Cambridge A Level Mathematics 9709 — 2025 May/June Paper 3 · Variant 5

1 Solve the equation 3 4 - 2x = 5 ( 6x - 1 ) . Give your answer correct to 3 significant figures. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Use law of logarithm of a product or quotient M1 ln 5 + ln 6x−1 or ln 34−2x – ln 5 or ln 34−2x − ln 6x−1 Allow logs to any base but must be consistent throughout the equation. Use law of logarithm of a power twice M1 (4 – 2x)ln 3 and (x – 1)ln 6. Omission of bracket(s) is an accuracy error if not corrected later. Can have M0M1, i.e. go wrong with product but powers dealt with correctly. E.g. (4 – 2x)ln 3 = ln 5×(x – 1)ln 6, or (4 – 2x)ln 3 = (x – 1)ln 30 or 33x×2x written as 3xln 3×xln 2. Obtain a correct equation in any form, A1 May see (4 – 2x)ln 3 written as (2 – x)ln 9. e.g. (4 – 2x)ln 3 = ln 5 + (x – 1)ln 6 or 1.10(4 – 2x) = 1.61 + 1.79(x – 1) Obtain x = 1.15 A1 Must be 3 s.f. If no working seen, no marks available. 4

More questions on Logarithmic and exponential functions

2 Solve the equation 3 cot i - 4 cosec 2 i + 5 = 0 for - r G i G r . [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Reduce to an equation in a single trig function by use of correct trig formula(s) M1 E.g. cosec2 = cot2 +1. Obtain correct simplified solvable equation in one trig function with all terms on A1 Other equations may be possible. one side, any order, including θ, e.g. one of 4cot2 θ – 3cot θ – 1 = 0 tan2 + 3tan − 4 = 0 34sin 4 − 49sin 2 + 16 = 0 34cos 4− 19cos 2 + 1 = 0 34 2 10 −1 5 *For this equation, it is not necessary to have all terms on = 3 9 cos 2+ 3 cos2 = 0 * 34 sin ( 2− tan ( 3 ) ) one side. Solve their equation correctly, by formula, factorisation or calculator, to obtain M1 M0M1 is available if only sign slip(s) in trig formula(s). two values for a trig function, e.g. one of Incorrect solvable equation from use of correct trig formula(s) can score M1M1. 1 cot θ = 1 and – 0 tan θ = 1 and – 4 4 4 1 1 1 sin θ =  and  cos θ =  and  17 2 17 2 15  −1  5   3 cos 2θ = 0 and − *sin  2− tan    = 17   3   34 *For this equation, only one value is needed. Obtain two of answers, 14 π, − 34 π, –1.33, 1.82 A1 ISW 1 3 Accept 0.785 for 14 π and –2.36 for − 34 π. Second M1 can be implied by 4 π or − 4 π, and –1.33 or 1.82 AWRT –1.33, 1.82, 0.785, –2.36. Obtain all answers 14 π, − 34 π, –1.33, 1.82 and no others in the interval A1 ISW Accept 0.785 for 14 π and –2.36 for − 34 π. AWRT –1.33, 1.82, 0.785, –2.36. Ignore answers outside the given interval. 5

More questions on Trigonometry

4 Find the exact coordinates of the stationary point of the curve with equation y = 3x 3 ln x 4 , for x 2 0 . [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 Use the correct product rule *M1 4 d 3 3 d 4 Attempt ln x 3 x + 3 x ln x with their derivatives. (Note: may start with 12 x3 ln x) ( ) ( ) dx dx Obtain the correct derivative in any form e.g. 9 x 2 ln x 4 + 12 x 2 A1 2 4 3 4 x 3 2 2 E.g. 9 x ln x + 3 x  4 . (If starting with 12 x3 ln x, should get e.g. 36 x ln x + 12 x ) x d 4 4 May see ln x = . ( ) d x x 4 − DM1 E.g. ax4 = eb, or cx3 = ed or x = e f. Equate to zero and eliminate ln, e.g. x4 = e 3 Allow this mark even if in decimals. Allow SCB1 if incorrect sign in product rule resulting in 4 x4 = e 3 , OE. 1 − A1 ISW Obtain x = e 3 only or exact simplified equivalent 1 4  − 4  x =  e 3  scores A0.   1 − x =  e 3 scores A0. Answers with no working score no marks. 4 A1 ISW Obtain y = − only or exact simplified equivalent 4 e −1 − 3 y = 3e lne scores A0. Answers with no working score no marks. 5

