Cambridge A Level Mathematics 9709 — 2023 May/June Paper 1 · Variant 2

3 (a) Express 4x2 −24x + p in the form a x + b 2 + c, where a and b are integers and c is to be given in terms of the constant p. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence or otherwise find the set of values of p for which the equation 4x2 −24x + p = 0 has no real roots. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) 2 4 3 seen or 4 and 3 x a b    B1 OE Award marks for the correct expression or their values a, b and c. Condone 4 ( 3) 36    x p = 0 and 4 9 4        p . –36 + p or p – 36 seen or c = p – 36 B1 2 3(b) 36 0   p leading to 36  p or 2 24 4 4 0 36    p p or 36 p B1 Allow (36, )  or 36 p  . Consider final answer only. 1

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4 Solve the equation 8x6 + 215x3 −27 = 0. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 6 3 [8 215 27 0] x x    3 3 8 1 27 0 x x    OR 2 215 215 4.8. 27 215 47089 or 2.8 2.8       If a substitution is used then the correct coefficients must be retained. Condone substitution of 3  x x . 1 , 27 8  A1 Both correct values seen. SC: if M0 scored SC B1 is available for sight of 1 8 and 27  OE 1 or 0.5 , 3 2  A1 SC: if M0SCB1 scored then SCB1 is available for the correct answers and no others. Do not ISW if answers given as a range. 3

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6 A x 1 rad O B x P The diagram shows a sector OAB of a circle with centre O. Angle AOB = 1 radians and OP = AP = x. (a) Show that the arc length AB is 2x1 cos 1. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the area of the shaded region APB in terms of x and 1. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) 1 cos 2 OA x   or   sin π 2 sin OA x     or   2 2 2 2 2 cos π 2 OA x x x      or 2 2 2 2 cos x r x rx     or other valid method. 2 OA, OA or 2 OA (allow p, a or r for OA) containing only terms with x and  but not just 2 cos   OA x . Do not condone sin π 2  until missing brackets recovered or cos(180 2  ) until it becomes 2 cos etc. OA = 2 cos x leading to Arc length = 2 cos   x DB1 AG Complete correct method showing all necessary working. Condone 2 cos   x . 2 If B0 but www then SCB1 for OA = 2 cos x leading to Arc length = 2 cos   x . 6(b) Sector area =   2 1 2 cos 2   x θ M1 OE Using sector formula with a correct OA. Condone 2  cos for 2 cos and missing brackets. Triangle area = 1 2 cos sin 2   x x OR   2 1 sin π 2 2 x   M1 Using a correct triangle formula for the correct triangle. Condone missing brackets and 180 for π. [Area APB =] Their sector area – their triangle area M1 Both expressions must be areas involving terms with x2 and  only. Condone missing brackets and 180 for π for the triangle. Condone calling the sector a segment. [Area APB =]   2 1 2 cos 2     x   2 1 sin π 2 2 x   2 2 [ (2 cos     x 1 sin2 ) 2  or 2 cos (2 cos    x sin ) ] A1 OE A correct expression. Mark the first unsimplified result of subtraction and ISW any incorrect ‘simplifications’. 4

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9 The second term of a geometric progression is 16 and the sum to infinity is 100. (a) Find the two possible values of the first term. 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(b) Show that the nth term of one of the two possible geometric progressions is equal to 4n−2 multiplied by the nth term of the other geometric progression. 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Mark scheme: 9(a) 16 , 100 1 a ar r          leading to   16 and 100 1 a a r r    These can be implied by a correct equation in one variable.   100 1 16 r r   leading to   2 100 100 16 0 r r    *M1 Using their two expressions and rearranging to get a 3-term quadratic expression with all of the terms on one side. Condone sign errors only.    5 4 5 1 0    r r OR 2 25 25 4.25.4 2.25   leading to 4 1 or 5 5 r       DM1 Condone    5 4 5 1   r r following 2 100 100 16   r r . a = 20, a = 80 A1 SC: if DM0 scored SCB1 is available for sight of 20 and 80. Alternative Method for Question 9(a) 16, 100 1 a ar r          leading to 16 100 and 100 a r r a    B1 Rearranging two algebraic statements to give  r . These can be implied by a correct equation in one variable. 2 1600 100 a a   leading to   2 100 1600 0 a a    *M1 Using their two expressions and rearranging to get a 3-term quadratic expression with all of the terms on one side. Condone sign errors and 160 instead of 1600 only.    20 80 0    a a OR 2 100 100 4.1600 2   DM1 a = 20, a = 80 A1 SC: if DM0 scored SCB1 is available for sight of 20 and 80. 4 Question Answer Marks Guidance 9(b) 4 1 , 5 5  r B1 OE SOI   1 1 4 1 [ ] 20 80 5 5                   n n n n u their their v their their B1FT 2 expressions for the nth term FT their values from part (a) if r less than 1. Method 1 for final 2 marks 1 1 1 20 4 5           n n M1 Correctly separating the numerator and denominator of their 1 4 5        n or one correct step towards the solution eg 2 1 4 80 5     n n n u . 1 1 1 2 2 1 1 1 80 4 4 80 4 4 5 5                           n n n n n n n u v A1 AG Given result clearly shown Method 2 for final 2 marks 1 1 1 20 0.8 1 4 4 80 0.2        n n n M1 Dividing two nth terms of the correct format and simplifying their terms in r. = 1 1 2 4 4 4      n n A1 AG 4

