Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 3 · Variant 2

1 + x 1 The equation of a curve is y = for x > −1 Show that the gradient of the curve is always 2. 1 + 2x negative. [3]

Mark scheme: 1 Use correct quotient or product rule M1 Obtain correct derivative in any form A1 Justify the given statement A1 [3] 2 ( ) 2 ( ) 2

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2 Solve the equation 2 3x −1 = 3x, giving your answers correct to 3 significant figures. [4]

Mark scheme: x2 EITHER: State or imply non-modular equation 2 2 ( 3 x − 1) 2 = ( 3 ) 2 , or pair of equations x M1 2 ( 3 x − 1) = ±3 x 2 Obtain 3x = 2 and 3 = (or 3x+1 = 2) A1 3 OR: Obtain 3x = 2 by solving an equation or by inspection B1 x 2 Obtain 3 = (or 3x+1 = 2) by solving an equation or by inspection B1 3 Use correct method for solving an equation of the form 3x = a (or 3x+1 = a), where a > 0 M1 Obtain final answers 0.631 and –0.369 A1 [4] 1 1 1 ∫

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4 ln x 3 Find the exact value of dx. [5] x 1

Mark scheme: 1 3 EITHER: Integrate by parts and reach kx 2 ln x − m ∫ x 2 . x dx M1* 1 1 Obtain 2 x 2 ln x − 2 ∫ 1 d x , or equivalent A1 x 2 1 1 Integrate again and obtain 2 x 2 ln x − 4 x 2 , or equivalent A1 Substitute limits x = 1 and x = 4, having integrated twice M1(dep*) Obtain answer 4(ln 4 − )1 , or exact equivalent A1 1 1 u u OR1: Using u = ln x, or equivalent, integrate by parts and reach kue 2 − m ∫ e 2 d u M1* 1 1 u u Obtain 2u e 2 − 2 ∫ e 2 d u , or equivalent A1 1 1 u u Integrate again and obtain 2ue 2 − 4e 2 , or equivalent A1 Substitute limits u = 0 and u = ln4, having integrated twice M1(dep*) Obtain answer 4 ln 4 −,4 or exact equivalent A1 1 OR2: Using u = x , or equivalent, integrate and obtain ku ln u − m ∫ u . u d u M1* Obtain 4u ln u − 4 ∫ 1du , or equivalent A1 Integrate again and obtain 4u ln u − 4u , or equivalent A1 Substitute limits u = 1 and u = 2, having integrated twice or quoted ∫ ln u du as u ln u ± u M1(dep*) Obtain answer 8 ln 2 − 4 , or exact equivalent A1 x ln x ± x x ln x ± x OR3: Integrate by parts and reach I = + k ∫ dx M1* x x x x ln x − x 1 1 1 I − Obtain I = + 2 2 ∫ dx A1 x x Integrate and obtain I = 2 x ln x − 4 x , or equivalent A1 Substitute limits x = 1 and x = 4, having integrated twice M1(dep*) Obtain answer 4 ln 4 −,4 or exact equivalent A1 [5] GCE A LEVEL – October/November 2013 9709 32

More questions on Integration

4 The parametric equations of a curve are x = e−t cost, y = e−t sin t. dy Show that = tan t −1 . [6] dx 4

Mark scheme: 4 Use correct product or quotient rule at least once M1* d x − t − t d y − t − t Obtain = e sin t − e cos t or = e cos t − e sin t , or equivalent A1 d t d t d y d y d x Use = ÷ M1 d x d t d t d y sin t − cos t Obtain = , or equivalent A1 d x sin t + cos t d y EITHER: Express in terms of tan t only M1(dep*) d x 1  Show expression is identical to tan − t π  A1  4  1  t M1 OR: Express tan − t π  in terms of tan  4  d y Show expression is identical to A1 [6] d x

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5 (i) Prove that cot + tan 2 cosec 2 . [3] 1 3 1 (ii) Hence show that cosec 2 d = ln 3. [4] 1 2 6

Mark scheme: 5 (i) Use Pythagoras M1 Use the sin2A formula M1 Obtain the given result A1 [3] (ii) Integrate and obtain a k ln sin θ or m ln cosθ term, or obtain integral of the form p ln tan θ M1* 1 1 1 Obtain indefinite integral ln sin θ − ln cos θ , or equivalent, or ln tan θ A1 2 2 2 Substitute limits correctly M1(dep)* Obtain the given answer correctly having shown appropriate working A1 [4] 2 2 2 ( )