More questions on Differentiation

6 The parametric equations of a curve are 2 x = and y = tan t3 , cos t3 for 0 G t G 2r . dy (a) Show that can be written as A cosec t3 , where A is a constant to be found. 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(b) Find an equation of the normal to the curve at the point where t = 1 r . Give your answer in the 12 form y = mx + c , where the constants m and c are exact. 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Mark scheme: 6(a) dy 2 B1 OE Obtain = 3sec 3t dt −1 d x M1 dx −2 d Attempt chain rule on 2 ( cos3t ) for or attempt to differentiate 2sec 3t Attempt = −2 ( cos3t ) ( cos3t ) with their derivative. d t dt dt d x A1 −2 cos3t ) sin3t . Obtain = 6sec 3t tan 3t OE, e.g. 6 ( d t Allow unsimplified, e.g. (−1)  2  (−1)  3 for 6. dy dy dt dy dt M1 2 1 Use =  with their and their Expect, for example, 3sec 3t  dx dt dx dt dx 6sec 3t tan 3t 2 1 or 3sec 3t  −2 if correct. 6 ( cos3t ) sin 3t dy 1 A1 WWW Obtain = cosec3t Must be in this form of answer given in the question. dx 2 Not required to state A = 12 . d y d x d y Allow slips in notation for , and . d t d t d x 6(a) Alternative Method for Question 6(a) 2 B1 2 x Convert to Cartesian form e.g. 1 + y = 4 d 2 d y B1 Correct use of implicit differentiation e.g. y = 2 y d y d x dy x B1 OE Obtain 2 y = dx 2 dy 2sec3t M1 Convert to parametric form e.g. 2tan3t = dx 2 dy 1 A1 WWW Obtain = cosec3t Must be in this form of answer given in the question. dx 2 5 6(b) 1 B1 Must be exact. Obtain x = 2 2 and y = 1 when t = 12 π 1 dy M1 Expect gradient of normal = − 2 if correct. Substitute t = 12 π into −1 their dx Allow if in decimals. Allow a small slip, but not with their coefficient A If their coefficient A is dealt with incorrectly M0, but allow second M1. 2 M1 E.g. y – 1 = − 2 x − 2 2 if correct or find c in equation Form equation of the normal with their (x, y), found using x = and ( ) cos3t of line. dy y = tan 3t , and −1 their Allow M1 even if decimals. dx M0 if using gradient of tangent. A1 CAO Obtain equation of normal y = − 2 x + 5 Require y = mx + c and exact m and c 2 Accept y = − x + 5. 2 4

