question one 3 marks for each part-question, to a maximum

Scholarship Calculus – page 1 of 14

Sample Assessment Schedule Scholarship Calculus Evidence Statement

QUESTION ONE 3 marks for each part-question, to a maximum of 8 per question. In all questions, marks are independent, and marks are allocated for answers consistent with earlier errors, so long as the question has not been simplified by the error.

(a) The point (if it exists) where

aex

2ex −1= 1 can be found:

aex = 2ex −1

(2 − a)ex = 1

e−x = 2 − a

x = − ln(2 − a) = ln 12 − a

So the point exists when 0 < a < 2 .

Note that 2ex −1< 0 when x < ln 1

2 and aex > 0 always; so for any positive a,

aex

2ex −1< 1 for

x < ln 1

2.

Also, if 0 < a < 2 , then

aex

2ex −1< 1 for x > − ln(2 − a) ; positive solutions only exist for these values of a.

Question One (a)

1 mark: 21ln<x

1 mark: )2ln( ax −−>

1 mark: condition, if 20 << a

Note: watch for errors with < and >.


(b) The total surface area is T = nA AA + nB AB = nA 2k2 − 2k +1 + nB

32

k2 − k +1( ) , where An and

Bn are positive integers.

dTdk

=nA (4k − 2)

2 2k2 − 2k +1+ nB

32

(2k −1)

= (2k −1)nA

2k2 − 2k +1+ nB

32

⎛

⎝⎜

⎞

⎠⎟

The second term is always positive, so the only critical point is at k = 1

2 .

Using the 1DT:

dTdk k=0

= −nA − nB3

2< 0 and

dTdk k=1

= nA + nB3

2> 0 , so we have a minimum.

Question One (b)

1 mark: correct derivative kTdd

1 mark: 5.0=k

1 mark: use test to show minimum

Note: MEI if An and Bn missing.

(c) Using separation of variables:

dydx

= ym+1

y−m−1∫ dy = dx∫y−m

−m= x + C [as m ≠ 0]

ym = 1m(k − x)

y = 1m(k − x)m

Using the chain rule,

ddx

yn( ) = ddy

yn( ) dydx

= nyn−1ym+1 = nyn+m , as required.

It is possible to show that the functions satisfy the property directly, but this work is not required.

Question One (c)

1 mark: integrate

1 mark: in form =y

1 mark: show ( ) mnnx yy +=dd

Note: various forms of =y possible.


QUESTION TWO 3 marks for each part-question, to a maximum of 8 per question. In all questions, marks are independent, and marks are allocated for answers consistent with earlier errors, so long as the question has not been simplified by the error.

a) We aim to find

dxdt x=3

and we see that dVdt

= 0.015× 25π = 0.375π . Also

dVdx

= −25π + π x2 .

dxdt

= dxdV

dVdt

= dVdt

÷ dVdx

= 0.375π(x2 − 25)π

= 0.375x2 − 25

.

dxdt x=3

= 0.3759 − 25

= −0.0234375 metres per hour. The water is rising at (approximately) 23.4 mm

per hour.

Question Two (a)

1 mark: π375.0dd =tV

1 mark: 2dd 25 xxV ππ +−=

1 mark: 23.4 mm per hour

Note: units required in final answer.

(b)(i) Differentiating:

ddx

A 1− x( ) 32

2 + 3 x( ) + C⎛⎝⎜

⎞⎠⎟= A 1− x( ) 3

2 32 x

− 34 x

A 1− x( ) 12

2 + 3 x( )= 3

4 xA 1− x( ) 1

2

2 1− x( ) − 2 + 3 x( )⎡⎣⎢

⎤⎦⎥

= 34 x

A 1− x( ) 12

2 − 2 x − 2 − 3 x⎡⎣

⎤⎦

= 34 x

A 1− x( ) 12

−5 x( )= − 15

4A 1− x

and so A = − 4

15, and the Fundamental Theorem of Calculus gives the integral as required.

