Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 1 · Variant 2

1 y 8 7 6 5 4 3 2 1 0 0 1 r rr 3 r 2rr x −1 2 2 The diagram shows the curve with equation y = a sin ( bx) + c for 0 G x G 2 r, where a, b and c are positive constants. (a) State the values of a, b and c. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) For these values of a, b and c, determine the number of solutions in the interval 0 G x G 2 r for each of the following equations: (i) a sin ( bx) + c = 7 - x [1] .................................................................................................................................................... .................................................................................................................................................... (ii) a sin ( bx) + c = 2 r ( x - 1) . [1] .................................................................................................................................................... ....................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) a = 4 B1 Allow 4sin ( 2 x ) + 3 if values of a, b and c are not stated. b = 2 B1 c = 3 B1 3 1(b)(i) 5 B1 Ignore attempts at finding solutions. 1 1(b)(ii) 1 B1 Ignore attempts at finding solutions. 1

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2 The first term of an arithmetic progression is -20 and the common difference is 5. (a) Find the sum of the first 20 terms of the progression. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ It is given that the sum of the first 2k terms is 10 times the sum of the first k terms. (b) Find the value of k. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 20 ( 2 −20 + ( 20 − 1)  5 ) or 20 ( −20 + 75 ) M1 Correct use of either S20 formula with a = — 20 and d = 5. 2 2 550 A1 2 2(b) 2 k k M1* Correct use of Sn formula with a = −20 , d = 5 and either k or ( −40 + ( 2k − 1)  5 ) or ( −40 + ( k − 1)  5 ) 2k. 2 2 n This mark can be awarded for clear use of ( a + l ) when 2 correct values of a and d are used.  −40k + 10k 2 − 5k = −200k + 25k 2 − 25k   15k 2 − 180k = 0 DM1 Equating their S2k to 10 × their Sk and reaching a 2-term   quadratic or 2 term linear equation if k has been cancelled. Condone errors in simplification. k = 12 A1 Condone extra solution k = 0. 3

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3 The equation of a curve is y = 2x 2 - 3 . Two points A and B with x-coordinates 2 and ( 2+ h) respectively lie on the curve. (a) Find and simplify an expression for the gradient of the chord AB in terms of h. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Explain how the gradient of the curve at the point A can be deduced from the answer to part (a), and state the value of this gradient. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a)  2 ( 2 + h ) 2 − 3 B1 SOI  f ( 2 + h ) = 2 M1 2 their − their 5 − 5 2 ( 2 + h ) − 3 2 ( 2 + h ) − 3 ( ( )  2 h 2 + 8h   )  =  can be implied by the ( 2 + h ) − 2  h  ( 2 + h ) − 2 simplified expression or the correct answer. 2 Their 5 must come from 2 ( 2 ) − 3. 2h + 8 or 2 ( h + 4 ) A1 3 3(b) h → 0 , or chord [AB] → tangent [at A] B1 Either of these statements or any sight of h = 0. 8 B1FT Could come from anywhere except wrong working. Either correct or FT their linear expression from (a). 2

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4 Find the term independent of x in the expansion of each of the following: 6 3 (a) e x + 2 o [2] x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 6 3 3 (b) ( 4x - 5 ) e x + 2 o . [4] x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 6 4  3  2 6 6  5  3  2 B1 OE 15 or   x  2  or x   3  May be in a list. 2  x  2  x  6 Allow  . 4 135 B1 Correct term must be identified if in a list. Allow 135x0. 2 4(b) 6 3  3 3 6 6.5.4  3 3 B1 OE 20 or   x  2  or x   3  May be in a list. 3  x  3!  x  540 B1 1 = Identifying term. x 3 x 3 This can be implied by sight of 2160 as part of the constant term. 4  540 −5 135 M1 4  their 540 −5 their 135 1485 A1 Allow 1485x0. 4