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6 O B C r A In the diagram, A is a point on the circumference of a circle with centre O and radius r. A circular arc with centre A meets the circumference at B and C. The angle OAB is radians. The shaded region is bounded by the circumference of the circle and the arc with centre A joining B and C. The area of the shaded region is equal to half the area of the circle. 2 sin 2 − (i) Show that cos 2 = . [5] 4 (ii) Use the iterative formula 2 sin 2 n − 1 n+1 = 2 cos−1 , 4 n with initial value 1 = 1, to determine correct to 2 decimal places, showing the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) State or imply AB = 2r cosθ or AB 2 = 2 r 2 − 2 r 2 cos (π − 2θ ) B1 Use correct formula to express the area of sector ABC in terms of r and θ M1 Use correct area formulae to express the area of a segment in terms of r and θ M1 State a correct equation in r and θ in any form A1 Obtain the given answer A1 [5] [SR: If the complete equation is approached by adding two sectors to the shaded area above BO and OC give the first M1 as on the scheme, and the second M1 for using correct area formulae for a triangle AOB or AOC, and a sector AOB or AOC.] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 0.95 A1 Show sufficient iterations to 4 d.p. to justify 0.95 to 2 d.p., or show there is a sign change in the interval (0.945, 0.955) A1 [3] GCE A LEVEL – October/November 2013 9709 32 A Bx + C

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2x2 −7x −1 7 Let f x = . x −2 x2 + 3 (i) Express f x in partial fractions. [5] (ii) Hence obtain the expansion of f x in ascending powers of x, up to and including the term in x2. [5]

Mark scheme: A Bx + C 7 (i) State or imply partial fractions are of the form + 2 B1 x − 2 x + 3 Use a relevant method to determine a constant M1 Obtain one of the values A = –1, B = 3, C = –1 A1 Obtain a second value A1 Obtain the third value A1 [5] (ii) Use correct method to obtain the first two terms of the expansions of ( x − 2 )−1 , − 1 − 1 − 1 1 x  , (x 2 + 3) −1 or + 1 1 x 2  M1  2   3  Substitute correct unsimplified expansions up to the term in x2 into each partial fraction A1 +A1 Multiply out fully by Bx + C, where BC ≠ 0 M1 1 5 17 2 Obtain final answer + x + x , or equivalent A1 [5] 6 4 72 − 1  [Symbolic binomial coefficients, e.g.   are not sufficient for the M1. The f.t. is  1  on A, B, C.] 2 −1 2 −1 [In the case of an attempt to expand (2 x − 7 x − 1)( x − 2 ) (x + 3) , give M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final answer.] [If B or C omitted from the form of partial fractions, give B0M1A0A0A0 in (i); M1A1 A1 in (ii)]

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8 Throughout this question the use of a calculator is not permitted. (a) The complex numbers u and v satisfy the equations u + 2v = 2i and iu + v = 3. Solve the equations for u and v, giving both answers in the form x + iy, where x and y are real. [5] (b) On an Argand diagram, sketch the locus representing complex numbers satisfying + i = 1 and the locus representing complex numbers w satisfying arg w −2 = 3 . Find the least value 4 of −w for points on these loci. [5]

Mark scheme: 8 (a) EITHER: Solve for u or for v M1 i2 − 6 5 Obtain u = or v = , or equivalent A1 l − i2 l − i2 Either: Multiply a numerator and denominator by conjugate of denominator, or equivalent Or: Set u or v equal to x + iy, obtain two equations by equating real and imaginary parts and solve for x or for y M1 OR: Using a + ib and c +id for u and v, equate real and imaginary parts and obtain four equations in a, b, c and d M1 Obtain b + 2d = 2, a + 2c = 0, a + d = 0 and –b + c = 3, or equivalent A1 Solve for one unknown M1 Obtain final answer u = –2 –2i, or equivalent A1 Obtain final answer v = l + 2i, or equivalent A1 [5] (b) Show a circle with centre –i B1 Show a circle with radius l B1 3 Show correct half line from 2 at an angle of π to the real axis B1 4 Use a correct method for finding the least value of the modulus M1 3 Obtain final answer −,1 or equivalent, e.g. 1.12 (allow 1.1) A1 [5] 2 GCE A LEVEL – October/November 2013 9709 32

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9 C D B A O The diagram shows three points A, B and C whose position vectors with respect to the origin O are 2 0 3 −−→ −−→ −−→ given by OA = −1 , OB = 3 and OC = 0 . The point D lies on BC, between B and C, and is 2 1 4 such that CD = 2DB. (i) Find the equation of the plane ABC, giving your answer in the form ax + by + c = d. [6] (ii) Find the position vector of D. [1] (iii) Show that the length of the perpendicular from A to OD is 1 65 . [4] 3 [Question 10 is printed on the next page.]

