PRE-LEAVING CERTIFICATE EXAMINATION, 2018 MARKING SCHEME - School … · 2019-04-08 · page 1 of...

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PRE-LEAVING CERTIFICATE EXAMINATION, 2018

MARKING SCHEME

MATHEMATICS

HIGHER LEVEL

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OVERVIEW OF MARKING SCHEME

Scale label A B C D E

No of categories 2 3 4 5 6

5 mark scale 0, 5 0, 2, 5 0, 2, 4, 5

10 mark scale 0, 10 0, 5, 10 0, 3, 7, 10 0, 2, 5, 8, 10

15 mark scale 0, 15 0, 7, 15 0, 5, 10, 15 0, 5, 7, 10, 15

20 mark scale 0, 20 0, 10, 20 0, 7, 13, 20 0, 5, 10, 15, 20

25 mark scale 0, 25 0, 12, 25 0, 8, 17, 25 0, 6, 12, 19, 25 0, 5, 10, 15, 20, 25

A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the body of the scheme, where necessary. Marking scales – level descriptors A-scales (two categories) incorrect response (no credit) correct response (full credit)

B-scales (three categories) response of no substantial merit (no credit) partially correct response (partial credit) correct response (full credit)

C-scales (four categories) response of no substantial merit (no credit) response with some merit (low partial credit) almost correct response (high partial credit) correct response (full credit)

D-scales (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (middle partial credit) almost correct response (high partial credit) correct response (full credit)

E-scales (six categories) response of no substantial merit (no credit) response with some merit (low partial credit) response almost half-right (lower middle partial credit) response more than half-right (upper middle partial credit) almost correct response (high partial credit) correct response (full credit)

Marking categories for all questions are shown throughout the solutions. In certain cases, typically involving rounding or omission of units, a mark that is one mark below the full-credit mark may also be awarded. Such cases are flagged with an asterisk. Thus, for example, Scale 10C* indicates that 9 marks may be awarded.

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SUMMARY OF MARK ALLOCATIONS TO BE APPLIED

Section A Section B

Question 1 Question 7 (a) 10C (a) 10C (b) (i) 5C (b) 10C (ii) 10D (c) 10C

(d) 10C Question 2 (e) 10C (a) 10C (f) 10C (b) (i) 10C (g) 5A (ii) 5C Question 8 Question 3 (a) 10A (a) 10D (b) 10C (b) (i) 10D (c) 15D (ii) 5B (d) 5B (e) 5B Question 4 (a) 15D (b) 10C Question 9 (a) (i) 10C Question 5 (ii) 15D (a) 15D (b) 15C (b) 10D Question 6 (a) (i) 5C (ii) 5C (b) 5C (c) 10D

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PAPER 1

Detailed marking notes Note: The model solutions for each question are not intended to be exhaustive – there may be other correct solutions. Any Examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact Examcraft.

Q1 Model Solutions – 25 Marks Marking Notes

(a) 2 (8 − 343) 2 (2 − 7)(4 + 14 + 49) Scale 10C (0, 3, 7, 10)

Low Partial Credit • Any correct factor High Partial Credit • Correct HCF with cubes identified • One error in difference of cubes

(b) (i)

log ( + 2) + log ( − 2) = 5 log ( + 2)( − 2) = 5 ( + 2)( − 2) = 32 = 36 = 6

Scale 5C (0, 2, 4, 5) Low Partial Credit • Any correct step High Partial Credit • Gives solution as ±6 • Error in expanding brackets but continues to

end correctly

(ii) 2 + 2 = 3 2 + 22 − 3 = 0

2 − 3(2 ) + 2 = 0 − 3 + 2 = 0 ( − 2)( − 1) = 0 2 = 2 2 = 1 = 1 = 0

Scale 10D (0, 2, 5, 8, 10) Low Partial Credit • Any correct step Mid Partial Credit • Eliminates fraction correctly High Partial Credit • Forms correct quadratic • Fails to give = 0 as a solution

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(a)

