Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 1 · Variant 2

1 (a) Express 9x 2 - 36 x + 8 in the form p ( x + q) 2 + r , where p, q and r are constants. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the set of values of the constant k for which the equation 9x 2 - 36 x + 8 = k has no real roots. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the exact roots of the equation 9x 2 - 36 x + 8 =-15 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 9 {(x − 2) 2 }−28 B2,1,0 B2 for all 3 correct values. B1 for any two correct. Allow p = 9, q = −2, r = −28 . If contradictory, give preference to the expression. Condone correct values in an expression of the correct form, even if missing the square on the bracket. Allow B1 for 9{(x − 2) 2 + k }. 2 1(b) k −28 B1FT k  theirr. 1 1(c) 2 M1 2 2 2 13 36  36 − 4 9 23 Solving as far as ' ( x + q ) = '. 9 ( x − their 2 ) − their 28 = −15  ( x − 2 ) = or x = 9 2  9 Allow sign errors only in the rearrangement, or rearranging and using the quadratic formula (see quadratics guidance). 13 6  13 13 36  468 A1 ISW if decimals given after the exact answers. x = 2  or or 2  or oe 3 3 9 18 2

More questions on Quadratics

2 3 62 Find the term independent of x in the expansion of b2 x - l . [3] x .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 6 2 2  3  6 − 2 B1 6 6   (2 x ) −   may be replaced by  or 15. 2  x  2 4 6 6 2 2  3  6 − 2 May be seen in a list. or   2  ( x ) −  Condone missing brackets only if recovered later. 2  2 x  Identifying the relevant term of the form: M1 May still be part of a list if, e.g., underlining or 6 − 2 arrow. 2 2  3  k  (2 x ) −    p   x  k can be any constant or any .   6 − 2  q  6 2 2  3  or k 2  ( x ) −   May be implied by correct final answer.  2 x  If constants evaluated, then condone errors for this mark. = 4860 A1 0 Accept 4860x SC B2 for – 4860 3

More questions on Series

3 (a) The graph of y = f ( x) is transformed to the graph of y = f ( 3 x) + 2 . Describe fully the two transformations which have been combined to give the resulting graph. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) A different graph has equation y = g ( x) . This graph is stretched by scale factor 3 in the y-direction and then reflected in the y-axis. Write down the equation of the transformed graph in terms of the function g. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 0 B1 Condone ‘shift’ and vector written as co-ordinates. Translation  , or Allow any mention of the y-axis but not just ‘up’. 2 Translation 2 ‘parallel to the y-axis’ or ‘in the y-direction’ or ‘vertically’. 1 B2,1,0 B2 for 3 correct components. B1 for 2. {Stretch} {factor } {parallel to x-axis or in x-direction or horizontally} Allow ‘the y-axis is invariant’. 3 3 3(b)  y =  3g ( − x ) B2,1,0 B1 for 3 outside the function and nowhere else, or – inside the function and nowhere else. So –3g(x),  x  3(–gx), g(–3x), g   , 3g(x) +2, –3g(–x) and  − 3  3g ( −x3 ) all score B1. But 3g ( 3x ) , − g ( −x ) score B0. SC B1 for 3f(–x). 2

More questions on Functions

dy 3 2 4 The equation of a curve is such that = kx + , where k is a constant. The curve passes through the dx x 2 point S (2, 20) and the gradient of the curve at S is 65. 2 (a) Find the value of k. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The coordinates of a point T on the curve are (1, t). Find the value of t. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) [ k =] 4 B1 1 4(b)  k 4  2 x −+2 1  B2,1,0FT FT their k from 4(a) or the letter k.  x +    + c  B2 for both correct components and no other x-terms.  4   −+2 1  B1 for one correct. their k 4 2 M1 Substituting x = 2 and y = 20 into their integrated 20 =  2 − + c 4 2 expression (defined by having at least one correct power and including a '+ 'c ). Allow one slip. c = 5 A1  t =  4 A1 5

