PEM 2020 Mathematics Advanced Trial HSC Marking Criteria PDF

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Summary

Due to the implementation of the new syllabus, past paper resources that cover the new content is almost impossible to find. HOWEVER, I have collated 35+ past papers from different schools for their 2020 trials :)...

Description

PEM 2020 Advanced Mathematics TRIAL EXAMINATION SUGGESTED SOLUTIONS Section 1 Multiple Choice Answer Key Question Answer D 1 B 2 C 3 B 4 B 5 A 6 D 7 A 8 C 9 D 10

Page 1 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Section II Question 11 Criteria

Marks

• Provides correct solution

2

• Differentiates correctly, no or incorrect substitution

1

Sample Answer:

f ( x ) = log e ( x 2 +1) f ' ( x) = x = −1

2x x 2 +1

f ' ( − 1) =

−2

( −1)2 +1

= −1

Question 12 Criteria

Marks

• Provides correct solution

3

• Differentiates the exponential expression but provides a partially correct product rule

2

• Only uses the product rule

1

Sample Answer:

d x sin x d e = e x sinx  ( x sin x ) ( ) dx dx xsin x =e  (sin x + x cos x )

Page 2 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 13 Criteria

Marks

• Provides correct solution

2

• Indicates one correct translation, horizontal or vertical

1

Sample Answer:

The tangent is shifted left 1 unit and down 2 units: y = 2 − 3 ( x + 1) − 2 y = −3 x − 3

Question 14 Criteria

Marks

• Provides correct solution

2

• Uses ln f ( x) with no constant adjustment

1

Sample Answer:

2 2 −3 =−  2 − 3x 3 2 − 3x Anti-derivative is −

2 ln 2 − 3 x + c 3

Page 3 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 15 Criteria

Marks

• Provides correct solution, with correct rounding

3

• Calculates correct value of 

2

• Attempts to use sine rule

1

Sample Answer:

By Sine Rule: sin sin 37 = 13 8 13sin 37 sin  =  0.97794... 8   77 56 ' 43.82 '' = 77 57 ' to nearest minute

Page 4 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 16 (a) Criteria

Marks

• Provides correct solution

2

• Indicates a graph that shows two of the indicated characteristics

1

Sample Answer:

Question 16 (b) Criteria • Provides correct solution

Marks 1

Sample Answer:

Minimum of y = |1+ f ( x) | occurs at x = a  minimum is |1 + f ( a)|

Page 5 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 17 (a) Criteria • Provides correct solution

Marks 1

Sample Answer:

P ( 3, then 4) = P ( 3)  P ( 4) 1 1 =  6 5 1 = 30

Question 17 (b) Criteria • Provides correct solution

Marks 1

Sample Answer:

Using part (a ), for each first number there is a corresponding number that will add to 7,  P (sum is 7 ) = 6 

1 1 = 30 5

Question 17 (c) Criteria

Marks

• Provides correct solution

2

• Attempts to use conditional probability formula or list a suitable sample space

1

Sample Answer:

Using P ( A | B) = P ( sum is 7|2

nd

P ( A B ) , P ( B)

1 1 ball is 4) = 30 = 1 6 5

Page 6 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 18 (a) Criteria

Marks

• Provides correct solution

2

• Attempts to use the chain rule

1

Sample Answer: d 1 sin 2 x ) = 2 ( sin x )  cos x ( dx = 2sin x cos x

Question 18 (b) Criteria

Marks

• Provides correct solution

3

• Correctly simplifies integrand, then incorrectly integrates

2

• Attempts to simplify the integrand

1

Sample Answer: 

4

 (sin x +cos x)

 2

dx =

0

4

 (sin

2

x +cos 2 x + 2sin xcos x ) dx

0



=

4

 (1 + 2sin x cos x) dx 0 

= [x + sin 2 x ] 0 4

using (a )

    2 =  +  sin  − 0 + (sin 0 ) 4  4   1 = + 4 2 2

(



) 