Question Two (b)(i)

1 mark: differentiate

1 mark: common factor xA −1

1 mark: 154−=A

Note: decimal value of A acceptable.


(b)(ii) The area under the curve )(xgy = is

g(x) dx0

1

∫ = − 415

1− x( ) 32

2 + 3 x( )⎡

⎣⎢

⎤

⎦⎥

0

1

= 815

.

The area beneath the dotted curve is the same as the area between y = g(x) and y = 1 ; that is,

1− 8

15= 7

15.

So the area between the curves is

115

.

Question Two (b)(ii)

1 mark: area under curve 158=

1 mark: use symmetry for related area

1 mark: area between curves 151=

Note: errors can carry from (b)(i).


QUESTION THREE 3 marks for each part-question, to a maximum of 8 per question. In all questions, marks are independent, and marks are allocated for answers consistent with earlier errors, so long as the question has not been simplified by the error.

(a) Applying angle sum formula:

cos7π12

⎛⎝⎜

⎞⎠⎟= cos π

3+ π4

⎛⎝⎜

⎞⎠⎟= cos π

3⎛⎝⎜

⎞⎠⎟cos

π4

⎛⎝⎜

⎞⎠⎟− sin π

3⎛⎝⎜

⎞⎠⎟sin

π4

⎛⎝⎜

⎞⎠⎟= 121

2− 321

2= 1− 3

2 2= 2 − 6

4

Question Three (a)

1 mark: useful split 43127 πππ +=

1 mark: use of trig identities

1 mark: 462− or other exact form

Note: other splits possible.

(b) First,

cosθ = 1

20 6( )2 +1

= 1

2401= 1

49.

Then cos

θ2=

1+ 149

2= 25

49= 5

7. Now

cos

θ4=

1+ 57

2= 6

7, and

sinθ

4=

1− 57

2= 1

7, so

tanθ4=

17

67

= 16

.

Question Three (b)

1 mark: exact 491cos =θ , any form

1 mark: use both identities given

1 mark: exact & simplified 61

4tan =θ

Note: various methods possible.


(c)

Fact A: The 2nd Derivative Test (2DT) only applies at critical points where 0)( =′ xf , and other points where )(xf ′ is undefined can be maxima and minima.

Fact B: The 2DT is inconclusive if 0)()( =′′=′ xfxf .

Other facts or examples:

example: xxf =)( has )0(f ′ and )0(f ′′ undefined, but )(xf has a minimum at 0=x .

example: 3)( xxf = has critical point 0=x since 23)( xxf =′ and xxf 6)( =′′ so 0)0( =′′f , so the test is inconclusive.

example: 4)( xxf = has critical point 0=x since 34)( xxf =′ and 212)( xxf =′′ so 0)0( =′′f , so the test is inconclusive.

Fact: (Full description of 2DT) If f is a function and c is a point for which 0)( =′ cf and )(cf ′′ is defined, then f has a maximum at c if 0)( <′′ cf and a minimum at c if 0)( >′′ cf .

Compare: The 1st derivative test can be applied at all critical points, including those where )(cf ′ is undefined, or where 0)( =′ cf but )(cf ′′ is undefined or zero.

Fact: The 2DT can be hard to apply if the second derivative is hard to find; more work is required.

Question Three (c)

1 mark: Fact A

1 mark: Fact B

1 mark: other facts or examples

Note: loss of marks for more than one incorrect statement.


QUESTION FOUR 3 marks for each part-question, to a maximum of 8 per question. In all questions, marks are independent, and marks are allocated for answers consistent with earlier errors, so long as the question has not been simplified by the error.

(a) First we find the value of Γ( 1

2) and work from there:

Γ( 12)Γ(1− 1

2) = π

sin 12π

Γ( 12)2 = π

Γ( 12) = π

Γ( 52) = 3

2Γ( 3

2) = 3

212Γ( 1

2) = 3

4π

Question Four (a)

1 mark: ( ) ( ) ππ21sin2

121 1 =−ΓΓ

1 mark: ( ) π=Γ 21

1 mark: ( ) π43

25 =Γ

Note: answer must be exact, not π43± .