More questions on Integration

5 y O a b x 1 The equation of a curve is y = 4x 2 - x . The curve has a maximum point when x = a and crosses the x-axis at the point with coordinates (b, 0), where b 2 0 . The shaded region is bounded by the curve, the line x = a and the x-axis (see diagram). (a) Find the value of a. 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(b) Find the exact area of the shaded region. 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Mark scheme: 1 B15(a)  dy  − 2 = 2 x − 1    dx  1 2 2 x −−=1 1 2 1 0  x = 2 M1 d y Setting their of the form k x −−2 1 equal to 0 and d x 2 solving as far as x 1 = d , condoning sign errors only. Can be implied by correct final answer but not if clearly following wrong working.  a = 4 A1 Alternative Method for Question 5(a): y = 4 x − ( x ) 2 B1 Recognising the quadratic in .x −b −4 M1 Condone sign errors only. Max at x = = = 2 2 2 a −2 Allow z = x , − ( z − 2) + 4  max at 2. 2 = 2 or − ( x − 2) 2 + 4  x 1  a = 4 A1 3 5(b)  x = or b =  16 B1 SOI It may be found in 5(a), but must be seen in 5(b). Condone extra ‘solution’ x = 0. 3  4   1 2  B2, 1, 0 B2 for both correct components and no other x terms. 2  12   4 x − x dx = −  x  +  x  B1 for one correct term. ( )   2    32  Allow any correct unsimplified form. 3 3  8 2 1 2   8 2 1 2  M1 Substituting their a (from part (a)) and their b (from [Area =]   16 −  16  −   4 −  4  1  3 2   3 2  2 an attempt to solve 4 x − x = 0 ) into an integrated expression (defined by having at least one correct power) and subtracting. If correct limits and integration, then minimum 128 40 acceptable working is − . 3 3 If incorrect limits or integration, then full substitution of every term must be seen. Note: needs 0  a  b, otherwise M0, but allow limits applied either way round. Allow missing brackets if recovered.  128 40  88 1 DB1 88  = −  = or 29 Must be exact. Allow − if it becomes 88.  3 3  3 3 3 3 Do not ISW if a further area is added or subtracted. Dependent upon B1B2 scored earlier. 5

More questions on Integration

6 (a) Sketch the graph of y = 3 sin x + 2 for 0 G x G 2 r . [2] y 3 0 1 r r r 2 r x 2 2 (b) Determine the number of solutions in the interval 0 G x G 2 r of each of the following equations. (i) 3 sinx + 2 = x [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) 3 sinx + 2 = 5 - x [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (c) Solve the equation 3 sin x + 2 = 5 cos 2 x - 1 for 0 G x G 2 r . 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Mark scheme: 6(a) B2, 1, 0 B2 Fully correct curve, including 2, 5 and − 1 clearly labelled or implied on the y-axis and the curve 3 1 and horizontal only at ( ( 2 π, 5 ) 2 π, − 1) (approximately). Condone the x-axis intercepts and the stationary points not being in exactly the correct positions and y-axis not to scale. B1 for one full cycle from 0 to 2π starting on the positive y-axis and initially increasing. Note: Starting at the origin or finishing at (2π, 0) scores B0. 2 6(b)(i) 1 B1 Ignore any working out seen or graphs drawn. 1 6(b)(ii) 3 B1 1 6(c) 3sin x + 2 = 5 1 − sin 2 x − 1 M1* For replacing cos 2 x with 1 − sin 2 x. ( ) 2 DM1 For simplifying to a 3-term quadratic. The terms 5sin x + 3sin x − 2  = 0 don’t all need to be on the same side. Condone sign errors only. Candidates who miss out the ‘ −’1 can score the M1DM1 only. sin x = 0.4 and −1 A1 x = 0.41, 2.73, 32 π AWRT A2, 1, 0 A2 for three correct values and no others in the range 0 x 2π. or 0.13π, 0.87π, 1.5π A1 for 2 correct values. 59 391 3 or π, π, π Ignore  symbol if seen with answers. 450 450 2 Ignore answers outside the domain. Allow 4.71 or equivalent fractions instead of 32 π SC A1 for 23.6, 156.4 AWRT and 270 or 32 π only. SC A1 for 0.4, 2.7 and 4.7 or 32 π only. 5

