Cambridge A Level Mathematics 9709 — 2007 Oct/Nov Paper 3 · Variant 1

1 1 Find the exact value of the constant k for which dx = 1. [4] 2x −1 1

Mark scheme: 1 Obtain indefinite integral of the form aln(2x –1), where a = 12 , 1, or 2 M1 Use limits and obtain equation 1 ln ( 2 k − )1 = 1 A1 2 Use correct method for solving an equation of the form aln(2k –1) = 1, where a = 1 , 1, or 2, for k M1 2 Obtain answer k = 1 e( 2 + )1 , or exact equivalent A1 [4] 2 2 2

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2 The polynomial x4 + 3x2 + a, where a is a constant, is denoted by p(x). It is given that x2 + x + 2 is a factor of p(x). Find the value of a and the other quadratic factor of p(x). [4]

Mark scheme: 2 EITHER: Attempt division by x 2 + x + 2 reaching a partial quotient of x 2 + kx M1 Complete the division and obtain quotient x 2 −x + 2 A1 Equate constant remainder to zero and solve for a M1 Obtain answer a = 4 A1 OR: Calling the unknown factor x 2 + bx + c , obtain an equation in b and/or c, or state without working two coefficients with the correct moduli M1 Obtain factor x 2 −x + 2 A1 Use a = 2c to find a M1 Obtain answer a = 4 A1 [4] 1 ∫

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3 Use integration by parts to show that 4 ln x dx = 6 ln 2 −2. [4] 2

Mark scheme: 1 3 Using 1 and ln x as parts reach xln x ± ∫ x. x dx M1* Obtain indefinite integral xln x –x A1 Substitute correct limits correctly M1(dep*) Obtain given answer A1 [4]

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4 The curve with equation y = e−x sin x has one stationary point for which 0 ≤x ≤π. (i) Find the x-coordinate of this point. [4] (ii) Determine whether this point is a maximum or a minimum point. [2]

Mark scheme: 4 (i) Use correct product or quotient rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1 Obtain answer x = 1 π or 0.785 with no errors seen A1 [4] 4 (ii) Use an appropriate method for determining the nature of a stationary point M1 Show the point is a maximum point with no errors seen A1 [2] [SR: for the answer 45° deduct final A1 in part (i), and deduct A1 in part (ii) if this value in degrees is used in the exponential.]

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5 (i) Show that the equation tan(45◦+ x) −tan x = 2 can be written in the form tan2x + 2 tan x −1 = 0. [3] (ii) Hence solve the equation tan(45◦+ x) −tan x = 2, giving all solutions in the interval 0◦≤x ≤180◦. [4]

Mark scheme: 5 (i) Use correct tan(A + B) formula to obtain an equation in tan x M1* Use tan 45° = 1 M1(dep*) Obtain the given answer A1 [3] (ii) Make reasonable attempt to solve the given quadratic for one value of tan x M1 Obtain tan x = − 1± 2 , or equivalent in the form ( a ± b /) c (accept 0.4, −2.4) A1 Obtain answer x = 22.5° A1 Obtain second answer x = 112.5 and no others in the range A1 [4] [Ignore answers outside the range.] [Treat answers in radians as a MR and deduct one mark from the marks for the angles.] GCE A/AS LEVEL – October/November 2007 9709 03

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6 (i) By sketching a suitable pair of graphs, show that the equation 2 −x = ln x has only one root. [2] (ii) Verify by calculation that this root lies between 1.4 and 1.7. [2] (iii) Show that this root also satisfies the equation x = 13(4 + x −2 ln x). [1] (iv) Use the iterative formula xn+1 = 13(4 + xn −2 ln xn), with initial value x1 = 1.5, to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Make a recognisable sketch of an appropriate graph, e.g. y = ln x B1 Sketch an appropriate second graph, e.g. y = 2 –x, correctly and justify the given statement B1 [2] (ii) Consider sign of 2 –x –ln x when x = 1.4 and x = 1.7, or equivalent M1 Complete the argument with correct calculations A1 [2] (iii) Rearrange the equation x = 13 ( 4 + x − 2ln x ) as 2 –x = ln x, or vice versa B1 [1] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.56 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.555, 1.565) A1 [3]

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7 The number of insects in a population t days after the start of observations is denoted by N. The variation in the number of insects is modelled by a differential equation of the form dN = kN cos(0.02t), dt where k is a constant and N is taken to be a continuous variable. It is given that N = 125 when t = 0. (i) Solve the differential equation, obtaining a relation between N, k and t. [5] (ii) Given also that N = 166 when t = 30, find the value of k. [2] (iii) Obtain an expression for N in terms of t, and find the least value of N predicted by this model. [3]

Mark scheme: 7 (i) Separate variables correctly and attempt integration of both sides M1* Obtain term ln N, or equivalent A1 k Obtain term sin(0.02t ) , or equivalent A1 .002 Use t = 0, N = 125 to evaluate a constant, or as limits, in a solution containing terms of the form aln N and bsin(0.02t), or equivalent M1 Obtain any correct form of solution, e.g. ln N = 50ksin(0.02t) + ln 125 A1 [5] (ii) Substituting N = 166 and t = 30, evaluate k M1(dep*) Obtain k = 0.0100479…(accept k = 0.01) A1 [2] (iii) Rearrange and obtain N = 125exp(0.502sin(0.02t)), or equivalent B1 Set sin(0.02t) = −1 in the expression for N, or equivalent M1 Obtain least value 75.6 (accept answers in the interval [75, 76]) A1 [3] [For the B1, accept 0.5 following k = 0.01, and allow 4.8 or better for ln 125.]

