Cambridge A Level Mathematics 9709 — 2022 May/June Paper 1 · Variant 2

2 The second and third terms of a geometric progression are 10 and 8 respectively. Find the sum to infinity. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 r = 0.8 B1 OE a = 12.5 B1 OE   12.5 1 0.8 S   M1 Using 1 a r  with ‘their a’ and ‘their r’ but must be 1 r . 125 1 , 62 or 62.5 2 2 S A1 1 12 2 1 5 or similar does not get A1. 4

More questions on Series

4 The first, second and third terms of an arithmetic progression are k, 6k and k + 6 respectively. (a) Find the value of the constant k. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the sum of the first 30 terms of the progression. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) or 2d = 6 leading to d = 3, ⸫ 6k – 3 = k B1 OE A correct equation in k only. Can be implied by correct final answer. 6 10 k  or 0.6 B1 OE 2 Question Answer Marks Guidance 4(b) d = 3 B1 Correct value of d can be implied by a correct final answer. Working may be seen in part (a) but must be used in (b).   30 30 2 ‘ ’ 29 ‘ ’ 2 S their k their d     M1 It needs to be clear that the candidate is using a correct sum formula. There is no requirement to check the candidates working for d but it must be clearly identified. S30 = 1323 A1 ISW if corrected to 1320. 3

More questions on Series

5 The equation of a curve is y = 4x2 −kx + 12k2 and the equation of a line is y = x −a, where k and a are constants. (a) Given that the curve and the line intersect at the points with x-coordinates 0 and 34, find the values of k and a. 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(b) Given instead that a = −72, find the values of k for which the line is a tangent to the curve. 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Mark scheme: 5(a) 2 2 1 4 0 0 0 2 k a      M1 Equating the equations of curve and line and substituting 0 Condone slight errors e.g. ± sign errors. 2 2 3 3 1 3 4 4 4 2 4 k k a            M1 Equating the equations of curve and line and substituting 3 4 x  . Condone slight errors e.g. ± sign errors. k = 2, 2 a  A1 A1 WWW Alternative method for question 5(a)   3 0 0 4 x x          or   4 3 0 x x   [⟹ 2 4 3 0] x x   *M1 Use 0, 3 4 to form a quadratic equation. Do not allow   3 0 0 4 x x          .     2 2 2 2 1 1 4 leading to 4 1 0 2 2 x kx k x a x k x k a          DM1 Equating the equations of curve and line and rearranging so that terms are all on same side. Condone slight errors e.g. ± sign errors. k = 2, 2 a  A1 A1 WWW Alternative method for question 5(a) 3 0 4 b a    and 3 0 4 c a   *M1 Using sum and product of roots. Condone ± sign errors. 1 3 4 4 k  and 2 1 2 0 4 k a   DM1 Equating the equations of curve and line and equating to 3 4 and 0. k = 2, 2 a  A1 A1 WWW 4 Question Answer Marks Guidance 5(b)   2 2 2 2 1 7 1 7 4 4 0 2 2 2 2 x kx k x x kx x k           *M1 OE Substitute 7 2 a  and rearrange so that terms are all on same side, condone ± sign errors. Watch for multiples.   2 2 1 7 1 4 4 2 2 k k          *DM1 Use of b2 – 4ac with the coefficients from their 3-term quadratic. Both coefficients ‘b’ and ‘c’ must consist of two components. 2 7 2 57 k k    A1 OE    3 7 19 k k   or other valid method DM1 Factorising or use of the formula or completing the square. Must be evidence of an attempt to solve for this mark. Dependent upon both previous method marks. k = 3, k = 19 7  A1 OE e.g. AWRT 2.71. No ISW if inequalities used. SC: If second DM1 not scored, SC B1 available for correct final answers. Alternative method for question 5(b) 8 1 x k   and 2 2 1 7 4 2 2 x kx k x     *M1 Equating gradients and equating line and curve.   2 2 1 7 4 8 1 (8 1) 2 2 x x x x x       or 2 2 1 1 1 1 7 4( ) 8 8 2 8 2 k k k k k              *DM1 Forming an equation in x or k only. 2 28 8 3 x x   or 2 7 2 57 k k   A1 OE A correct 3 term quadratic in x or k only.    14 3 2 1 x x   or    3 7 19 k k   or other valid method DM1 OE Factorising or use of the formula or completing the square. Must be evidence of an attempt to solve for this mark. Dependent upon both previous method marks. Question Answer Marks Guidance 5(b) k = 3, k = 19 7  A1 OE e.g. AWRT 2.71. No ISW if inequalities used. SC: If second DM1 not scored, SC B1 available for correct final answers. 5

