Cambridge A Level Mathematics 9709 — 2012 May/June Paper 3 · Variant 3

1 1 Expand in ascending powers of x, up to and including the term in x2, simplifying the √(4 + 3x) coefficients. [4]

Mark scheme: 1 EITHER: Obtain a correct unsimplified version of the x or x2 term of the expansion of − 12 3 − 12 ( 4 + 3 x ) or 1( + 4 x ) M1 1 State correct first term B1 2 3 27 2 Obtain the next two terms − x + x A1 + A1 16 256 − 32 OR: Differentiate and evaluate f(0) and f ′(0), where f ′( x ) = k ( 4 + 3 x ) M1 1 State correct first term B1 2 3 27 2 Obtain the next two terms − x + x A1 + A1 [4] 16 256 − 12  [Symbolic coefficients, e.g.   are not sufficient for the M or B mark.]  2 

More questions on Series

2 Solve the equation 2 ln x ln 3, giving your answer correct to 3 significant figures. [4] ln(2x + 3) = +

Mark scheme: 2 Use law of the logarithm of a power and a product or quotient and remove logarithms M1 2 x + 3 Obtain a correct equation in any form, e.g. = 3 A1 x 2 Solve 3-term quadratic obtaining at least one root M1 Obtain final answer 1.39 only A1 [4] dx dy

More questions on Logarithmic and exponential functions

3 The parametric equations of a curve are x sin 2θ y cos 2θ 2 sin θ. = −θ, = + dy 2 cos θ Show that . [5] dx = 1 2 sin θ +

Mark scheme: d x dy 3 Obtain = 2 cos 2θ − 1 or = −2 sin 2θ + 2 cos θ , or equivalent B1 d θ dθ dy dy dx Use = ÷ M1 d x dθ dθ dy − 2 sin 2θ + 2 cos θ Obtain = , or equivalent A1 dx 2 cos 2θ − 1 At any stage use correct double angle formulae throughout M1 Obtain the given answer following full and correct working A1 [5]

More questions on Differentiation

e2x 4 The curve with equation y has one stationary point. = x3 (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 quotient or product rule M1 2 e 2 x 3e 2 x Obtain correct derivative in any form, e.g. − A1 x 3 x 4 Equate derivative to zero and solve a 2-term equation for non-zero x M1 3 Obtain x = correctly A1 [4] 2 (ii) Carry out a method for determining the nature of a stationary point, e.g. test derivative either side M1 Show point is a minimum with no errors seen A1 [2] GCE AS/A LEVEL – May/June 2012 9709 33

More questions on Differentiation

5 In a certain chemical process a substance A reacts with another substance B. The masses in grams of A and B present at time t seconds after the start of the process are x and y respectively. It is given that dy 70. and x When t 0, y = 5e−3t. = = dt = −0.6xy (i) Form a differential equation in y and t. Solve this differential equation and obtain an expression for y in terms of t. [6] (ii) The percentage of the initial mass of B remaining at time t is denoted by p. Find the exact value approached by p as t becomes large. [2]

Mark scheme: 5 (i) Substitute for x, separate variables correctly and attempt integration of both sides M1 Obtain term ln y, or equivalent A1 Obtain term e−,t3 or equivalent A1 Evaluate a constant, or use t = 0, y = 70 as limits in a solution containing terms a ln y and be−3 t M1 Obtain correct solution in any form, e.g. ln y − ln 70 = e −t3 − 1 A1 Rearrange and obtain y = 70exp(e −t3 − )1 , or equivalent A1 [6] (ii) Using answer to part (i), either express p in terms of t or use e −t3 → 0 to find the limiting value of y M1 100 Obtain answer from correct exact work A1 [2] e

More questions on Differential equations

6 It is given that tan 3x k tan x, where k is a constant and tan x = ≠0. (i) By first expanding show that tan(2x + x), tan2x k (3k −1) = −3. [4] (ii) Hence solve the equation tan 3x k tan x when k 4, giving all solutions in the interval = = [3] x 0◦< < 180◦. 2. [1] (iii) Show that the equation tan 3x k tan x has no root in the interval x k 0◦< = < 180◦when =