(b) The equation has real coefficients, so c = 2 − 3i is also a root. Since all the terms are

even powers of x, so are −c = − 2 − 3i and −c = − 2 + 3i .

These roots give factors of the polynomial

x − 2 − 3i( ) x − 2 + 3i( ) x + 2 + 3i( ) x + 2 − 3i( )= x2 − 2 2x + 5( ) x2 + 2 2x + 5( ) = x4 + 2x2 + 25

By observation, we find that (x4 + 2x2 + 25)(x2 − k) = 0 is a factorisation of the original equation.

The roots are c ,−c,−c , k and − k .

It is possible to find the last two roots by inspection, by substituting x2 = k .

Question Four (b)

1 mark: i32 −=c is also a root

1 mark: c− and c− are also roots

1 mark: other roots are k±

Note: all roots must be found for each part.


(c) Some quick geometry gives that the distance between two roots is 2 and the centre is at

C = a + bi = 1+ 1

2( ) + 1+ 12( ) i .

Shifting the points to be centred on the origin, the new points are the roots of x8 + r8 = 0 .

From the diagram we find r = 1

2( )2 + 1+ 12( )2 = 1

2+1+ 2 + 1

2= 2 + 2

x8 + 2 + 2( )4 = 0

Now shift the centre from the origin to C = 1+ 1

2( ) + 1+ 12( ) i .

p(z) = z − (a + bi)( )8 + r8 = z − 1+ 1

2( ) − 1+ 12( )i⎛

⎝⎞⎠

8+ 2 + 2( )4 .

In the original question, as well as the obvious n = 8 , we have a + bi = 1+ 1

2( ) + 1+ 12( ) i = (1+ i) 1+ 1

2( )

and q = 2 + 2( )4 = 68 + 48 2 .

Question Four (c)

1 mark: finding centre

1 mark: finding r or 2r

1 mark: )(zp in required form

Note: no marks for finding the roots as complex numbers iα and writing

))()(()( 321 ααα −−−= zzzzp .


QUESTION FIVE 4 marks for each part-question. In all questions, marks are independent,

and marks are allocated for answers consistent with earlier errors, so long as

the question has not been simplified by the error.

(a) The curves intersect when 242 )95( xxxk −=− . Making a substitution of 2xu = it is a quadratic.

uukku −=− 295 . When this quadratic has a repeated root, the curves intersect at a common tangent.

09)15(2 =++− kuku

The discriminant is kk 36)15( 2 −+=Δ ; when this is zero, there is a repeated root.

01262503611025

2

2

=+−=−++

kkkkk

Using the quadratic formula again, 251or 1

502426

5010077626 =±=−±=k .

Alternatively, equate slopes xxkx 2410 3 −= to find 0=x (not required, gives 0=k ) or kx 5.25.02 += .

Substituting kx 5.25.02 += into 242 )95( xxxk −=− , we find the same values of k above.

Question Five (a)

1 mark: set two equations equal

1 mark: substitution 2xu = or equate slopes

1 mark: set discriminant to zero, or find kx 5.25.02 +=

1 mark: find both values of 1=k and

04.0251 ==k


(b) EITHER

The substitution should be another line perpendicular to the major axis 02 =+ yx , to form the minor axis. So let yxv −= 2 .

Now rearrange to 1728

724

725 22

=++ yxyx.

2

2

22

2

2

2

22

2

2

2

2

2 4444)2()2(dy

dxy

dx

cy

cxy

cx

dyx

cyx +−+++=−++

Equating terms:

72541

22 =+dc

and 72444

22 =−dc

and 72814

22 =+dc

.

From the first, 22

47251

dc−= ; substituting into the third, we find

72814

7254 22 =+⎟

⎠⎞⎜

⎝⎛ −

dd.

0157212

2 =−d

, so 902 =d . Then 904

7251

2 −=c

, so 402 =c .

