4E Northbrook AM P2 2026 Mark Scheme

2026 Additional Mathematics Prelim Marking Scheme 1 5tan 2 1cot tan 2 5 2 55 A A A = = = = M1 M1 A1 2(a) ( ) B BB BA BABA 2tan31 tantan3 tantan1 tantantan + −= + −=−       += += B B B B B B 2 2 2 cos sin31cos sin2 tan31 tan2 B B BB BB BB B B B 2 22 22 2 sin21 2sin sin3sin1 cossin2 sin3cos cos cos sin2 += +−= += M1 (sub in tan B) M1 (change tan to sin cos ) M1 ( use trigo identity) A1 2(b)(i) 2 2 cos 2 3 cos 1 2cos 1 3 cos 1 2cos 3 cos 0 cos (2cos 3) 0 3cos 0 or cos 2 3 11, ,2 2 6 6 xx xx xx xx xx xx     =− − = − −= −= == == M1 (use double angle formulae) M1 (factorise) A1 A1 2(b)(ii) , 2 2 6 , 3 x x   = = B1 3(a) 24 At max. speed, 0 when 2, 24(2) 0 48 0 48 dv ptdt at p p p =− == −= −= = M1 M1

2 2 48 12 max speed 48(2) 12(2) 48cm/s v t t=− = − = A1 3(b) At instantaneous rest, v = 0, 248 12 0 12 (4 ) 0 0 or 4 tt tt tt −= −= == The person changes direction, moves in the opposite direction at t = 4. So the total distance travelled by the person in the interval t = 0 to t = 7 is not obtained by finding the value of s when t = 7. M1 B1 3(c) 2 23 23 (48 12 ) 48 12 = 23 = 24 4 s t t dt tt c t t c =− −+ −+  When t = 0, s = 0, 0.c= 2324 4s t t = − 23 23 When 4, 24(4) 4(4) = 128 cm When 7, 24(7) 4(7) = 196 cm Total distance travelled = 128 128 196 = 452 cm t s t s = =− = =− − ++ M1 (obtain equation) M1 M1 A1 4(a) 0168422 =−+−+ yxyx 0164)4(2)2( 2222 =−−++−− yx 222 6)4()2( =++− yx )4,2( − C and 6=r M1 A1 A1 4(b) New centre is (2,10) Equation is 22( 2) ( 10) 36xy− + − = Distance between the 2 centres = 14 units Sum of 2 radius = 12 units Since distance between the 2 centre of circle is more than the sum of 2 radius, the 2 circles did not touch each other. M1 A1 M1 A1

5(a) 32 2 2 Plot against . Gradient Vertical-intercept d av bv d av bv d vv a b =+ =+ = = M1 A1 A1 (for a) A1 (for b) 5(b)(i) B1 (all points plotted) B1 (straight line drawn) B1 (axis labelled) 5(b)(ii) 9.39 ln ln ln ln ln ln ln Vertical-axis intercept 9.39 11968.09933 12000 (to 3 s.f.) 9.59 9.41Gradient, 8 0.8 0.025 kt o kt o kt o o o o o V V e V V e Ve V kt e kt V InV Ve k = = =+ =+ =+ == = =  −= − = [Accept 0.0025] M1 A1 M1 A1 5(b)(iii) It refers to the value of the watch on 1st January 2016. (Accept initial value of the watch.) B1