Mark scheme: 6 (i) Use tan (A + B) and tan 2A formulae to obtain an equation in tan x M1 Obtain a correct equation in tan x in any form A1 Obtain an expression of the form a tan 2 x = b M1 Obtain the given answer A1 [4] (ii) Substitute k = 4 in the given expression and solve for x M1 Obtain answer, e.g. x = 16.8° A1 Obtain second answer, e.g. x = 163.2°, and no others in the given interval A1 [3] [Ignore answers outside the given interval. Treat answers in radians as a misread and deduct A1 from the marks for the angles.] (iii) Substitute k = 2, show tan 2 x < 0 and justify given statement correctly B1 [1]

More questions on Trigonometry

7 y R x O p2 The diagram shows part of the curve y for x where x is in radians. The shaded region = cos(√x) ≥0, between the curve, the axes and the line x p2, where p 0, is denoted by R. The area of R is equal = > to 1. p2 3 cos p (i) Use the substitution x u2 to find dx. Hence show that sin p −2 . [6] = ã 0 cos(√x) = 2p 3 cos pn (ii) Use the iterative formula sin−1 −2 , with initial value p1 1, to find the value of pn+1 = 2pn = p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 7 (i) Substitute for x and dx throughout the integral M1 Obtain ∫ 2u cos u d u A1 Integrate by parts and obtain answer of the form au sin u + b cos u , where ab ≠ 0 M1 Obtain 2u sin u + 2 cos u A1 Use limits u = 0, u = p correctly and equate result to 1 M1 Obtain the given answer A1 [6] (ii) Use the iterative formula correctly at least once M1 Obtain final answer p = 1.25 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 (1.245, 1.255) A1 [3] GCE AS/A LEVEL – May/June 2012 9709 33 B C

More questions on Integration

9 The lines l and m have equations r 3i k 2j and r 4i 4j 2k bj = −2j + + λ(−i + + k) = + + + µ(ai + −k) respectively, where a and b are constants. (i) Given that l and m intersect, show that 2a 4. −b = [4] (ii) Given also that l and m are perpendicular, find the values of a and b. [4] (iii) When a and b have these values, find the position vector of the point of intersection of l and m. [2]

Mark scheme: 9 (i) Express general point of l or m in component form, i.e. (3 − λ , − 2 + 2 λ , 1 + λ ) or ( 4 + a µ , 4 + b µ , 2 − µ ) B1 Equate components and eliminate either λ or µ from a pair of equations M1 Eliminate the other parameter and obtain an equation in a and b M1 Obtain the given answer A1 [4] (ii) Using the correct process equate the scalar product of the direction vectors to zero M1* Obtain − a + 2b − 1 = 0 , or equivalent A1 Solve simultaneous equations for a or for b M1(dep*) Obtain a = 3, b = 2 A1 [4] (iii) Substitute found values in component equations and solve for λ or for µ M1 Obtain answer i + 2j + 3k from either λ = 2 or from µ = −1 A1 [2] GCE AS/A LEVEL – May/June 2012 9709 33

More questions on Vectors

10 (a) The complex numbers u and w satisfy the equations u 4i and uw 5. −w = = Solve the equations for u and w, giving all answers in the form x iy, where x and y are real. + [5] (b) (i) On a sketch of an Argand diagram, shade the region whose points represent complex numbers satisfying the inequalities arg 4π and Re where Re |ß −2 + 2i| ≤2, ß ≤−1 ß ≥1, ß denotes the real part of [5] ß. (ii) Calculate the greatest possible value of Re for points lying in the shaded region. [1] ß

Mark scheme: 10 (a) EITHER: Eliminate u or w and obtain an equation in w or in u M1 Obtain a quadratic in u or w, e.g. u 2 − i4 u − 5 = 0 or w 2 + i4 w − 5 = 0 A1 Solve a 3-term quadratic for u or for w M1 OR1: Having squared the first equation, eliminate u or w and obtain an equation in w or u M1 Obtain a 2-term quadratic in u or w, e.g. u2 = –3 + 4i A1 Solve a 2-term quadratic for u or for w M1 OR2: Using u = a + ib, w = c + id, equate real and imaginary parts and obtain 4 equations in a, b, c and d M1 Obtain 4 correct equations A1 Solve for a and b, or for c and d M1 Obtain answer u = 1 + 2i, w = 1 – 2i A1 Obtain answer u = −1 + 2i, w = −1 − 2i and no other A1 [5] (b) (i) Show point representing 2 − 2i in relatively correct position B1 Show a circle with centre 2 – 2i and radius 2 B1 1 Show line for arg z = − π B1 4 Show line for Re z = 1 B1 Shade the relevant region B1 [5] (ii) State answer 2 + 2 , or equivalent (accept 3.41) B1 [1]

More questions on Complex numbers