Mathematics 9709/33 October/November 2020: question paper and mark scheme

Mark scheme

1 Make a recognisable sketch graph of y = 2 x − 3 and the line B1 Need to see correct V at x = 3, roughly symmetrical, y = 2 – 5x x = 3 stated, domain at least (– 2, 5). Find x-coordinate of intersection with y = 2 – 5x M1 Find point of intersection with y = 2|x – 3| or solve 2 – 5x with 2(x – 3) or –2(x – 3) 4 A1 Obtain x = − 3 4 A1 Do not accept x < –1.33 State final answer x < − [Do not condone ⩽ for < in the final answer.] 3 Alternative method for question 1 State or imply non-modular inequality/equality B1 Two correct linear equations only (2 – 5x)2 >, ⩾, =, 22(x – 3)2, or corresponding quadratic equation, or pair of linear equations (2 – 5x) >, ⩾ , =, ± 2(x – 3) Make reasonable attempt at solving a 3-term quadratic, or solve M1 21x2 + 4x – 32 = (3x + 4)(7x – 8) = 0 one linear equation, or linear inequality for x 2 – 5x or –(2 – 5x) with 2(x – 3) or –2(x – 3) 4 A1 Obtain critical value x = − 3 4 A1 Do not accept x < –1.33 State final answer x < − [Do not condone ⩽ for < in the final answer.] 3 4

Mark scheme

2 Show a circle with centre the origin and radius 2 B1 Show the point representing 1 – i B1 Show a circle with centre 1 – i and radius 1 B1 FT The FT is on the position of 1 – i. Shade the appropriate region B1 FT The FT is on the position of 1 – i. Shaded region outside circle with centre the origin and radius 2 and inside circle with centre ±1 ± i and radius 1 4

Mark scheme

3 dx dy B1 State or imply = 2sin 2θ or = 2 + 2 cos 2θ dθ dθ dy dy dx M1 Use = ÷ dx dθ dθ dy 2 + 2cos 2θ A1 OE Obtain correct answer = dx 2sin 2θ Use correct double angle formulae M1 dy A1 AG. Must have simplified numerator in terms of cosθ. Obtain the given answer correctly = cot θ dx Alternative method for question 3 Start by using both correct double angle formulae M1 e.g. x = 3 – (2cos2θ − 1), y = 2θ + 2sinθcosθ dx dy B1 or dθ dθ dy = ( ( 2 2 + 2 cos θ − sin θ 2 )) M1 A1 dx 4cos θ sin θ dy A1 AG Simplify to given answer correctly = cot θ dx 3 Alternative method for question 3 dx dy B1 Set = 2θ. State = sin t or = 1 + cos t dt dt dy dy dx M1 Use = ÷ dx dt dt dy 1 + cos t A1 OE Obtain correct answer = dx sin t Use correct double angle formulae M1 dy A1 Obtain the given answer correctly = cot θ dx 5

Mark scheme

4 State or imply log10 10 = 1 B1 log10 10–1 = − 1 Use law of the logarithm of a power, product or quotient M1 Obtain a correct equation in any form, free of logs A1 e.g. (2x + 1)/(x + 1)2 = 10–1 or 10(2x + 1)/(x + 1)2 = 100 or 1 or x2 + 2x + 1 = 20x + 10 Reduce to x 2 − 18 x − 9 = 0 , or equivalent A1 Solve a 3-term quadratic M1 Obtain final answers x = 18.487 and x = –0.487 A1 Must be 3 d.p. Do not allow rejection. 6

Mark scheme

5(a) Sketch a relevant graph, e.g. y = cosec x B1 π π cosec x, U shaped, roughly symmetrical about x = , y = 1 2 2 π 5π and domain at least , . 6 6 1 − x B1 Exponential graph needs y(0) = 2, negative gradient, always Sketch a second relevant graph, e.g. y = 1 + e 2 , and justify the increasing, and y(π) > 1 given statement Needs to mark intersections with dots, crosses, or say roots at points of intersection, or equivalent 2 5(b) Use the iterative formula correctly at least twice M1 2, 2.3217, 2.2760, 2.2824… Need to see 2 iterations and following value inserted correctly Obtain final answer 2.28 A1 Must be supported by iterations Show sufficient iterations to at least 4 d.p. to justify 2.28 to A1 2 d.p., or show there is a sign change in the interval (2.275, 2.285) 3

