Cambridge A Level Mathematics 9709 — 2025 Feb/March Paper 1 · Variant 2

1 A curve has equation y = 5 + 3x - 2x 2 and a straight line has equation y = kx + 13 , where k is a constant. Find the set of values of k for which the curve and the line do not meet. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 2 2 B1 OE [kx + 13 = 5 + 3x − 2 x ] 2 x + ( k − 3) x + 8  = 0 Eliminate y to obtain a three-term quadratic. Use of b 2 − 4ac  0 or b 2 − 4ac = 0 with their coefficients of their new M1 OE quadratic equation. Condone  errors only. Use of ‘ ’0 scores M0, unless recovered. –5 and 11 A1 Identification of correct critical values, may only be seen in their final answer. −5 k  11 A1 CWO Do not allow ‘or’. A0 if ⩽ sign or signs used. 4

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2 y M P O x 2 5 The diagram shows the curve with equation y = 2 x - + 3 . The curve crosses the x-axis at the point x P (1, 0) and M is a minimum point. (a) Find the gradient of the curve at P. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the coordinates of M. Give each coordinate correct to 3 significant figures. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 5 B1 OE 4 x + x 2  d y  B1 FT Correct use of x = 1 in their two-term differentiated = 9    d x  expression, defined as an expression with one correct power. 2 2(b)  5  M1 to zero, where Their  4 x + 2  = 0 and valid method as far as ' x = ...' Equate their derivative of the form Ax B2  x  x A, B ≠ 0, and solve. If no working is seen, this can be implied by a correct answer for x. x =−1.08 A1 AWRT y = 9.96 A1 AWRT 3

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3 43 (a) Find the complete expansion of b2x - l . [4] x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 2 3 4 (b) Hence determine the coefficient of x2 in the expansion of `x + 5bj2x - l . [2] x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 4  3  4 B1 May be seen in a full expansion. ( 2 x ) and   4 81  x  This can be implied by 16x and + 4 unless they are x 3 clearly using + throughout. x 2 3 B1 Correct combination of numerical coefficients for the 3  −3  2  −3   −3  4 ( 2 x )   + 6 ( 2 x )   + 4 ( 2 x )   middle three terms.  x   x   x  Can be implied by a correct full expansion.  16 x 4    + k1 x 2 + k 2  x 0  + k3 x −2  +81x −4  M1 OE Powers now simplified correctly with their k1 , k 2 , k3  0. 16 x 4 − 96 x 2 + 216 − 216 x −2 + 81x − 4 A1 OE 4 3(b) Use of (their 216) + 5×(their –96) only, to arrive at the coefficient of x 2 M1 Other terms may be seen. –264 A1 Accept −264 x 2 as the final answer. 2

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4 A B 4 cm 4 cm C D 6 cm 0.8 rad 6 cm O The diagram shows a triangle OAB where OA = OB = 10 cm and angle AOB = 0.8 radians. Points C and D on OA and OB respectively are such that the arc CD is part of a circle with centre O and radius 6 cm. The shaded region is bounded by the arc CD and the line segments CA, AB and BD. (a) Find the perimeter of the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the area of the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 6  0.8 B1 45.8 Accept  12π. 360 2 2 2 M1 Allow angles correctly converted to degrees for this mark. AB =10 + 10 − 2  10  10cos0.8 or AB = 2 (10sin0.4 ) or 0.8 rad = 45.8, OABˆ = OBAˆ = 67. 1. 10sin0.8 AB =   π − 0.8 = 1.17  π − 0.8  sin    2   2  This mark can be implied by AWRT 7.8. 20.6 A1 AWRT 3 4(b) 1 2 B1 [Area of sector =]  6  0.8 2 1 2 M1 OE [Area of triangle =]  10  sin0.8 or 10sin0.4 10cos0.4 2 Allow use of their value of  or 12  in degrees. or other complete method. 21.5 A1 AWRT 3