8 ( ) 1

( ) ( ) = ( ) = 8 − 2 = 0 = 4 84 ( ) 1

70

Scale 10C (0, 3, 7, 10) Low Partial Credit • Any correct step towards finding general term High Partial Credit • One error in dealing with indices • Correct equation but error in solving • Finds but fails to find term

(b) (i)

*Long division alternate method. ( + + )( + ) = + + − 4, + ( + ) + ( + ) + = + + − 4,

(i) + = = −

(ii) + ( − ) =

(iii) ( − ) = −4

Scale 10C (0, 3, 7, 10) Low Partial Credit • Any correct step in expansion/division High Partial Credit • Equates 2 coefficients correctly • Correct long divison but coefficents not

equated correctly

(ii) From (i) ( − ) =

From (iii) = −4 − = −4 4 = −

Scale 5C (0, 2, 4, 5) Low Partial Credit • Any correct step High Partial Credit • Isolates ( − )

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(a)

Step 1: Prove true for = 1[ ( + )] = [ ( (1 ) + (1 )]∴ = 1 Step 2: Assume true for = [ ( + )] = [ (cos( ) + ( )] Step 3: Prove true for = + 1 [ ( + )]= [ (cos(k + 1)θ + ( + 1) ] LHS: [ ( + )]= ( + ) (+ ) [ ( + )]= (+ )( + ) = ( + +− ) = [cos( + ) + ( + )] = [cos(k + 1)θ + ( + 1) ] ∴ ∈ ℕ

Scale 10D (0, 2, 5, 8, 10) Low Partial Credit • Step 1 correct Mid Partial Credit • Step 1 correct and step 2 partially correct High Partial Credit • Step 1 and 2 fully correct but no statement

(b) (i) = (1) + −√3 = 2

= √31 = tan √31 = 3 ∴ = 53 = 2 cos 53 + 53

Scale 10D (0, 2, 5, 8, 10) Low Partial Credit • Any correct step Mid Partial Credit • Finds modulus and High Partial Credit • Finds modulus and

(ii) = 2 cos 53 + 53

= 64 cos(10 ) + (10 ) = 64

Scale 5B (0, 2, 5) Partial Credit • Any correct step

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(a)

Function Injective Surjective Bijective Yes Yes Yes No No No Yes Yes Yes

Scale 15D (0, 5, 7, 10, 15) Low Partial Credit • Any one entry correct Mid Partial Credit • One function fullly correct • Two functions partially correct High Partial Credit • Two functions fully correct • Three functions partially correct

(b)

ℎ ∘ ( ) = 2(4 − 3) − 1 ℎ ∘ ( ) = 8 − 7 ∘ ℎ ∘ ( ) = (8 − 7) + 2(8 − 7) ∘ ℎ ∘ ( ) = 64 − 112 + 49 + 16 − 14 ∘ ℎ ∘ = 64 − 96 + 35

Scale 10C (0, 3, 7, 10) Low Partial Credit • Any correct step High Partial Credit • ℎ ∘ ( ) fully correct • ∘ ℎ ∘ ( ) not expanded/error in expansion

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(a)

( ) = 3 − 4 − 5 ( + ℎ) = 3 − 4( + ℎ) − 5( + ℎ) ( + ℎ) = 3 − 4 − 4ℎ − 5 − 10 ℎ − 5ℎ ( + ℎ) − ( ) = −4ℎ − 10 ℎ − 5ℎ ( + ℎ) − ( )ℎ = −4 − 10 − 5ℎ

lim→ ( + ℎ) − ( )ℎ = −4 − 10 − 5(0) lim→ ( + ℎ) − ( )ℎ = −4 − 10

Scale 15D (0, 5, 7, 10, 15) Low Partial Credit • Any correct step Mid Partial Credit • ( + ℎ) − ( ) correct and stops • ( + ℎ) − ( ) incorrect but continues to end

correctly High Partial Credit • ( ) ( ) fully correct • Candidate forgets limit • Fully correct with LHS of equation missing or

incorrect

(b)

