Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 1 · Variant 1

2 A line has equation y = 2cx + 3 and a curve has equation y = cx2 + 3x −c, where c is a constant. Showing all necessary working, determine which of the following statements is correct. A The line and curve intersect only for a particular set of values of c. B The line and curve intersect for all values of c. C The line and curve do not intersect for any values of c. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 2 2 M1 Forming a 3-term quadratic, all terms on one side. cx + 3 x − c = 2cx + 3 leading to cx + ( 3 − 2c ) x − ( c + 3)  = 0 2 2 M1 2nd M1 for b 2 − 4ac correct for their a, b, c b − 4ac = ( 3 − 2c ) + 4c ( c + 3) i.e. no sign errors. = 8c 2 + 9 A1 > 0 [for all values of c] leading to B [Intersects for all values of c] A1 WWW 4

More questions on Quadratics

3 x x x The diagram shows a cubical closed container made of a thin elastic material which is filled with water and frozen. During the freezing process the length, xcm, of each edge of the container increases at the constant rate of 0.01cm per minute. The volume of the container at time t minutes is V cm3. Find the rate of increase of V when x = 20. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 dV 2 B1 SOI = 3 x dx dV  dV dx  2 M1 Correct use of chain rule with x = 20 substituted into =  = 3  20  0.01   dt  dx dt  dV . dx 12 A1 3

More questions on Differentiation

5 (a) Show that the equation 5 2 4 sin x + + = 0 tan x sin x may be expressed in the form a cos2x + b cos x + c = 0, where a, b and c are integers to be found. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 5 2 (b) Hence solve the equation 4 sin x + + = 0 for 0Å ≤x ≤360Å. [3] tan x sin x ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 2 *M1 Multiply by sin x (or writing as a single fraction) and 4sin x + 5cos x + 2  = 0 sin x using tan x = . cos x 2 DM1 Correctly obtaining a quadratic in cos x (allow sign 4 1 − cos x + 5cos x + 2  = 0  ( ) errors). 4cos 2 x − 5cos x − 6 = 0 A1 Condone missing x. Must be = 0 unless 0 appears on RHS earlier. 3 5(b) (4cos x + 3)(cos x − 2)  = 0  M1 Or use of formula or completing square. 138.6º, 221.4º A1 B1 FT FT on 360º ‒ 1st solution from quadratic in cos x. Use of radians (2.42) A0 but allow B1 FT for 2π: 1st solution if use of radians is clear. SC If M0 scored SC B1 B1 for correct final answer(s). If extra incorrect solutions in the range 0 → 360º are given award A1 B0. 3

More questions on Trigonometry

6 O r A B r C The diagram shows a motif formed by the major arc AB of a circle with radius r and centre O, and the minor arc AOB of a circle, also with radius r but with centre C. The point C lies on the circle with centre O. (a) Given that angle ACB = kπ radians, state the value of the fraction k. 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(b) State the perimeter of the shaded motif in terms of π and r. 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(c) Find the area of the shaded motif, giving your answer in terms of π, r and 3. 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Mark scheme: 6(a) 2 B1 2π k = Allow ACB = . 3 3 1 6(b) Perimeter of shaded area = 2πr B1 1 6(c) 1 2 4π *M1 2 2 Major sector OAB = r  Expect 3 πr . Finds area of any relevant sector or 2 3 triangle. Can be embedded in segment formula.  1 2 π 1 2 π  *M1 One or both segments = 2   r  − r sin   2 3 2 3   2 π 2 3  A1 = 2  r − r     6 4  2 2  1 2 r 2 3  DM1 Shaded area = πr − 2  πr −    3  6 4  πr 2 r 2 3 A1 = + 3 2 6(c) Alternative method for Question 6(c) 1 2 1 *M1 1 2 Sector CAOB = 2  r their π Expect [2] × π r . 2 3 6 Can be embedded in segment formula.  1 2 π 1 2 π  *M1 One or both segments = 2   r  − r sin   2 3 2 3   2 π 2 3  A1 = 2  r − r     6 4  1 2  2 π 2 3  DM1 2 Shaded area = π r −  πr + 2  r − r    3  6 4  πr 2 r 2 3 A1 = + 3 2 6(c) Alternative method for Question 6(c) 1 2 π M1 3 Area of rhombus AOBC = 2  r sin Expect [2]  . 2 3 4 Can be embedded in segment formula.  1 2 π 1 2 π  M1 One or both segments = 2   r  − r sin   2 3 2 3   2 π 2 3  A1 = 2  r − r     6 4  2  3 2  2 π 2 3   DM1 Shaded area = πr −  r − 4  r − r     2  6 4  πr 2 r 2 3 A1 = + 3 2 5

