

x=140⋅Cosπ6−

t10

⎛⎝⎜

⎞⎠⎟−70 3

whent=5π3

x=18.8m OR x= 14⋅Sinπ6−

t10

⎛⎝⎜

⎞⎠⎟⋅dt=18.8m

0

5π3

∫

short answer 2011

• Calculator free

y=kx⋅e2x dydx

=k(2x+1)⋅e2x d2ydx2 =4k(x+1)⋅e2x

⇒ 4k(x+1)−2k(2x+1)+ 5kx=15x+6⇒ 4kx+ 4k−4kx−2k+5kx=15x+6k=3

Many students did not use the product rule to find dy/dx.

z=2cis

−π3

⎛⎝⎜

⎞⎠⎟

2cis3π4

⎛⎝⎜

⎞⎠⎟

Arg z( ) =−π3

−3π4

=−13π12

=11π12

F =0∑considerhorizontal onlyT1sin30 =T2 sin60

T1 =T2 3

T2 =T1

3

⇒ T2 ≤ 98

∴T1 ≤ 3 × 98

F = 0 vertical only∑T1 cos30 +T2 cos60 ≥ mg

98 3 ×3

2+ 98 ×

1

2≥ mg

m ≤196

g= 20 kg

y=cosec2 πx6

⎛⎝⎜

⎞⎠⎟

43=cosec2 πx

6⎛⎝⎜

⎞⎠⎟

±32

=sinπx6

⎛⎝⎜

⎞⎠⎟

πx6

=π3,2π3

,4π6

,5π6

,...

x=2,4,8,10

pts⇒ 2,43

⎛⎝⎜

⎞⎠⎟, 4,

43

⎛⎝⎜

⎞⎠⎟, 8,

43

⎛⎝⎜

⎞⎠⎟, 10,

43

⎛⎝⎜

⎞⎠⎟

Many students miss the last two points that come from the negative sine angle.

multiple choice 2011

z=x−iyz =x+ iy1z=

1x+ iy

×x−iyx−iy

1z=

x−iyx2 + y2

Same angle but closer, as it will be divided by magnitude of z. C

z3 =27cis−π2

⎛⎝⎜

⎞⎠⎟

z=3cis−π6

⎛⎝⎜

⎞⎠⎟

other solns±2π3

P4 ,P8 ,P12

A Perpendicular vectorsB Equal vectorsC Perpendicular diagonalsD Diagonals same lengthE Vectors equal and opposite direction

−dx

dy=2 y −1( )x −1( )

−dy

dx=

x −1( )2 y −1( )

dy

dx+

x −1( )2 y −1( )

= 0

Quite a difficult question. Only 25% correct.

When x=0, gradient is constant and positive. Omit EWhen y=0, gradient is constant and negative. Omit CWhen y=x, m is negative. Omit DWhen y=-x m is positive. Omit BLeaves A

t =0, xo =5

xn+1 =xn +h⋅f ' tn( )

⇒ x1 =xo +h⋅f ' t0( )x1 =5−0.5⋅1.0 =4.5

t=0.5, xo =4.5

x2 =4.5−0.5⋅109.5

=3.97

Could also use CAS program to find the answer.Be careful with the negative gradient.

a=v⋅dvdx

if v=2 1−x2

dvdx

=−2x

1−x2

a=v⋅dvdx

=2 1−x2

1×

−2x

1−x2

=−4x

extended answer 2011

y=x

Question 1

let z=x+ iyz =x−iy

⇒ x+ iy=a x−iy( )lety=x

⇒ x+ ix=a x−ix( )LHS=RHSiff a=i

Area =2(1/4 circle - triangle)

A=214π 1( )

2 −12⋅1⋅1⎛

⎝⎜⎞⎠⎟=π2−1

v= a⋅dt∫v(t) =−9.8tk+c

whent=0,v=35i +5 j +24.5k=c

v(t) =35i +5 j + 24.5−9.8t( ) k

r t( ) = v⋅dt∫whent=0,r(t) =0⇒ c=0

r(t) =35ti +5tj + 24.5t−4.9t2( ) k

Not very well done for a easy question. Many answer lacking care and details. Marks were lost for omitting the tilda’s on the vectors.

when dist k =0⇒ 24.5 =4.9tt=5.0sec

r(5) =175i +25 j +0k

asholeis200i

position fromhole=25i +25 j

25i +25 j =35m

F =ma∑T −3gsinθ =3a

2gsin 2θ( )−T =2a

2gsin 2θ( )−T =2a

T −3gsinθ =3a

⇒ T =2gsin 2θ( )−2a=3a+ 3gsinθ⇒ T =2gsin 2θ( )−3gsinθ =5a

but sin 2x( ) =2sin x( )cos x( )

4gsin θ( )cos θ( )−3gsinθ =5a

a=gsin θ( )

54cos θ( )−3( )

⇒ a = 0

4cos θ( ) − 3 = 0

θ = cos−13

4⎛⎝⎜

⎞⎠⎟= 41.40

F =ma∑5a=2gsin2θ −3gsinθ −μN5a=2gsin60−3gsin30−μ3gcosθ

5a=2g32

−3g12−μ3g 3

210a=2g 3 −3g−μ3g 3

a=0.2ms−2

Conc.=mass

volume=

x10 +10t

volume starts at 10 and increases by 10 every min

dx

dt=Input−Output

dxdt

=20e−0.2t −10x

10 +10tdxdt

=20e−0.2t −x

1+ t

⇒dxdt

+x

1+ t=20e−0.2t

must show dx/dt=Input-Output

dx

dt=20 t2 + 7t+ 31( )e−0.2t

t+1( )2 −

600t+1( )

2

Use CAS don’t put x as variable instead of t

dx

dt=20 t2 + 7t+ 31( )e−0.2t

t+1( )2 −

600t+1( )