More questions on Differentiation

8 (a) By sketching a suitable pair of graphs, show that the equation sec 2x =- 2x - 1 has exactly one 2 root in the interval 0 G x G 1 r . [2] 2 (b) Show by calculation that this root lies between 0.8 and 1.2. 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(c) Show that, if a sequence of real values given by the iterative formula 1 -1 - 2 x = cos e o n + 1 2 4x + 1 n converges, then it converges to the root of the equation in part (a). 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(d) Use this iterative formula to calculate this root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 8(a) Sketch y = sec 2x for 0 ⩽ x ⩽ 12 π M1 Need 1 or –1 and 14 π or 12 π. Ignore regions outside 0 ⩽ x ⩽ 12 π. Sketch y = –2x – 12 for 0 ⩽ x ⩽ 12 π and justify the given statement A1 Need a dot at the intersection of graphs, or dotted line parallel to the y-axis from where graphs cross to the x-axis, or state only one point of intersection OE. Do not allow, e.g. ‘only one root’. Ignore regions outside 0 ⩽ x ⩽ 12 π. Diagram for reference 2 8(b) Calculate the values of a relevant expression or pair of expressions at x = 0.8 and M1 M1 two values attempted and at least one correct in x = 1.2 f(x) = sec 2x +2 x + 12 : f(0.8) = –32.1 < 0, f(1.2) = 1.54 > 0. Can use smaller interval provided it contains root M1 four values attempted and at least three correct M0 if working in degrees 1 when comparing sec 2 x and –2 x – 2 : 1 Using 0.8: sec2 x = –34.2 –2 x – = –2.1 , 2 so –34.2 < –2.1. 1 Using 1.2: sec 2 x = –1.36 –2 x – = –2.9, 2 so –1.36 > –2.9. M1 two values attempted and at least one correct in 1 −1 −2 f(x) = cos − x : 2 4 x + 1 f(0.8) = 0.234 > 0, f(1.2) = – 0.239 < 0. M1 four values attempted (must see 0.8 and 1.2 explicitly, not just embedded) and at least three correct when 1 −1 −2 comparing x and 2 cos 4 x + 1: 1 −1 −2 Using 0.8: cos = 1.03, so 0.8 < 1.03. 2 4 x + 1 1 −1 −2 Using x = 1.2 cos = 0.961 , so 1.2 > 0.961 2 4 x + 1 Complete the argument correctly with correct calculated values A1 If accurate to only 1sf, M1A0. < 0 and > 0 or change of sign is sufficient For A1, answers must be correct to at least 2 sf 2 8(c) 1 −1 −2 1 −1 −2 M1 Need consistent variable, could be xn or xn+1. Express xn +1 = cos as x = cos 2 4 xn + 1 2 4 x + 1 1 −1 −2 1 A1 AG Rearrange x = cos to sec2 x = −2 x − with full and correct working, 2 4 x + 1 2 −2 Full working should include cos2x = or no slips allowed 4 x + 1 1 1 1 −2 x − = or cos2 x = . 2 cos2x 1 −2 x − 2 Alternative Method for Question 8(c) 1 1 −1 −2 M1 −2 1 Rearrange sec2 x = −2 x − to x = cos after full and correct working, Full working should include = or 2 2 4 x + 1 4 x + 1 sec2x no slips allowed −2 1 = cos 2x or cos2 x = . 4 x + 1 1 −2 x − 2 1 −1 −2 1 −1 −2 A1 AG Express x = cos as xn +1 = cos 2 4 x + 1 2 4 xn + 1 2 8(d) Use the iterative formula correctly at least twice (consecutive) even if only 3 M1 M0 if first value is not between 0.8 and 1.2 inclusive. decimal places M0 if working in degrees. Obtain final answer [x = or α = ] 0.992 A1 A0 if state nx = 0.992. Show sufficient iterations to at least 5 d.p. to justify 0.992 to 3 d.p. or show there A1 Iterations for each starting value: is a sign change in the interval (0.9915, 0.9925) 0.8, 1.03356, 0.98547, 0.99373, 0.99226, 0.99252, 0.99247, 0.99248 0.9, 1.1030, 0.98938, 0.99303, 0.99238, 0.99250, 0.99248, 0.99248 1, 0.99116, 0.99271, 0.99244, 0.99249 1.1, 0.97510, 0.99560, 0.99193, 0.999258, 0.99246, 0.99248 1.2, 0.96143, 0.99813, 0.99148, 0.99265, 0.99245, 0.99248 3

More questions on Numerical solution of equations

12x 2 + 55 x - 2 9 (a) Express in partial fractions. 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Mark scheme: 9(a) B C B1 Dx + E F P Qx + R State or imply the form A + + + and + B0 3 x − 2 x + 6 3 x − 2 x + 6 3 x − 2 x + 6 Values for A, B and C do not need to be substituted into this form to gain full However, can recover all marks from these marks S T + B0 and can only gain maximum M1A1 3 x − 2 x + 6 Use a correct method for finding a constant M1 Obtain one of A = 4, B = 6 and C = –5 A1 Allow maximum M1A1 for one or more ‘correct’ values or F = − 5 or P = 6 S T after B0, even if from + or S = 6 or T = –5 3 x − 2 x + 6. D = 12 E = − 2 F = − 5 P = 6 Q = 4 R = 19 S = 6 T = –5 Dx + E B Qx + R C A1 Obtain a second value from = A + or = A + 3 x − 2 3 x − 2 x + 6 x + 6 Obtain a third value A1 Alternative Method for Question 9(a) Divide numerator by denominator and reach quotient of 4 and remainder of M1 Or by inspection Px + Q P or Q ≠ 0 −9 x + 46 −9 x + 46 A1 Obtain 4 + or 4 + 2 ( 3 x − 2 )( x + 6 ) 3 x + 16 x –12 D E B1 State or imply their remainder is of form + 3 x − 2 x + 6 Values for D and E do not need to be substituted into this form to gain full marks 9(a) Obtain one of D = 6, E = –5 A1 Obtain a second value A1 5 9(b) Use the correct method to find the first two unsimplified terms of the expansion M1 E.g. –2−1 – 2−2 (3x), or 6−1 − 6−2(x), or 1 + 32 x or 1 − 16 .x of ( 3 x − 2 ) −1 or ( x + 6 ) −1 or (1 − 32 x ) −1 or (1 + 16 x ) −1 Obtain correct unsimplified expansions up to the term in x2 of each partial A1FT The FT is on B and C, fraction B  3  3  2  C  1  1  2  B = 6 C = − 5 A = 4 A1FT e.g.  1 + x +  x   +  1 − x +  x   OE. −2  2  2   6  6  6   1 157 1463 2 A1 OE − x − x Do not ISW, e.g. multiplication through by 216. 6 36 216 Allow terms in any order. 4