More questions on Quadratics

6 y y = 2x + 2 1 y = 5x 2 x O 1 The diagram shows the curve with equation y = 5x 2 and the line with equation y = 2x + 2. Find the exact area of the shaded region which is bounded by the line and the curve. 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Mark scheme: 6 Line meets curve when: 1 2 2 2 5 x x   leading to   1 2 2 — 5 2 0 x x   or 2 4 8 4 25 x x x    leading to   2 4 17 4 0 x x    or 2 25 y x  leading to   2 2 25 50 0 y y    M1 Equating line and curve and rearranging so that terms are all on same side, condone sign errors, and making a valid attempt to solve by factorising, using the formula or completing the square. Factors are:(2 1 2 x -1)( 1 2 x -2), (4x-1)(x-4) and (2y-5)(y-10). 1 , 4 4 x x   A1 SC: If M1 not scored, SC B1 available for correct answers, could just be seen as limits. Area = 1 2 5 (2 2) dx x x    = 1 2 5 2 2 x x    dx *M1 Intention to integrate and subtract areas. Condone missing brackets and/or subtraction wrong way around. 4 3 2 2 1 4 10 2 3 x x x            = 10 10 1 1 1 8 16 8 3 3 8 16 2                          DM1 Integrating( 3 2 kx seen) and substituting ‘their points of intersection’ (but limits need to be found, not assumed to be 0 and something else). 45 16 or 13 216 or 2.8125 A1 OE exact answer. Condone 45 16  if corrected to 45 16 . A0 for inclusion of π. SC: If *M1 DM0 scored, SC B1 available for correct answer. Question Answer Marks Guidance 6 Alternative method for question 6 Line meets curve when: 1 2 2 2 5 x x   ⇒   1 2 2 — 5 2 0 x x   or 2 4 8 4 25 x x x    ⇒   2 4 17 4 0 x x    or 2 25 y x  ⇒   2 2 25 50 0 y y    M1 Equating line and curve and rearranging so that terms are all on same side, condone sign errors, and making a valid attempt to solve by factorising, using the formula or completing the square. Factors are:(2 1 2 x -1)( 1 2 x -2), (4x-1)(x-4) and (2y-5)(y-10). 1 , 4 4 x x   A1 SC: If M1 not scored, SC B1 available for correct answers, could just be seen as limits. Area = 1 2 5 dx x   { (2 2) x dx   or area of trapezium} *M1 Intention to integrate and subtract areas. Or integrate curve and subtract area of trapezium.   4 3 4 2 2 1 1 4 4 10 1 15 2 sum of ‘ values’ ‘ ’ 3 2 4 x x x or their y their                       10 10 1 1 1 1 5 15 8 16 8 10 3 3 8 16 2 2 2 4 or                                                    DM1 Integrating ( 3 2 kx seen) and substituting ‘their points of intersection’ (but limits need to be found, not assumed to be 0 and something) or a trapezium using the correct formula (‘their 15 4 ’ must be ‘their 4’ – ‘their 1 4 ’ but not 0). 45 16 or 13 216 or 2.8125 A1 OE exact answer. Condone 45 16  if corrected to 45 16 . A0 for inclusion of π. SC: If *M1 DM0 scored, SC B1 available for correct answer. 5

More questions on Integration

7 B 2 cm A 10 cm 16π O P C The diagram shows a sector OBAC of a circle with centre O and radius 10cm. The point P lies on OC and BP is perpendicular to OC. Angle AOC = 16π and the length of the arc AB is 2cm. (a) Find the angle BOC. 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(b) Hence find the area of the shaded region BPC giving your answer correct to 3 significant figures. 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Mark scheme: 7(a) ˆ AOB      2 10 B1 OE Sight of 0.2 from s r  but 10 2  is not enough. ISW if 2 π 10 5  . [B ˆO C 5π 6 ] 30   or 1 π 0.2 6  B1 OE e.g. 0.724c AWRT or 41.5 degrees AWRT. But not 5π 2 3 10  – fraction within a fraction. ISW incorrect simplifications. Alternative method for question 7(a) OR [Arc AC =]   10π 10π or Arc 2 or 7.2 6 6 BC   B1 AWRT. Sight of 10π 6 or 5.2 or 7.2. [B ˆO C 5π 6 ] 30   or 1 π 0.2 6  B1 OE e.g. 0.724c AWRT or 41.5 degrees AWRT. But not 5π 2 3 10  – fraction within a fraction. ISW incorrect simplifications. 2 Question Answer Marks Guidance 7(b) [BP] = 5π 6 10sin 30        and [OP ] = 5π 6 10cos 30        [= 6.6208…] and [= 7.494…] OR [BP] = 5π 6 10sin 30        and [O ˆB P] = 5π 3 15        [= 6.6208…] and [= 0.84719…] M1 OE Any correct method for both lengths, for their angle BOC (which may have been incorrectly ‘simplified’ but not 0.2) or length BP and O ˆB P. May be seen as part of 1 sin 2 ab C . Sight of correct method enough. Can be implied by the next A1. Area of ∆OBP = 1 2  5π 6 10sin 30        5π 6 10cos 30         or 1 5π 6 5π 3 10 10sin sin 2 30 15                        [=24.809] A1 OE Can be implied by any answer in range (24.7, 24.9) or a final answer in the range (11.3, 11.5) WWW. [Sector BOC] = 2 1 10 2 their   5π 6 30        5π 6 50 36.1799... 30                M1 Use of 2 1 2 r  with their angle BOC (may have been incorrectly ‘simplified’ but not 0.2). Area of region BPC = 11.4 A1 CAO 4

