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: