2023 JC1 Promo Practice Paper B (solutions) VJC

2023 PROMO PRACTICE PAPER B Solutions Qn Solution Notes 1i 5 3yx= has a stationary point at ( )0,0 hence gradient of it’s graph at the origin should be 0. Graph of 3ln 3yx=+ extends down to infinity as x approaches 0. Equation of asymptote: 0x= and coordinates of x- intercept: ( )0.368,0 need to be stated (required by question) 1ii Graphs of 3ln 3yx=+ and 5 3yx= intersect at 0.395x= and 3.02x= . 5 33ln 3 0.395 3.02 xx x +    2 ( ) ( ) 22 d3 2 1 d yx y xy x− = −−− Differentiating w.r.t. x ( ) 2 22 2 d d d d3 6 2 2 2 d d d d y y y yx y x y y xx x x x − + − = + When 0x= , 1y= ( ) ( )( )dd0 1 2 0 1 0dd yy xx− =  = ( ) ( )( )( ) ( )( ) 22 22 dd0 1 0 2 0 0 2 2 0 0 2dd yy xx− + − = +  =−  the Maclaurin’s series for y is ( ) 2 2 210 2! 1 y x x x −= + + + = − + Differentiate (1) immediately using product rule. No need to make d d y x the subject before differentiation.

2023 PROMO PRACTICE PAPER B Solutions 3i ( ) ( ) 2 22 2 2 2 2 2 2 2 2 2 2 2 2 2 e sin d 11e sin e cos d22 1 1 1 1e sin e cos e sin d2 2 2 2 1 1 1e sin e cos e sin d2 4 4 5 1 1e sin d e sin e cos4 2 4 21e sin d e sin e cos55 1 e 2sin cos5 x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x D x x x x C x x C =− = − − −  = − − = − + = − + = − +       3ii Let ( ) 2f ( ) e 2sin cosxx x x=− . Observe that ( ) ( )( ) 22e 2sin 1 cos 1 f ( 1)xy x x x+= + − + = + is a translation of f ( )yx= by 1 unit in the negative x- direction. Hence, the gradient of the curve f ( 1)yx=+ at 12x =− is the gradient of the curve f ( )yx= at 2x = , which is given by f 2y =  . From part (i), 2f ( ) 5e sin xxx = . Hence, the required gradient is f 5e2  = .

2023 PROMO PRACTICE PAPER B Solutions 4i Asymptotes: ( ) ( ) ( ) 22 23 94 23 32 33 22 3 5 3 13 or 2 2 2 2 yx yx xy y x y x −− = −− = −= + = − =− + 4ii ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 22 22 22 22 12 48 48 6 3 0 12 4 6 3 48 0 12 4 4 48 6 9 9 3 48 0 12 2 3 12 23 112 y y ax ax a y y a x x a y y a x x a a y a x a yx a − + + − − = − + − − + = − + − + − + − − + = − + − = −− += which is an ellipse with centre ( )3,2 and vertices at ( )3,2 a+ and ( )3,2 a− For curve C and D to not intersect, 39aa   Since a is a positive constant,  :0 9aa    Modified mark allocation from 2 to 4.

2023 PROMO PRACTICE PAPER B Solutions 5i ( )( ) ( ) ( ) ( ) ( )( ) ( ) 2 122 2 12 21 222 2 24 2 244 2 24 1 122 1 2 1 2 121 1 1 12 2 2! 2 1 112 2 4 1 12 2 2 4 1 3 3 2 4 8 x xxx xx xxx xxx xxxx xx − − − + = + −− = + −   −−   = + + − − + − +        = + + + +  = + + + + +  = + + + Refer to MF26 and apply the standard series ( ) ( ) 2111 2! n nnx nx x −+ = + + + 5ii For expansion to be valid, 22 1122 xx−    ( ) ( )( )   2 2 2 2 2 2 2 since 0 2 2 0 : 2 2 x x x x x xx xx   =  − +   −   Note that the standard series ( ) ( ) 2111 2! n nnx nx x −+ = + + + is only valid when 1x  . Since the above standard series is applied to 12 1 2 x − −  , hence the expansion is only valid if 2 12 x− . 5iii ( )( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 4 2 22 3 22 3 3 3 22 322 222 d 1 d 1 3 3 d 2 d 2 4 8 2 2 1 2 6 12 482 4 2 2 2 6 12 482 6 1262 48 12 44 x x x x x x x x x x xx x x x x x x x x xxxx xx − −    + = + + +   −    − − + − = + + − − + + = + + − − = + + − = + +

2023 PROMO PRACTICE PAPER B Solutions 6 g( ) 2 ax byx xc +== + Vertical asym : 3 322 cxc=− =  =− Horizontal asym : 242 aya= =−  =− y-intercept : 4 43 byb c= =−  = 44g( ) 23 xyx x −+ = = − 1f1 2yx =− ( )11f 2 1 f22y x x   = + − =       ( ) ( )1f 2 f2y x x== The graph of 1f1 2yx =− is translated 2 units in the negative x-direction and then stretched parallel to the x-axis by factor 1 2 with y-axis invariant. ( ) ( ) ( ) ( ) ( ) 1 4 4f1 2 2 3 4 2 41f 2 2 2 3 44 21 4 2 4f 2 2 1 84 41 xx x xx x x x xx x x x −+ −= − − + + = +− −−= + −−= + −−= + Replace x by x+2 Replace x by 2x

