Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 1 · Variant 2

1 Points A and B have coordinates 5, 2 and 10, −1 respectively. (a) Find the equation of the perpendicular bisector of AB. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the circle with centre A which passes through B. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a)  10 + 5 2 − 1    15 1   B1 Accept unsimplified. Mid-point AB is , = ,        2 2    2 2   −−1 2 −3 5 M1 Change in y Gradient of AB = = Gradient perpendicular = For use of , condone inconsistent order of x and y, 10 − 5 5 3 Change in x and m1m2 = –1. 1 A1 OE ISW y − 2 5  1 5  15   5 =  y − =  x −   Any correct version e.g. y = x − 12 or 5 x − 3 y = 36 . 15 3  2 3  2   3 x − 2 3 1(b) [Radius =] 34 or 5.8 AWRT or [(radius)2 =] 34 B1 Sight of 34 or 34. Condone confusion of r and r 2. ( x − 5 ) 2 + ( y − 2 ) 2 B1 Sight of ( x − 5 ) 2 + ( y − 2 ) 2 ( x − 5 ) 2 + ( y − 2 ) 2 = 34 B1 CAO ISW Alternative method for Question 1(b) x 2 + y 2 − 10 x − 4 y B1 2 + y 2 − 10 x − 4 y + c = 0.  c =  5 or  c =  − 5 B1 Substitution of (10, –1) into x x 2 + y 2 − 10 x − 4 y − 5 = 0 B1 3

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2 The first, second and third terms of an arithmetic progression are a, 2a and a2 respectively, where a is a positive constant. Find the sum of the first 50 terms of the progression. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 2 a − a = a 2 − 2a B1 OE An unsimplified correct equation in a or d only, e.g. a 2 + a = 4 a . Can be implied by correct values for a or d. a = 3 or d = 3 B1 Condone ‘extra’ solution of a = 0 or d = 0. a = 3 and d = 3 B1 SOI 50 M1 May be done using 50th term (=150). Their a and d must be S50 = ( 2  their a + 49  their d ) numerical. 2 3825 A1 ISW SC B2 for 1275 a or 1275d 5

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3 (a) Find the set of values of k for which the equation 8x2 + kx + 2 = 0 has no real roots. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Solve the equation 8 cos21 −10 cos 1 + 2 = 0 for 0Å ≤1 ≤180Å. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) k 2 − 4 8 2 [  0] M1 Use of b2 – 4 a c but not just in the quadratic formula. −8 < k < 8 or −8 < k , k < 8 or k  8 or (–8, 8) A1 Condone ‘ − 8 < k or k < 8’, ‘ − 8 < k and k < 8’ but not 64 . 2 3(b) 2 ( 4cos− 1)( cos− 1) or ( 4cos− 1)( cos− 1) M1 OE Or use of formula or completing the square. Allow use of replacement variable. 2 A1 OE For both answers. cos= , cos= 1 2 8 SC: If M0, SC B1 available for sight of cos= and 1 8 [θ =] 0°, 75.5° A1 AWRT ISW rejection of 0°. For both answers and no others in the range 0  180, must be in degrees. SC: If M0 B1 scored, SC B1 available for correct answers. 2 SC: If M1 A0 scored, SC B1 available for cos= and θ = 8 75.5° only, WWW. 3

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4 A geometric progression is such that the third term is 1764 and the sum of the second and third terms is 3444. Find the 50th term. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 a r2 = 1764 and a r + a r2 = 3444 or a r = 1680 or B1 Two correct algebraic statements. a 1 − r 3 ( ) − a = 3444 1 − r Attempt to solve as far as r = or a = M1 Any valid method, e.g. 1764  1680 or from 20 r 2 − 41r + 21 OE (condone solving using a calculator). 1764 21 A1 1764 r = = or 1.05 [ a = 1600] Note: r = www implies B1 and M1. 1680 20 3444 − 1764 17 500 A1 AWRT e.g. 17 474.1….. 4

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6 The equation of a curve is y = 4x2 + 20x + 6. (a) Express the equation in the form y = a x + b 2 + c, where a, b and c are constants. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 4x2 + 20x + 6 = 45. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Sketch the graph of y = 4x2 + 20x + 6 showing the coordinates of the stationary point. You are not required to indicate where the curve crosses the x- and y-axes. [3]

Mark scheme: 6(a) 2 There is no requirement for the candidate to list a , b and c.  5  y = 4  x +  − 19 Look at values in their final expression, condone omission of 2 ,  2  and award marks as follows: B1 a = 4 B1 5 b = OE 2 B1 c = −19 3 6(b)   5  2    5  2  *M1 Equate their quadratic completed square form from 6(a) to 45 or  Their 4  x +  − 19  = 45    x +  = 16  re-start and use completing the square.    2      2   Solve as far as x = DM1 Any valid method leading to two answers. 13 A1 SC: If M0 or M1 DM0 awarded, B1 available for correct final  x = 3 , − answers. 2 2 3 6(c) Quadratic curve that is the right way up (must be seen either side B1 No axes required, ignore any axes even if incorrect. of stationary point) Stationary point stated using any valid method or correctly B1 FT FT their values from 6(a) as long as their expression is of the labelled on their diagram. 2  5  Expect  − , −19  . B1 FT form p ( qx + r ) + s .  2  Condone if stated correctly but plotted incorrectly. 3

