Answers - 3E NV AM EOY 2021

3E AM EOY 2021 Marking Scheme Qn Marks Marker’s Report 1 2 22 53 5 7 31(3 1)( 10) 10 x x A Bx C xx x x        M1 (correct form) Generally well done across the level. Common mistakes: 1. Wrong form 2. 2 22 53 5 7 5 4 3 31(3 1)( 10) 10 x x x xx x x        2253 5 7 ( 10) ( )(3 1)x x A x Bx C x       M1 (DB) 5, 4, 3A B C   B1,B1,B1 2 22 53 5 7 5 4 3 31(3 1)( 10) 10 x x x xx x x        A1 (conclusion) 2(a) 22 1x by y a a  B1 (linearization) Badly done. Candidates were unable to linearize the equation. Many mistook 2x y as the horizontal axis (X) 2y as the vertical axis (Y) Vertical-intercept= 1 a , so a can be found B1 (statement) Gradient = b a so b can now be found B1 (statement) (b)(i) Scale; Points; Line B1, B1, B1 Well done. Only about 4 candidates (surprisingly from 3E1) did not follow the given scale. (ii) k = gradient using two points M1 Generally well done 0.0414k  A1 (iii) ln m = 4.3 to 4.5 M1 Well done 73.7m to 90 A1 (iv) 4.4 ln 4.4 0.0413792 e t M1 Badly done. Many candidates computed the value of t rather than using the graph to estimate the value of t. Many did not answer the question by stating explicitly the year. 16.751 17 yearst  M1 Year 2021 + 17 = 2038 A1

3(a)   92 1 3 9 52 rr rTx r x      M1 Generally well done. Candidates are able to simplify the general term to its simplest form. For the mathematical reasoning, many used the word “integer” rather than whole number.     9 18 59 (2) 5 rrr xr  M1 (simplify to simplest form) For independent term, 18 5 0r 3.6r  M1 Since r is not a whole number, it is not possible to get a term independent of x. A1 (b)(i) 26561 8748 5103xx B3 Well done. However, a handful of scripts from 3E2 left this as the final answer: 29 12 7xx (candidates divided throughout by 729) (ii)   3 5 3 4 8 3 17013 2 xTx      8748(2) 5103( 7) 9 1701 68526      M1 A1 Badly done. 85% of the candidates were unaware that the T 4 contains the x3 term. Hence, did not obtain the correct answer. 4(i) 1ABm  B1 (gradient of AB) Well done. 1m  B1 (gradient of bisector)  8.5,13.5ABM  B1 (midpoint)

13.5 1( 8.5)yx   M1 5yx A1 (ii) (0,5)Centre B1 Generally well done. Radius = 13 units B1 22 ( 5) 169xy   or 213 or 22 10 144 0x y y    B1 (iii) Distance of C from centre = 7.07 or 52 or 50 units M1 (DB dep on their centre) Well done. Interestingly, candidates failed to see that 169 13 Since distance of C from centre is less than the radius, the point C lies inside the circle. A1 (iv) Centre(-18,5) B1 Badly done. Candidates were unable to figure out the centre of the reflected circle. 22( 18) ( 5) 169xy    or 213 22 36 10 11 0x y x y     B1 or B2 if centre not shown If centre is wrong, max 1 out of 2 marks 5(i) Let 32( ) 2 11 22 13f x x x x    Well done. Quite a handful of candidates used long division. 32( 1) 2( 1) 11( 1) 22( 1) 13 0f          Therefore, 1x is a factor. B1 (ii) By long division, comparing coefficient or synthetic division, M1 Long division was the most commonly used method followed by Synthetic Division. 3 2 22 11 22 13 ( 1)(2 9 13)x x x x x x       M1 Candidates can write this step clearly in their working. To cut the x-axis, 2( 1)(2 9 13) 0x x x   