More questions on Circular measure

7 The sum of the first two terms of a geometric progression is 15 and the sum to infinity is 125 . The 7 common ratio of the progression is negative. Find the third term of the progression. [7] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 − r 27 a (1 + r ) = 15 B1 a ( ) Accept = 15 for first B1. 1 − r a 125 B1 = 1 − r 7 125 M1 Eliminate a. (1 − r )(1 + r ) = 15 7 2 105 M1 1 − r = 125 2 4 2 A1 2 2 r = leading to r = − Condone or ± . 25 5 5 5 125 7 A1 Ignore 2nd answer. a =  = 25 7 5 4 A1 CAO 3rd term = 25  = 4 25 7 Alternative method for Question 7 a (1 + r ) = 15 B1 a 125 B1 = 1 − r 7 7(15-15r) = (125 – 125r)(1 – r2) M1 125r3 – 125r2 – 20r +20 = 0 M1 −2 2 A1 Condone extra ‘answer’ of r = 1. r = [1, ] 5 5 a = 25 A1 Ignore 2nd answer. 3rd term = 4 A1 CAO 7 Alternative method for Question 7 a (1 + r ) = 15 B1 a 125 B1 = 1 − r 7 a 125 M1 Eliminate r. =  15  7 1 −  − 1   a  7a2 – 250a + 1875 [= 0] M1  75  A1  75  a = 25, Condone extra ‘answer’ of r = .      7   7  −2 A1 Ignore 2nd answer. r = 5 4 A1 CAO 3rd term = 25  = 4 25 7

More questions on Series

8 y y = 2x −3 2 + 1 A B y = 2 2x −3 4 x O The diagram shows the curves with equations y = 2 2x −3 4 and y = 2x −3 2 + 1 meeting at points A and B. (a) By using the substitution u = 2x −3 find, by calculation, the coordinates of A and B. 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(b) Find the exact area of the shaded region. 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Mark scheme: 8(a) 4 2 4 2 B1 u = 2 x − 3 leading to 2u = u + 1 leading to 2u − u −=1 0 2 2 M1 Factors or formula or completing square must be 2u + 1 u − 1 = 0  ( )( ) shown. u = 1 leading to 2 x−=3 1 leading to x = 1 or 2 A1 (1, 2), (2, 2) A1 Special case: If B1 M0 scored then SC B2 can be awarded for correct coordinates or SC B1 for correct x values only. Special case 2(2x – 3)4 = (2x – 3)2 + 1 32x4 – 192x3 + 428x2 – 420x + 152 = 0 x = 1, 2 finding both from a correct quartic SC B1 (1, 2), (2, 2) SC DB1 Special case: Trial and improvement without quartic. Both x values correct B1, both coordinates correct B2. 4 8(b)   ( 2 x − 3 ) 3   2 ( 2 x − 3 ) 5  B1 B1 Integrate the 2 functions.  + x  −   3  2 5  2      1   1   1  1   M1 Apply their limits 1 → 2 (must be shown) to an + 2 −− + 1     −  −−   integral.  6   6   5  5   Some evidence of substitution. Minimum (13 − 5) – ( 1 + 1 ) or equivalent. 6 6 5 5 Allow 1 sign error for 1st M1. 4 2 M1 Subtract (at some point) the 2 areas. − Must subtract areas and not just integrals. 3 5 14 A1 Special case: If M0 for substitution of limits can award SC B1 for correct answer. 15 14 Condone − if corrected. 15 If subtraction is the wrong way round award B1 B1 M1 M1 A0. y 2 dx or x dy scores 0 /5. π y dx used. Award B1 B1 M1 M1 A0. 8(b) Alternative method for Question 8(b) u = 2x – 3 B2,1,0  u 2 + 1 − 2u 4 du ( ) 1  1 3   2 5    u + u  −  u   2  3   5   1   1 2   −1 2   M1 Applies limits –1 → 1. + 1 − − −+1       2   3 5   3 5   M1 Subtract (at some point) the 2 areas. 1  14 14  A1  +  2  15 15  14 15 5