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sin 1 cos 1 tan21 + 1 7 (a) Prove the identity +  [3] sin 1 + cos 1 sin 1 −cos 1 tan21 −1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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........................................................................................................................................................ ........................................................................................................................................................ sin 1 cos 1 (b) Hence find the exact solutions of the equation + = 2 for 0 ≤1 ≤π. sin 1 + cos 1 sin 1 −cos 1 [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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Mark scheme: 2 + cos 2  *M1 Sight of a correct common denominator, either in one or two7(a) sin( sin− cos) + cos( sin+ cos)  sin  = 2 2  fractions, condone missing brackets if recovered. In the numerator ( sin+ cos)( sin− cos)  sin − cos  condone sign errors only. sin 2  cos 2  DM1 Divide throughout by cos 2 . + cos 2  cos 2  sin 2  cos 2  − cos 2  cos 2  tan 2 + 1 A1 AG tan 2 − 1 Alternative method for Question 7(a) sin 2  *M1 sin 2  + 1 and multiply top and bottom by Replace tan2 with cos 2   sin 2 + cos 2  cos 2  cos 2   = or the equivalent step   sin 2  cos 2  sin 2 − cos 2    cos 2 . Condone sign errors. − 1 cos 2  Sight of convincing use of partial fractions DM1 sinθ cosθ A1 + AG sinθ + cosθ sinθ − cosθ 3 Note: M1 DM1 A1 for working on both sides at the same time and finishing at the same correct expression. M1 DM1 for starting separately and finishing at the same correct expression and A1 if there is a final conclusion e.g. QED. Do not allow cross multiplication. Condone use of s, c and t and omission of . 7(b) tan 2 + 1 2 2 *M1 Equate expression from (a) to 2 and clear fraction. = 2  tan + 1 = 2 tan − 1 2 ( ) tan − 1 tan=  3 DM1 Simplify as far as tan= . May be implied by a correct final answer in degrees or radians. Alternative method for first two marks of Question 7(b) sin 2 + cos 2  2 2 *M1 Equate expression to 2, clear fraction and use trig identities to = 2  1 = 2 sin − 2 1 − sin  2 2 ( ) form an equation in sin or cos only. sin − cos  3 1 DM1 Simplify as far as sin= , or cos= . sin=  or cos=  4 4 1 2 A1 A1 for either correct answer then A1FT For their second value = π , π being − (their first) and no others in range 0   , both 3 3 A1 FT values must be exact and in radians. SC: B1 for = 60,120 or 0.333, 0.667 AWRT. or 1.05, 2.09 AWRT. 4

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dy 18 The equation of a curve is such that = 3x 2 −3x−12. The curve passes through the point 3, 5 . dx (a) Find the equation of the curve. 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(b) Find the x-coordinate of the stationary point. 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(c) State the set of values of x for which y increases as x increases. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a)  3   1  B1 B1 Marks can be awarded for correct unsimplified expressions, 1  3 x 2   3 x 2   32 12  mark each for contents of { } ISW.  y =    +  −   + c   = 2 x − 6 x  3 1        2   2  3 1 M1 Correct use of (3,5) in an integrated expression (defined by at least 5 = 2  3 2 −6 3 2 + c one correct power) including + c. 3 1 A1 Condone c = 5 as their final line if either y = or f(x) = seen y = 2 x 2 − 6 x 2 + 5 elsewhere in the solution, but coefficients must not contain unresolved double fractions. 4 8(b) 1 − 1 M1 Setting given differential to 0. 3 x 2 − 3 x 2 = 0 [x=] 1 A1 CAO WWW Condone extra solution of —1 only if it is rejected. 2 8(c) x>1 or x> “their 8(b)” B1FT Allow ⩾ 1

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10 R O 23π 2.5 m P 56π 2.24 m A B S The diagram shows a cross-section RASB of the body of an aircraft. The cross-section consists of a sector OARB of a circle of radius 2.5m, with centre O, a sector PASB of another circle of radius 2.24m with centre P and a quadrilateral OAPB. Angle AOB = 23π and angle APB = 56π. (a) Find the perimeter of the cross-section RASB, giving your answer correct to 2 decimal places. 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(b) Find the difference in area of the two triangles AOB and APB, giving your answer correct to 2 decimal places. 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(c) Find the area of the cross-section RASB, giving your answer correct to 1 decimal place. 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Mark scheme: 10(a) 4 5 10 28 B1 For either arc correct. Arc ARB could be AR+RB. 2.5  + 2.24  [= 10.47[2] + 5.86[4] or + ] 3 6 3 15 M1 For adding two (or three) arc lengths using different radii and angles and nothing else. SOI 26 π A1 AWRT 16.34 or Condone 16.33 only. 5 3 10(b) 1 2 2 M1 For either AOB or APB (AB = 4.33, h= 1.25, 0.58) or any other Area AOB =  2.5 sin [=2.706] valid method. 2 3 1 2 5 Area APB =  2.24 sin [=1.254] 2 6 [Difference =] 1.45 A1 AWRT Condone 1.46 only. 2 10(c) 1 2 4 B1 For either sector area correct Area AOB =  2.5  [=13.09] 2 3 1 2 5 Area APB =  2.24  [=6.57] 2 6 [Area of cross section =] M1 Adding two sector areas from different sectors and ‘ their 10(b) ’ 1 2 4 1 2 5 and nothing else. SOI  2.5  +  2.24  + “ their 10 ( b )” 2 3 2 6 =  13.09 + 6.57 + “their 10 ( b )”  21.1 A1 CAO Condone slight inaccuracies in intermediate working if the correct answer is arrived at. 3

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