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Cambridge AS & A Level Mathematics (9709), Paper 3 Pure Mathematics 3, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w22_qp_339709_w22_ms_339709/33/O/N/22
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
Solve the equation ln 2x − 1 = 2 ln x + 1 − ln x. Give your answer correct to 3 decimal places. [4] _ 1 + 2x
Mark scheme
1 Use law for the logarithm of a product, quotient or power M1 Remove logarithms and state a correct equation, e.g. x ( 2 x − 1) = ( x + 1) A1 2 Solve a 3-term quadratic obtaining at least one root M1 Obtain answer 3.303 only A1 4
Question
Expand in ascending powers of x, up to and including the term in x2 , simplifying the 1 − 2x coefficients. [5] 1π Find the exact value of Ó 4
Mark scheme
2 1 B1 State a correct unsimplified term in x or x 2 of the expansion of either (1 + 2 x ) 2 1 or (1 − 2 x ) 2 − 1 B1 State correct unsimplified expansion of (1 + 2 x ) 2 up to the term in x 2 1 B1 State correct unsimplified expansion of (1 − 2 x ) 2 up to the term in x 2 − Obtain sufficient terms of the product of the expansions M1 Obtain final answer 1 + 2 x + 2 x 2 A1 Alternative method for question 2 1 B1 ( ) − State that the expression equals (1 + 2 x ) 1 − 4 x 2 2 and state a term of the expansion 1 B1 + B1 ( ) − State correct unsimplified expansion of 1 − 4 x 2 2 up to the term in x 2 Obtain sufficient terms of the product of (1 + 2x) and the expansion M1 Obtain final answer 1 + 2 x + 2 x 2 A1 5
Question
x sec2 x dx. [5] 0
Mark scheme
3 Commence integration by parts and reach x tan x ± tan x.1dx *M1 Use a correct method to integrate tan x M1 Obtain integral x tan x − lnsec x , or equivalent A1 Use limits correctly, having integrated twice DM1 1 1 A1 Obtain answer π − ln 2 , or exact equivalent 4 2 5
Question
The parametric equations of a curve are x = 2t − tan t, y = ln sin 2t, for 0 < t < 12 π. dy Show that = cot t. [5] dx
Mark scheme
4 dx dy cos 2t B1 State or imply = 2 − sec 2t or =2 dt dt sin 2t dy dy dx M1 Use = ÷ dx dt dt Obtain correct answer in any form A1 Use double angle formula to express derivative in terms of cos x and sin x M1 Obtain the given answer correctly A1 AG 5
Question
(a) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z + 2 ≤ 2 and Im z ≥ 1. [4] (b) Find the greatest value of arg z for points in the shaded region. [2]
Mark scheme
5(a) Show a circle with centre – 2 B1 Show a circle with radius 2 and centre not the origin B1 Show the line y = 1 B1 Shade the correct region B1 4 5(b) Identify the correct point and carry out a correct method to find the argument M1 11 A1 2.88 radians or 165°. Obtain answer π 12 2
Question
Solve the quadratic equation 1 − 3iz2 − 2 + iz + i = 0, giving your answers in the form x + iy, where x and y are real. [6]
Mark scheme
6 Use quadratic formula to solve for z M1 SC M1: For substitution of x + iy and multiplying out. Use i 2 = −1 throughout M1 SC M1: Use i 2 = −1 throughout. Obtain correct answer in any form A1 SC A1: For two correct equations x 2 − y 2 + 6 xy − 2 x + y = 0 and ( ) −3 x2 − y 2 + 2 xy − x − 2 y + 1 = 0 . Multiply numerator and denominator by (1 + 3i), or equivalent M1 1 1 A1 Obtain final answer, e.g. − + i 2 2 2 1 A1 Obtain second final answer, e.g. + i 5 5 6
Question
(a) Show that the equation 5 sec x + tan x = 4 can be expressed as R cos x + ! = 5, where R > 0 and 0Å < ! < 90Å. Give the exact value of R and the value of ! correct to 2 decimal places. [4] (b) Hence solve the equation 5 sec 2x + tan 2x = 4, for 0Å < x < 180Å. [4] x3
Mark scheme
7(a) Rearrange and obtain 4cos x − sin x = 5 B1 State R = 17 B1 Use trig formulae to find α M1 Obtain α =14 .04° A1 4 7(b) 5 B1 FT FT their R. Evaluate cos −1 17 Carry out a correct method to find a value of x in the given interval M1 Obtain answer, e.g. 21.6 ° A1 Obtain a second answer, e.g. 144.4 ° and no other in the interval A1 Treat answers in radians as a misread. Ignore answers outside the given interval. 4
