MI 9758 2025 Prelim P1 Solutions

PU3 MATHEMATICS – Paper 1 Qn Solution 1 [4] Since is sufficiently small, sin .    Method 1: Binomial Series (MF27) ( ) ( ) ( )( ) 1 1 1 2 2 2 1 1 1 3 1 3 11 1f ( ) 3 3 131 3 1 1 13 1 1 ...3 1 1 11 ...3 3 9 1 1 1 3 3 121 2 (shown)92 3 3 ! 7          − − − − −  − =− =−  =+  =+   = + + +  = + + + + −−− − −    −−       + 1 1 1, and .3 9 27p q r = = = Method 2: Repeated Differentiation ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 22 33 f3 f 3 1 3 f 2 3 1 2 3        − −− −− −  − − − = −  − − − = − Sub 0 : = ( ) ( ) ( ) ( ) ( ) 1 2 3 f 0 3 f 0 3 0 1 3 1 9 2 2f 0 2 3 0 7 − − − ==  = − =  = − = ( ) ( ) 2 2 Using MF27, f ... 2! 11 2 11 1f ... (shown)3 9 27 27 39 xx xx    + + +  + + + 1 1 1, and .3 9 27p q r = = =

PU3 MATHEMATICS – Paper 1 Qn Solution 2(i) [3] From the question, the equation of the curve is 2 2 x px qy x ++= − . Since the vertical asymptote is 2 2.xr=  = C passes through the point ( )0, 1.5− . 2 00 1.5 32 0 2 x px q qyq x + + + +=  − =  =−− Hence, the equation of C is 2 3 ----- (1)2 x pxy x ++= − C has asymptotes with equations 1yx=− and 2x= . 1 , where is a constant.2 ay x a x = − + −  The equation of C is ( )( ) 212 321 -- (2)2 2 2 x x aa x x ay x y x x x − − + − + += − + =  =− − − Comparing (2) and (1), p = –3. 2(ii) [1] 2 3 3 7For f( ) , from graph,22 2 2.5 or 4. xxx x xx −+= −    4 2.5 7 2y=

PU3 MATHEMATICS – Paper 1 Qn Solution 3(a) [2] 1 2 1 2 2 2 2 2 d 4 1 ( ) d2 1 ( 4) 12 2 4 2 4 x x x x x x x x C C − − =   −=+    = − + −   3(b) [3] ( ) 2 3 dLet ln 2 d d1 d3 vu x x x ux vxx == == ( ) ( ) ( ) ( ) 2 33 32 33 ln 2 d 1ln 2 d33 11 ln 2 d33 11 ln 239 x x x xx xx x x x x x x x x C    =−       = − + = − +    3(c) [2] ( ) 2 2 2 2 sin 2 d1 cos 2cos sin d1 cos ln 1 cos ln 1 cos x xx xx xx xC xC + −=− + =− + + =− + +  

PU3 MATHEMATICS – Paper 1 Qn Solution 4(i) [3] Let x, y and z be the usual selling price of Blend A, B and C coffee beans respectively. 176 (2) 0.85 0.90 0.95 161.80 (2) 0.93 0.95 0.98 169.56 (3) x y z x y z x y z + + = −−−− + + = −−−− + + = −−−− From GC: Hence x = $32, y =$44, z = $100 4(ii) [3] Price difference: 169.56 161.80 7.76−= To make it more attractive: ( ) ( ) Additional loyalty discount price differe nce 32 44 7.76100 100 10.2 11 xx x x      +        = Note: x% = 100 x

PU3 MATHEMATICS – Paper 1 Qn Solution 5(i) [4] ( ) ( ) ( ) 2 3 2 2 2 11 22 31 22 31 22 322 Let 3 d 3 3 3 1 3 d2 1 3312 22 1 33 1 3 3 3 (sho d 2 d 2 wn d d )3 ux x x x x x u u u u u u xx xx u C u u C x x C u u −   = +  = + = + −= =−   = − +          = − + = + − + +     Or ( ) ( ) ( ) ( ) ( ) 1 2 2 3 2 3 2 3 2 11 22 31 22 31 22 22 dLet 3 2 d d 3 3 d 2 3 d 3 2 3d 2 1 3 d2 1 3312 22 1 3 3 3 (shown)3 uu x x x x x x u u u u u u uu u u u u u x u u C x x C − = +  = + − =  − =  − =  =−   = − +          = + − + + −        Continue to replace x in terms of u if it can not be done easily with 1 step