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10 The equation of a circle is x −a 2 + y −3 2 = 20. The line y = 12x + 6 is a tangent to the circle at the point P. (a) Show that one possible value of a is 4 and find the other possible value. 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(b) For a = 4, find the equation of the normal to the circle at P. 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(c) For a = 4, find the equations of the two tangents to the circle which are parallel to the normal found in (b). 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Mark scheme: 10(a)   2 2 1 6 3 20 2            x a x or using 2 12   x y *M1 Obtaining an unsimplified equation in x or y only.     2 2 5 3 2 11 0 4      x a x a A1 OE e.g.   2 2 5 4 3 2 4 44 x a x a     Rearranging to get a correct 3-term quadratic on one side. Condone terms not grouped together.   2 2 5 54 4 133 24      y y a a .      2 2 5 3 2 4 11 0 4     a a DM1 OE Using   2 4 on 3 term quadratic 0   b ac their . Method 1 for final 2 marks Using a = 4:    2 3 8 5 5 0    A1 Clearly substituting a = 4. 16  a B1 Condone no method shown for this value. Method 2 for final 2 marks 2 12 64 0     a a ⇒    4 16 0    a a ⇒ 4  a A1 AG Full method clearly shown. 16  a B1 Condone no method shown for this value. 5 If M0, SCB1 available for substituting 4,  a finding P(2, 7) and verifying that CP2 = 20. Question Answer Marks Guidance 10(b) Centre (4, 3) identified or used or the point P is (2, 7) B1 ⸫ gradient of normal 2  B1 SOI Forming normal equation using their gradient (not 0.5) and their centre or P M1 Condone use of   4, 3   .   3 2 4    y x or   7 2 2    y x A1 OE Condone  f x . 4 Question Answer Marks Guidance 10(c) Method 1 for Question 10(c) Diameter:   1 3 4 2 y x    1 leading to 1 2 y x         Or     2 4 2 3 0 dy x y dx     1 leading to 1 2 y x         *M1 Using gradient 1 2 with their centre. By implicit differentiation.   2 2 1 4 1 3 20 2           x x 2 5 10 0 4         x x DM1 Obtaining an unsimplified equation in x or y only. 2 [ 6 5 0]    y y . x = 0 or 8, y = 1 or 5 [(0, 1) and (8, 5)] A1 Correct co-ordinates for both points. Condone no method shown for solution. Equations are   1 2 and 5 2 8 y x y x     A1 2 1 and 2 21 x y x y     . Method 2 for Question 10(c) Coordinates of points at which tangents meet curve are (4+4, 3+2) = (8, 5) and (4 – 4, 3 – 2) = (0, 1) *M1 A1 Vector approach using their centre and gradient = 0.5 . Condone answers only with no working. Equations are   5 2 8    y x and 1 2  y x DM1 A1 Forming equations of tangents using their (0, 1) and (8, 5). Method 3 for Question 10(c)     2 2 4 2 3 20      x x c     2 2 5 4 4 3 4 0 x c x c           *M1 Obtaining an unsimplified equation in x only using equation of circle with 2   y x c .        2 2 4 4 20 3 4 0      c c [leading to 2 4 32 120 16 100 0       c c c ] DM1 Using   2 4 0   b ac . Question Answer Marks Guidance 10(c)   2 4 88 84 0    c c [leading to 2 22 21 0    c c ] A1 21 and 1 or 2 21 c c y x     and 2 1   y x A1 Condone no method shown for solution. 4

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