( ) = + 2 + ( ) = 3 + 4 + (2) = 3(2) + 4 (2) + + 8 + 12 = 0 ( + 6)( + 2) = 0 = −6,−2

Scale 10D (0, 2, 5, 8, 10) Low Partial Credit • Any correct differential Mid Partial Credit • ( ) correct • ( )incorrect and subs in = 2 High Partial Credit • Correct quadratic set up • Incorrect quadratic but solved to end correctly

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(a) (i)

(3 − 2 + 4)

− + 4 |10 = 4

Scale 5C (0, 2, 4, 5) Low Partial Credit • Any correct step High Partial Credit • Fully correct but final solution incorrect

(ii) (sin 3 cos 2 )

12 (sin 5 + sin )

= 12 −cos(5 )5 − cos( ) 20

12 − cos 5 25 − cos 2 − −cos 5(0)5 − cos 0

= 1

Scale 5C (0, 2, 4, 5) Low Partial Credit • Correct identity used High Partial Credit • Fully correct but final solution incorrect

(b) ( − ) ( − ) 13 + 12 +

Scale 5C (0, 2, 4, 5) Low Partial Credit • Any correct step High Partial Credit • Fully correct but final solution incorrect • Omits constant of integration

(c) ( ) = 3 − 2 + 4 + 2 14 − 1 34 − 23 + 2 + 2 13 34 (4) − 23 (4) + 2(4) + 2(4)− 34 (1) − 23 (1) + 2(1) + 2(1) 13 5683 − 4912 2474

Scale 10D (0, 2, 5, 8, 10) Low Partial Credit • Any correct step Mid Partial Credit • Integrals correct • One incorrect integration but continues to end

correctly High Partial Credit • Fully correct integrals but error in evaluating

answer

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(a)

− = ( − ) 98 − 21 = ( − 21) ( ) = 98℃

Scale 10C (0, 3, 7, 10) Low Partial Credit • Any correct substitution High Partial Credit • Fully correct substitution

(b)

63 − 21 = (98 − 21) ( ) 42 = (77) ( ) 4277 = ( ) ln 4277−5 = = 0 ∙ 12123

Scale 10C (0, 3, 7, 10) Low Partial Credit • Any correct substitution High Partial Credit • Fully correct substitution with 1 error in

solving • 1 error in substitution but solved to end

correctly

(c) − 21 = (98 − 21) ∙ ( ) = 47℃



correctly

(d) 22 − 21 = (98 − 21) ∙ 177 = ∙

ln 177−0 ∙ 12123 =

= 35 ∙ 83


solving • 1 error in substitution but solved to end correctly

(e) 63 − 26 = (98 − 26) ∙ 3772 = .

ln 3772−0.12123 =

= 5 ∙ 49 Takes longer than 5 minutes to cool to 63℃

Scale 10C (0, 3, 7, 10) Low Partial Credit • Any correct substitution • Chooses unsuitable temperature < 26℃ High Partial Credit • Fully correct substitution with 1 error in


correctly

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(f) Scale 10C (0, 3, 7, 10) Low Partial Credit • One correct point High Partial Credit • 3 or more points plotted correctly • Points connected with straight lines

(g) 53℃ 55℃ Scale 5A (0, 5)

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(a)

Scale 10A (0, 10)

(b)

Pattern 1 2 3 4 5Number of Tiles 6 10 16 24 34

Nature: Quadratic pattern

Scale 10C (0, 3, 7, 10) Low Partial Credit • Any correct entry High Partial Credit • 4 correct entries

(c) 2 = 2 ∴ = 1 = 1:(1) + (1) + = 6 = 2:2 + (2) + = 10 ∴ = 1, = 4 = + + 4

Scale 15D (0, 5, 7, 10, 15) Low Partial Credit • Works towards 2nd difference Mid Partial Credit • One equation set correctly High Partial Credit • Both equations set correctly error in solving • Incorrect equations but solved correctly