Mark scheme: 9 (i) EITHER: Obtain a vector parallel to the plane, e.g. AB = − 2 i + 4 j − k B1 Use scalar product to obtain an equation in a, b, c, e.g. − 2a + 4b − c = 0 , 3a − 3b + 3c = 0 , or a + b + 2c = 0 M1 Obtain two correct equations in a, b, c A1 Solve to obtain ratio a : b : c M1 Obtain a : b : c = 3 : 1 : −2 , or equivalent A1 Obtain equation 3x + y – 2z = 1, or equivalent A1 OR1: Substitute for two points, e.g. A and B, and obtain 2a − b + 2c = d and 3b + c = d B1 Substitute for another point, e.g. C, to obtain a third equation and eliminate one unknown entirely from the three equations M1 Obtain two correct equations in three unknowns, e.g. in a, b, c A1 Solve to obtain their ratio, e.g. a : b : c M1 Obtain a : b : c = 3 : 1 : −2 , a : c : d = 3 : −2 : 1 , a : b : d = 3 : 1 : 1 or b : c : d = −1 : −2 : 1 A1 Obtain equation 3 x + y − 2 z = 1 , or equivalent A1 OR2: Obtain a vector parallel to the plane, e.g. BC = 3i − 3 j + 3k B1 Obtain a second such vector and calculate their vector product e.g. (− 2i + 4 j − k ) × (3i − 3 j + 3k ) M1 Obtain two correct components of the product A1 Obtain correct answer, e.g. 9i + 3j – 6k A1 Substitute in 9 x + 3 y − 6 z = d to find d M1 Obtain equation 9 x + 3 y − 6 z = 3 , or equivalent A1 OR3: Obtain a vector parallel to the plane, e.g. AC = i + j + 2k B1 Obtain a second such vector and form correctly a 2-parameter equation for the plane M1 Obtain a correct equation, e.g. r = 3i + 4k + λ (− 2i + 4 j − k ) + µ (i + j + 2k ) A1 State three correct equations in x , y , z , λ, µ A1 Eliminate λ and µ M1 Obtain equation 3 x + y − 2 z = 1 , or equivalent A1 [6] (ii) Obtain answer i + 2j + 2k, or equivalent B1 [1] GCE A LEVEL – October/November 2013 9709 32 OA.OD (iii) EITHER: Use to find projection ON of OA onto OD M1 OD 4 Obtain ON = A1 3 Use Pythagoras in triangle OAN to find AN M1 Obtain the given answer A1 OR1: Calculate the vector product of OA and OD M1 Obtain answer 6i + 2j – 5k A1 Divide the modulus of the vector product by the modulus of OD M1 Obtain the given answer A1 OR2: Taking general point P of OD to have position vector λ (i + 2 j + 2k ) , form an equation in λ by either equating the scalar product of AP and OP to zero, or using Pythagoras in triangle OPA, or setting the derivative of AP to zero M1 4 Solve and obtain λ = A1 9 4 Carry out method to calculate AP when λ = M1 9 Obtain the given answer A1 OR3: Use a relevant scalar product to find the cosine of AOD or ADO M1 4 5 Obtain cos AOD = or cos ADO = , or equivalent A1 9 3 10 Use trig to find the length of the perpendicular M1 Obtain the given answer A1 OR4: Use cosine formula in triangle AOD to find cos AOD or cos ADO M1 8 10 Obtain cos AOD = or cos ADO = , or equivalent A1 18 6 10 Use trig to find the length of the perpendicular M1 Obtain the given answer A1 [4] 3

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10 h 60° C A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi- vertical angle is 60 , as shown in the diagram. At time t = 0, the tank is full and the depth of water is H. At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to h, where h is the depth of water at time t. The tank becomes empty when t = 60. (i) Show that h and t satisfy a differential equation of the form dh −3 = −Ah 2, dt where A is a positive constant. [4] (ii) Solve the differential equation given in part (i) and obtain an expression for t in terms of h and H. [6] [1] (iii) Find the time at which the depth reaches 12H. [The volume V of a cone of vertical height h and base radius r is given by V = 1 r2h.] 3

Mark scheme: 10 (i) State or imply V = πh 3 B1 d V State or imply = − k h B1 d t d V d V d h Use = . , or equivalent M1 d t d h d t Obtain the given equation A1 [4] d V [The M1 is only available if is in terms of h and has been obtained by a d h correct method.] d V [Allow B1 for = k h but withhold the final A1 until the polarity of the constant d t k has been justified.] 3π GCE A LEVEL – October/November 2013 9709 32 (ii) Separate variables and integrate at least one side M1 5 2 2 Obtain terms h and –At, or equivalent A1 5 5 Use t = ,0 h = H in a solution containing terms of the form ah 2 and bt + c M1 5 Use t = 60, h = 0 in a solution containing terms of the form ah 2 and bt + c M1 5 5 5 2 2 1 2 2 2 Obtain a correct solution in any form, e.g. h = H t + H A1 5 150 5  5    h  2  (ii) Obtain final answer t = 60 1 −   , or equivalent A1 [6]   H     1 (iii) Substitute h = H and obtain answer t = 49.4 B1 [1] 2

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