More questions on Integration

d2y 10 A curve has a stationary point at 2, −10 and is such that = 6x. dx2 dy (a) Find dx. 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(b) Find the equation of the curve. 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(c) Find the coordinates of the other stationary point and determine its nature. 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(d) Find the equation of the tangent to the curve at the point where the curve crosses the y-axis. 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Mark scheme: 10(a) dy 2 B1 = 3 x  + c  dx 3  2 2 + c = 0 M1 dy Substitute x = 2 and = 0 into an integral (c must be dx present). dy 2 A1 = 3 x −12 dx 3 10(b) 3 B1 FT FT on their non-zero c (dependent on c being found at y = x − 12 x  +k  some stage). −10 = 23 − 12  2 + k M1 Substitute x = 2, y = ‒10 (k present). 3 A1 3 y = x − 12 x + 6 Must be y = (unless y = x − 12 x + k stated earlier). 3 10(c) 3 x 2 − 12 = 0 [leading to x =−2 ] M1 dy Set their two term = 0 . Expect x =−2 . dx Ignore x = 2 given in addition. 3 A1 y = ( −2 ) − 12 −( 2 ) + 6 = 22 leading to (‒2, 22) d 2 y A1 dy When x = −2, 2  0 (or ‒12) hence Maximum Can be from correct conclusion from sign diagram dx dx dy if calculated correctly. dx Do not allow concave downward for final A1. Can be awarded if the only error is incorrect or missing y-coordinate. 3 10(d) dy M1 d y At x = 0, = −12, y = 6 Both required. FT on their and y. dx d x y − 6 = −12 x A1 OE 2

More questions on Differentiation

11 y x −4 2 + y + 1 2 = 40 A x O B The diagram shows the circle with equation x −4 2 + y + 1 2 = 40. Parallel tangents, each with gradient 1, touch the circle at points A and B. (a) Find the equation of the line AB, giving the answer in the form y = mx + c. 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(b) Find the coordinates of A, giving each coordinate in surd form. 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(c) Find the equation of the tangent at A, giving the answer in the form y = mx + c, where c is in surd form. 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Mark scheme: 11(a) Gradient of AB = ‒1 B1 SOI Centre of circle = (4, ‒1) B1 SOI Equation of AB is y + 1 = −1 ( x − 4 ) leading to y = −+x 3 B1 FT FT their centre with gradient ‒1. 3 11(b) ( x − 4 )2 + ( −+x 3 + 1) 2 = 40 *M1 Substitute their AB into circle equation. 2 2 DM1 Forming and solving 3-term quadratic. leading to x − 8 x − 4 2 ( x − 4 ) = 40 OR [2]( ) 8  64 + 16 16  256 + 64 or 2 4 A1 OE. No fractions. x = 4  20 4 − 20 , −+1 20 A1 OE ( ) Special case: If M1 M0 scored then SCB2 can be awarded for correct coordinates or SCB1 for correct x values only. Ignore other coordinate 4 M1 OE −+1 20 = 1 x − their 4 − 2011(c) y − their ( ) (  ) A1 y = x −+5 2 20 or y = x −+5 80 or y = x −+5 4 5 2

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