(d) = (18) + (18) + 4 = 346 Scale 5B (0, 2, 5) Low Partial Credit • Any correct substitution

(e) + + 4 = 310 + − 306 = 0 ( − 17)( + 18) = 0 = 17

Scale 5B (0, 2, 5) Low Partial Credit • Any correct substitution

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(a) (i)

105 ∙ 2 = 100(1 + ) = 105 ∙ 2100 − 1

= 0 ∙ 423336



correctly

(ii)

= 550(1 ∙ 00423336) + 550(1 ∙ 00423336)+ ⋯550(1 ∙ 00423336) = 550[(1 ∙ 00423336)+ (1 ∙ 00423336) +. . . +(1∙ 00423336) ] = 550 1 ∙ 00423336(1 − (1 ∙ 00423336)1 − 1 ∙ 00423336 = €332,876

Scale 15D (0, 5, 7, 10, 15) Low Partial Credit • Any correct term of sequence • Identifies a geometric series and stops Mid Partial Credit • Correct series identified • Series set up with 1 error High Partial Credit • Error in calculation • Fully correct but error in power

(b) = 25 × 12 = 300 ℎ = 104 ∙ 2100 − 1 = 0 ∙ 343437929% = 332,876(0 ∙ 00343437929)(1 ∙ 00343437929)(1 ∙ 00343437929) − 1 = €1779 ∙ 41



correctly

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SUMMARY OF MARK ALLOCATIONS TO BE APPLIED

Section A Section B

Question 1 Question 7 (a) (i) 5B (a) 5B (ii) 5B (b) 5B (b) 15D (c) 5C (d) 10C (e) 15C Question 2 (f) 10C (a) (i) 5B (g) 10B (ii) 5B (b) 15D Question 8 (a) (i) 5B Question 3 (ii) 10B (a) (i) 5B (iii) 10C (ii) 5B (iv) 10C (b) 5C (b) (i) 5B (c) (i) 5B (ii) 10C (ii) 5B Question 9 Question 4 (a) 5B (a) 10C (b) 5B (b) 5A (c) 10B (c) 5B (d) 5B (d) 5A (e) 15D Question 5 (a) 15D (b) 10D Question 6 (a) (i) 10D (ii) 5B (b) 10C

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PAPER 2

Detailed marking notes Note: The model solutions for each question are not intended to be exhaustive – there may be other correct solutions. Any Examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact Examcraft.


(a) (i)

Line 1 and Line 3 × = −1

Scale 5B (0, 2, 5) Partial Credit • Identifies lines no reason • States perpendicular slopes multiply to give −1

(ii) Line 2 and Line 4 =

Scale 5B (0, 2, 5) Partial Credit • Identifies lines no reason • States parallel slopes are equal

(b) (1,0) 2 + 3 = 2 2 + 3 + = 0 |2(1) + 3(0) + |√2 + 3 = 2

| + 2| = 2√13 = −2 ± 2√13 2 + 3 + (−2 + 2√13) = 0

Or 2 + 3 + (−2 − 2√13) = 0

Scale 15D (0, 5, 7, 10, 15) Low Partial Credit • Finds a point • Writes equation of all parallel lines Mid Partial Credit • Correct substitution into perpendicular

distance formula

High Partial Credit • Solves

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(a) (i)

Scale 5B (0, 2, 5) Partial Credit • Any correct centre plotted • One circle plotted

(ii) Distance between (4, 2) and (5, 3) = √2 √18 − √8 = √2

Scale 5B (0, 2, 5) Partial Credit • Finds distance between centres

(b) = 2 − 2(2 − ) + 2 − 24(2 − ) + 12 + 25 = 0 4 + 28 − 15 = 0 ∴ = 0 ∙ 5, = −7 ∙ 5 (1 ∙ 5, 0 ∙ 5)(9 ∙ 5, −7 ∙ 5) (9 ∙ 5 − 1 ∙ 5) + (−7 ∙ 5 − 0 ∙ 5) 8√2

