2025 A Level H2 Math 9758 Paper 1 SUGGESTED MS

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/01 Syllabus 9758 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (a) 2 35ua=+ B1 c.a.o. 3 9 20ua=+ B1 FT from 2u (b) 4 3(9 20) 5 27 65u a a= + + = + M1 s.o.i. 3 5 9 20 27 65 110a a a a+ + + + + + = M1 Know to sum 1u to 4u s.o.i. 140 20 2aa=  = A1 o.e. [5] 2 (a) Let the first term be a and common difference be d. ( )20 2 19 652 ad+= ( )28 2 27 65 302 ad+ − =− M1 Attempt to use AP sum formula s.o.i. 10(2 19 ) 65 20 190 65 ad ad += += 14(2 27 ) 35 28 378 35 ad ad += += M1 Simplify to either equation By GC, 8a= and 1 2d =− . First term 8= and common difference 1 2=− . A1 Conclude, o.e. (b) ( 2) 1 ( 1) e ee kn kn kn n u u + + +== M1 Find 1n n u u + o.e. As the ratio of subsequent terms is a constant, the sequence is geometric. A1 Ratio is a constant o.e. [5]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/01 Syllabus 9758 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 3 (a) 3 i 3 i 3 i 3 i z ++= −+ M1 Attempt to multiply by conjugate s.o.i. 2 2 3 i 1 3 i3 1 2 2z += = ++ M1 Find z o.e. 22 13 122z = + =    A1 ( ) ( ) 1 πarg tan 3 3z −== A1 Radians only Alternatively: 3i3i 3i 3i ++ = − − M1 Uses 11 22 zz zz = s.o.i. 1z = A1 ( ) ( ) 3iarg arg 3 i arg 3 i 3i  + = + − − − M1 Uses ( ) ( ) arg arg arg z w zw   =− ( ) ( ) ( )( ) 11 πarg tan 3 tan 3 3z −−= − − = A1 Radians only (b) Plotting of A correctly with modulus and argument B1 Plotting of B correctly relative to A B1 Since B is obtained by rotating A by π 2 radians in the anti-clockwise direction, the complex number represented by A is multiplied by i. Hence, the complex number represented by B is iz. B1 Alternatively (using exponential form): Since B is obtained by rotating A by π 2 radians in the anti-clockwise direction, the complex number represented by B is ππ π πi i i62 6 2e e e i z  + =  = . B1 [7]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/01 Syllabus 9758 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 4 (a) 1 213 1 (3 10) ( 1)d 2 d1 x x xy xx − + − +  + = + M1 Know to use product rule s.o.i. 33 3 1031 6 6 3 10 3 421 1 2 ( 1) 2 ( 1) xx x x xx x xx ++− + − − −+= = = + ++ A1 Simplify x-coordinate of turning point 4 3= A1 o.e. (b) 3 1 2 32 2 d2 ( 1) 3 4 d 3 d d2 ( 1) 2 ( 1) 32 d d yxx x yyxx xx + = −   + + + = M1 Reasonable attempt to differentiate wrt x When 4 3x= , d 0d y x = . 3 2 2 4d2 1 3 0.420848... 0.421 (3 s.f.)3d y xx  + =  = = M1 Value seen Hence, the stationary point at 4 3x= is a minimum. A1 M0A0 [6]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/01 Syllabus 9758 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 5 (a) Let exu= . 2 2 1 22 1 0 0 u u uuu uu + −−− −    M1 Join fraction ( )( )e 2 e 1( 2)( 1) 00 e xx x uu u −+−+    The critical value is ln 2x= . A1 Find critical value Hence ln 2x . A1 (b) ( ) ( ) 1 0 ln 2 1 0 ln 2 e 2e 1 d e 2e 1 d e 2e 1 d xx x x x x x xx − −− −− = − − − + − −   M2 M1 each integral with integrand and correct limits ln 2 0 e 2e e 2ex x x x xx−−   = − − + + + −    M1 Correct integration 2ln 4 4 e e= − + + A1 o.e. [7] ln 2 + –