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5 An arithmetic progression has first term 5 and common difference 6. For this progression, find the sum of all the terms that lie between 150 and 400. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 Attempt to solve either: M1 Attempt to determine positions of first and last terms involved 5 + ( n − 1) 6 = 150 or 5 + ( n − 1) 6 = 400 151 401 A1 OE and Can be implied by 25 or 26 and 66 or 67. 6 6 Correct use of Sn formula with their 66 and their 25 M1 S 66 or 67 − S 25 or 26 M1 S66 – S25 A1 11275 A1 Alternative Method for Question 5: Attempt to find new a and l for reduced series M1 155 and 395 A1 their 395 = their 155 + ( n − 1) 6 M1 Attempt to find n, which results in 40, 41 or 42. n = 41 A1 CWO their 41 M1 OE Stheir 41 = ( their 155 + their 395 ) 2 11275 A1 6

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6 y r x O The diagram shows a circle C of radius r, where x 2 0 and y 2 0 for all points on C. The least distance between any point on C and the x-axis is 8 units, and the least distance between any point on C and the y-axis is 5 units. (a) State the coordinates of the centre of the circle in terms of r. [1] ............................................................................................................................................................ (b) Given that the distance between the origin and the centre of the circle is 15 units, find the value of r. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) The point on the circle furthest from the origin is denoted by P. Find the gradient of the tangent to the circle at P. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) ( r + 5, r + 8 ) B1 OE Allow x = r + 5, y = r + 8. If values are stated without reference to x and y, take the first value to be their x. 1 6(b) 2 2 2 B1 FT OE their ( r + 5 ) + their ( r + 8 ) = 15 Following their answers to (a), which must both contain r.  r + 17 )( r − 4 ) = 0  M1 Or other valid method of solution for their three-term  r 2 + 13r − 68  = 0    ( quadratic.  r = 4 A1 CWO r = −4 r = 4 scores A0. Special Case: After B1M0, r = 4 scores SCB1, but after B1M0, r = −4 r = 4 scores B0. 3 only.6(c) their ( r + 8 ) ( their r from ( b ) ) + 8 M1 r  0 from (a) with their r from (b), or their ( r + 5 ) ( their r from ( b ) ) + 5 3 A1 FT their ( r + 5 ) ( their r from ( b ) ) + − OE, i.e. − or − 8. 4 their ( r + 8 ) ( their r from ( b ) ) + 5 2

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2 4 2 2 8 sin i - 5 sin i 7 (a) Show that 3 tan i + 5 sin i / . [3] 1 - sin 2 i ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence solve the equation 3 tan 2 i + 5 sin 2 i = 9 for 0° 1 i 1 270° . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) 2 sin 2  M1 Use of tan  = 2 cos  Relevant use of cos 2 = 1 − sin 2  at least once M1 8sin 2 − 5sin 4  A1 AG 2 All necessary detail needed. 1 − sin  3 7(b) Attempt to solve their 5sin 4 − 17sin 2 + 9 = 0 using a “correct” method *M1 Allow  errors in arriving at their quadratic in sin 2 . This can be implied by either sin 2 =  or 0.6559.  2.744 A1 Condone inclusion of sin= 1.65 for this mark. 17 − 109 sin=  0.810 or 10 17  109 Allow . 10 This mark can be implied by correct values. Their 54.1, and 180 − their 54.1 or 180 + their 54.1 DM1 A correct method for obtaining a second angle within the range 0  their 54.1  90. 54.1, 125.9, 234.1 A1 AWRT A0 for additional values between 0 and 270. 4

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8 A geometric progression is such that its second term is - 120 and its sum to infinity is 160. (a) Find the common ratio. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The first nine terms of the progression are now removed. Find the sum to infinity of the remaining terms of the progression. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) a B1 ar = −120 and = 160 1 − r 120 1 a M1 Elimination of either a or r. −  = 160 or = 160 r 1 − r 120 Condone  errors for this mark. 1 + a 2 2 A1 OE 4r − 4r − 3 = 0 or a − 160a − 19200  = 0 Rearrange to arrive at a three-term quadratic. 1 A1 r = −only2 4 8(b)  a =  240 B1 FT −120  ( their r ) , where −1 r 1, r  0.  1 9  M1 With (their 240) and ( their r ) as long as −1 r 1, r  0. 240  1 −−( )   2  Condone reversed subtraction. 160 −  1  1 −−   2  Alternative Method 1 for first two marks of Question 8(b)  a =  240 B1 FT −120  ( their r ) , where −1 r 1, r  0. 1 9 M1 Correctly using the 10th term as ‘a’ and the sum to infinity. 240 −( ) 2 With (their 240) and ( their r ) as long as −1 r 1, r  0.  1  1 −−   2  Alternative Method 2 for first two marks of Question 8(b) 15 B1 FT 8 [10th term =] − , 0.46875 OE −120  ( theirr ) , where −1 r 1, r  0. 32 15 M1 With ( their a and r ) , where −1 r 1, r  0. − 32  1  1 −−   2  5 A1 5 − or − 0.3125 A0 for −0.313 without sight of − or −0.3125. 16 16 3 Condone use of r = to provide a second solution. 2 3

