Mathematics 9709/31 October/November 2019: question paper and mark scheme

Mark scheme

1 State 1 + e 2 y = e x B1 Make y the subject M1 Rearrange to e 2 y = ... and use logs A1 OE 1 ( Obtain answer y = ln e x − 1 2 ) 3

Mark scheme

2 State or imply non-modular inequality B1 12 x 2 + 44 x + 7 < 0 ( 2 x − 3) > 4 ( x + 1) , or corresponding quadratic equation, or pair of linear equations 2 2 2 ( 2 x − 3) = ±4 ( x + 1) Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations M1 Correct method seen, or implied by correct for x answers 7 1 A1 Obtain critical values x = − and x = − 2 6 7 1 A1 State final answer − < x < − 2 6 Alternative method for question 2 7 B1 Obtain critical value x = − from a graphical method, or by inspection, or by solving a 2 linear equation or an inequality 1 B2 Obtain critical value x = − similarly 6 7 1 B1 State final answer − <x<− 2 6 4

Mark scheme

3 dx B1 State = 2 + 2cos 2t dt Use the chain rule to find the derivative of y M1 dy 2sin 2t A1 OE Obtain = dt 1 − cos 2t dy dy dx M1 Use = ÷ dx dt dt dy A1 AG Obtain = cosec2t correctly dx 5

Mark scheme

4(i) dN B1 OE State = ke −0.02t N and show k = – 0.01 dt ( −10 = k × 1× 1000 ) 1 4(ii) Separate variables correctly and integrate at least one side B1 1 ∫ N dN = ∫ −0.01e −0.02 t dt Obtain term ln N B1 OE Obtain term 0.5e −0.02t B1 OE Use N = 1000, t = 0 to evaluate a constant, or as limits, in a solution with terms a ln N M1 and be −0.02t , where ab ≠ 0 Obtain correct solution in any form A1 1 ln1000 − = 6.41 ( e.g. ln N − ln1000 = 0.5 e −0.02t − 1) 2 Substitute N = 800 and obtain t = 29.6 A1 6 4(iii) 1000 B1 Accept 606 or 607 or 606.5 State that N approaches e 1

Mark scheme

5(i) Use correct product rule M1 Obtain correct derivative in any form A1 dy e−2 x = −2e−2 x ln ( x − 1) + dx x −1 Equate derivative to zero and derive A1 AG 1 1 x =1+ e 2( x−1) or p = 1 + 2( p−1) 3 5(ii) Calculate values of a relevant expression or pair of relevant expressions at x = 2.2 and M1 x = 2.6 1 f ( x ) = ln ( x − 1) − ⇒ f ( 2.2 ) = −0.234, f ( 2.6 ) = 0.317 2 ( x − 1) e −2 x f ( x ) = 2e −2 x ln ( x − 1) + ⇒ f ( 2.2 ) = 0.005… , f ( 2.6 ) = −0.0017... x −1 Complete the argument correctly with correct calculated values A1 2 5(iii)  1  M1 Use the iterative process pn +1 = 1 + exp  correctly at least once  2 ( p − 1)   n  Obtain final answer 2.42 A1 Show sufficient iterations to 4 d.p. to justify 2.42 to 2 d.p., or show there is a sign A1 change in the interval (2.415, 2.425) 3

Mark scheme

6(i) Use correct quotient rule M1 dy A1 AG Obtain = − cosec 2 x correctly dx 2 6(ii) Integrate by parts and reach ax cot x + b ∫ cot x dx *M1 Obtain − x cot x + ∫ cot x dx A1 OE State ± ln sin x as integral of cot x M1 Obtain complete integral − x cot x + ln sin x A1 OE Use correct limits correctly DM1 π 1 0+0+ − ln 4 2 1 A1 AG Obtain ( π + ln 4 ) following full and exact working 4 6

