2025 A Level H2 Math 9758 Paper 2 SUGGESTED MS

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/02 Syllabus 9758 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 Section A: Pure Mathematics [40 marks] 1 (a) 8 11 5 114 5 14 7 112 9 9 4 110 c t b c t b c t b + + = + + = + + = M2 M1 for any correct equation M1 for all three 7, 3, 5c t b= = = A1 (b) Let x, y and z denote the number of cars, trains and boats Sam has. 7 3 5 113x y z+ + = M1 Set up equation Process of Guess and Check, Trial and Error, Algebraic manipulation, inequalities, o.e. M1 Any valid method seen and reason 6 cars, 7 boats, 10 trains A1 [6]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/02 Syllabus 9758 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 2 (a) 1 2 3 6 3 3 6 0 6 PQ −           = − =                 M1 Find PQ  s.o.i. 34 34 44 3433 68 QR PQ QR PQ =        = = =             M1 Attempt to find QR  o.e. 4 4 1 5 4 4 6 10 8 8 6 14 OR OQ OR                − =  = + =                       A1 (b) 1 1 2 6 2 8 66 SQ cc −           = − − =           −       M1 Find SQ  s.o.i. 23 0 8 3 0 66 6 24 36 6 0 6 66 11 SQ PQ c c cc         =   =      −    + + − = =  =  A1 (c) 1 2 1 2 3 5 11 0 11 PS −−           = − − = −                 M1 Find PS  s.o.i. 1 3 1 3 5 3 5 3 cos 11 6 11 6                 −  = −                      M1 Reasonable attempt to use dot product s.o.i. 54cos 52.7 (1 d.p.) or 0.920 rad (3 s.f.) 7938 =  =  A1 [8]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/02 Syllabus 9758 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 3 (a) 2 2 2 d 1 d sin 3 sin 3dd d sin 3 d yy y x xx y x y y x x− =  = = M1 Variable separable s.o.i. 1 cos3 3 x cy− =− + M1 Correct integration When πx= , 1y= . 141 33 cc− = +  =− M1 Find arbitrary constant o.e. 1 cos3 4 3 f ( )3 3 4 cos3 x xyx− =− −  = + A1 (bi) B2 B1 for correct maximum, minimum and y- intercept B1 for correct period and shape 1c= or 3 5c= B1 o.e. (bii) π ,3 kxk=  B2 B1 for equation B1 condition for k (biii) 3f( π) 4 cos3x x+= − B1 FT from (a) (c) 2 2 2 dd 2 sin 3 3 cos3dd yy y x y xxx =+ M2 M1 for attempt to use product rule M1 for implicit differentiation ( ) ( ) ( ) ( ) 2 22 2 2 2 2 3 2 d 3cos3 2 sin 3 sin 3d d 3cos3 2sin 3d y x y y x y xx y x y x yx =+ =+ A1 [13] y x O (0, 0.6) f ( )yx= 1

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/02 Syllabus 9758 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 4 (a) Full capacity of bowl 21 π(15) (45 15) 2250π3= − = M1 Time taken to fill the bowl 2250π 22510π== seconds A1 (b) 2d 2 1 π (45 ) π ( 1)d 3 3 V h h hh = − + − M1 Product rule s.o.i. ( )11π 2(45 ) π (90 3 )33 h h h h h= − − = − M1 Simplify s.o.i. ( ) d d d d d d 1d10π π(12) 90 3(12)3d V V h t h t h t = = −  M1 Substitution d5 d 108 h t = cm/s A1 c.a.o. (c) 2 d 10π 10π dd 5π V t V t tt V t c =  = =+  M1 Integrate correctly When 0t = , 0V = 0 0 0 cc= +  = M1 Find arbitrary constant When 972πV = , 2972π 5π 13.94274005 13.9tt=  = = seconds (3 s.f.) A1 (d) When 972πV = , 2 2 31 π (45 ) 972π 45 29163 h h h h− =  − = M1 Form equation 9h= or 43.456h= (reject) or 7.4558h=− (reject) M1 Must reject for M1 ( ) d d d d d d 972 1 d10π π(9) 90 3(9)5 3 d V V h t h t h t = = −  M1 Substitution d 0.7377111136 0.738d h t == cm/s (3 s.f.) A1 M0A0 [13]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/02 Syllabus 9758 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 Section B: Probability and Statistics [60 marks] 5 (a) P(RB or RY or BY or BR or YR or YB) 3 5 3 7 5 7 2!15 14 15 14 15 14 =  +  +   M2 M1 Any correct case considered M1 All cases 71 105= A1 (b) P(nth counter is red) 12 1 3 36 15 16 445 1 n n n   −=  = −  − M1 Any equivalent equation or method seen By GC, 6n= . A1 [5] 6 (a) ( )P 7 0.05X = B1 o.e. (b) ( )P 7 0.05 7P 0.05 1.8 X Z  = − = M1 Attempt to standardise s.o.i. 7 1.6448536261.8 9.960736527 9.96 (3 s.f.)   − =−  = = A1 (c) ( ) ( ) ( ) P 7 P 7 0.38 0.38P 7 0.4 1 0.05 XY Y   =  = = − M1 0.4 s.o.i. 7P 0.4 2.8Z − = M1 Attempt to standardise s.o.i. 7 0.25334710112.8 6.29062811692 6.29 (3 s.f.)   − =  = = A1 (d) ( )2P 7 0.800000107Y = M1 By GC, 3.934087593 3.93a== (3 s.f.) A1 [8]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/02 Syllabus 9758 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 7 (a) Number of ways 8! 2!2!= M1 10080= A1 (b) Number of ways to arrange the 5 odd numbers 5! 2!2!= M1 Choosing an even number and slotting in between the odd numbers 34 11    =       M1 Total number of ways 345! 2! 720112!2!    =    =       A1 (c) Number of ways 3!6! 2!2!= M1 3!6! s.o.i. 1080= A1 (d) Required probability 1 3= B1 FT from (c) if working is shown [8]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/02 Syllabus 9758 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 8 (a) The value of the correlation coefficient is not close to 1, showing that the relationship between Long jump and High jump have a weak positive linear correlation. M1 Make reference to value of r clearly This is also seen in the scatter diagram where there is little to no linear relationship seen in the data. Hence, it suggests that there is no linear relationship. A1 AG (bi) B1 Must mark end- points or B0 As x increases, t increases at an increasing rate, suggesting that x and t have a non-linear relationship but perhaps a quadratic one. B1 Non-linear o.e. (bii) (A) For t ax b=+ , 0.958r = (3 s.f.) For 2t cx d=+ , 0.990r = (3 s.f.) M1 Either one seen As the magnitude of the value of r for the model 2t cx d=+ is closer to 1, and the scatter diagram shows a non -linear correlation between x and t, 2t cx d=+ is the better model to fit the given data. A1 A0 if ‘magnitude’ was not stated o.e. (bii) (B) 21.25 0.876tx=+ B2 B1 for each term (biii) (A) When 11x= , 152.088469 152t == seconds (3 s.f.) B1 FT from (b)(ii)(B) (biii) (B) The value of r is very close to 1, showing that there is a strong positive linear correlation between t and 2x . B1 The value of 11x= lies within the data range of  2.1, 15 . B1 Must quote the interval  2.1, 15 [11] t x (2.1, 11) (15, 300)

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/02 Syllabus 9758 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jord