MI Prelims 2025 H2 Math Paper 2 Solutions

PU3 MATHEMATICS – Paper 2 Qn Solution 1(a) [3] Method 1 ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 , 1, f f 1 1 , f 2 1 2 x x x x ab y x y x y x a b a b → − → + = ⎯⎯⎯⎯ → = − ⎯⎯⎯→ =  − +  Method 2 ( ) ( ) ( ) ( ) ( ) 1 2 2 2 1 1 1 , , f f 2 f 2 , 2 2 2 1 xxxx a b a b yx b y a y x x →−→= ⎯⎯⎯→ = ⎯⎯⎯⎯⎯ → =     +       − P becomes 11 , 022Pp + , Q becomes 11 , 022Qq + and R becomes 1 , 2Rr . 1(b) (i) [1] The required area or II=− . Transformation: Reflection of the graph about the x-axis. The area in question remains the same. 1(b) (ii) [1] 11The required area or 22II=− . Transformations: Scaling of the graph parallel to the y-axis by scale factor 1 2 . The area in question is halved. Translation of the graph in the negative x-direction by 3 units. The area in question remains the same. 1(c) [1]  00 f ( ) f ( ) ff ( ) ()d 0 0 q q xq r xx r = = − =−  =− 

PU3 MATHEMATICS – Paper 2 2(i) [3] 2 3 2 ,4 d 1 d ,2dd 2 2 1 d 2 d 4 x t y t xy ttt t x t t t y = = + == == Note: tp= ( ) ( ) ( ) 2 3 2 2 3 22 The eqn of tangent to at : 44 4 3 4 , 4P p pC y p p x p y p x p − + − =− + = + 2(ii) [2] The eqn of tangent of C at P passes through ( )0,1 ( ) ( ) 3 22 2 At 0,1 : 1 4 0 3 4 33 1 (reject 1 0) pp p pp = − + = = −  ( ) When 1 1, 5 1,5 p xy P = == P (1,5) (0,1)

PU3 MATHEMATICS – Paper 2 2 (iii) [2] 3(a) [4] 2 6 i ... (1) i 3 i 3 ... * (2) zw z w w z z+ + = + − =  = − ( ) ( )( ) ( ) ( ) ( ) Let i. Then, * Substitute (2) into (1): 2 * i 3 6 i 1 i 2 * 9 i 1 i i 2 i 9 i i i 2 2 i 9 i 3 i 9 i i.z z z z zz a b a b a b a b a b a b a b a b z a b= + = + + − = + + + = + + + + − = + + + − + − = + − + − = + − Comparing corresponding real and imaginary parts: 39 1 Solving the equations simultaneously: 4 and 3 4 3i ab ab ab z −= −= ==  = + ( ) From (2), i 4 3i 3 6 4iw= + − =− + 4 3i, 6 4i.zw = + =− + 3(b) [4] Method 1: Use conjugate root and factorisation All the coefficients of f(x) are real. Thus, 2i− is also a root of f ( ) 0x = . A quadratic factor of f ( )x is: ( ) ( ) ( ) ( ) ( ) 2 2 2 2 i 2 i 2 i 2i 45 2ixx x x x xx − + = − − − +    −−      = − − =−  +  x y 4 O (4, 260)

PU3 MATHEMATICS – Paper 2 ( )( ) 3 2 2f ( ) 5 4 5 , where is a real constant. x x mx nx x c x x c = + + + = + − + Comparing constant term: 5 5 1cc=  = f ( ) 0 1 or 2 ixx =  =−   The coordinates of P is ( )1, 0− . Method 2: Apply Factor Theorem by substituting the given root directly. 2i+ is a root of f ( ) 0x = . f( 2 i 0)+ = ( ) ( ) ( ) ( ) ( ) 32 2 i 2 i 2 i 5 0 2 11i 3 4i 2 5 i 0 7 3 2 11 4 i 0 mn m n n m n m n + + + + + + = + + + + + + = + + + + + = Comparing corresponding real and imaginary parts: 7 3 2 0 3 2 7 11 4 0 4 11 Solving the equations simultaneously: 3 and 1 m n m n m n m n mn + + =  + =− + + =  + =− =− = 32f ( ) 3 5x x x x = − + + Using GC to solve f ( ) 0x = , 1 or 2 ix=−  . ------------------------------------------------------------- Alternatively (instead of GC), All the coefficients of f(x) are real. Thus, 2i− is also a root of f ( ) 0x = . A quadratic factor of f ( )x is: ( ) ( ) ( ) ( ) ( ) 2 2 2 2 i 2 i 2 i 2 i 2i 45 x x x x x xx − + − − = − − − +           = − − = − + ( )( ) 3 2 2f ( ) 3 5 1 4 5 . x x x x x x x = − + + = + − + f ( ) 0 1 or 2 ixx =  =− 