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2x + 1 1 5 The function f is defined by f ( x) = for x 1 . 2x - 1 2 (a) (i) State the value of f ( - 1 ) . [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) y 4 2 0 x −4 −2 2 4 −2 −4 The diagram shows the graph of y = f ( x) . Sketch the graph of y = f -1 ( x) on this diagram. Show any relevant mirror line. [2] (iii) Find an expression for f -1 ( x) and state the domain of the function f -1 . 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The function g is defined by g ( x) = 3x + 2 for x d R . (b) Solve the equation f ( x) = gf b 1 l. 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Mark scheme: 5(a)(i) 1 B1 Condone 0.333.   f ( − 1) = 3 1 y5(a)(ii) 4 B1 For showing the correct mirror line. 2 B1 For correct shape: the curves should intersect in the first square in the third quadrant. To the left of the point of intersection, the −4 −2 2 4 reflection is below the original and crosses the x-axis. To the right of the point of intersection, the reflection is to the right the −2 original. −4 2 5(a)(iii) 2 x + 1 M1* Equating y to the given function and clearing of fractions. = y  2 x + 1 = y ( 2 x − 1) x and y may be interchanged at this stage. 2 x − 1 2 xy − 2 x = y + 1 DM1 Condone  errors during simplification. x + 1 −−x 1 A1 Allow ‘ f −’or1 ‘y =’ but NOT ‘x =’, nor fractions within , 2 ( x − 1) 2 − 2 x fractions. [Domain of f −1 is] x  1 B1 Accept — ∞ < x <1 or (— ∞, 1), condone [— ∞, 1). Alternative Method for Question 5(a)(iii) 2 2 M1* Equating y to the given function after division by 2 x − 1. y = 1 +  y −=1 Isolating the term in .x 2 x − 1 2 x − 1 x and y may be interchanged at this stage. 2 DM1 Condone  errors during simplification. 2 x = +1 y − 1 1 1 A1 OE + 1 x − 1 2 Allow ‘ f −’or ‘y =’ but NOT ‘x =’, nor fractions within fractions. [Domain of f −1 is] x  1 B1 Accept — ∞ < x <1 or (— ∞, 1), condone [— ∞, 1). 4 5(b)  1 B1 gf = − 7    4  2 x + 1 M1 2 x + 1  1  = −7 Equating to their gf   . 2 x − 1 2 x − 1  4  A1 OE  x = 3 8 Alternative solution for Question 5(b)  1 B1 gf = − 7    4  x = f −1 ( −7 ) M1 −1  1   x = f  their gf      4   A1 OE  x = 3 8 3

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6 C D 2r cm 2r cm B E 2θ rad A F θ rad θ rad r cm r cm O The diagram shows a metal plate OABCDEF consisting of sectors of two circles, each with centre O. The radii of sectors AOB and EOF are r cm and the radius of sector COD is 2r cm. Angle AOB = angle EOF = i radians and angle COD = 2i radians. It is given that the perimeter of the plate is 14 cm and the area of the plate is 10 cm 2. Given that r 2 3 and i 1 3 , find the values of r and i. 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Mark scheme: 6 [Perimeter =] r + r+ r + 2 r  2+ r + r+ r  = 4r + 6r B1 1 2 1 2 1 2 2 B1 [Area =] r + ( 2r )  2+ r   = 5r    2 2 2 4r + 6r= 14 and 5r 2= 10 M1* ar + br= 14 and cr 2= 10 where a, b and c are constants  0. Terms may be uncollected. EITHER 2 14 − 4 r  10  DM1 Eliminate  to get an equation in r. 5r = 10 or 4 r + 6 r  2  = 14 6 r  5r    2r 2 − 7 r + 6 = 0   ( r − 2 )( 2 r − 3 ) = 0 DM1 Factorise or other accepted method for solving their 3-term   quadratic. OR  14  2  10   10  DM1 Eliminate r to get an equation in . 5   = 10 or 4   + 6  = 14  4 + 6  5   5  [ 182 − 25+ 8 = 0 ] ( 9− 8)( 2− 1) = 0 DM1 Factorise or other accepted method for solving their 3-term quadratic. Then r = 2 and = 0.5 B1 3 8 Condone extra answers r = and = . 2 9 6