Mark scheme

7(i) Express general point of l or m in component form B1 e.g. (a + λ, 2 – 2λ, 3 + 3λ) or (2 + 2µ, 1 – µ, 2 + µ) Equate at least two pairs of corresponding components and solve for λ or for µ M1 Obtain either λ = – 2 or µ = – 5 A1 1 2 or λ = a or µ = a − 1 3 3 1 1 or λ = ( a − 4 ) or µ = ( 3a − 7 ) 5 5 Obtain a = – 6 A1 4 7(ii) Use scalar product to obtain a relevant equation in a, b and c, e.g. a – 2b + 3c = 0 B1 Obtain a second equation, e.g. 2a – b + c = 0 and solve for one ratio M1 Obtain a : b : c = 1 : 5 : 3 A1 OE Substitute a relevant point and values of a, b, c in general equation and find d M1 Obtain correct answer x + 5y + 3z = 13 A1FT OE. The FT is on a from part (i), if used Alternative method for question 7(ii) Attempt to calculate vector product of relevant vectors, M1 e.g. ( i − 2 j + 3k ) . ( 2i − j + k ) Obtain two correct components A1 Obtain correct answer, e.g. i + 5j + 3k A1 Substitute a relevant point and find d M1 Obtain correct answer x + 5y + 3z = 13 A1FT OE. The FT is on a from part (i), if used 7(ii) Alternative method for question 7(ii) Using a relevant point and relevant vectors, form a 2-parameter equation for the plane M1 State a correct equation, e.g. r = 2i + j + 2k +λ(i – 2j + 3k) +µ(2i – j + k) A1FT State three correct equations in x, y, z, λ and µ A1FT Eliminate λ and µ M1 Obtain correct answer x + 5y + 3z = 13 A1FT OE. The FT is on a from part (i), if used 5

Mark scheme

8(i) A B C B1 State or imply the form + 2+ x x x+2 Use a correct method for finding a constant M1 Obtain one of A = – 1, B = 3, C = 2 A1 Obtain a second value A1 Obtain the third value A1 Ax + B C Allow in the form + x 2 x+2 5 8(ii) 3 B1FT + The FT is on A, B, C; or on A, D, E. Integrate and obtain terms ln x − + 2ln ( x + 2 ) B1FT + x B1FT b M1 3 Substitute limits correctly in an integral with terms a ln x , and c ln ( x + 2 ) , where − ln 4 − + 2ln 6 ( + ln1) + 3 − 2ln 3 x 4 abc ≠ 0 9 A1 AG – work to combine or simplify logs is Obtain following full and exact working required 4 5

Mark scheme

9(i) Use cos(A + B) formula to express cos3x in terms of trig functions of 2x and x M1 Use double angle formulae and Pythagoras to obtain an expression in terms of cos x M1 only Obtain a correct expression in terms of cos x in any form A1 Obtain cos3 x ≡ 4cos3 x − 3cos x A1 AG 4 9(ii) Use identity and solve cubic 4cos 3 x = −1 for x M1 cos x = −0.6299... Obtain answer 2.25 and no other in the interval A1 Accept 0.717π M1A0 for 129.0° 2 9(iii) 1 3 B1 + Obtain indefinite integral sin 3x + sin x B1 12 4 Substitute limits in an indefinite integral of the form a sin 3 x + b sin x , where ab ≠ 0 M1 1 1 π 1 π π  sin π + 3sin − sin − 3sin  4 3 3 3 2 6 Obtain answer 1 24 ( ) 9 3 − 11 , or exact equivalent A1 Alternative method for question 9(iii) B1 + ∫ cos x (1 − sin x ) dx = sin x − 3 sin x ( +C ) 2 1 3 B1 Substitute limits in an indefinite integral of the form a sin x + b sin 3 x where ab ≠ 0 M1  3 1 1 3 1   − − +   2 2 4 2 24  Obtain answer 1 24 ( ) 9 3 − 11 , or exact equivalent A1 4

Mark scheme

10(a) Square a + ib and equate real and imaginary parts to – 3 and −2 10 respectively *M1 Obtain a 2 − b 2 = −3 and 2ab = −2 10 A1 Eliminate one unknown and find an equation in the other DM1 Obtain a 4 + 3a 2 − 10 = 0 , or b 4 − 3b 2 − 10 = 0 , or horizontal 3-term equivalent A1 Obtain answers ± ( 2 − 5i ) , or exact equivalent A1 5 10(b) Show point representing 3 + i in relatively correct position B1 Show a circle with radius 3 and centre not at the origin B1 1 B1 Show correct half line from the origin at π to the real axis 4 Show horizontal line y = 2 B1 Shade the correct region B1 Im(z) shaded Im(z) = 2 i π 4 3 Re(z) 5