Cambridge A Level Mathematics 9709 — 2022 Feb/March Paper 1 · Variant 2

2 A curve has equation y = x2 + 2cx + 4 and a straight line has equation y = 4x + c, where c is a constant. Find the set of values of c for which the curve and line intersect at two distinct points. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 [ ] 2 2 2 4 4 leading to 2 4 4 0 + + = + + − + − = x cx x c x cx x c *M1 Equate ys and move terms to one side of equation. ( ) ( ) 2 2 4 2 4 4 4 − = − − − b ac c c DM1 Use of discriminant with their correct coefficients. 2 2 4 16 16 16 4 4 12   − + − + = −   c c c c c A1 ( ) ( ) 2 4 0 leading to 4 3 0 − > − > b ac c c M1 Correctly apply ‘> 0’ considering both regions. 0, 3 < > c c A1 Must be in terms of c. SC B1 instead of M1A1 for c ⩽ 0, c ⩾ 3 5

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4 The first term of a geometric progression and the first term of an arithmetic progression are both equal to a. The third term of the geometric progression is equal to the second term of the arithmetic progression. The fifth term of the geometric progression is equal to the sixth term of the arithmetic progression. Given that the terms are all positive and not all equal, find the sum of the first twenty terms of the arithmetic progression in terms of a. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 + ar a d B1 4 5 = + ar a d B1 ( ) ( ) 2 2 4 2 5 leading to 5 = + + = + a r a a d a ad a d *M1 Eliminating r or complete elimination of a and d. 2 3 0 leading to 3   − = =   ad d d a OR [ ] 2 leading to 3 = = r d a A1 [ ] 20 20 2 19 3 2 = + × S a a DM1 Use of formula with their d in terms of a. 590a A1 6

More questions on Series

6 y y = 3x −20 x + 1 2 + y −2 2 = 85 A C x O B The circle with equation x + 1 2 + y −2 2 = 85 and the straight line with equation y = 3x −20 are shown in the diagram. The line intersects the circle at A and B, and the centre of the circle is at C. (a) Find, by calculation, the coordinates of A and B. 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(b) Find an equation of the circle which has its centre at C and for which the line with equation y = 3x −20 is a tangent to the circle. 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Mark scheme: 6(a) ( ) ( ) 2 2 1 3 22 85 + + − = x x M1 OE. Substitute equation of line into equation of circle. [ ] 2 10 130 400 0 − + = x x A1 Correct 3-term quadratic [ ]( )( ) 10 8 5 leading to 8 or 5 − − = x x x A1 Dependent on factors or formula or completing of square seen. (8, 4), (5, ‒5) A1 If M1A1A0A0 scored, then SC B1 for correct final answer only. 4 6(b) Mid-point of AB = ( ) 1 1 2 2 6 ,- M1 Any valid method Use of C = (‒1, 2) B1 SOI ( ) ( ) 2 2 2 1 1 2 2 1 6 2 = −− + + r M1 Attempt to find r2. Expect 2 1 2 62 = r . Equation of circle is ( ) ( ) 2 2 1 2 1 2 62 + + − = x y A1 OE. 4

More questions on Coordinate geometry

7 (a) Show that sin 1 + 2 cos 1 −sin 1 −2 cos 1  4 [4] cos 1 −2 sin 1 cos 1 + 2 sin 1 5 cos21 −4. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation sin 1 + 2 cos 1 −sin 1 −2 cos 1 = 5 for 0Å < 1 < 180Å. [3] cos 1 −2 sin 1 cos 1 + 2 sin 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) ( )( ) ( )( ) ( )( ) sin 2cos cos 2sin sin 2cos cos 2sin cos 2sin cos 2sin θ θ θ θ θ θ θ θ θ θ θ θ + + − − − − + *M1 Obtain an expression with a common denominator ( ) 2 2 2 2 2 2 5sin cos 2cos 2sin 5sin cos 2sin 2cos cos 4sin θ θ θ θ θ θ θ θ θ θ + + − − − − ( ) 2 2 2 2 4 cos sin cos 4sin θ θ θ θ + = − A1 ( ) 2 2 4 cos 4 1 cos θ θ − − DM1 Use 2 2 cos sin 1 θ θ + = twice 2 4 5cos 4 θ − A1 AG 4 7(b) 2 2 4 5 leading to 25cos 24 5cos 4 θ θ = = − ( ) 24 leading to cos 0.9798 25 θ = = ±    M1 Make cos θ the subject 11.5 or 1 68.5 θ = ° ° A1 A1 FT FT on 180º ‒ 1st solution 3

More questions on Trigonometry

8 y A B x O x −2 2 + y2 = 8 The diagram shows the circle with equation x −2 2 + y2 = 8. The chord AB of the circle intersects the positive y-axis at A and is parallel to the x-axis. (a) Find, by calculation, the coordinates of A and B. 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(b) Find the volume of revolution when the shaded segment, bounded by the circle and the chord AB, is rotated through 360Å about the x-axis. 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Mark scheme: 8(a) ( ) ( ) 2 2 2 8 leading to 2 leading to 0, 2 − + = = = y y A B1 Substitute 2 = y their into circle ( ) 2 leading to 2 4 8 − + = x M1 Expect x = 4. B = (4, 2) A1 3 8(b) Attempt to find [ ] ( ) ( ) 2 π 8 2 d − −  x x *M1 [ ] ( ) [ ] 3 3 2 2 π 8 or π 8 2 4 3 3       −  −  − − +               x x x x x x A1 [ ] [ ] 16 64 π 32 or π 32 32 16 3 3       − − − +             DM1 Apply limits 0 → their 4. Volume of cylinder = 2 π 2 4 16π × × = B1 FT OR from 2 π 2 d  x with their limits from (a). FT on their A and B 2 2 3 3 Volume of revolution = 26 π 16π 10 π é ù - = ê ú ë û A1 Accept 33.5 5

More questions on Integration

10 D E A 5 C 8 B The diagram shows a circle with centre A of radius 5cm and a circle with centre B of radius 8cm. The circles touch at the point C so that ACB is a straight line. The tangent at the point D on the smaller circle intersects the larger circle at E and passes through B. (a) Find the perimeter of the shaded region. 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(b) Find the area of the shaded region. 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Mark scheme: 10(a) 12 5 12 tan or cos or sin 5 13 13 = = = A A A M1 5 12 5 OR tan or cos or sin 12 13 13 = = = B B B A = 1.176 B = 0.3948 A1 Allow 1.18 or 67.4°, Allow 0.395 or 22.6°. May be implied by π 1.176 2 − DE = 4 B1 If trigonometry used accept AWRT 4.00 Arcs = 5 1.176 8 0.3948 × × their and their M1 Or corresponding calculations in degrees. [Perimeter = 5.880 + 3.158 + 4 =] 13.0 A1 Accept 13. If DE is outside the given range this mark cannot be awarded. 5 10(b) Area of triangle = 1 2 × 5 × their12 [ = 30] B1 FT Area of sectors = 2 1 1 2 2 2 5 1.176 8 0.3948 × × + × × their their M1 Or corresponding calculations in degrees [Area = 30 ‒ 14.70 ‒12.63 =] 2.67 A1 Allow 2.66 to 2.67 3

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