PU3 MATHEMATICS – Paper 1 5(ii) [3] 3 3 2 1 3 d 3 x x x− +     3 0 0 1 3 3 3 2 0 3 2 0 3 2 1 2 33 dd 33 d d3 3 original c 3 3 Part of original curve reflected in the -axis ve r u x xx xx xx xx x x x x − − − =+ ++ =+ ++   ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 03 332 2 2 2 10 33 3 9 3 3 9 3 27 9 3 64 9 4 12 9 12 3 9 3 3 3 9 3 8 18 12 12 9 12 3 3 9 3 10 6 3 3 12 6 3 10 6 3 3 4(3) 6 3 10 6 3 6 3 6 3 10 18 3 x x x x −    =− + − + + + − +       =− − − − + − − −      =− − − − + − − −    =− − + +  =− − + +     =− − + +    =− + Use GC: ( ) 3 2 3t f 3 Le xx x = + ( ) 3 2 3t f 3 Le xx x = + 10 x−   , we have ( )f0 x  03 x , we have ( )f0 x      3 3 3 1 0 2 2 0 3 Reflection in -ax 3 Part of original c is 3 ur d d ve 3 x x x x x x x − + + + −  

PU3 MATHEMATICS – Paper 1 Qn Solution 6(i) [3] 2 2 Area : 1 2 2 22 2 4 xSx xSx y y    =   +  =+ ( ) 2 2 22 22 Perimeter: : 125 2 2 2 222 2 25 2 = 25 24 25 24 325 2 (shown)4 xSx x x xx xS x x x x x x x y x y S         =   + +   = − − + − − = + − − = − − 6(ii) [4] ( ) 22 2 2 325 2 4 d3 25 4 d2 dAt stationary point, 0 d 325 4 0 2 50 8 3 0 8 3 50 50 83 d3 25 4 d2 d3 4 = 8.71 02d is maximum area S x x x S xxx S x xx xx x x S xxx S x S         = − − = − − = − − = − − = += = + = − − = − − − 

PU3 MATHEMATICS – Paper 1 Qn Solution 7(i) [2] ( ) 0 1 2 2 GP: 500, 0.6 500 0.6 500 300 0.6 0.6 500 500 0.6 180 (verified) ar u u u == = =  = =   =  = 7 (ii) [2] ( ) ( ) ( ) 2 1 1 12500 500 0.6 500 0.6 ... 500 0.6 500 1 0.6 0.6 ... 0.6 1 1 0.6 500 1 0.6 1250 1 0.6 (shown) n n n n n n n n S S S S + + = +  +  + +  = + + + +  − = − =− Week n New case, nu Total Cumulative, nS 0 500 500 1 1500 0.6 1500 500 0.6+ 2 2500 0.6 12500 500 0.6 500 0.6+  +  … … … n 500 0.6 n 12500 500 0.6 500 0.6 ... 500 0.6 n +  +  + +  7 (iii) [1] Method 1: GC ( ) 11250 1 0.6 1240n+− From GC n ( ) 11250 1 0.6 n+− 8 1237.4 9 1242.4 > 1240 Method 2: Solving Inequality ( ) ( ) ( ) 1 1 1 1250 1 0.6 1250 1 0.6 1240 12400.6 1 1250 1ln 125 1 8.45 l 0.6 9 n n n n n c S n n n + + + =− − −  −   =  ( )ln 0.6 < 0, handle the inequality sign accordingly

PU3 MATHEMATICS – Paper 1 7 (iv) [2] ( ) 1 1 1250 1 0.6 , 0.6 0, 1250 n n n n S nS + + =− → → → Since the maximum cumulative infection is 1250 < 1280, the patient capacity will be sufficient. Qn Solution 8 (i) [2] Method 1 12sin 1e ln 2sinxy y x − −=  = Differentiating both sides wrt x: 2 2 1 d 2 d 1 d1 2 (shown)d y yx x yxy x = − −= Method 2 12sine xy − = 1 1 2sin 2 2 2sin 2 d2 ed 1 d1 2e d d1 2 (shown)d x x y x x yx x yxy x − − = − −= −= 8 (ii) [4] ( ) 2 22 d14 d yxy x −=  Differentiating both sides wrt x: ( ) ( ) 22 2 2 d d d d1 2 2 8 d d d d y y y yx x y x x x x    −  + − =        ---------------------------------------------------------- Alternatively, ( ) ( ) ( ) ( ) 1 2 1 2 2 1 2 22 2 2 d12 d Differentiating both sides wrt : d 1 d d1 1 2 2 d 2 d d yxy x x y y yx x x x x x − −= − + − − = ----------------------------------------------------------

PU3 MATHEMATICS – Paper 1 ( ) ( ) ( )( ) 12sin 0 2 22 22 22 When 0, e 1 dd1 0 2(1) 2dd dd1 0 2 2 0 2 8(1)(2) 4dd x yy yy xx yy xx − = =  = − =  = −  + =  = -------------------------------------------------------------------- Alternatively, ( ) ( ) ( ) 1 2 1 22 22 2 22 d 1 d1 0 1 0 0 (2) 2(2) 4d 2 d yy xx − − + − =  = -------------------------------------------------------------------- 2 2 The Maclaurin series for : 1 2 4 ... 2! 1 2 2 ...yx y x xyx  = + + +   = + + + 8 (iii) [2] ( ) 1 0.1 0.1 2sin 0 2 0 d d 0.00571667 0.005717 e 1 2 2 (4 s.f.) xx x x xxx − = = ++