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4 −3i 8 (a) The complex number is given by = 1 −2i. (i) Express in the form x + iy, where x and y are real. [2] (ii) Find the modulus and argument of . [2] (b) Find the two square roots of the complex number 5 −12i, giving your answers in the form x + iy, where x and y are real. [6]

Mark scheme: 8 (a) (i) EITHER: Carry out multiplication of numerator and denominator by 1 + 2i, or equivalent M1 Obtain answer 2 + i, or any equivalent of the form (a + ib)/c A1 OR1: Obtain two equations in x and y, and solve for x or for y M1 Obtain answer 2 + i, or equivalent A1 OR2: Using the correct processes express z in polar form M1 Obtain answer 2 + i, or equivalent A1 [2] (ii) State that the modulus of z is 5 or 2.24 B1 State that the argument of z is 0.464 or 26.6° B1 [2] (b) EITHER: Square x + iy and equate real and imaginary parts to 5 and −12 respectively M1 Obtain x 2 −y 2 = 5 and 2xy = −12 A1 Eliminate one variable and obtain an equation in the other M1 Obtain x 4 −x5 2 − 36 = 0 or y 4 + 5 y 2 − 36 = 0 , or 3-term equivalent A1 Obtain answer 3 –2i A1 Obtain second answer –3 + 2i and no others A1 [SR: Allow a solution with 2xy = 12 to earn the second A1 and thus a maximum of 3/6.] OR: Convert 5 –12i to polar form (R, θ) M1 Use the fact that a square root has the polar form ( R , 1 θ ) M1 2 Obtain one root in polar form, e.g. ( 13 , − .0588) or ( 13 , −337.°) A1 + A1 Obtain answer 3 –2i A1 Obtain answer –3 + 2i and no others A1 [6] GCE A/AS LEVEL – October/November 2007 9709 03 A B C

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2 −x + 8x2 9 (i) Express in partial fractions. [5] (1 −x)(1 + 2x)(2 + x) 2 −x + 8x2 (ii) Hence obtain the expansionof inascending powers of x, up to and including (1 −x)(1 + 2x)(2 + x) the term in x2. [5]

Mark scheme: A B C 9 (i) State or imply the form + + B1 1 − x 1 + 2 x 2 + x Use any relevant method to determine a constant M1 Obtain A = 1, B = 2 and C = −4 A1 + A1 + A1 [5] (ii) Use correct method to obtain the first two terms of the expansion of 1( −x ) −1 , 1( + 2 x ) −1 , ( 2 + x ) −1 , or 1( + 1 x ) −1 M1 2 Obtain complete unsimplified expansions up to x 2 of each partial fraction A1√ + A1√ + A1√ Combine expansions and obtain answer 1 − 2 x + 17 x 2 A1 [5] 2 − 1 [Binomial coefficients such as   are not sufficient for the M1. The f.t. is on A, B, C.]  2  [Apply this scheme to attempts to expand ( 2 − x + 8 x 2 )(1 − x ) −1 1( + 2 x ) −1 ( 2 + x ) −1 , giving M1A1A1A1 for the expansions, and A1 for the final answer.] [Allow Maclaurin, giving M1A1√A1√ for f(0) = 1 and f ′(0) = −2, A1√ for f ″(0) = 17 and A1 for the final answer (f.t. is on A, B,C).]

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10 The straight line l has equation r = i + 6j −3k + s(i −2j + 2k). The plane p has equation (r −3i).(2i −3j + 6k) = 0. The line l intersects the plane p at the point A. (i) Find the position vector of A. [3] (ii) Find the acute angle between l and p. [4] (iii) Find a vector equation for the line which lies in p, passes through A and is perpendicular to l. [5]

Mark scheme: 10 (i) Substitute for r and expand the given scalar product, or correct equivalent, to obtain an equation in s M1 Solve a linear equation formed from a scalar product for s M1 Obtain s = 2 and position vector 3i + 2j + k for A A1 [3] (ii) State or imply a normal vector of p is 2i –3j + 6k, or equivalent B1 Use the correct process for evaluating a relevant scalar product, e.g. (i – 2j + 2k).(2i –3j + 6k) M1 Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse sine or cosine of the result M1 Obtain final answer 72.2° or 1.26 radians A1 [4] (iii) EITHER: Taking the direction vector of the line to be ai + bj + ck, state equation 2a –3b + 6c = 0 B1 State equation a –2b + 2c = 0 B1 Solve to find one ratio, e.g. a : b M1 Obtain ratio a : b : c = 6 : 2: −1, or equivalent A1 State answer r = 3i + 2j + k + λ(6i + 2j –k), or equivalent A1√ OR1: Attempt to calculate the vector product of a direction vector for the line l and a normal vector of the plane p, e.g.(i –2j + 2k)×(2i –3j + 6k) M2 Obtain two correct components of the product A1 Obtain answer –6i –2j + k, or equivalent A1 State answer r = 3i + 2j + k + λ(−6i –2j + k), or equivalent A1√ OR2: Obtain the equation of the plane containing A and perpendicular to the line l M1 State answer x –2y + 2z = 1, or equivalent A1√ Find position vector of a second point B on the line of intersection of this plane with the plane p, e.g. 9i + 4j M1 Obtain a direction vector for this line of intersection, e.g. 6i + 2j – k A1 State answer r = 3i + 2j + k + λ(6i + 2j – k), or equivalent A1 [5] [The f.t. is on A.]

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