Page 7 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 19 Criteria

Marks

• Provides correct solution

2

• Attempts to equate appropriate area formula

1

Sample Answer: 30 =



c

6

  52 =

1 2 R  R 2 = 300 2 6

 R = 10 3 cm

Question 20 Criteria

Marks

• Provides correct solution

3

• Equates the appropriate areas without determining a

2

• Attempts to calculate the appropriate areas

1

Sample Answer:

Since the areas are equal, but one region is above and the other below, a

0 =  (9 x − x 3 )dx 0 a

 9 x2 x4  = −  4 0  2 9a 2 a 4 − 2 4 2  18 a = a4 =

 a 2 = 18 since a  0  a = 3 2, since a  3

Page 8 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 21 (a) Criteria

Marks

• Provides correct solution

2

• Provides correct centre or radius or states both correctly but does not provide correct equation

1

Sample Answer: Centre (2,5 ) , radius 5  ( x − 2 ) + ( y − 5 ) = 25 2

2

Question 21 (b) Criteria

Marks

• Provides correct solution

2

• Substitutes point P and forms an inequality

1

Sample Answer:

Since P (6, a ) is inside the circle,

( 6 − 2) + ( a − 5) 2  (a − 5 )  9 2

2

 25

 − 3 a − 5 3 2  a  8

Page 9 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 22 Criteria

Marks

• Provides correct solution, with appropriate rounding

3

• Calculates distance using cosine rule

2

• Indicates an appropriate diagram or attempts to use cosine rule

1

Sample Answer:

By the Cosine Rule, x 2 =50 2 +90 2 −2 50 90cos (45 +50

)

= 11384.40168...  x = 106.6977, i.e. x = 107 m, correct to nearest metre

Page 10 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 23 Criteria

Marks

• Provides correct solution

3

• Calculates values of A and B , doesn’t calculate range

2

• Attempts to form simultaneous equations

1

Sample Answer:

 A At P : Asin   + B = −3  + B = −3 2 6   At Q : Asin   + B =1  A + B =1 2  A By subtraction: =1 − (−3 )  A = 8  B =1 −8 = −7 2  f ( x ) = 8sin x − 7  R f is [−15,1]

Question 24 (a)

Criteria

Marks

• Provides correct solution

2

• Attempts to prove the identity

1

Sample Answer:

1 1 + 1 − sin x 1 + sin x 1 + sin x + 1 − sin x  (1− sin x )(1 + sin x )

LHS =

2 1 − sin 2 x 2  cos2 x  2sec 2 x = RHS 

Page 11 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 24 (b) Criteria

Marks

• Provides correct solution

3

• Correctly forms integral and uses part (a)

2

• Attempts to form an integral to indicate area

1

Sample Answer: 

3 1 −1 Area =   − 0  1 −sin x 1 +sin



3 1 1  =  + x 1 +sin 0  1 −sin



 dx  x  dx x 

3

=  2sec2 x dx , using (a ) 0

=  2 tan x0 3 

= 2 tan



3 = 2 3 u2

− 2 tan 0

Page 12 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 25 (a)

Criteria • Provides correct solution

Marks 1

Sample Answer:

Question 25 (b) Criteria • Provides correct solution

Marks 1

Sample Answer:

There are 3 points of intersection.

Page 13 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 25 (c)

Criteria

Marks

• Provides correct solution

2

• Attempts to rearrange the equation or draw cos ( 2x )

1

Sample Answer:

The equation rearranges to e −x = − cos ( 2 x ) , so the graph changes to:

There are now 4 solutions, so there are 4 solutions of the equation.