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7 (a) By expressing - 2x 2 + 8x + 11 in the form - a ( x - b) 2 + c , where a, b and c are positive integers, find the coordinates of the vertex of the graph with equation y =- 2x 2 + 8 x + 11. 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(b) y O x The diagram shows part of the curve with equation y =- 2x 2 + 8 x + 11 and the line with equation y = 8x + 9 . Find the area of the shaded region. 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Mark scheme: 7(a) 2 2 M1* p  0. −2 ( x  p )  q or −2 ( x  p )  q ( ) 2 2 DM1 −2 ( x − 2 )  q or −2 ( x − 2 )  q ( ) 2 A1 Accept x = 2, y = 19 or 2, 19. −2 ( x − 2 ) + 19 and (2, 19) 3 7(b) Method 1  x = 1 B1* Both x co-ordinates for the points of intersection. Subtract and attempt to integrate M1* 2 2 3 B1* Both terms correct.  −2 x + 2 dx  − x + 2 x  ( )  3  2   2  M1 Apply their limits, one positive and one negative, obtained  − + 2  −  − 2  from equating the line and the curve to their integrated  3   3  expression. 8 2 DB1 AWRT 2.67 WWW. = , 2 8 8 3 3 Condone −→ . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. Method 2  x = 1 B1* Both x co-ordinates for the points of intersection. Attempt to integrate and subtract M1* The second integral can be replaced with what is clearly their area of a trapezium.  −2 x 3 8 2   8 2  B1* OE  + x + 11x  −  x + 9 x  All terms correct.  3 2   2  1 The second integral can be replaced by (1 + 17 )  2 OE. 2 7(b)  −2   2   M1 Apply their limits, one positive and one negative, obtained −  4 + 9 ) − ( 4 − 9 ) from equating the line and the curve, to their integrated  + 4 + 11 −  + 4 − 11  (  3   3   expressions. If the trapezium has been used, the second integral can be replaced by their 18. 8 2 DB1 AWRT 2.67 WWW. = , 2 8 8 3 3 Condone −→ . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. Method 3  x = 1 B1* Both x co-ordinates for the points of intersection. Subtract and attempt to integrate M1* 2 3 8 2 B1* All terms correct. − ( x − 2 ) − x + 10 x 3 2  2  M1 Apply their limits, one positive and one negative, obtained − 4 + 10   − (18 − 4 − 10 ) from equating the line and the curve, to their integrated  3  expression. 8 2 DB1 AWRT 2.67 WWW. = , 2 3 3 7(b) Method 4  x = 1 B1* Both x co-ordinates for the points of intersection. Attempt to integrate and subtract M1* The second integral can be replaced with what is clearly their area of a trapezium.  2 3   8 2  B1* All terms correct.  − ( x − 2 ) + 19 x  −  x + 9 x  1  3   2  The second integral can be replaced with (1 + 17 )  2 OE. 2  2   M1 Apply their limits, one positive and one negative, obtained 18 − 19 )  (−  4 + 9 ) − ( 4 − 9 ) from equating the line and the curve, to their integrated  + 19  − (  3   expression. If the trapezium has been used the second integral can be replaced with their 18 OE. 8 2 DB1 AWRT 2.67 WWW. = , 2 8 8 3 3 Condone −→ . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. 5

