Mathematics 9709/22 October/November 2025: question paper and mark scheme

Mark scheme

1 Apply relevant logarithm property *M1 3x + 5 4 A1 Allow recovery from an incorrect log statement. Obtain correct equation without logarithms = e or equivalent x−2 Solve equation to find exact value of x DM1 Must be in correct form, condone one sign slip. 2e 4 + 5 A1 Or exact equivalent. Obtain e4 − 3 4

Mark scheme

2 Attempt identity involving tan 2 θ and sec 2 θ to obtain quadratic equation in secθ or *M1 cosθ Obtain 2sec2 θ + 3secθ − 20 = 0 or 20cos2 θ − 3cosθ − 2 = 0 A1 Solve three-term quadratic equation to obtain at least one value of θ DM1 Obtain at least two values of ±66.4 and ±104.5 A1 Or greater accuracy 66.421…, 104.477… Obtain all four values and no others between –180 and 180 A1 Or greater accuracy. 5

Mark scheme

3(a) Solve 3x − 4 = 2x + 5 to obtain x = 9 B1 Or inequality. Attempt solution of linear equation where 3x and 2x have different signs M1 Or inequality. Obtain x = − 15 A1 Conclude − 15 x 9 A1 1 Now using ⩽ signs, allow x − and x 9. 5 Alternative Method for Question 3(a) State or imply non-modular equation (3x − 4)2 = (2 x + 5) 2 B1 Or inequality. Attempt solution of three-term quadratic equation or inequality M1 5x2 − 44 x − 9 = 0 Obtain values − 15 and 9 A1 Allow if one critical value is rejected. Conclude − 15 ≤ x ≤ 9 A1 1 Now using ⩽ signs, allow x − and x 9. 5 4 3(b) State 70.01N = 9 B1 FT Or inequality; following their positive upper limit from (a). Use logarithms to find value of N M1 Allow non-integer value 112.915… so far, or exact equivalent. If using trial and improvement, must show N = 113 > 9 and N = 112 < 9. Conclude with single integer 112 A1 3

Mark scheme

4(a) Carry out division as far as x 2 ± 10 x M1 Or equivalent method, e.g. inspection. Obtain quotient x 2 − 10 x + 17 A1 WWW Confirm remainder is –11 A1 WWW AG – necessary detail needed. 3 4(b) State or imply that equation is ( x 2 + 3)( x 2 − 10 x + 17) = 0 B1 FT Following their quotient from part (a). Attempt solution of their three-term quadratic quotient from (a) to give two exact M1 roots Obtain 5 ± 8 or 5 ± 2 2 and no other solutions A1 Do not ISW. 16 ± 2 2 gets A0. 3

Mark scheme

5(a) Differentiate to obtain −8sin 2x + 8cos x B1 Attempt to solve equation of form k1 sin 2 x + k2 cos x = 0 using correct identity *M1 Must have attempted differentiation. Obtain at least sin x = 12 A1 Attempt to find both coordinates of at least one maximum point DM1 Obtain ( 16 π, 6) and ( 56 π, 6) and no others between 0 and π A1 Angles must be in radians. 5 5(b) Integrate to obtain expression of form k1 sin 2 x + k2 cos x *M1 Obtain correct 2sin 2x − 8cos x A1 Condone poor notation for this mark. Apply their x-limits correctly to evaluate area under curve between A and B *DM1 Condone use of degrees for this mark. Obtain 6 3 or exact unsimplified equivalent A1 Carry out correct process to find area of shaded region DDM1 The y-coordinates of both points must be the same. Rectangle – their 6 3. Obtain 6 ( 56 π − 16 π ) − 6 3, and hence 4π − 6 3 A1 Or exact equivalent. 6

Mark scheme

6(a) Integrate to obtain expression of form k1e 2 x + k2 e − x *M1 Obtain correct 14 e2 x − 14 e− x A1 Apply limits and arrange at least as far as 12 e2a = ... or e 2 a = ... DM1 Confirm a = 12 ln(10 + 12 e−a + 12 e−4a ) A1 AG – necessary detail needed. 4 6(b) Consider sign of a − 12 ln(10 + 12 e−a + 12 e−4a ) or equivalent for values 1.0 and 1.2 M1 May use an intermediate equation from part (a). Example 1: consider 12 ln(10 + 12 e−a + 12 e−4a ) for values 1.0 and 1.2. 1 1 1 Example 2: consider e2a − e− a − e−4a − 5 2 4 4 for values 1.0 and 1.2. Obtain –0.16… and 0.041… or equivalents and justify conclusion A1 Example 1: 1.0 < 1.1608... and 1.2 > 1.1589... (allow to 2dp and truncation), so 1.0 < α < 1.2 OE. Example 2: f (1.0 ) = −1.40... f (1.2 ) = 0.43... and change of sign so 1.0 < α < 1.2 OE. 2 6(c) Use iteration process correctly at least once M1 Obtain final answer 1.159 A1 Required to precisely 4 significant figures. Show sufficient iterations to justify answer or show a sign change in the interval A1 [1.1585, 1.1595] 3

Mark scheme

7(a) Attempt use of product rule for differentiating 5x 2 y *M1 Must have evidence of implicit differentiation. dy A1 Obtain 10 xy + 5 x2 dx dy B1 Differentiate 4e2 y and obtain 8e2 y dx dy dy dy 7 − 10 xy A1 OE Obtain 10 xy + 5x 2 + 8e2 y − 7 = 0 and hence = 2 dx dx dx 5 x + 8e2 y Attempt to find relevant value of x and substitute that and y = 0 to find gradient DM1 x=2 Obtain x = 2 and gradient 14 A1 OE 6 7(b) Attempt to show that denominator of derivative cannot be zero M1 dy Must not be using when y = 0. dx State that both terms are non-negative or equivalent and conclude appropriately A1 FT Following their denominator of form k1 x 2 + k2 e 2 y , provided that k1k2 > 0. 2