More questions on Trigonometry

7 The coordinates of the points P and Q are (1, 1) and (7, 11) respectively. The line segment PQ forms a diameter of a circle. (a) Find the equation of the circle. 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(b) Find the equation of the tangent to the circle at the point Q. 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(c) The other point on the circle with x-coordinate 7 is R. Find the coordinates of the point of intersection of the tangent at Q with the tangent at R. 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Mark scheme: 7(a) [Centre is] ( 4, 6 ) B1 SOI 2 2 B1 SOI 3 + 5 2 1 1 2 Finding PQ or PQ or 2 PQ or ( 2 PQ ) . or 32 + 5 2 or 62 + 102 Allow 5.83 or 5.832 or 11.7 or 11.7 2 AWRT. or 6 2 + 10 2 oe Allow unsimplified, e.g. ( 4 − 1) 2 + ( 6 − 1) 2 . ( x − their 4 ) 2 + ( y − their 6 ) 2 = (their radius) 2 M1 Using what they think that the centre and radius are, correctly, in the equation of a circle. Do not allow P or Q as the centre or the diameter being used as the radius. ( x − 4 ) 2 + ( y − 6 ) 2 = 34 or x 2 + y 2 − 8 x − 12 y + 18 = 0 A1 May be done by other equivalent method. 2 Do not allow 34 or 5.832 for the final answer. ( ) 4 7(b) 11 − 1 5 B1 Allow unsimplified. [Gradient of PQ =] or oe 7 − 1 3 3 M1 Use of m1m2 = −1 with their gradient PQ. [Gradient of tangent =] − 5 y − 11 3 3 76 A1 ISW = − oe e.g. y = − x + x − 7 5 5 5 76 3 Condone stopping at c = if y = − x + c stated 5 5 earlier. Alternative Method for Question 7(b): dy B1 Or rearranging to form ' y = ' and differentiating 2 ( x − 4 ) + 2 ( y − 6 ) = 0 dx using the chain rule. dy dy 3 M1 Substituting x = 7 and y = 11 into a correct Substitute x = 7, y = 11  6 + 10 = 0  = − dx dx 5 d y differential equation and rearranging to give d x allowing sign errors only. y − 11 3 3 76 A1 = − oe e.g. y = − x + x − 7 5 5 5 3 7(c) y-coordinate of R is 1 B1 y − 1 3 3 16 B1 [Equation of tangent at R is] = oe eg y = x − x − 7 5 5 5 3 3 M1 Using their stated line and the equation from part (b) ( x − 7 ) + 1 = − ( x − 7 ) + 11  x = … or y = … to reach a value of x or y. 5 5 Do not allow if the gradients are the same. 46 A1 x = oe and y = 6 3 Alternate Method for Question 7(c): y-coordinate of R is 1 B1 Can be implied by correct y-coordinate of point of intersection. y-coordinate of point of intersection is 6 B1 Then either 3 M1 Using y = 6 in their equation of tangent at R. 6 − 11 = − ( x − 7 ) 5 Or ( x − 4) 2 = 34 + ( x − 7) 2 + 25 (M1) Using the right-angled triangle formed by the centre, Q and the point of intersection of the tangents. 46  46  A1 Allow 15.3 AWRT. x = ,6 oe   3  3  4

More questions on Coordinate geometry

8 The first three terms of a geometric progression are a, b and c respectively, where a, b and c are positive constants. The first three terms of an arithmetic progression are a, b and -3c respectively. (a) Show that a 2 - 10ac + 9c 2 = 0 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ It is now given that a = 9 and c takes the smaller of its two possible values. (b) (i) Find the sum to infinity of the geometric progression. 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(ii) Find the sum of the first 20 terms of the arithmetic progression. 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Mark scheme: 8(a) − 3c − b = b − a B1 For one correct equation connecting a, b and c only. or 2b = a − 3c Allow a + 2 ( b − a ) = −3c. b c or = a b or b 2 = ac oe 2 M1 Forming an equation in a and c only using both the 4ac = ( a − 3c ) or 2 ac = a − 3c oe AP and GP. Condone sign errors only. 2 2 2 2 A1 AG ⇒ a − 6ac + 9c = 4ac ⇒ a − 10 ac + 9c = 0 Convincing proof needed, e.g. ( a − 3c ) 2 expanded. 3 8(b)(i) 2 90  5184  90  72  M1 Factorising or other accepted method for solving. 81 − 90c + 9c = 0 ⇒ (9 c − 9 )( c − 1) = 0 or  =  See quadratics guidance but condone missing the 18  18  factor 9. Allow presence of a. c = 1 B1 Condone extra ‘solution’ c = 9. B0 if any error seen leading to c = 1 but all further marks are still available in this part and (b)(ii). 2 their 1 1 M1* Correctly finding a value for r from their c. r = ⇒ r = , 1 1 1 9 3 May be implied by r = or r =  or r = − . 3 3 3 2  1  or b = their 1  9 ⇒ b = 3 ⇒ r =    3  9 DM1 Use of correct sum to infinity formula with 9 and S= 1 1 − 3 their r  1. Ignore any extra working with a negative value of r. 27 A1 Do not condone extra ‘answer’. S= oe This mark is not dependent upon the first M1. 2 5 8(b)(ii) d = −6 B1 SOI 20 M1 Use of a correct sum formula with n = 20 , a = 9 and S20 = ( 2 +9 ( 20 − 1) −( 6 ) ) 2 their non-zero d stated in this part (can be implied by quoting the correct formula then substituting). But allow use of 6 without stating either the formula or ‘d =’. S20 = −960 A1 3