Mark scheme

6(a) State R = 15 B1 Use trig formulae to find α M1 sin α 3 3 = with no error seen or tan α = quoted then cos α 6 6 allow M1 Obtain α = 50.77 A1 Must be 2 d.p. If radians 0.89 A0 MR 3 6(b) 2.5 B1 FT The FT is on incorrect R. Evaluate β = cos −1 (49.797° to 4 d.p.) 15 x = β − α [–2.9° and –301.7°] 3 x M1 x Use correct method to find a value of in the interval Needs to use 3 3 Obtain answer rounding to x = 301.6° to 301.8° A1 Obtain second answer rounding to x = 2.9(0)° to 2.9(2)° and no A1 others in the interval 4

Mark scheme

7(a) Substitute −1 + 5 i in the equation and attempt expansions of M1 All working must be seen. x2 and x3 Allow M1 if small errors in 1 − 2 5i − 5 or 1 − 5i − 5i − 5 and 4 − 2 5i + 10 or 4 − 4 5i + 2 5i + 10 Use i2 = –1 correctly at least once M1 1 – 5 or 4 + 10 seen Complete the verification correctly A1 ( ) ( ) ( ) 2 14 − 2 5i + −4 − 2 5i + 6 −1 + 5i − 18 = 0 3 7(b) State second root −1 − 5i B1 Carry out a complete method for finding a quadratic factor with M1 zeros −1 + 5i and −1 − 5i Obtain x2 + 2x + 6 A1 3 A1 OE Obtain root x = 2 Alternative method for question 7(b) State second root −1 − 5i B1 (x + 1 – 5i ) (x + 1 + 5i ) (2x + a) = 2x3 + x2 + 6x – 18 M1 (1 – 5 i) (1 + 5 i) a = –18 A1 3 A1 OE 6a = –18 a = –3 leading to x = 2 7(b) Alternative method for question 7(b) State second root −1 − 5i B1 POR = 6 SOR = – 2 M1 Obtain x2 + 2x + 6 A1 3 A1 OE Obtain root x = 2 Alternative method for question 7(b) State second root −1 − 5i B1 ( )( ) POR −1 − 5i −1 + 5i a = 9 M1 A1 3 A1 OE Obtain root x = 2 Alternative method for question 7(b) State second root −1 − 5i B1 1 M1 A1 ( ) ( ) SOR −1 − 5i + −1 + 5i + a = − 2 3 A1 OE Obtain root x = 2 4

Mark scheme

8 Separate variables correctly and attempt integration of at least B1 1 1 − 2 x2 one side dy = dx y x Obtain term ln y B1 Obtain terms ln x − x 2 B1 Use x = 1, y = 1 to evaluate a constant, or as limits, in a solution M1 The 2 terms of required form must be from correct working containing at least 2 terms of the form a ln y , b ln x and cx 2 e.g. ln y = lnx – x2 + 1 Obtain correct solution in any form A1 Rearrange and obtain y = xe1− x 2 A1 OE 6

Mark scheme

9(a) A B C B1 State or imply the form + + 1 − x 2 + 3x ( 2 + 3x )2 Use a correct method for finding a coefficient M1 Obtain one of A = 1, B = –1 , C = 6 A1 Obtain a second value A1 Obtain the third value A1 A Dx + E In the form + , where A = 1, D = – 3 1 − x ( 2 + 3 x )2 and E = 4 can score B1 M1 A1 A1 A1 as above. 5 9(b) Use a correct method to find the first two terms of the expansion M1 Symbolic coefficients are not sufficient for the M1 −1 −2 −1 −1 3 −2 3 1 + ( −1)( − x ) + ( −1)( −2 )( − x )2 of (1 − x ) , ( 2 + 3 x ) , 1 + x , ( 2 + 3 x ) or 1 + x A A=1 2 2 2 3x 3x 2 1 + ( −1) + ( −1)( −2 ) B 2 2 B=1 2 2 3x 3x 2 1 + ( −2 ) + ( −2 )( −3) C 2 2 C=6 4 2 Obtain correct un-simplified expansions up to the term in A1 FT of each partial fraction + (1 + x + x ) + − 12 + 34 x − 89 x 2 2 A1 FT 6 18 81 + + − x + x 2 [The FT is on A, B, C] 4 4 8 A1 FT 1 6 3 18 9 81 2 1 − + + 1 + − x + 1 − + x 2 4 4 4 8 8 11 A1 Allow unsimplified fractions Obtain final answer 2 − x + 10 x 2 , or equivalent 4 3x 3x 2 ( Dx + E ) ( ) 2 ( )( ) 2 1 + −2 + −2 −3 D = –3, E = 4 4 2 The FT is on A, D, E. 5