1x or 22 9 13 0xx   I do not agree, since 2 4 23b ac  < 0, therefore there is only 1 real root so the graph only cuts the x-axis at one point. DB1 (discriminant) A1 (reasoning) Careless mistake was seen in the value of the discriminant of -68. 6(a) 21xy  or 21xy  o.e M1 (Eqn 1) Common error when forming equation 315 5 5 3 5(25) x x y y xy     0.5 2 1.5xy or 43xy o.e M1 (Eqn 2) Quite alright for most students 1, 1xy  A1,A1 (b) 760 (25 7 5)(4 5 2) BC  Or 380 (25 7 5)(2 5 1) BC  M1 There are a few students who wrote formula as area of triangle = BC x height 760 114 5 190   Or 380 57 5 95   M1 (Expansion & simplifying) Mistakes in expansion commonly seen. 760 114 5 190 114 5 190 114 5 190   Or 380 57 5 95 57 5 95 57 5 95   DB1 (rationalization) 3 5 5 A1 (simplest form) Alternatively: 380(25 7 5) 25 7 5 (25 7 5)(25 7 5) Area     M1 (rationalization) (25 7 5)(2 5 1) (2 5 1)(2 5 1) BC   o.e M1 (rationalization) 50 5 25 70 7 5 19BC  o.e M1(expansion)

57 5 95 3 5 519BC    A1(simplest form) 7(a) 23 8 3 0uu   M1 Majority obtained the right quadratic equation (3 1)( 3) 0uu   M1 13 ( ) 3 x rejected or 33 1 x x   A1,A1 (deduct 1 if -3 is not rejected) A number of students rejected the answer 13 3 x  after putting log on both sides (b) 44 23 12log ( 4) log 1log 4 3x x      44 4 4 12log ( 4) log 1 log 4 3 log (2 3) x x      Or 4 44 4 log (2 3) 2log ( 4) log 1 og 4 3 xx l     M1(change of base) Common error 44 23 4 2 3 4 12log ( 4) log 1log 4 3 log 41 log 4 log 3 x x x x         4 4 4 2log ( 4) log (2 3) log 1 3xx     M1 A few students has this as first step - that is they did not show how the base is changed to 4. 44 48log log2 3 3 x x   OR 4 3( 4)log 12(2 3) x x   M1 (quotient and product law) 48 2 3 3 x x   OR 4 423 x x   M1 Remove log base 4 on both sides 3( 4) 8(2 3)xx   OR 4 4(2 3)xx  

36 13x A1 8(i) 2.5ACm  M1 Careless mistakes were seen when solving simultaneous equation. Most of the students are able to tackle this question. Equation of AC: 2.5 5.9yx M1 2 5(2.5 5.9) 15xx   DB1 (substitution/eli mination) 1, 3.4xy  M1 (x or y values) C(-1,3.4) A1 (coordinate form) (ii) (1.5,9.65)ACM  Badly done. Common error – Students consider the mid point of AB = Mid point of CD. (7.5,0)B M1 7.5 0, (1.5,9.65)22BD xyM  M1 (midpoint) ( 4.5,19.3)D  A1 Alternatively : Equation of AD : 2 17.55yx  M1 Common error – Students consider solving simultaneous equation using eqn AD and BD Equation of CD : 19 14135 7yx  M1 ( 4.5,19.3)D  A1 (iii) Area = 7.5 1 5 7.51 | |0 3.4 9 02   DB1 (shoelace method) Students are able to apply the shoelace method correctly. Students who did not obtain correct answer is due to wrong coordinates of C and/or B in earlier part.

= 0.5((7.5)(3.4) ( 1)( 9)) (7.5( 9) 5(3.4)) 59.5       units2 A1 9(i) For minimum graph, 0a B1 A number of students gave incomplete set of 3 conditions. A few scripts with this error:   2 93 4 0 => 4ac ac    For graph to lie above the x axis,   23 4 0 ac   M1 9 4ac ; 49ac ; 9 4a c ; 9 4c a A1 (ii) Any suitable pairs B0 or B2 Not so well done. Students with incomplete set of conditions but able to give a correct pair of values were awarded the marks. 10(i) y 1 x Shape correct (B1); Followed by y-intercept (B1) Generally fine

(ii) 15 23 x x  M1 (index form) Generally fine 123 (5 )xx M1 Equation of line: 23yx A1 No of solution = 2 B1 Badly done. Majority gave 1 solution. Mark is awarded only if the shape of curve and line are correct. 11(i) 24 (3 1) 1q px x p x q      M1 (equate) Careless mistakes for few 24 (3 1) 1q px x p x q      . Generally fine. 24 (2 1) 1 0x p x    M1 2(