5(b)(iv) When 7, lnV = 9.57 V=14328.42 14300 t =  M1 for their line at t = 7 A1 6(a) 22 2y p px x= + − + 2 2 2 2 2 2 2 4 ( ) 4(1)( 2) 3 8 For all real values of , 0 30 3 8 8 Since discriminant 0 b ac p p p p p p p − = − − + =− −  − − − −  No real roots M1 M1 M1 A1 6(b) ( ) ( ) ( ) 2 2 3 1 1 2 3 1 0 k x x k x k x kx + + − = + − + − + = ( ) ( )( ) ( )( ) 2 2 4 3 1 0 4 12 0 2 6 0 kk kk kk − − +  − −  + −  2 and 6kk−  Since 30 3 k k + − 3 2 and 6kk−  −  M1 M1 M1 A1 7(a) 33 24)(3 xyx +− ]8)[(3 33 xyx +−= ])2()2)(()][(2)[(3 22 xxyxyxxyx +−−−+−= )3)(3(3 22 yxyx +−= M1 M1 A1 7(b)(i) 32 1( ) 02 1 1 14( ) ( ) ( ) 3 02 2 2 1 1 1 34 2 2 2 14 f ab ab ab = + + + = + =− + =− 32 ( 1) 18 4( 1) ( 1) ( 1) 3 18 19 f ab ab −= − + − + − + = −= 19 19 2 14 11, 8 ab bb ba =+ + + =− =− = M1 M1 M1 M1 M1 (substitution) A1 (for both answer)

7(b)(ii) 2 Long divisi 3 on or obse ( r[ v ) a t 5 i o o ( n 1 ] ( ) (2 ) 2 3 (2 1)(2 1) 3) 0 1 r 2 f x x x x x x x xx = − + − = − − + = = =− M1 M1 A1 8(a) ( ) 3 22 22 22 42 d1 tan 2d6 1 tan 2 (sec 2 )(2)2 tan 2 sec 2 (sec 2 1)(sec 2 ) sec 2 sec 2 xx xx xx xx xx   = = =− =− M1 – differentiate trigo M1 A1 (use of identity) 8(b) 2 2 9cos3 5sindy xxdx =− ( ) 9cos3 5sin 19 sin 3 5cos3 3sin 3 5cos dy x x dxdx x x c x x c =− = + +  = + +  3 ,82 dyx dx == 38 5 c c += = 3sin 3 5cos 5dy xxdx = + + 3sin 3 5cos 5 13 cos3 5sin 53 cos3 5sin 5 y x x dx x x x d x x x d = + + =− + + + =− + + +  When 3 15,22xy == , 15 3 3 3 cos3( ) 5sin( ) 5( )2 2 2 2 5 d d    =− + + + = cos3 5sin 5 5y x x x=− + + + M1 (integrate either trigo correctly) M1 (attempt to sub in the 2 values) M1 for c = 5 M1 M1 (attempt to sub in x and y) A1

9(a) 9 9 3 93 3 32 3 2 33 2 2 log (25 10 ) log 5 log (1 ) 25 10log log (1 )5 log (5 2 ) log (1 )log 3 log (5 2 ) log (1 ) 5 2 (1 ) 40 2 (reject) 2 xx x x x x xx xx x x x − − = − − =− − =− − = − − = − −= = =− M1 M1 (Change of base) M1 (removing log) A1 9(b) 2 1 1 2 1 1 2 2 2 15 2 15 2 15 2 15 0 (2 5)( 3) 0 5 32 ( ) 3 ln 3 xx xx xx x x ee e e e e e ee Let y e yy yy y or y reject e x ++=+ =+ =+ = − − = + − = =− = = = M1 M1 M1 A1 10(a) ( ) ( ) ( ) ( ) ( ) 1 2 2 3 2 2 3 2 2 33 22 3 2 3(2 3) 26 3(2 3) 263 22 2 3 26 2 3 4 3 26 2 3 27 2 3 9 6 k k k x dx x x k k k k −=   − =  −= − − − = −= −= =  M1 (integrate correctly) M1 (sub in values) A1 10(b)(i) ( ) 2 3 2 0 2 dy xdx x =− − = = x < 2 x = 2 x > 2 dy dx Negative 0 Negative slope It is a point of inflexion. M1 A1 M1 A1

10(b)(ii) ( ) 2 3 1 2 3 1gradient of normal = 3 dy dx =− − =− 1 3 10 (18)3 6 1 63 y x c c c yx =+ =+ =− =− M1 M1 M1 (sub in 18 and 0) A1