More questions on Circular measure

8 The equation of a circle is x2 + y2 + ax + by −12 = 0. The points A 1, 1 and B 2, −6 lie on the circle. (a) Find the values of a and b and hence find the coordinates of the centre of the circle. 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(b) Find the equation of the tangent to the circle at the point A, giving your answer in the form px + qy = k, where p, q and k are integers. 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Mark scheme: 8(a) 1 1 12 0 10 a b a b          4 36 2 6 12 0 2 6 28 a b a b         correct values for a and b. 4, 6 a b   B1 Centre is   , 2, 3 2 2 their a theirb           B1 FT Or 2, 3 x y   4 Question Answer Marks Guidance 8(b) Gradient of AC is 1 [ 1 their y their x    1 3 1 2   1 3 4 ] 1 2 3    *M1 Using their centre correctly. Gradient of tangent is 1 3 4 4 3 their          A1 FT Use of 1 2 1 m m  to obtain the gradient of the tangent. Equation:   3 1 ‘ ’ 1 4 y their x    or y 3 7 4 4 x   DM1 Using   1,1 with their gradient of the tangent at A. 3 4 7 x y   or 4 3 7 y x   . or integer multiples of these A1 Alternative method for question 8(b) 2 2 4 6 0 dy dy x y dx dx     *M1 Implicit differentiation with at least one y term differentiated correctly. 6 8 6 8 dy dy dx dx   A1 Equation:   3 1 ‘ ’ 1 4 y their x    or y 3 7 4 4 x   DM1 Using   1,1 with their gradient of the tangent at A. 3 4 7 x y   or 4 3 7 y x   . or integer multiples of these A1 Alternative method for question 8(b)   1 2 2 1{25 ( 2) } 2 4 2 dy x x dx       *M1 Rearranging to form y  and differentiating using the chain rule.   1 2 1 6 (25 9) 6 2 8 dy dx      A1 Question Answer Marks Guidance 8(b) Equation:   3 1 ‘ ’ 1 4 y their x    or y 3 7 4 4 x   DM1 Using   1,1 with their gradient of the tangent at A. 3 4 7 x y   or 4 3 7 y x   . or integer multiples of these A1 4

More questions on Coordinate geometry

1 9 The equation of a curve is y = 3x + 1 −4 3x + 1 2 for x > −13. dy d2y (a) Find and . 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(b) Find the coordinates of the stationary point of the curve and determine its nature. 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Mark scheme: 9(a)  1 1 2 2 d 1 3 4 (3 1) 3 3 6(3 1) d 2 y x x x                          1 differentiation of   1 2 4 3 1 x   . Accept unsimplified.     2 3 3 2 2 2 d 1 6 3 1 3 [ 9 3 1 2 d y x x x               ] B1 WWW. Accept unsimplified. Do not award if d d y x is incorrect. 3 9(b)   1 2 d 0 leading to 3 6 3 1 0 d y x x      M1 Setting their d d y x = 0.   1 2 3 1 2 3 1 4 x x    leading to 1 x  A1 CAO – do not ISW for a second answer. y = −4 [coordinates (1, −4)] A1 Condone inclusion of second value from a second answer.   2 3 2 2 d 9 3 1 1 d y x    = 9 or 8 > 0 so minimum A1 Some evidence of substitution needed but 2 2 d d y x . Do not award if 2 2 d d y x is incorrect or wrongly evaluated. Accept correct consideration of gradients either side of 1 x  . 4

More questions on Differentiation