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/01 Syllabus 9758 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 6 (a) d 1 d dd 1 dd MM kM kt M t M k tM =−  =− =−  M1 Variable separable s.o.i. ln e e , ekt c kt c M kt c M A A − + − =− + = = = M1 Remove modulus after integration When 0t = , 0MM= . 0e ktMM −= M1 Finds arbitrary constant When 5730t = , 0 1 2MM= . (5730) 5730 00 11 ee22 kkMM −−=  = By GC, 0.00012096809 0.000121 (3 s.f.)k == A1 o.e. (b) When 00.2MM= , 000.2 e e 0.2 kt ktMM −−=  = M1 By GC, 13304.64846 13300 (nearest100)t== A1 [6]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/01 Syllabus 9758 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 7 (a) d1 ( ln ) ln 1d y x x x x xxx − =  + − M1 A1 M1 Attempt to use product rule A1 Differentiate correctly ln x= AG (b) Volume generated ( ) e 2 1 π ln d xx=  M1 Correct integral ( ) ee2 1 1 1π ln π 2 ln dx x x x x x =  −    M1 Attempt to use by parts integration   e 1π(e 0) π ln x x x= − − − M1 Use result from (a) ( ) 3πe 2π units=− A1 Exact value (c) Volume generated ( ) 1 22 0 π(e) (1) π e d y x=−  M2 M1 for cylinder M1 for integral 313.18 units= (2 d.p.) A1 Alternatively (using Shell Method in H2 FM): Volume generated e 1 2π ln dx x x=  M2 M1 for limits M1 for integrand 313.18 units= (2 d.p.) A1 [9]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/01 Syllabus 9758 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 8 (a) Let N denote the foot of perpendicular of P to 1π . 32 11 03 ON  −       =+           , for some   M1 Find a line that N lies on 3 2 2 1 1 2 33    −+       +  =          M1 Substitute back into plane 16 4 1 9 2 2   − + + + + =  = M1 o.e. 2 1.5 1.5 ON − =  or ( 2, 1.5, 1.5)N − A1 o.e. (b) Let 'P denote the reflection of P in 1π . ' 2 OP OPON +=   M1 Attempt to use ratio theorem s.o.i. 4 3 1 ' 3 1 2 3 0 3 OP − − −           = − =                 A1 (c) Normal vector of 2π 2 2 13 1 2 4 5 3 6 −           = −  =                B1 c.a.o. Let the acute angle between both planes be  . 2 13 2 13 1 4 1 4 cos 3 6 3 6  −−               =                      M1 Dot product formula s.o.i. 4cos 85.9 (1 d.p.) or 1.50 rad (3 s.f.) 3094 =  =  A1 [9]

General Certificate of Education Advanced Level 2025 Higher 2 Mathematics 9758/01 Syllabus 9758 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 9 (a) 1sin sin dcos d y px y px yyp x −=  = = M1 Differentiate wrt x d secd cos yp pyxy== A1 AG Divide by cos y (b) 2 2 dd sec tandd yy p y yxx = M1 Differentiate correctly ( ) ( ) ( ) 2 2 22 2 2 2 3 d sec tan secd sec tan tan 1 tan tan tan y p y y p yx p y y p y y p y y = = = + = + Use identity and simplify (not needed but used to make working simpler) ( ) 3 2 2 2 2 3 d sec 3tan secd y p y y yx =+ M1 Any equivalent expression When 0x= , 0y= , d d y px = , 2 2 d 0d y x = and 3 3 3 d d y px = . M2 M1 for any two M1 for all correct 3 3 ...6 py px x= + + A1 o.e. (c) 1 22 dsin d 1 ypy px x px −=  = − M1 Find d d y x Letting 3p= , 22 d 3 1 1 d d 3 d 1 9 1 9 yy xx xx =  = −− . 3 32 2 1 1 d 3 1 27 3 ... 3 ...3 d 6 3 219 x x xxx  = + + = + + −  M1 Differentiate