So the conic section is in the form 190)2(

40)2( 22

=−++ yxyx

On the major axis, yx 2−= , so substituting: 728820 222 =+− yyy . We get ( )518518 ,2),( −±=yx

with the length of the major axis 72 .

On the minor axis, xy 2= , so substituting: 723285 222 =++ yxx . We get ( )5858 2,),( ±=yx with

the length of the minor axis 16 .

1 mark: substitution yxv −= 2

1 mark: find 402 =c and 902 =d

1 mark: substitute major axis line to find length 72

1 mark: substitute major axis line to find length 16


(b) OR

There are four critical paths: along the top, along the bottom, and crossing between these two.

Because there are several critical paths, all must be ‘crashed’ in order to reduce the project completion time.

We can crash all four paths by spending more resources in two places; the best choice for this looks to be the longer tasks at the end of the project, which take 9 or more days.

If the time to complete the critical paths is reduced by more than 9 days, then the sub-critical path across the middle of the diagram becomes critical, and would also need to be reduced.

Because all of the times are estimated, it might become apparent as the project progresses which of the critical paths is taking longer to finish. Rather than planning to shorten the length of a particular task before the project starts, it might be better to wait until the first tasks are completed (those with 1, 3, 4, 7, and 8 day times) before crashing the critical paths.

1 mark for the first statement

1 mark for the second statement

1 mark for another meaningful statement about shortening the project

1 mark for another meaningful statement about shortening the project

Other statements possible, as determined by the Panel Leader


QUESTION SIX 4 marks for each part-question. In all questions, marks are independent, and marks are allocated for answers consistent with earlier errors, so long as the question has not been simplified by the error.

(a)

dBdx

= I−2x3

+ 2(d − x)3

⎛

⎝⎜

⎞

⎠⎟ = 2I

1(d − x)3

− 1x3

⎛

⎝⎜

⎞

⎠⎟ = 2I

x3 − (d − x)3

x3(d − x)3= 2I

(2x − d )(x2 − dx + d 2 )x3(d − x)3

= 0

The only real root is x = d / 2 ; this gives the two points equidistant from the foci, the points on the minor axis. The function B(x) is decreasing for x < d / 2 and increasing for x > d / 2 .

Question Six (a)

1 mark: find xBdd

1 mark: solve 0dd =xB to get dx 2

1=

1 mark: describe one point

1 mark: describe both points*

*Note: can be in a diagram.

(b) EITHER

Working with matrices (useful, but not required)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+−−→

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+−−+−→

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−++++−

0220i20i0

111

0220i22i20

111

i21iii2i322

111

kkk

i

kkk

i

kkki

i

When 2=k , the system is consistent:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+−+−0000i20i20

111 i so 1=y , and i1+−−= xz (infinitely many solutions).

When i=k , the system is inconsistent and has no solution.

1 mark: eliminate x from two equations

1 mark: eliminate to final form

1 mark: test 2=k and find solutions exist

1 mark: test i=k and find no solution


(b) OR

First we need a particular solution to the problem; any from the table below is sufficient.

Since 32395170 =×+×+× , )9,1,0(),,( =pnm is a solution.

Now the inequalities are ⎪⎩

⎪⎨⎧

≥−−≥+−

≥−

039021

02

yxyx

yx and we sketch them.

There are seven points with integer values of x and y in the feasible region as shown. They are:

x y m n p

0 0 0 1 9

1 0 2 0 6

1 1 1 2 5

1 2 0 4 4

2 1 3 1 2

2 2 2 3 1

2 3 1 5 0

Note that the points on the lines have one of 0,, =pnm . The yx, values will change if a different starting point is chosen, but the shape of the feasible region and the other ),,( pnm solutions will not change.

1 mark: one solution ),,( pnm

1 mark: three inequalities

1 mark: find integer points in feasible region

1 mark: list all seven solutions ),,( pnm

question one 3 marks for each part-question, to a maximum

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