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] Replace x with 3x Replace x with x – 2 Replace x with x – 6 Replace x with 3x O y x 1 2y= 2x= ( )1, 0 13, 4  − 0x= 1 f ( )y x=

5 © Millennia Institute 9758/01/PU1/EOY/24 Solution Qn Solution 5(c) [3] Centre: (a, 0) Radius: b – a Qn Solution 6(i) [2] ( ) 2 . Asymptote: 1. Turning point at 0, 2 .x bx byx xa −+= = −− Vertical asymptote is 1x a a=  = ( )At 0, 2 : 2 2 1 b b− − =  =− 6(ii) [4] 2 2 2 1 1 + .11 Asymptotes: 1 and 1. xxyx xx x y x −+= = −−− = = − ( ),0a O g( )yx= ba− y x b c O y x 1yx=− 1x= ( )0, 2− ( )2, 2 2 22 1 xxy x −+= −

6 © Millennia Institute 9758/01/PU1/EOY/24 Solution Qn Solution 6(iii) [3] 2 22 2 ln(5 )1 xx xx −+  + −− . 2 ln(5 ) is the additional curve to be ske tched. Asymptote : 5 0 5 yx xx = + − − =  = From graph, for 2 22 2 ln(5 )1 xx xx −+  + −− , 1 1.34 or 3.15 5 (3 s.f.)xx    Note: The graph of 2 ln(5 )yx= + − is to be sketched on the graph in part (ii) but is sketched separately here for clarity of illustration. O y x 1yx=− 1x= 5x= Intersection Points: ( )1.34,3.30 and ( )3.15, 2.62 ( )2 ln 5yx= + −

7 © Millennia Institute 9758/01/PU1/EOY/24 Solution Qn Solution 7(i) [2] Rg = ( )3, . Df = )0, . Since ( ) ) fgR 3, 0, D=    = , fg exists. 7(ii) [2] ( ) ( ) ( ) 2 2 e + 3 5( e + 3 2) 5 (e + f f f 1) gg x x x xx  = − − =− == ( )fg gD D , or = = −  7(iii) [2] Method 1: GC/Graph y = fg(x) fgR = ( ),4− Method 2: Mapping ( ) ( ) ( ) g fg fg gf R D , 3, ,4 R= −  ⎯⎯ →  ⎯⎯ → − = fgR = ( ),4− Note: ( ) 2 2 2(e + 1) 5 (e 2e 1) 5 e e 4 fg 2 x xx xx x −+ =− + + + =− − + = As , 0, fg( ) 4 4 is horizontal asymptote xxe x y → → → =

8 © Millennia Institute 9758/01/PU1/EOY/24 Solution 7(iv) [1] Method 1: Horizontal Line Test Since there exists a horizontal line 3y= that intersects the graph of ( )fyx= more than once, f is not one-one and 1f− does not exist. Method 2: Counterexamples 2 2 f (0) 5 (0 2) 1 f (4) 5 (4 2) 1 = − − = = − − = Since f (0) f (4) 1== , f is not one-one and 1f− does not exist. 7(v) [1] For 1f − to exists under domain ),k  , f must be a one-one function. Hence, the smallest value of 2k= . 7(vi) [3] ) ( ) ( ) 2 2 2 Let f ( ) 5 ( 2) , for 2, 52 52 2 5 2 5 Since 2, 2 5 reject 2 5 as 2 yx y x x yx yx xy xy x x y y x = = − −    − =− − − + = − − = − + =  −  = + − − −  ( ) 1f 2 5xx− = + − (1 ff ,5DR− = = −

9 © Millennia Institute 9758/01/PU1/EOY/24 Solution Qn Solution 8(i) [4] Let nT be the nth term of an AP. 8 4 2,,T T T of AP are first 3 terms (consecutive terms) of a GP. 8 4 2 7 , 3 ,T a d T a d T a d= + = + = + 42 84 3 73 TTr TT a d a d a d a d == ++= ++ ( ) 2 2 2 2 2 2 ( 3 ) ( )( 7 ) 6 9 8 7 2 2 0 20 0 (rejected) or 0 a d a d a d a ad d a ad d ad d d a d d a d + = + + + + = + + −= −= = − = Since the terms of the arithmetic series are increasing, 0d  0ad−= (shown)ad= 8(ii) [2]       20 2 ( 1)2 20 2 (20 1)2 20630 2 (20 1) ( )2 630 10(2 19 ) 630 210 3 n nS a n d S a d a a a d aa a a = + − = + − = + − = =+ = = 8(iii) [3] First term of GP = 8T of AP = ( )7 3 7 3 24ad+ = + = Common ratio of GP r = 3 ad ad + + = ( ) 33 3 3 3 + + = 1 2 10 10 124 1 2 11 2 3069 61 (or 47 )64 64 S − = − =

10 © Millennia Institute 9758/01/PU1/EOY/24 Solution Qn Solution 8(iv) [2] From part (iii), r = 1 2 Since 1 12r = , the geometric series converges. 24 4811 2 S == − Qn Solution 9(i) [2] 0 0 0 0 15 15 12 6 6 PR OR OP            = − = − = −                20 0 20 0 15 15 9 6 3 PQ OQ OP            = − = − = −                9(ii) [2] A normal vector to plane 0 20 15 15 63 45 120 300 3 15 8 20 PQR PR PQ=        = −  −           =  =  Equation of plane : 3 0 3 3 8 0 8 8 240 (shown) 20 12 20 20 PQR                =  =                      rr 9(iii) [2] Normal to plane is parallel to -axis 0 Normal to plane // 0 1 OAB z OAB   