Scale 15D (0, 5, 7, 10, 15) Low Partial Credit • Any correct substitution into non linear

equation Mid Partial Credit • Incorrect quadratic solved correctly • Correct quadratic solved with errors High Partial Credit • Two correct points found • Two incorrect points found but continues to

find distance between them

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(a) (i)

19 14 = 11,628 Scale 5B (0, 2, 5) Partial Credit • 20 15 • Uses 20 or 15 in a combination

(ii) 20 15 − 18 13 = 6,936 Scale 5B (0, 2, 5) Partial Credit • Writes 18 or 13 • Fails to subtract from total

(b) ( ∩ ) = 0 ∙ 2 ( ) = 0 ∙ 46 ( | ) = 0 ∙ 20 ∙ 46 = 1023

Scale 5C (0, 2, 4, 5) Low Partial Credit • ( ∩ ) or ( ) correct High Partial Credit • ( ∩ ) and ( ) correct

(c) (i)

25 35 = 0 ∙ 0009675


(ii)

115 25 35 × 35 = 0 ∙ 088


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(a) Scale 10C (0, 3, 7, 10) Low Partial Credit • Two correct points plotted High Partial Credit • 8 points plotted correctly • All points correct but axis not labelled

(b) = 0 ∙ 883732 Scale 5A (0, 5)

(c) Strong positive correlation between hours of sunshine and maximum temperature reached in the towns.

Scale 5B (0, 2, 5) Partial Credit • Partially correct

(d) =≅ 16 ∙ 5℃ ± 1℃ Scale 5A (0, 5)

0

10

20

30

0 5 10 15 20

Max

Tem

p

Number of hours of sunshine

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(a)

Given: || || ℎ| | = | | To Prove: | | = | | Construction: Draw ’|| , cutting at ’ and at ’ Draw ’ ’|| , cutting at ’ and at ’ Proof: | ’ ’| = | | Opposite sides in

a parallelogram are equal |∠ | = |∠ | Alternate angles| | = | | Given |∠ | = |∠ | Vertically opposite angles |∠ | = |∠ | Angles in a triangle ∴ ∆ ≡ ∆ ASA ∴ | | = | |

But | | = | | | |= | | Opposite sides in a parallelogram are equal ∴ | | = | |

Scale 15D (0, 5, 7, 10, 15) Low Partial Credit • Any correct step Mid Partial Credit • Given/To Prove/Construction 2 correct High Partial Credit • Given/To Prove/Construction partially correct

but fully correct proof • Given/To Prove/Construction fully correct but

proof only partially correct

(b)

Scale 10D (0, 2, 5, 8, 10) Low Partial Credit • Any correct step Mid Partial Credit • One perpendicular bisector correct High Partial Credit • One median correct

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(a) (i)

Let ( , ) and ( , ) be two points on the unit circle. Using the Cosine Rule: | | = 1 + 1 − 2(1)(1) cos( − )= 1 + 1 − 2 cos( − )| | = 2 − 2 cos( − )…

Using the distance formula: | | = (cos − cos ) + (sin − sin ) | | = (cos − cos ) + (sin − sin ) = cos − 2 cos cos + cos + sin− 2sin sin + sin = cos + sin + cos + sin− 2 cos cos − 2sin sin = 1 + 1 − 2 cos cos − 2sin sin

(by Formula 1)∴ | | = 2 − 2(cos cos + sin sin )… = 2 − 2cos( − ) = 2 − 2(cos cos + sin sin )∴ cos( − ) = cos cos + sin sin

Scale 10D (0, 2, 5, 8, 10) Low Partial Credit • Any correct step Mid Partial Credit • Distance and Cosine rule partially correct High Partial Credit • Distance fully correct with partially correct

Cosine • Cosine fully correct with partially correct

distance

(ii) 2 = cos( + ) cos( + ) = − 2 = −


(b) 3 = cos √32

3 = 30°, 330° 3 = 30°, 390°, 750°, 330°, 690°, 1050° = 10°, 130°, 250°, 110°, 230°, 350°

