Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 3 · Variant 1

x2 1 Find the exact coordinates of the points on the curve y = at which the gradient of the tangent 1 −3x is equal to 8. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Use correct quotient or product rule *M1 Obtain correct derivative in any form A1 2  2  (1 − 3 x )2 x − x ( −3) 2 x − 3 x e.g. 2  = 2  or (1 − 3 x )   ( 1 − 3 x )  2 −2 −1 3x (1 − 3x ) + 2 x (1 − 3x ) . Equate derivative to 8 and solve for x DM1 75 x 2 − 50 x + 8 = (15 x − 4 )( 5 x − 2 ) . 2 4 A1 Exact values required. Obtain answers x = and 5 15 4 16 A1 Allow A1 for one correct point. Obtain answers y = − and 5 45 5

More questions on Differentiation

3 + 2i 4 The complex number u is defined by u = a −5i, where a is real. (a) Express u in the Cartesian form x + iy, where x and y are in terms of a. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Given that arg u = 14π, find the value of a. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Multiply numerator and denominator by a + 5i M1 OE Use i2 = –1 M1 At least once. 3a − 10 2 a + 15 A1 Obtain answer + i a 2 + 25 a 2 + 25 Alternative Method for Question 4(a) Multiply x + iy by a – 5i and use i2 = –1 M1 Compare real and imaginary parts M1 3 = ax + 5y, 2 = ay – 5x. 3a − 10 2 a + 15 A1 Obtain answer + i a 2 + 25 a 2 + 25 3 4(b) State or imply Im(a) ÷ Re(a) = 1 M1 Or Im(a) = Re(a) or equivalent for their u. Obtain answer a = 25 A1 2

More questions on Complex numbers

5 (a) Given that sin x + 16π −sin x −16π = cos x + 13π −cos x −13π , find the exact value of tan x. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the exact roots of the equation sin x + 16π −sin x −16π = cos x + 13π −cos x −13π for 0 ≤x ≤2π. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Use correct trig formulae and obtain an equation in sin x and cos x *M1 Allow one sign error. Obtain a correct equation in any form A1 e.g 2cos x sin π6 = −2sin x sin π3 . Substitute exact trig ratios and obtain an expression for tan x DM1 Allow one sign error. 1 A1 Or exact equivalent. Obtain answer tan x = − 3 4 5(b) 5π B1 Obtain answer, e.g. x = 6 11π B1FT FT first answer +  (provided 0 ⩽ first answer ⩽ ). Obtain second answer, e.g. x = and no others in the interval 6 Or FT first answer –  (provided  ⩽ first answer ⩽ 2). Ignore any answers outside interval. 2

More questions on Trigonometry

6 The parametric equations of a curve are x = t + 3, y = ln t, for t > 0. dy (a) Obtain a simplified expression for in terms of t. [3] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the exact coordinates of the point on the curve at which the gradient of the normal is −2. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) State correct derivative of x or y with respect to t B1 dx 1 − 12 dy 1 = t , = . dt 2 dt t dy dy dt M1 Use correct chain rule. Use =  dx dt dx − dy 2 A1 2 t 2 Obtain answer = . Or simplified equivalent e.g 2t 1 or dx t t 3 6(b) dy 1 M1 State or imply their = dx 2 Obtain t = 4 A1 Or equivalent. Obtain answer (7, ln 16) A1 Or exact equivalent. Can state the two components separately. 3

More questions on Differentiation

7 The variables x and 1 satisfy the differential equation x dx = x2 + 3. tan 1 d1 It is given that x = 1 when 1 = 0. Solve the differential equation, obtaining an expression for x2 in terms of 1. [7] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 7 Separate variables correctly B1 x  tan=d  x dx . 2 + 3 Condone missing integral signs or missing dx, d. Can be implied by later work. Obtain term –ln(cos ) B1 Or equivalent e.g. ln(sec ). 2 M1 Obtain term of the form a ln x + 3 ( ) 1 2 A1 Obtain term ln x + 3 ( ) 2 Use x = 1, 𝜃 = 0 to evaluate a constant or as limits in a solution containing M1 If they have rearranged then the constant must be of the 2 correct form. terms of the form a ln x + 3 and b ln(cos) ( ) Obtain correct answer in any form A1 1 2 x + 3 = − lncos+ ln2 . 2 ln ( ) 2 4 A1 Or equivalent e.g. x 2 = 4sec 2 x − 3 . Obtain final answer x = − 3 cos 2 lns removed. 7

