Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 1 · Variant 3

1 Solve the equation 8 sin21 + 6 cos 1 + 1 = 0 for 0Å < 1 < 180Å. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 8cos 2− 6cos− 9 = 0 . 8 1 − cos 2 + 6cos+ 1  = 0  M1 Expect ( ) (4cos+ 3)(2cos− 3)  = 0  A1 Factors or formula or completing square must be shown.  → cos = −0.75 →  =  138.6° only, A1 AWRT, ignore solutions outside the given range, answer in radians A0. 3

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5 y 8, 12 12 10 8 6 4 2 x O 2 4 6 8 10 12 The diagram shows a curve which has a maximum point at 8, 12 and a minimum point at 8, 0 . The curve is the result of applying a combination of two transformations to a circle. The first transformation @ A 7 applied is a translation of . The second transformation applied is a stretch in the y-direction. −3 (a) State the scale factor of the stretch. [1] ........................................................................................................................................................ ........................................................................................................................................................ (b) State the radius of the original circle. [1] ........................................................................................................................................................ ........................................................................................................................................................ (c) State the coordinates of the centre of the circle after the translation has been completed but before the stretch is applied. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) State the coordinates of the centre of the original circle. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 3 B1 Ignore any description. 1 5(b) 2 B1 Ignore any description. 1 5(c) (8, 2) B1 B1 Ignore any description. Allow vector notation and absence of brackets. 2 5(d) (1, 5) B1 FT FT each coordinate, (their8 – 7, their2 + 3) Allow vector notation and absence of brackets. B1 FT 2

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6 It is given that ! = cos−1 8 . 17 1 1 Find, without using the trigonometric functions on your calculator, the exact value of + sin ! tan !. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 6 2 *M1 Or Pythagoras seen (may quote 8, 15, 17 triple). 2 2  8  Use of sin + cos = 1 eg sin=  1 −   17  15 A1 sinα = 17 15 A1 tan= 8 1 1 17 8 DM1 Dealing with reciprocals and addition of fractions + = + correctly. sin tan 15 15 5 A1 Correct answer with no working shown scores 0. = oe 15 3 Extra answers from sinα = − are allowed. 17 5

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8 A r 5 3r P Q r B The diagram shows two identical circles intersecting at points A and B and with centres at P and Q. The radius of each circle is r and the distance PQ is 3r.5 (a) Find the perimeter of the shaded region in terms of r. 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(b) Find the area of the shaded region in terms of r. 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Mark scheme: 8(a) 5 *M1 May use cosine rule to find APB. Stating APQ or r −1 6  −1 5  APB as an incorrect multiple of  is M0. APQ = cos = cos   r  6  = 0.5857 A1 Accept 0.586 or 33.6° or APB (1.171 or 67.1o). Perimeter = 4 r their 0.5857 = 2.34r or 0.745π r or (293/125)r DM1 A1 Must use a numerical value of their angle. 4 8(b) Use of sector formula: Sector APB = ½r 2  ( 2  their 0.5857 ) or Sector APC M1 Any sector with their appropriate angle. It must be clear the appropriate numerical angle is being used. (C is on PQ so PC = r) = ½r 2  ( their 0.5857 ) Use of appropriate formula for area of triangle and correct combination with M1 e.g. Area APB = ½r 2  sin ( 2  their 0.5857 ) . the sector to find the area of a half segment, one segment or both segments Shaded area [ = 2  0.1250 r 2 ] = 0.250 r 2 A1 2 1 2 2 or 0.0796π r , allow r or 0.25 r . 4 3

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9 The first term of a geometric progression is 216 and the fourth term is 64. (a) Find the sum to infinity of the progression. 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The second term of the geometric progression is equal to the second term of an arithmetic progression. The third term of the geometric progression is equal to the fifth term of the same arithmetic progression. (b) Find the sum of the first 21 terms of the arithmetic progression. 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Mark scheme: 9(a) 216 r 3 = 64 →=r 2/3 B1 Allow decimal to 3sf (AWRT). 216 M1 A1 M1 depends on their |r| < 1. S = = 648 cao 1 − their⅔ 3 9(b)  2 B1 FT SOI, may be implied in the use of 96 = 144 + 3d 216  = 144 → 144 = a + d and finding .a Mis-reads not condoned in 9(b).  3  2 B1 FT SOI, may be implied in the use of 96 = 144 + 3d  2 = 216   96 → 96 = a + 4d and finding .a  3  Solve simultaneously *M1 No working may be seen. d = −16, a = 160 A1 Both required. 21 DM1 A1 S 21 = 320 + 20 ( −16 ) = 0 Or use of 21( a+u21). 2 2 6