2

& x(t) =600t+1

−100e−0.2t t+6( )

t+1

ifdxdt

+x

t+1=20e−0.2t

⇒20 t2 + 7t+ 31( )e−0.2t

t+1( )2 −

600t+1( )

2 +1

t+1⎛⎝⎜

⎞⎠⎟

600t+1

−100e−0.2t t+6( )

t+1⎛

⎝⎜⎞

⎠⎟

⇒20t2 +140t+620( )e−0.2t

t+1( )2 + −

e−0.2t 100t+600( )t+1( )

2

⎛

⎝⎜

⎞

⎠⎟

⇒20t2 + 40t+20( )e−0.2t

t+1( )2 =

20 t2 +2t+1( )e−0.2t

t+1( )2

=20e−0.2t

Initial cond.

x(0)=600−600 =0thisneedstobeshown.

Must have turning point. I would have end point, it needed to be shown in the correct square. Marks were lost if the graph did not show correct concavity.

=600

t +1( )2 −100e−0.2t t + 6( )

t +1( )2

0

10

∫ ⋅dt = 51.6g

short answer 2010

AB=−a+b

= −i −2 j −2k( )+ −i + 3 j + 4k( )

= −2i + j +2k( )

a⋅b=a bcosθ−2 +2 + 4 =3⋅3cosθ

⇒ cosθ =49

Must show step 2 to get mark.

Area=12

a⋅b⋅sinθ

if cosθ =49

sinθ =659

Area=123⋅3⋅

659

=652

units2

Many students were unable to start this problem. Either ignoring the instruction hence, or unable to find sin(x) given cos(x)=4/9.

I would consider an easy question but only 50% correct

Many students got the answer as 1/6 instead of-1/6

Note there is a typo in first part but does not change the answer

Implicit Differentiation

Need to identify the Difference of Perfect Squares.

area=− x2 −1( ) x+1−1

1

∫ ⋅dx

area=− x−1( ) x+1( ) x+1−1

1

∫ ⋅dx

area= x−1( ) x+1( )32

1

−1

∫ ⋅dx

letu=x+1 and x−1=u−2terms: x=−1⇒ u=0

x=1⇒ u=2

area= u−2( ) u( )32

2

0

∫ ⋅du

area= u52 −2u

32

⎛

⎝⎜⎞

⎠⎟2

0

∫ ⋅du

area=27

u72 −

45

u52

⎡

⎣⎢

⎤

⎦⎥2

0

area= 0( )−27

2( )72 −

45

2( )52

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎤

⎦⎥

area=45

2( )52 −

27

2( )72 =

256 27

−128 2

5

area=32 235

multiple choice 2010

Not A, C and D

B has the same co-efficient for x2 and y2. Asymptote y=x

Only E will have non-perpendicular asymptote.

Graph the function. There are 4 intersections. Only option is E

There is a complex coefficient so the conjugate root theorem does not apply. Minimum would be 3.

Gradient is independent of t: NOT AGradient for positive V is negative: NOT B or DAs V increases Abs(V) increases NOT EMust be C

V =e−t

dVdt

=−e−t =−V

∴C

OR

F =ma=md

12

v2⎛⎝⎜

⎞⎠⎟

dx

⇒ m×12

v2 = F(x)⋅dx∫⇒ v2 =

2m

F(x)⋅dx∫

v=2m

F(x)⋅dx∫ ⇒ D

extended answer 2010

m=80

4− 3fromprevious

F =0∑mgsinθ + μN =20gmgsinθ + μmgcosθ =20g

μ =20−msin30

mcos30=0.07735

F =ma∑ma=mgsinθ −μNma=mgsinθ −μmgcosθa=gsinθ −μgcosθa=9.8sin30− 0.077( ) 9.8( )cos30 =4.244ms−2

v=u+at

v=3×4.244 =12.7ms−1

π +12,0

⎛

⎝⎜⎞

⎠⎟−π +12,0

⎛

⎝⎜⎞

⎠⎟

Note : ArSin(1)=π2

y=Arsin(2x2 −1)

⇒ x2 =12

siny+1( )

Area=π x2 ⋅dya

b

∫

Area=π 12

siny+1( )⋅dy−π2

0

∫

y=Sin−1 2x2 −1( )

let u=2x2 −1⇒dudx

=4x

y=Sin−1 u( )⇒dydu

=1

1−u2

dydx

=dydu

dudx

=4x

1−u2

=4x

1− 2x2 −1( )2=

4x

4x2 1−x2( )

=2x

1−x2( )

and a=1

Use CAS show: z1 =−10−10i

z1 =10 2

z3 =20012 cis

−3π4

⎛⎝⎜

⎞⎠⎟

z= 20012 cis

−3π4

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

13

z=20016 cis

−π4

⎛⎝⎜

⎞⎠⎟

also 20016 cis

−π4

+2π3

⎛⎝⎜

⎞⎠⎟=200

16 cis

5π12

⎛⎝⎜

⎞⎠⎟

20016 cis

−π4

+4π3

⎛⎝⎜

⎞⎠⎟=200

16 cis

13π12

⎛⎝⎜

⎞⎠⎟=200

16 cis

−11π12

⎛⎝⎜

⎞⎠⎟

u =2 10cis −α( )

w= 10cis π / 2−α( )

⇒ u=2 10cis α( )

iw= 10cis π −α( ), rotatebyπ / 2

uiw

=2 10cis α( )10cis π −α( )

=2cis 2α −π( )

z1 =u+w( )u

iw

Arg z1( ) =Arg(u+w)+ Arguiw

⎛⎝⎜

⎞⎠⎟

⇒−3π4

=Arg(u+w)+ 2α −π

∴Arg(u+w) =π4−2α