Page 14 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 26 (a)

Criteria

Marks

• Provides correct solution

2

• Attempts to calculate the cumulative distribution function

1

Sample Answer:

FX ( x ) = P ( X  x ) = P ( X = 1) + P ( X = 2 ) +

+

(

= x)

1 1 1 + + + x 2 4 2 x 1  1    1−  2   2   = 1 1− 2 1 = 1− x 2 = 1− 2− x

=

Question 26 (b) Criteria

Marks

• Provides correct solution

2

• Attempts to use part (a) or calculate directly

1

Sample Answer:

P ( 2  X  5) = FX ( 5) − FX ( 2) = (1 − 2 − 5 ) − (1 − 2 − 2 ) 1 1 − 4 32 7 = 32 =

Page 15 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Or, P ( 2  X  5) = P ( 3  X  5 ) = P ( X = 3) + P ( X = 4 ) + P ( X = 5 ) =

1 1 1 7 3 + 4 + 5 = 2 2 2 32

Question 26 (c) Criteria • Provides correct solution

Marks 1

Sample Answer: P ( X  4) = 1− P ( X  4) = 1− FX ( 4) = 1− ( 1− 2−4 ) =

1 16

Question 27 (a) Criteria

Marks

• Provides correct solution

3

• Determines the coordinates of the two stationary points, or equivalent merit

2

• Determines the x values of the two stationary points, or equivalent merit

1

Sample Answer:

f ( x) = − x3 + 9 x2 − 24 x + 16 f ' ( x) = −3 x 2 +18 x − 24 = −3 ( x 2 − 6 x + 8 ) = −3 ( x − 2 )( x − 4 ) f ' ( x) = 0 when x = 2 or 4 f '' ( x ) = −6 x + 18 : f '' (2 ) = 6  minimum turning point at (2, −4 ) f '' (4 ) = −6  maximum turning point at (4,0 )

Page 16 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 27 (b) Criteria

Marks

• Provides correct solution with stationary points and intercepts labelled

2

• Sketches correctly, no or little labelling

1

Sample Answer:

Question 27 (c) Criteria • Provides correct solution

Marks 1

Sample Answer:

Global minimum is -20 and occurs when x = 6 .

Page 17 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 28 (a) Criteria

Marks

• Provides correct solution

2

• Attempts to use the correct equation, stating that the integral equals one, or equivalent merit

1

Sample Answer: 



−

1

f ( x ) dx = 1  k  (x 3 − x 5 )dx = 1 0 1

1 1 1  x 4 x 6 Since  ( x − x ) dx =  −  = − =  4 6  0 4 6 12 0 1

3

then

5

k =1  k = 12 12

Question 28 (b) Criteria

Marks

• Provides correct solution

4

• Attempts to show the medians are equivalent

3

• Successfully finds the median of either model

2

• Attempts to find the median of either model

1

Sample Answer:

Let x = M be the median of model g. Then: M

M

M 1 1 =  g ( x ) dx =  2 x dx = x 2  = M 2, i.e. M 2 = 0 2 0 2 0 M

Consider



M

f ( x) dx = 12 ( x3 − x5 ) dx = 3 x4 − 2 x 6  = 3 M 4 − 2 M 6 M 0

0

0 M

2

3

1 1 1 1 Since M 2 = ,  f ( x ) dx = 3   − 2   = 2 0 2  2 2 Hence x = M is also the m edian of modelf , i.e. the two models have the same median.

Page 18 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 29 (a) Criteria

Marks

• Provides correct solution

2

• Attempts to calculate the area function

1

Sample Answer:

h (a + b ), where h =10sin , a =10, b =10 +2 10 cos  2 10sin  A ( ) = (10 + ( 10+ 20cos ) ) 2 = 5sin ( 20 + 20cos )

Cross-sectional area =

A ( ) = 100sin  (1 + cos  )

Page 19 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 29 (b) Criteria

Marks

• Provides correct solution

4

• Finds the correct value of  , but does not test nature

3

• Finds the correct derivative function and attempts to find stationary points

2

• Attempts to find the derivative function

1

Sample Answer:

A ' ( ) = 100 cos  (1 + cos  ) + sin  ( − sin  ) 

(

= 100 cos  + cos 2  − (1 − cos 2  )

)

= 100 (2cos2  + cos  −1 ) = 100 (cos  +1 )(2cos  −1 )  A ' ( ) = 0 if cos = − 1 or cos =

1 2

 1  2 3 A '' ( ) = 100 (−4cos  sin  − sin  ) = −100sin  (4cos  + 1 )

Since  is acute, cos  =

 A ''  3

  = −150 3  0 

Hence, the cross-sectional area is a maximum when  =

 3

Page 20 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 30 Criteria

Marks

• Provides correct solution

2

• Provides correct method with incorrect table factor.