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8 The equation of a circle is x 2 + y 2 + px + 2y + q = 0 , where p and q are constants. (a) Express the equation in the form ( x - a) 2 + ( y - b) 2 = r 2 , where a is to be given in terms of p and r2 is to be given in terms of p and q. 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The line with equation x + 2y = 10 is the tangent to the circle at the point A (4, 3). (b) (i) Find the equation of the normal to the circle at the point A. 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(ii) Find the values of p and q. 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Mark scheme: 2 8(a) B1* 1 1     1   − p , −1 . Allow a = − p and b = −1, or centre is   x −− p + ( y −−( 1) ) 2 OE     2 2   2     2 2 DB1   1   2  1  − P  x −− p   + ( y −−( 1) ) = −+q 1 +   OE   2    2  2 8(b)(i) 1 B1 OE [Gradient of tangent =] − SOI 2 [Gradient of normal =] 2 M1 Use of m1m2= −1with their numeric tangent gradient. y − 3 A1 OE = 2  y = 2 x − 5  ISW x − 4 Allow y = 2 x + c, 3 = 2 +c4  c = −5. 3 8(b)(ii) Method 1 for the first two marks:  1  M1*   p  −−1 3 = 2  − p − 4  or −=1 − p − 5 Using their stated centre or  ,  1  in their equation of the  2   2  normal. p = −4 A1 Method 2 for the first two marks: 1 M1* Using their normal equation and their stated centre or −=1 2 x −=5 x 2 − p = 2 2   p   ,  1  .  2  p = −4 A1 Method 3 for the first two marks: dy dy  dy  M1* 2 x + 2 y + p + 2 = 0  p = −−8 8   dx dx  dx   dy 1  A1 = −  p = −4    dx 2  8(b)(ii) Method 1 for the last 3 marks: r 2 = ( 4 − 2 ) 2 + ( 3 −−( 1) ) 2  = 20  M1* Using (4, 3) and their centre or   their p ,  1  to find r2 or r.  2  1 2 DM1 OE −+q 1 + p = 20 Using their expression for r2 (from (a)) equated to their 20. 4 q = −15 A1 Method 2 for the last 3 marks: 2 − 2 − 10  10  M1* Using ( 2, −1) and x + 2 y − 10 = 0 (distance from a point to a r = =   5  5  line). 2 DM1 OE 1 2  10  −+q 1 + p = 2    10  4  5  Using their expression for r2 equated to their   .  5  q = −15 A1 Method 3 for the last 3 marks: 4 2 + 32 + 4 p + 6 + q = 0   4 p + q + 31 = 0  M1* Substituting ( 4,3 ) into their circle equation. OR   1   2 2  1  2 − p  4 −− p   + ( 3 −−( 1) ) = −+q 1 +     2    2  4 ( −4 ) + q + 31 = 0 DM1 Substituting their p = −4. q = −15 A1 8(b)(ii) Alternative Method for Question 8(b)(ii) 4 2 + 32 + 4 p + 6 + q = 0 M1* Substituting ( 4, 3 ) into their circle equation, or 2 2 replacing y with 2 x − 5 from the normal equation, or x + ( 2 x − 5 ) + px + 2 ( 2 x − 5 ) + q = 0 with x = 4 2 10 − x 2  10 − x   10 − x  replacing y with from the tangent equation, or x + + px + 2 + q = 0 with x = 4 2      2   2  y + 5 2 replacing x with from the normal equation, or  y + 5  2  y + 5  2 + y + p + 2 y + q = 0 with y = 3      2   2  replacing x with 10 − 2 y from the tangent equation, and using (10 − 2 y ) 2 + y 2 + p (10 − 2 y ) + 2 y + q = 0 with y = 3 either x = 4 or y = 3 to form an equation in p and q.  Each of these  4 p + q + 31 = 0 5 2 2 5 M1* Solving the tangent and circle equations simultaneously to form x + ( p − 6 ) x + 35 + q = 0  ( p − 6) −4  ( 35 + q ) = 0 a quadratic equation in either x or y. 4 4 OR Then using b 2 − 4 ac = 0 on their quadratic to form an equation 2 2 in p and q. 5 y − y ( 38 + 2 p ) + 100 + 10 p + q = 0  (38 + 2 p) −4 5 (100 + 10 p + q ) = 0 Each of these  p 2 − 12 p − 139 − 5q = 0   Solving the equations simultaneously to find p or q DM1 p = −4 A1 q = −15 A1 5