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d 2 y 6 5 1 9 A curve is such that = - . It is given that the curve has a stationary point at b , 9l. dx 2 x 4 x 3 2 d 2 y (a) Use the expression for to determine whether the stationary point is a maximum or a dx 2 minimum point. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the equation of the curve. [7] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 9(a) 6 5 M1 1 and evaluate second derivative. − Substitute x = 2  1  4  1  3      2   2  96 − 40  = 56   0  Minimum A1 CWO Evidence and conclusion. d 2 y SC B1 for 2  0 without sight of 96 − 40 or 56. d x 2 9(b)  6 −3  5 −2  B1 B1 OE  x  − x   + c1   −3  −2  B1 for each correct { }. −3 1 into two terms of an integrated expression −2 M1 Substitute x = 2 6  1  5  1  0 = −     + c1 (at least one correct power), now with c1, and equate to 0 to −3  2  −2  2  find c1. c1 = 6 A1 k1 x −2 + k 2 x −1 + k3 x  +c 2  M1 d y Integration of their to produce at least two terms with d x correct powers, k1 , k 2  0. 1 5  1  M1 OE 9 = − + 6 + c2 1   2 , 9 ) into integrated expression (at least two  1   1   2  Substitute ( 2 2      2   2  correct powers) to find c2. −2 5 −1 A1 OE y = x − x + 6 x + 7 Condone their final answer being c2 = 7 if a completely 2 correct simplified expression for the equation containing c 2 has been stated previously. 7

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10 y Q B A P x O The diagram shows the curve with equation 1 y = 4 ( 3x + 4) 2 - 2x - 6 for values of x such that 0 G x G 7 . The tangent to the curve at the point P (7, 0) meets the y-axis at the point Q. Region A is bounded by the curve and the two axes. Region B is bounded by the curve, the line segment PQ and the y-axis. (a) Find the area of region A. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the area of region B. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 10(a)  2 4 3   2 x 2  B1 B1 B1 for each correct { }.   ( 3 x + 4 ) 2  − − 6 x   3 3  2   8 3 2   8 3  M1 Correct use of 7 and 0 in an expression with at least two 2 − 7 −6 7 − 21 + 4 ) 4 )  A =   (   ( 2  terms with two correct powers.  9   9  13 A1 4 10(b)  − 1  B1 2  3 − 2 3 x + 4 )  2 (     dy  *M1 dy y − 0 =  their with x = 7  ( x − 7 ) Using x = 7, their value for and then any form of the  dx  dx equation of a straight line using ( 7, 0 ) . Either:  28  DM1 Use x = 0 in their equation of PQ.  Q is   0,   5  1  28  DM1 [Area of OPQ =]  7  their  2  5   1  28   33 A1 FT Only FT, following B0M1DM1DM1, from their (a) if 7 = 98  their  − (  their 13 from ( a ) )  their 13 from ( a ) )  and the area is > 0.  2  5   5 ( 5 Or:     DM1 their − 4( x − 7) dx        5    4  x 2   DM1 Evaluating their  −  – 7 x   with limits 7 and 0  5 2      10(b) 33 A1 FT Only FT, following B0M1DM1DM1, from their (a) if   = ( their area of OPQ ) − ( their 13 from ( a ) )   5 ( their 13 from ( a ) )  98 and the area is > 0. 5 5

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