Scale 10C (0, 3, 7, 10) Low Partial Credit • One correct angle found High Partial Credit • One incorrect angle • Incorrect range used but angles correct

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(a)

= 24m Scale 5B (0, 2, 5) Partial Credit • Partially correct

(b) = [2m, 46m] Scale 5B (0, 2, 5) Partial Credit • Partially correct

(c) = 6 Or 6 = 2 ∴ = 3


(d) Joan. The loading platform is not the midway line so it is a cosine graph

Scale 10C (0, 3, 7, 10) Low Partial Credit • States Joan • States Tom with correct explanation High Partial Credit • States Joan with partially correct answer

(e) (iv) ( ) = 24 − 22 Scale 15C (0, 5, 10, 15) Low Partial Credit • Correct formula with no values filled in • Incorrect formula with partially correct

substitution High Partial Credit • Incorrect formula with all values filled in • Correct formula with partially correct

substitution

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(f)

1·5 0 1·5 3 4·5 6

Scale 10C (0, 3, 7, 10) Low Partial Credit • Any correct point on graph • Partially correct graph for incorrect ( ) in part

(e) High Partial Credit • Correct graph for incorrect ( ) in part (e)

(g) (2) = 24 − 22 3 (2)

= 35m


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Q8 Model Solutions –50 Marks Marking Notes

(a) (i)

6103010 = 6103people Scale 10B (0, 5, 10) Partial Credit • Partially correct

(ii) Decrease in number in both categories overall. Changes in industry, economy picking up. More emigration. Increased employment.

Scale 5B (0, 2, 5) Partial Credit • Correct explanantion with incorrect reason • Incorrect explanation with correct reason

(iii) 4266510 = 4267people

6103118,000 × 100 = 5 ∙ 2%

426796,000 × 100 = 4 ∙ 4%

Scale 10C (0, 3, 7, 10) Low Partial Credit • One percentage calculated High Partial Credit • Two percentages with no reason or incorrect

reason

(iv)

Or other suitable display.

Scale 10C (0, 3, 7, 10) Low Partial Credit • One correct month High Partial Credit • Two correct months

(b) (i)

: = 0 ∙ 75 : ≠ 0 ∙ 75

Scale 5B (0, 2, 5) Partial Credit • One correct

(ii) = 360504

360504 − 1 ∙ 96 360504 1 − 360504504 << 360504 + 1 ∙ 96 360504 1 − 360504504

0 ∙ 675 < < 0 ∙ 754 0·59 is not within this interval so therefore we reject the null hypothesis

Scale 10C (0, 3, 7, 10) Low Partial Credit • Calculates • Partially sets up interval correctly High Partial Credit • Interval set up with one error • Correct interval but incorrect conclusion

0200400600800

1000

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(a)

cos( ) = 5

∴ = 5cos( ) sin( ) = 5

∴ = 5sin( )


(b) = 216 = 12 Scale 5B (0, 2, 5) Partial Credit • Partially correct

(c) (9) = + (5 sin( )) 81 = + 25 sin ( ) = 81 − 25 (12 )

Scale 10B (0, 5, 10) Partial Credit • Any correct substitution

(d) 5 (12 ) + 81 − 25 (12 ) Scale 5B (0, 2, 5) Partial Credit • Partially correct

(e) 10 = 5 (12 ) + 81 − 25 (12 ) 10 − 5 (12 ) = 81 − 25 (12 ) 100 − 100 (12 ) + 25 (12 )= 81 − 25 (12 ) 44 − 100 (12 ) = 0 (12 ) = 44100

12 = cos 44100

= 0 ∙ 0296 , = 0 ∙ 1371

Scale 15D (0, 5, 7, 10, 15) Low Partial Credit • Any step towards setting up correct equation Mid Partial Credit • Correct equation squared incorrectly • Incorrect equation squared correctly High Partial Credit • Mishandles angle/omits one answer

PRE-LEAVING CERTIFICATE EXAMINATION, 2018 MARKING SCHEME - School … · 2019-04-08 · page 1 of...

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