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