MI_9758_2024_Promo_PU1_Solutions

1 © Millennia Institute 9758/01/PU1/EOY/24 Solution Solution to 9758/01: Qn Solution 1(i) [2] 2 nS n kn=+ 2 1 2 ( 1) ( 1) 21 nS n k n n n kn k − = − + − = − + + − 1 22 22 ( 2 1 ) 21 21 n n nu S S n kn n n kn k n kn n n kn k nk −=− = + − − + + − = + − + − − + = − + 1(ii) [2] 1 1 2( 1) 1 2 2 1 23 2 1 (2 3 ) 2 1 2 3 2 n nn u n k nk nk u u n k n k n k n k − − = − − + = − − + = − + − = − + − − + = − + − + − = Since 1 2nnuu −−= which is a constant independent of n, the sequence is an arithmetic progression. 2 (i) [4] ( )( ) 2 2 2 2 32 24 32 204 3 2 2 8 04 56 04 16 04 xx x xx x x x x x xx x xx x −+ + −+ −+ − + − − + −− + +− + 4 1 or 6xx−  −  4− 1− 6 − − + +

2 © Millennia Institute 9758/01/PU1/EOY/24 Solution Qn Solution 2 (ii) [4] ( ) ( ) 22 3232 2 244 xxxx xx − − − +++   − − + Replace with : 4 1 or 6 4 1 or 6 1 4 or 6 xx xx xx xx − − − − −    −   − 3(i) [4] 32y ax bx cx d= + + + At (0,7): 7 d = 32 32 32 At ( 2, 11): 11 ( 2) ( 2) ( 2) 7 8 4 2 18 ..... (1) At (1,1): 1 (1) (1) (1) 7 6 ..... (2) At (3,19): 19 (3) (3) (3) 7 27 9 3 12 ..... (3) a b c a b c a b c abc a b c a b c − − − = − + − + − + − + − =− = + + + + + =− = + + + ++= Using GC, 2, 3 and 5a b c= =− =− 3(ii) [1] 32Equation of : 2 3 5 7C y x x x= − − + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 32 replace with 32 replace with 1 32 32 32 2 3 5 7 2 3 5 7 1 2 3 5 7 2 3 5 8 Equation of resulting curve: 2 3 5 8 xx yy y x x x y x x x y x x x y x x x y x x x − − = − − + ⎯⎯⎯⎯⎯→ = − − − − − + ⎯⎯⎯⎯⎯⎯ → − = − − − − − +  = − − − − − +  =− − + +

3 © Millennia Institute 9758/01/PU1/EOY/24 Solution Qn Solution 4(i) [2] Method 1 : 1: 3 : 1: 2 AC AB AC CB = = (1) (2) 12 2+ 3 += + = bac ab Method 2 : 1: 3 3 AC AB AC AB = = ( ) ( ) 3 3 33 2+ 3 OC OA OB OA− = − − = − = − + = c a b a baca abc 4(ii) [2] Method 1 1Area of triangle Area of triangle 3 11 32 11 i.e. 66 OAC OAB OA OB k = =   =  =ab Method 2 ( ) 1Area of triangle 2 1 2 + 23 1 2+6 1 26 11 i.e. 66 OAC OA OC k = =  = =  +  =  = aba a a b a a a b ab 4(iii) [3] ( ) ( ) ( ) ( ) ( ) ( ) 0 22 22 2+22 3 1 2 2 +3 1 4 2 23 1 43 1 4(1) (1) since , are unit vectors3 1 − = −  =− = − −  =− =− = aba b c a b a b a b a a + a b b a b b ab a b A B C 1 2 C A B O 1 2

4 © Millennia Institute 9758/01/PU1/EOY/24 Solution Qn Solution 5(a) [2] Sequence 1 2 2 19 x y+= ( ) 2 2 22 3 1 9 1 x y xy +=  + = ( ) 2 221xy− + = 1. Scaling parallel to x-axis by scale factor 1 3 . 2. Translation in the positive x-direction by 2 units. Sequence 2 2 2 19 x y+= ( ) 2 26 19 x y− += ( ) ( ) ( ) 2 2 22 2 2 2 36 19 32 19 21 x y x y xy − += − + =  − + = 1. Translation in the positive x-direction by 6 units. 2. Scaling parallel to x-axis by scale factor 1 3 . 5(b) [3]