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10 y y = 2x −1 A x O D x2 + y2 = 2 B The diagram shows the circle x2 + y2 = 2 and the straight line y = 2x −1 intersecting at the points A and B. The point D on the x-axis is such that AD is perpendicular to the x-axis. (a) Find the coordinates of A. 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(b) Find the volume of revolution when the shaded region is rotated through 360Å about the x-axis. π Give your answer in the form b c −d , where a, b, c and d are integers. 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(c) Find an exact expression for the perimeter of the shaded region. 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Mark scheme: 10(a) Or 5 y 2 + 2 y − 7  = 0 . x 2 + ( 2 x − 1) 2 − 2  = 0 → 5 x 2 − 4 x − 1  = 0 *M1 A1 ( 5 x + 1)( x − 1) = 0  or ( 5 y + 7 )( y − 1) = 0  DM1 May see factors or formula or completing square. x = 1, y = 1 or (1, 1) only A1 May be implied on the diagram. 4 10(b) *M1 A1 3  2 − y 2 dy . ) Attempt integration of y 2 , allow  ( 2 − x 2 dx = ()  2 x − x  ) ()  ( 3    3  DM1 Apply limits 1 → √2. ( 2)  1  2 − ()  2 2 −  −   )  3   3     A1  4 2 − 5 CAO, allow 2 8 − 5 , must be in given form. ( ) ( ) 3 3 4 10(c) 1  2 B1 Must be exact. Arc length = (2 2) or oe 8 4 Perimeter = 2 + their arc length B1 FT Must be exact, do not allow inverse trig functions. 2

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11 The coordinates of points A, B and C are A 5, −2 , B 10, 3 and C 2p, p , where p is a constant. (a) Given that AC and BC are equal in length, find the value of the fraction p. 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(b) It is now given instead that AC is perpendicular to BC and that p is an integer. (i) Find the value of p. 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(ii) Find the equation of the circle which passes through A, B and C, giving your answer in the form x2 + y2 + ax + by + c = 0, where a, b and c are constants. 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Mark scheme: 11(a) ( 5 − 2 p ) 2 + ( p + 2 ) 2 = (10 − 2 p ) 2 + ( 3 − p )2 M1 A1 Allow one sign error for M mark only. 25 − 20 p + 4 p 2 + p 2 + 4 p + 4 = 100 − 40 p + 4 p 2 + 9 − 6 p + p 2 A1 Allow 2.67 AWRT. 8 30 p = 80 → p = oe 3 3 11(b)(i) p + 2 p − 3 M1 Allow a sign error. m AC = mBC = 2 p − 5 2 p − 10 p + 2 p − 3 M1 Use of m1m2 = -1 with their mAC and mBC.  = −1 2 p − 5 2 p − 10 2 2 2 A1 p − p − 6 = − 4 p − 30 p + 50 → 5 p − 31 p + 44 ( = 0 ) ( ) 11 A1 Factors ( p − 4 )( 5 p − 11) , or formula or p = 4 (Ignore p = ) 5 completing square must be seen. 4 11(b)(ii) Mid-point of AB = (7½, ½) B1 SOI 2 2 2  50  2 2  5  2  *M1 2 1 2 2  50  5 + 5 = Or r = etc. r = 2½ + 2½ = = ( or r =  (2½ + 2½ )   )     4 4   4 2     2 2 50 DM1 Must use r 2 not r or d or d 2 Equation of circle is ( x − their 7½ ) + ( y − their ½ ) = their 4 x 2 + y 2 − 15 x − y + 44 = 0 A1 CAO 4

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