Mark scheme

10(a) Use correct product or quotient rule *M1 dy 1 1 − x 1 − x = − (2 − x) e 2 − e 2 dx 2 M1 requires at least one of derivatives correct Obtain correct derivative in any form A1 Equate derivative to zero and solve for x DM1 Obtain x = 4 A1 ISW Obtain y = –2e–2, or exact equivalent A1 5 10(b) Commence integration and reach *M1 Condone omission of dx 1 1 1 1 1 − x − x − x − x − x a (2 − x) e 2 +b e 2 dx −2 ( 2 − x ) e 2 + 4e 2 or 2 xe 2 1 − x 1 − x A1 OE Obtain −2 ( 2 − x ) e 2 − 2 e 2 dx 1 − x A1 OE Complete integration and obtain 2 xe 2 Use correct limits, x = 0 and x = 2, correctly, having integrated DM1 Ignore omission of zeros and allow max of 1 error twice Obtain answer 4e–1, or exact equivalent A1 ISW Alternative method for question 10(b) − x 1 *M1 A1 d 2 xe 2 1 1 = 2e− 2 x − xe− 2 x dx 1 − x A1 ∴ 2 xe 2 Use correct limits, x = 0 and x = 2, correctly, having integrated DM1 Ignore omission of zeros and allow max of 1 error twice Obtain answer 4e–1, or exact equivalent A1 ISW 5

Mark scheme

11(a) Express general point of at least one line correctly in component B1 form, i.e. (1 + aλ, 2 + 2λ, 1 – λ) or (2 + 2µ, 1 – µ, – 1 + µ) Equate at least two pairs of corresponding components and solve M1 May be implied for λ or for µ 1 + aλ = 2 + 2µ 2 + 2λ = 1 – µ 1 – λ = – 1 + µ Obtain λ = – 3 or µ = 5 A1 11 A1 Allow a = – 3.667 Obtain a = – 3 State that the point of intersection has position vector A1 Allow coordinate form (12, – 4, 4) 12i – 4j + 4k 5 11(b) Use correct process for finding the scalar product of direction M1 (a, 2, – 1) . (2, – 1, 1) = 2a – 2 – 1 or 2a – 3 vectors for the two lines Using the correct process for the moduli, divide the scalar *M1 product by the product of the moduli and equate the result to 1 ± 6 State a correct equation in a in any form, A1 2a − 2 − 1 1 e.g. =± ( 6 a2 + 5 ) 6 Solve for a DM1 Solve 3-term quadratic for a having expanded (2a – 3)2 to produce 3 terms e.g. 36(2a – 3)2 = 6(a2 + 5) 138a2 – 432a + 294 = 0 23a2 – 72a + 49 = 0 (23a – 49) (a – 1) = 0 Obtain a = 1 A1 49 A1 Allow a = 2.13 Obtain a = 23 11(b) Alternative method for question 11(b) cos(θ) = [|a2 + 22 + (–1)2|2 + | 22 + (–1)2 + 12|2 M1 Use of cosine rule. Must be correct vectors. – |(a – 2)2 + 32 + (–2)2|2] / [2|a2 + 22 + (–1)2|.|22 + (–1)2 +12|] 1 *M1 Allow M1* here for any two vectors Equate the result to ± A1 6 Solve for a DM1 Solve 3-term quadratic for a having expanded (2a – 3)2 to produce 3 terms e.g. 36(2a – 3)2 = 6(a2 + 5) 138a2 – 432a + 294 = 0 23a2 – 72a + 49 = 0 (23a – 49) (a – 1) = 0 Obtain a = 1 A1 49 A1 Allow a = 2.13 Obtain a = 23 6