1

Sample Answer: Annuity conditions are 24 periods, at

3% = 0.25% per period 12

Hence the annuity factor is 24.7028  monthly deposit is

$2800 = $113.35 24.7028

Question 31 (a) Criteria

Marks

• Provides correct solution

2

• Indicates no change without reason.

1

Sample Answer:

Since the 25th and 26th tip scores are in the $2.50-$5 bar, a tip of $6 is already above the median and so increasing the tip to $16 would not affect the median. Question 31 (b) Criteria

Marks

• Provides correct solution

2

• Argues that the mean increases, but does not calculate the increase.

1

Sample Answer:

Since the mean calculation uses the specific value of each tip, increasing a single tip from $6 to $16 must increase the mean by $10 = $0.20 . 50

Page 21 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 32 (a)

Criteria

Marks

• Provides correct solution

2

• Attempts to use a recursive relationship

1

Sample Answer:

    A1 =  5, 000,000 − R  1.03 = 5, 000,000 1.03 − R 1.03  payment of $R   made at start of year  A2 = ( A1 − R ) 1.03 = (5, 000,000 1.03 − R 1.03 − R )1.03 = 5, 000,000 1.032 − R  1.03 (1.03 + 1)

Question 32 (b) Criteria

Marks

• Provides correct solution

2

• Attempts to generalise from part (a)

1

Sample Answer:

Similarly, using the recursive rule An+ 1 = ( An − R ) 1.03, we get A3 = 5, 000,000 1.033 − R 1.03 (1.03 2 +1.03 +1 ), and more generally n n An = 5, 000,000 1.03 − R  1.03 (1.03 −1 +

= 5, 000,000 1.03 − R  1.03 (1 + 1.03 + n

+ +

+ 1) n −1

)

 1.03n − 1  = 5, 000,000 1.03 n − R 1.03   , using the sum to n-terms of a GP  1.03 −1  100 =5,000,000 1.03n − R 1.03 (1.03n − 1) 3 n 1.03 −1) ( n = 5,000,000 1.03 − 103R 3 Page 22 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 32 (c) Criteria

Marks

• Provides correct solution

2

• States A20 = 0 or equivalent merit

1

Sample Answer:

Maximum payments  A20 =0

(1.03

20

 5, 000, 000 1.0320 = 103R

−1)

3

15, 000, 000 1.03 = $326, 289.84 103 (1.0320 −1 ) 20

 R=

Therefore, the maximum annual payment is $326,289.84

Page 23 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 33 (a) Criteria

Marks

• Provides correct solution

1

Sample Answer:

Friday test marks

Monday test marks

Page 24 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 33 (b) Criteria • Provides correct solution

Marks 1

Sample Answer: y = 0.65 x + 22.50

Question 33 (c) Criteria • Provides correct solution

Marks 1

Sample Answer:

When x = 63:

y = 0.65  63 + 22.50 = 63.45  63 (to nearest whole mark)

Question 33 (d) Criteria • Provides correct solution

Marks 1

Sample Answer:

A student who appears below the line y = x performed better on Monday’s test.

Page 25 of 26

PEM

Advanced Mathematics

Trial HSC Marking Guidelines

Question 34 Criteria

Marks

• Provides correct solution

3

• Correctly determines outlier region above the mean, or equivalent merit

2

• Attempts to calculate IQR

1<...