More questions on Algebra

10 With respect to the origin O, the points A, B and C have position vectors given by OA = 2i - j - 6k , O B = b i - 2j + 3k and O C =- 4 i + 5 j - 2k . (a) It is given that AB = BC . Find the value of b. 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(b) A, B, C and D are the vertices of a rhombus. Find the position vector of D. 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(c) Calculate angle ABC. 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Mark scheme: 10(a) For reference A = (2, −1, − 6), B = (b, − 2, 3), C = (−4, 5, − 2) Allow column vectors throughout. Allow coordinates throughout except in 10(b). Allow their notation for vectors throughout Carry out a correct method for finding AB or BC M1 E.g. AB = ( b − 2 ) i + ( −+2 1) j + ( 3 −−6 ) k Allow if use BA for AB or CB for BC or BC = ( −−4 b ) i + ( 5 −−2 ) j + ( −−2 3 ) k. Correct method to form equation with their AB = their BC M1 May see ( b − 2 ) 2 + 12 + 9 2 = ( −−4 b ) 2 + 7 2 + 5 2 allow one further slip. Note that M0M1 is possible. Obtain [b =] − 13 A1 3 1110(b) Find OA + their BC or OC − their AB M1 E.g. ( 2 − 3 ) i + ( −+1 7 ) j + ( −−6 5 ) k OD = OC + BA M1 or ( −4i + 5 j − 2k ) − ( ( b − 2 ) i + ( −+2 1) j + ( 3 −−6 ) k ) OD = OC − BA M0 7 = ( −+4 3 ) i + ( 5 + 1) j + ( −−2 9 ) k. OD = OA + BC M1 May equate the midpoint of AC and BD to find OD: OA + OC OB + OD . OD = OA − BC M0 12 ( ) = 12 ( ) Incorrect order of vertices scores M0. Obtain  i + 6 j − 11k A1  – 53   – 53 i   OD =  − 53     Allow 6 but not 6 j .         −11 − 11k     OD = + 53 i − 6 j + 11k scores M1 A0. 5 OD = ( – 3 ,6, −11) scores M1 A0. 2 710(c) Carry out correct process for evaluating the scalar product of their  BA and M1 E.g. ( 3 − 113 ) + (1  7 ) + ( −−9 5 ) , their  BC 77 or − + 7 + 45 or 391. 9 9 Using the correct process for the moduli, divide the scalar product by the M1 391 391 product of the moduli for their pair of vectors and obtain cosine of angle (allow 9 9 unsimplified form as in above scalar product) E.g. cosine of angle = or AB BC AB AB  7 11   −  + (1  7 ) + ( −−9 5 )  3 3  = 49 121 + 1 + 81 + 49 + 25 9 9  391    9 391  = = 0.4968  787 787    9  AB = BC , so may be expressed differently. Obtain answer 60.2 or 1.05 A1 3

More questions on Vectors

11 The variables x and y satisfy the differential equation 2 d y 3 y ( x + 3 ) = e ( x - 2) . d x It is given that y = 0 when x = 0 . Solve the differential equation, and find the value of y when x = 2 . 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Mark scheme: 11 Separate variables correctly B1 1 Sight of sufficient for f(y) in f(y)dy = g(x)dx to obtain e 3 y B1. 1 −3 y B1 Obtain term − e 3 1 2 B1 Separate fractions and obtain term ln ( x + 3 ) 2 Separate fractions and obtain term of the form a tan −1 bx M1 2 −1 x A1 OE Obtain term − tan 3 3 Use x = 0, y = 0 to evaluate a constant or as limits in a solution containing terms M1 −1 mx of the form a e 3 y , b ln ( x 2 + 3 ) and c tan Obtain correct solution in any form relating x and y A1 1 −3 y 1 2 2 −1 x 1 1 E.g. − e = ln ( x + 3 ) − tan − − ln 3. 3 2 3 3 3 2 1 1 Constant − − ln3 may be – 0.883. 3 2 Obtain – 0.331 A1 OE AWRT. E.g. –0.33084… Note A0A1 is possible. 8

More questions on Differential equations