More questions on Differential equations

8 (a) By sketching a suitable pair of graphs, show that the equation x = ex −3 has only one root. [2] (b) Show by calculation that this root lies between 1 and 2. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Show that, if a sequence of values given by the iterative formula xn+1 = ln 3 + / xn ! converges, then it converges to the root of the equation in (a). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Use the iterative formula to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Sketch a relevant graph, e.g. y = ex – 3 B1 y = ex – 3 Should cut vertical axis at (0, –2) and have increasing gradient. 2 y= x Sketch a second relevant graph, e.g. y = x and justify the given statement B1 y = x should start at (0, 0) and have reducing grading y=ex - 3 2 Ignore anything outside 1st and 4th quadrants. For second B1 need to mark intersection with a dot, a cross, or say root at point of intersection, or equivalent. 2 8(b) Calculate the values of a relevant expression or pair of expressions at x = 1 M1 and x = 2 Complete the argument correctly with correct calculated values A1 e.g.1 −0.28..,1.41  4.39.. 1.28 > 0, –2.98 < 0. 2 8(c) State x = ln 3 + x and rearrange to the given equation x = e x − 3 B1 Or rearrange x = e x −to3 x = ln(3 + x ) and state ( ) iterative formula of xn +1 = ln(3 + xn ) . AG 1 8(d) Use the iterative process correctly at least once M1 Obtain final answer 1.43 A1 Show sufficient iterations to at least 4 d.p. to justify 1.43 to 2 d.p. or show A1 e.g. 1, 1.3864, 1.4297, 1.4341, … there is a sign change in the interval (1.425, 1.435) 1.5, 1.4210, 1.4332, 14.344, 1.4345, … Condone recovery and small differences in the final figure in the iteration 2, 1.4848, 1.4395, 1.4350, 1.4346, 1.4345, … 3

More questions on Logarithmic and exponential functions

9 y M x O 3 The diagram shows the curve y = xe−14x2 , for x ≥0, and its maximum point M. (a) Find the exact coordinates of M. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Using the substitution x = u, or otherwise, find by integration the exact area of the shaded region bounded by the curve, the x-axis and the line x = 3. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 9(a) Use the correct product rule *M1 Condone error in chain rule. 2 x Obtain correct derivative in any form A1 − d y x 2 − x42 4 e.g. = − e + e . d x 2 Equate derivative to zero and solve for x DM1  − 1  A1 Or exact equivalent. Obtain answer  2, 2e 2  Can state the components separately.     4 9(b) 1 − 12 B1 Or equivalent e.g. du = 2 xdx . State or imply dx = u du 1 2 2 Alternative substitution: u = − x . 4 Substitute for x and dx M1 u A1 OE 1 − 14 Obtain correct integral  e d u 2 1 1 − u − x 2 M1 u = 9 and u = 0 or x = 3 and x = 0. Use correct limits in an integral of the form a e 4 or ae 4 9 A1 Or exact equivalent. − Obtain answer 2 − 2e 4 Alternative Method for Question 9(b) 1 1 − x 2 − x 2 M1 Recognition used. xe 4 dx = ae 4  a negative A1 a = − 2 A1 1 − x 2 M1 x = 3 and x = 0. Use correct limits in an integral of the form ae 4 9 A1 Or exact equivalent. − Obtain answer 2 − 2e 4 5

More questions on Integration

11 G F M C B D k E j O A i In the diagram, OABCDEFG is a cuboid in which OA = 3 units, OC = 2 units and OD = 2 units. Unit vectors i, j and k are parallel to OA, OD and OC respectively. M is the midpoint of EF. (a) Find the position vector of M. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The position vector of P is i + j + 2k. (b) Calculate angle PAM. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the exact length of the perpendicular from P to the line passing through O and M. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a) Obtain 3i + 2j + k B1 Accept coordinates in place of position vector. 1 11(b) AM or AP correct soi B1 AM = 2j + k, or AP = –2i + j + 2k. Carry out correct process for evaluating the scalar product of AM and AP M1 or MA and PA: 0 + 2 + 2 . Using the correct process for the moduli, divide the scalar product by the M1 − 1  4  For their vectors.  = cos   . product of the moduli and obtain the inverse cosine of the result  3 5  Obtain answer 53.4° or 0.932c A1 4 11(c) Find PQ (or QP ) for a general point Q on the line passing through O and M, B1 FT e.g. PQ = - (i + j + 2k) + μ(3i + 2j + k). Follow their M. Calculate the scalar product of PQ and a direction vector for the line passing *M1 through O and M and equate to zero 1 A1 Solve and obtain correct solution e.g. = − 2 Carry out method to calculate PQ DM1 2 2 .5 + 0 + 1.5 . 10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 1 for Question 11(c) Find PQ (or QP ) for a general point Q on the line passing through O and M, B1 FT e.g. PQ = - (i + j + 2k) + μ(3i + 2j + k). Follow their M. Use a correct method to express PQ2 (or PQ) in terms of µ *M1 Obtain a correct equation in any form A1 e.g. PQ2 = (1 + 3μ)2 + (1 + 2μ)2 + (2 + μ)2 11(c) Carry out a complete method for finding its minimum DM1 1 e.g. 6(1 + 3) + 4(1 + 2) + 2( 2 + ) = 0, = − . 2 10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 2 for Question 11(c) Calling (0, 0, 0) A, state PA (or AP ) in component form, e.g. i + j + 2k B1 Use a scalar product to find the projection of PA (or AP ) on the line passing M1 through O and M 7 A1 OE Obtain correct answer 14 Use Pythagoras to find the perpendicular M1 2 2 2 2  7  d = AP − AQ = 1 + 1 + 2 −  .  14  10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 3 for Question 11(c) Calling (0, 0, 0) A, state PA (or AP ) in component form, e.g. i + j + 2k B1 Calculate the vector product of PA and a direction vector for the line passing M1 through O and M Obtain correct answer, e.g. 3i – 5j + k A1 11(c) Divide modulus of the product by that of the direction vector M1 2 2 2 3 + 5 + 1 e.g. . 32 + 2 2 + 12 10 A1 Or exact equivalent. Obtain answer 2 5

More questions on Vectors