2024 FMSS AMATH P2 PRELIM MS

1 Sec 4 Add Math Preliminary Exam 2024 P2 Marking Scheme Qn. Solution Marks AO 1 xx ee 2 1 )3(2 =− x ey 2 1 Let = , 2 2 11 22 2( 3) 2 6 0 (2 3)( 2) 0 3 or 22 3 (rej. as -ve) 22 1 ln 22 xx yy yy yy yy ee x −= − − = + − = =− = =− = = 2ln 2 1. 39 (3 s.f) Therefore, the equation has only one solution where =1.39 x x = = M1 (substitution) M1 (factorization) A1 (reject 3 2− ) (Did not award marks for students who squared both sides and could not justify why they rejected one answer when they ended up with 2 answers) AG1 3

2 2 3 2 3 2 3 5 2 12 5 2 12 2 12 55 2Grad of normal = 5 5Grad of tangent = 2 5 2 At 1, 2 12(1)55 2 At 1, 2 2 (1) 2 ------ (1) 5At 1, 2 5 3 (1) ------ (2)2 5 6 Substituting y ax b dy axdx yx yx yx dy dx x y y xy ab ab dyx dx a a a =+ = += =− + =− + − = = =− + = == =+ += == = = 5 into (1)6 5 26 7 6 b b = += = B1 (find dy dx ) B1 (grad of tangent) M1 (find y = 2) M1 (for either (1) or (2)) A1 (for a) A1 (for b) 2

3 3a 2 3ln(2 )xy x= ( ) 2 22 4 3 3(2) 6 ln 22 3 6 ln 2 3 (1 2ln 2 ) x x xdy x dx x x x x x xx  −= −= =− M1 (Apply quotient rule) M1 (Able to diff 21ln 2 2x xx== .) A1 1 3b 3 3 2 13 3 2 23 2 2 33 2 2 3 6 3ln 2ln 2 d 1 2 ln 2ln 2 d 2ln 2 ln 2 1d 2 ln 2 1 ln 2 1d 22 xx x Cx x x xx x Cx x x xx xCx x x xx xCx x x  − = +  − = + =− − + =− + +     M1 (reverse differentiation – must include + C) M1 (integrate 3 1 x ) A1(must include + C) (Whole question will only deduct once if they did not put + C) 2

4 4a f ( ) 1xx− 2 31x ax x− +  − 2 ( 1) 4 0x a x− + +  Since it is always positive for all real values of x, the graph of the curve 2 ( 1) 4y x a x= − + + lies entirely above the x-axis  the equation 2 ( 1) 4 0x a x− + + = has no real roots Discriminant, 0D 2( 1) 4(1)(4) 0a+ −  22( 1) 4 0a+ −  ( 5)( 3) 0aa+ −  53 a−   M1 (form inequality) (students who equated both eqns together will not get M1 unless they explain that there are no real roots and lead to D<0) M1 (Discriminant less than zero) M1 (Factorization) A1 2 4b f ( ) 4 yx ya    = =+ 2 34x ax a− + = + 2 10x ax a− − − = Since the line 4ya=+ is a tangent to the curve, this equation has equal real roots Discriminant, 0D= 2( ) 4(1)( 1) 0aa− − − − = 2 4 4 0aa+ + = ( 2)( 2) 0aa+ + = 2 a =− M1 (equate equations together) M1 (D = 0) A1 1 5a sin( ) sin cos cos sin cos sin sin cos sin( ) 53 = 88 1 = 4 A B A B A B A B A B A B − = − = − − − M1 (make cos sinAB the subject) A1 1 –5 3

5 5b 2 2sin 2 (sec tan ) 1 sin2(2sin cos ) cos cos 1 sin2(2sin cos ) cos 4sin 4sin     