PU3 MATHEMATICS – Paper 2 4(a) [1] 2 1 1 2Given that , and 3, 1n n nT T T T T++ = + = = 3 2 1Sub 1, 1 3 4n T T T= = + = + = 4 3 2Sub 2, 4 1 5n T T T= = + = + = 4(b) [2] 1 12 1 1 1 1 1 1 1 1 1 1 Given 1 11 (shown) n n n n n n n n n n n n n n n n n Tr T TTr TT TTr T Tr T r r + ++ + + + + + + + + + + =  = = += =+ =+ 4(c) [3] Since nr is convergent to L, it implies that as n→ , nrL→ and 1 nrL+ → . Hence, 1 111 1 as n n r L n rL + = +  = + → ( ) ( ) ( )( ) ( ) 2 2 10 1 1 4 1 1 21 15 2 1 5 1 5 (rej as 0)22 n LL L L Lr − − = − −  − − −= = +−= 4(d) [2] Method 1   1 930 2(5) ( 1)(2) 9302 ( 4) 930 n k k u n n nn =  + −  +  From GC n n(n+4) 28 896 29 957 Least n = 29 Method 2 ( ) 1 1 930 5 2( 1) 930 n k k n k u k = =  + −    From GC n ( ) 1 5 2( 1) n k k = +− 28 896 29 957 Least n = 29

PU3 MATHEMATICS – Paper 2 5(i) [3] 5(ii) [1] Every horizontal line , where ,y k k= cuts the graph of g( )yx= at most once. g is a one-one function. 1g− exists. 5(iii) [3] ( ) ( ) ( ) ( ) ( ) 1 1 3Let g . 1 13 13 33 g11 3g 1 xyx x y x x y x y yyxy yy xx x − − −== − − = − − = − −−=  =−− −= − ( ) ( ) ( ) ( )1 g gg From graph in part , R , 1 1, D R , 1 1, − = −    = = −   (i) 5(iv) [1] For x gD , ( ) 3g 1 xx x −= − and ( ) 1 3g 1 xx x − −= −  ( ) ( ) 1ggxx −= and hence g is self-inverse. 5(v) [2] 2 1 1Since g is self-inverse, g ( ) g g( ) gg ( ) .x x x x−−= = = 21 3g ( ) g ( ) 1 xx x x x − − =−  =− − 2 2 3 1 3 3 3 xx x x x x x x −=− − − =− + = = O y x

PU3 MATHEMATICS – Paper 2 6(i) [3] Let X be the waiting time to get pastries from the popular bakery of a randomly chosen customer. ( ) 2~ N , X  Given ( ) ( )P 5 P 25 0.10XX =  = . Method 1 5 25By symmetry, 15 2 +== ( )P 25 0.10 25 15P 0.10 10 1.28155 7.80305 X Z    = − = = = -------------------------------------------------------------------- Alternatively, ( )P 25 0.10X = Using GC, 7.80304 = -------------------------------------------------------------------- 15, 8 (to the nearest minute). = = Method 2 ( )P 25 0.10 25P 0.10 25 1.28155 1.28155 25 ... (1) X Z      = − = − = += ( )P 5 0.10 5P 0.10 5 1.28155 1.28155 5 ... (2) X Z      = − = − =− −= Solving (1) and (2) simultaneously, 15, 7.80305.== 15, 8 (to the nearest minute). = = y

PU3 MATHEMATICS – Paper 2 6(ii) [1] Consider ( ) 2~ N 15, 8 .X ( )P 0 0.0303.X = (or 0.02728 if we use 7.80305 = ) The waiting times cannot be negative but ( )P 0 is about 3%X  , which is not insignificant. Thus, the normal distribution is not a suitable model. Note: For normal distribution: ( )P 3 3 0.997 i.e. 99.7%.X   −   −  Hence, whenever the probability of obtaining unreasonable values of x exceeds 0.003, we can conclude that the normal distribution is not suitable 7(i) [1] The probability of obtaining a card of one colour depends solely on the colour and not any other factors. Or The probability of obtaining a card of one colour is exactly the same as the probability of obtaining another card of the same colour in the same deck. 7(ii) [4] Method 1 ( ) 2 10 21 12 3 1P 2 (shown) 22 CCG C= = = ( ) ( ) 2 10 12 12 3 10 3 12 3 9P1 22 12 6P0 22 11 CCG C CG C = = = = = = = ---------------------------------------------------- Alternatively, ( ) 1 9 6P 0 1 22 22 11G= = − − = ---------------------------------------------------- Method 2 ( ) 2 1 10 1P 2 (shown)12 11 1 3! 2 22!0G    = = =        ( ) 3! 2! 2 10 9 1P1 12 11 10 22G    = = =       ( ) 10 9 8 6P0 12 11 10 11G    = = =       ---------------------------------------------------- Alternatively, ( ) 1 9 6P 0 1 22 22 11G= = − − =

PU3 MATHEMATICS – Paper 2 7(iii) [2] G (4 pt) S (1 pt) Score Probability 0 3 3 12 6 22 11= 1 2 6 9 22 2 1 9 1 22 Probability Distribution of X x 3 6 9 ( )P Xx= 12 6 22 11= 9 22 1 22 7(iv) [1] ( ) 10 10 6 9 1E winnings (3) (6 10) (9) 4511 22 22=  +  +  = The expected winnings is $45. 8(i) (a) [3] ( ) ( )