Question
The curve with equation y = has a stationary point at x = p, where p > 0. ex − 1 (a) Show that p = 3 1 − e−p . [3] (b) Verify by calculation that p lies between 2.5 and 3. [2] (c) Use an iterative formula based on the equation in part (a) to determine p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
Mark scheme
8(a) Use quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative at x = p to zero and obtain the given equation A1 3 8(b) Evaluate a relevant expression or pair of relevant pair of expressions at p = 2.5 M1 and p = 3 Complete the argument with correct calculated values A1 2 8(c) ( Use the iterative formula pn+1 = 3 1 − e− pn ) correctly at least once M1 Obtain final answer p = 2.82 A1 Show sufficient iterations to 4 d.p.to justify 2.82 to 2 d.p., or show there is a sign A1 change in the interval (2.815, 2.825) 3
Question
With respect to the origin O, the position vectors of the points A, B and C are given by ` a ` a ` a OA = 5 , OB = 0 and OC = −3 . 2 1 −2 The midpoint of AC is M and the point N lies on BC, between B and C, and is such that BN = 2NC. (a) Find the position vectors of M and N . [3] (b) Find a vector equation for the line through M and N . [2] (c) Find the position vector of the point Q where the line through M and N intersects the line through A and B. [4]
Mark scheme
9(a) 2 B1 State OM = 1 0 Use a correct method to find ON M1 3 A1 Obtain answer −2 −1 3 9(b) Carry out a correct method to form a vector equation for MN M1 2 1 A1 OE Obtain a correct equation in any form, e.g. r = 1 + λ −3 0 −1 2 9(c) 1 −1 B1 State a correct vector equation for AB in any form, e.g. r = 0 + µ 5 1 1 Equate components of AB and MN and solve for λ or for µ M1 Obtain λ = – 3 or µ = 2 A1 −1 A1 Obtain position vector 10 , or equivalent, for Q 3 4
Question
A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time t minutes after filling begins the volume of water in the pool is V litres. The pool has a small leak and loses water at a rate of 0.01V litres per minute. dV The differential equation satisfied by V and t is of the form = a − bV . dt (a) Write down the values of the constants a and b. [1] (b) Solve the differential equation and find the value of t when V = 1000. [6] (c) Obtain an expression for V in terms of t and hence state what happens to V as t becomes large. [2] 5 − x + 6x2
Mark scheme
10(a) a = 30 and b = 0.01 B1 1 10(b) Separate variables and integrate one side M1 Obtain terms −100ln (30 − 0.01V ) and t, or equivalent A1 FT FT their a and b. + A1 FT Evaluate a constant, or use t = 0, V = 0 as limits, in a solution containing terms c M1 ln (30 − 0.01V ) and dt where cd ≠ 0 Obtain solution 100ln30 −100ln ( 30 − 0.01V ) = t , or equivalent A1 Substitute V = 1000 and obtain answer t = 40.5 A1 6 10(c) ( Obtain V = 3000 1 − e−0.01t ) B1 OE State that V approaches 3000 B1 2
Question
Let f x = . 3 − x 1 + 3x2 (a) Express f x in partial fractions. [5] 1 (b) Find the exact value of Ó f x dx, simplifying your answer. [5] 0 Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
11(a) A Bx + C B1 State or imply the form + 3− x 1 + 3x 2 Use a correct method to find a constant M1 Obtain one of A = 2, B = 0 and C = 1 A1 Obtain a second value A1 Obtain the third value A1 5 11(b) Integrate and obtain term −2ln ( 3 − x ) B1 FT Obtain term of the form b tan −1 ( 3x ) M1 1 A1 FT Obtain term tan −1 ( 3 x ) 3 Substitute limits correctly in an integral with terms a ln ( 3− x ) and M1 b tan −1 ( 3x ) , where ab ≠ 0 1 A1 Obtain answer 2 ln 3 + π , or equivalent 2 3 3 5
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Mathematics 9709/33 for October/November 2022 is out of 75 marks, with 1 hour 50 minutes to answer it.
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