More questions on Coordinate geometry

9 The equation of a curve is y = 1 k 2 x 2 - 2 kx + 2 and the equation of a line is y = kx + p , where k and p 2 are constants with 0 1 k 1 1. (a) It is given that one of the points of intersection of the curve and the line has coordinates b 5 , 1 l. 2 2 Find the values of k and p, and find the coordinates of the other point of intersection. 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(b) It is given instead that the line and the curve do not intersect. Find the set of possible values of p. 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Mark scheme: 9(a)  1 2 25 5 1  M1*  5 1   k  − 2 k  + 2 =  Using  ,  in the curve equation or equating the line and 2 4 2 2  2 2     OR  5 1 5 the curve and then using x = and p = − k .    1 2 25 5 5  1 5   2 2 2 k  − 2 k  + 2 = k  + − k   Simplify to get a three-term quadratic in k. Condone errors in   2 4 2 2  2 2   simplification. 25k 2 − 40k + 12  = 0  2 A1 OE k = 5 Condone inclusion of k = 6. 5 1  2  5  DM1*  5 1  =  their   + p  p = Using  ,  and their k in an equation in p. 2  5  2   2 2  Either the line (as shown) or 4 p 2 + 12 p + 5 = 0 are the most likely and solving for p. 1 A1 OE p = − 2 Condone inclusion of p = − 5. 2 2 2 6 5 2 DM1 Equating the line and curve using their k and p and simplify to x − x +  = 0   4 x − 60 x + 125  = 0   get a three-term quadratic [= 0]. 25 5 2  25 9  A1 A1 OE  ,  25 9  2 2  Accept x = , y = . 2 2 9(a) Alternative Method for Question 9(a)  1 2 25 5 5  M1* OE k  − 2 k  + 2 = k  + p   5  1   2 4 2 2 ,  Using   in the curve equation or equating the line and 2  2 2  4 p + 12 p + 5  = 0 5 1 2 the curve and then using x = and k = − p. 2 5 5 Simplify to get a three-term quadratic in p [= 0]. 1 A1 p = − OE Condone inclusion of p = − 5. 2 2 1  5   1  DM1*  5 1  =  k  +  their −   k = Using  ,  and their p in the line equation and solving for 2  2   2   2 2  k. 2 A1 OE k = 5 Condone inclusion of k = 6. 5 2 2 6 5 2 DM1 Equating the line and curve using their k and p and simplify to x − x +  = 0   4 x − 60 x + 125  = 0    get a three-term quadratic [= 0]. 25 5 2  25 9  A1 A1 OE  ,  25 9  2 2  Accept x = , y = . 2 2 7 9(b)  1 2 2  1 2 2 M1* Equate the original equations of the curve and the line and k x − 2 kx + 2 = kx + p  k x − 3kx + 2 − p   collect like terms; k and p must still be present.  2  2 2 1 2 DM1 Use of b 2 − 4 ac for their quadratic in x to give an expression in 9 k −4 k ( 2 − p ) 2 k and p. This expression can come from their equation in (a). 5 A1 p − 2 3

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10 A function f with domain x 2 0 is such that f l (x) = 8 ( 2x - 3 ) 3 - 10x 3 . It is given that the curve with equation y = f ( x) passes through the point (1, 0). (a) Find the equation of the normal to the curve at the point (1, 0). [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find f ( x) . 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It is given that the equation f l ( x) = 0 can be expressed in the form 125x 2 - 128 x + 192 = 0 . (c) Determine, making your reasoning clear, whether f is an increasing function, a decreasing function or neither. 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Mark scheme: 10(a) −18 B1 SOI 1 M1 Use of m1m2 = −from1 f ( x ) with x = 1. 18 y  − 0  1 A1 OE = ISW x − 1 18 3 10(b)    B1B1 B1 for each unsimplified {}. 5 4    Can be implied by equivalent simplified or partly simplified 1 1 1  3 . . 3 . x ) = 8 ( 2 x − 3 )  −10 x   + c   f ( versions. 4 5 2     3  3   4 5  3 − 6 x 3 + c   3 ( 2 x − 3 )   5 M1 Use of x = 1 and y = 0 in their integrated f ( x ) , defined as an 3 − 6 (1) 3 + c  0 = 3 − 6 + c  0 = 3 ( 2 (1) − 3 ) 4 expression with at least one correct power, which must contain + c. 4 5 A1 Only condone c = 3 as their final answer if all coefficients have  3 ( 2 x − 3 ) 3 − 6 x 3 + 3 previously been simplified in a correct statement.  f ( x ) or y = 4 10(c) b 2 − 4ac = 1282 −4 125  192 and stating “< 0” M1* b 2 − 4ac = −79616 can be accepted in place of working. OR use of the quadratic formula and stating “No solutions” OR completing the square for the given quadratic and stating positive or > 0. OR sketch of the given quadratic and stating positive. No turning points [in the original function.] DM1 Decreasing because f  ( any positive x value )  0 A1 WWW e.g. f ' (1) = −18. 3

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