BBSS 4E_AM Prelim P2 2021 Marking Scheme
Uploaded by hima · 11 June 2023
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Habits of Mind: Striving for Accuracy and Precision Bukit Batok Secondary School GCE ‘O’ Level Preliminary Examination 2021 Secondary 4 Express ADDITIONAL MATHEMATICS (Marking Scheme) 4049/02 1 (i) LHS = 1 – 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛𝑠𝑖𝑛 𝜃 1+ 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛𝑠𝑖𝑛 𝜃 = 1 – (1 − 2 𝜃 2 ) + 𝜃 2 𝜃 2 1 + (2 𝜃 2 − 1) + 𝜃 2 𝜃 2 = 2 𝜃 2 + 𝜃 2 𝜃 2 2 𝜃 2 + 𝜃 2 𝜃 2 = 𝜃 2 (𝜃 2 +𝜃 2 ) 𝜃 2 (𝜃 2 +𝜃 2 ) = 𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 2 = RHS M1 for converting at least one cos θ correctly M1 for converting sin θ correctly M1 for factorizing A1 (ii) 1 + sin θ = 2 cos θ ------ (1) 1 − 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛𝑠𝑖𝑛 𝜃 1+ 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛𝑠𝑖𝑛 𝜃 =𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 2 --------- (2) Sub (1) into (2): 2𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 − 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 𝜃 + 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 =𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 2 1 3 =𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 2 (since cos θ ≠ 0) Let β be a basic angle. tan β = 1 3 β ≈ 18.435o 𝜃 2 ≈ 18.435o θ ≈ 36.9o M1 for substitution M1 for simplifying A1 [7] 2 (i) x2 + y2 – 6x + 8y = 0 Centre of circle = (3, – 4) Radius of circle = √32 + (−4)2 = 5 units Diameter of circle = 10 units B1 for centre M1 for radius A1 for diameter (ii) tangent at R: 3x + 4y = 18 y = − 3 4 x + 9 2 Gradient of normal = 4 3 Normal will pass through centre of circle (3, – 4). M1 for making y the subject M1 for gradient of normal
2 BBSS/Prelim/2020/Sec 4NA/AMaths/Paper1 Habits of Mind: Striving for Accuracy and Precision Equation of normal: – 4 = 4 3 (3) + c c = – 8 Equation of normal: y = 4 3 x – 8 A1 (iii) Since the origin (0, 0) satisfies the equation, x2 + y2 – 6x + 8y = 0, Thus the origin lies on the circle. B1 [7] 3 (i) 2x3 + cx2 + dx – 6 = (x2 – 2x – 3) Q(x) + 23x + 21 2x3 + cx2 + dx – 6 = (x – 3) (x + 1) Q(x) + 23x + 21 When x = 3, 2(3)3 + c(3)2 + d(3) – 6 = 23(3) + 21 9c + 3d = 42 3c + d = 14 ----- (1) When x = – 1, 2(– 1)3 + c(– 1)2 + d(– 1) – 6 = 23(– 1) + 21 c – d = 6 ----- (2) (1)+ (2): 4c = 20 c = 5 Sub c = 5 into (1): d = 14 – 15 = – 1 M1 for factorising (x2 – 2x – 3) correctly M1 for sub x = 3 or sub x = – 1 correctly A1 for correct c A1 for correct d (ii) f(x) = 2x3 + 5x2 – x – 6 f(1) = 2(1)3 + 5(1)2 – (1) – 6 = 0 (x – 1) is a
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