2023 PROMO PRACTICE PAPER B Solutions Alternatively, 1f1 2yx =− ( ) ( )1f 2 1 f 12y x x= − = − ( )( ) ( )f 1 1 fy x x= + − = The graph of 1f1 2yx =− is stretched parallel to the x-axis by factor 1 2 with y-axis invariant and then translated 1 unit in the negative x-direction. ( ) ( ) ( ) ( ) ( ) ( ) 1 4 4f1 2 2 3 4 2 4f1 2 2 3 84 43 8 1 4f 4 1 3 84 41 xx x xx x x x xx x x x −+ −= − −+−= − −+= − − + += +− −−= + 7a 66 7 7 1sin 2 cos 2 d 2sin 2 cos 2 d2 1 cos 2 27 cos 2 14 x x x x x x x C x C =− − =− +  =− +  This is of the standard form ( ) ( )f f d n x x x  where ( )f cos2xx= 7bi ( ) 22 2 22 3 3 2dd2 2 2 3 ln 22 3 ln 2 ( 2 0)2 xx xxxx xC x C x =++ = + + = + + +   Replace x by 2x Replace x by x+1

2023 PROMO PRACTICE PAPER B Solutions 7bii ( ) ( )( ) ( )( ) ( )( ) ( )( ) 2 2 2 22 2 2 2 2 2 3 1 32 3 1 32 32 2 4 3 8 3 32 5 8 1 32 x x xx xx xx x x x xx xx xx + + + −++=−+ −+ + + − −= −+ −+= −+ biii ( )( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 2 0 2 2 2 0 2 2 0 2 22 0 2 21 0 1 1 10 16 2 d 32 5 8 12d 32 2 3 12d 32 2 3 12d 3 2 2 312 2ln 3 ln 2 tan2 22 3 1 22 2ln1 ln 6 tan2 22 312 2ln 3 ln 2 tan 02 2 xx x xx xx x xx x xxx x xx x x xxx − − − −+ −+ −+= −+ +=+  −+ = + + − + +  = − + + +   = + +  − + +     ( ) ( ) 1 1 3ln 6 2 tan 2 4ln 3 3ln 2 2 tan 2 ln 3 − −  = + − − =−

2023 PROMO PRACTICE PAPER B Solutions 7c 21 sin3x =  d2 sin cosd3 x  = When x = 0,  = 0; when x = 1 4 ,  = 3  . 1 4 0 d13 x xx− = 23 2 0 1 sin 23 sin cos dcos 3        = 23 0 2 sin d 33   = 3 0 2 1 cos 2 d233   −  = 3 0 11 sin 2233   − = 13 3433  −  8ai ( ) 33 1i1i 22*2 21 i 3 1 i 3 z −−−−= = = = − − ( ) ( ) ( ) ( ) 3 1i*arg arg 1 3i 3arg 1 i arg 1 3i 313 43 23 1 12 12 z   −−=  − = − − − −    = − − −       =−  1 1 1 1 * 2zz z = = = ( ) ( ) 1 23 *arg arg arg 12zzz  =− = =− or 1 12

2023 PROMO PRACTICE PAPER B Solutions 8aii 44 23 123 iii 3312 4 1 i2 i 2 i 3 4 1 1 1 1 1 e e e 442 11e e e e 4 a b a b zz z   −− + = = = =  =   = Therefore we have 2 1 1 1e 2 ln ln or ln 24 4 2 a aa=  =  = − 1 ii 3 1ee 3 b b  =  = 8b i 3 i 3u v v u− =  = − Then substituting i3wz=− into the other equation, ( )( ) ( ) 2 * 1 i i 3 7 4i * i 3 i 3i 7 4i * i 10 i 2 i i 10 i 2 i 10 i uu u u u u u u a a b a b a + − − = + + − − + = + + + = + + + = + − + = + Comparing the real and imaginary parts, we get 1a= and 2 10 2 10 8a b b b− =  − =  =− . Therefore 1 8iu=− and ( )i 1 8i 3 5 iv= − − = +

2023 PROMO PRACTICE PAPER B Solutions 9 2 1 2 xxy x  ++= + ( )( ) ( ) ( ) ( ) 2 2 22 2 2 1 1d 4 1 d 22 x x x xy x x x xx  + + − + + ++== ++ For C to have 2 stationary points, d 0d y x = has 2 real roots. For 2 4 1 0xx + + = to have 2 real roots, Discrimant > 0 ( ) ( ) 2 4 4 0 4 4 1 0   − − 10 or 4 1 4k  = Alternatively, 2 1 4 1 1222 xxyx xx  + + −= = + − +++ ( ) 2 d 4 1 d 2 y x x  −=− + For C to have 2 stationary points, d 0d y x = has 2 real roots. ( ) 2 d 4 100d 2 y x x  −=  − = + ( ) 2 412x   − + = for equation to have 2 real roots, 41 0  −  10 or 4 1 4k  = It is a show question. Clear steps on solving quadratic inequality (i.e. number line or equivalent) must be shown clearly. 0 + – + 0 + – + α α