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 −   = +  = + = + −= =−   = − +    