More questions on Series

4 1 9 The function f is defined by f ( x) = + for x 2 2 . ( 3x - 6) 2 ( 3 x - 6) 3 (a) Find an expression for fl( )x and hence determine whether f is an increasing function, a decreasing function or neither. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State whether f - 1 exists. Give a reason for your answer. 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The function g is defined by g ( )x = 4x - 3 for x 2 a . (c) Find the range of g in terms of the constant a. 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(d) Find the set of values of a for which the composite function fg exists. 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Mark scheme: 9(a) 4 −3 ( 2 ) 1 −3 ( 3 ) B1B1B1 B1 for the correct powers, B1 for ×3 in at least one + 3 4 term, B1 for all correct which can be unsimplified. ( 3 x − 6 ) ( 3 x − 6 ) −3 −4 or 4 −3 ( 2 )( 3x − 6 ) + 1 −3 ( 3)( 3 x − 6 ) Decreasing. B1* This mark is only available if f ' ( x ) is of the form ( − p )( 3x − 6 ) −3 + ( −q )( 3x − 6 ) −4 , where p and q are positive coefficients. 4 9(b) f −1 exists because f is a decreasing function DB1 Or f −1 exists because it is one-to-one, or passes the horizontal line test. 1 9(c) g ( x )  4 a − 3 B1 Allow ‘ y  ’ or ‘g ’ only. Accept 4a −3 y  or ( 4 a − 3 ,  ) . Condone ( 4 a − 3 ,   . Accept g ( x )  g ( a ) , but not if they make an error ‘simplifying’ it. 1 9(d) Either 4a −3 2 allow with x or a or  M1 Do not allow = or , unless they reach a correct inequality later. 5 5 A1 5 5  5   5  a  or a  oe Accept  a ,  a  ,  ,   or  ,   . 4 4 4 4  4   4  Or 3  ( 4 a − 3 ) − 6  0 allow with x or a or  M1 Do not allow = or , unless they reach a correct inequality later. 5 5 A1 5 5  5   5  a  or a  oe Accept  a ,  a  ,  ,   or  ,   . 4 4 4 4  4   4  2

More questions on Differentiation

10 B s r s C r A r s D The diagram shows a circle with centre A and radius r passing through points B, C and D. A larger circle of radius s has centre C and passes through B and D. The length BD is also s. (a) Show that s = 3 r . 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(b) Find an expression for the area of the shaded region. Give your answer in the form `a + b rj r2 , where a and b are constants to be found. 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Mark scheme: 10(a) 2 2 2 2 2π M1 Forming a correct equation. s = r + r − 2 r cos Allow use of degrees. 3 r s or = π 2π sin sin 6 3 π or s = 2 r  cos 6  r  2  s  2 2 or   +   = r  2   2   3r  2  s  2 2 or   +   = s oe  2   2  3 A1 AG ⇒ E.g. s = 2 r   s = 3 r Convincing proof required. E.g., for those using trig, 2 the exact value of the trig ratio substituted before the final answer is needed. 2 10(b) 2π B1 SOI, unless clearly associated with the wrong angle. [Angle BAD, BAC or CAD =] or 120 Condone 2.09 for this mark only. 3 1 2 2π B1 [Sector BAD =] r  oe 2 3 1 2 π B1* [Sector BCD = ]2 s  3 oe 3 2 1 2 DB1 Sector BCD simplified in terms of r. = πr or π r 6 2 1 2 2π r s B1* May be multiplied by 2 or by 3. Either  Area triangle ABD,ABC or ACD =  r  sin or  oe 2 3 2 2 1 2 π 3 2 Note: r  sin = r is B0DB0. 2 3 4 DB1 May be multiplied by 2 or by 3. 3 2 = r 4 1 2 π 3r s (B1* 1 2 2π 3 3 2 Or [Area triangle CBD =] s  sin or  oe Note: s  sin = r is B0DB0. 2 3 2 2 2 3 4 3 3 2 DB1) = r 4  3 π  2  12 3 − 4π  2 B1 Fully correct solution.  −  r or e.g.   r oe in the required exact form     3 π  2 6   24  Accept a = , b = − . 2 6 7

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