VJC 2023 promo prac paper (A) solutions

2023 PROMO PRACTICE PAPER A Solutions Qn Solution Comments 1i 1n nnuSS −= − ( ) ( )ln 1 ln 1 1nn= + − −+ 11ln ln 1n nn + = = +  1ii Observe that as 1, 0 ln1 0. nnu n→∞ → ∴ → = The sequence decreases and converges to 0. 2 2 2 10 3 10 3 03 3 31 24 03 xx xx x x x ≥− −+ ≥−  −+ ≥− Since 2 3 31 024x − +> for all real x values, 1 03 30 x x ≥− −> 3x< If we conclude that the numerator has no real roots, (i.e. 2 40b ac−< , it is not sufficient to conclude .. 1 03 x ≥− ..). What else do we need? 210 3 xx−+ is an expression, not an equation. There are no roots for an expression. Is it correct to only state that 2 3 31 024x − +≥ ? Which real number will be a solution to 2 3 31 024x − += ? 3i ( ) 32f x x ax bx c=+ ++ ( ) ( ) ( ) ( ) 32 f1 1 1 1 8 7 (1) a bc abc = + + += ++= ( ) ( ) ( ) ( ) 32 f 2 2 2 2 12 4 2 4 (2) a bc a bc = + + += + += ( ) ( ) ( ) ( ) 32 f 3 3 3 3 25 9 3 2 (1) a bc a bc = + + += + += − By GC, a = – 1.5, b = 1.5 and c = 7 Recall factor / remainder theorem from O-level Additional Math. This is an assumed knowledge that you should have. 3ii ( ) ( ) ( ) ( ) ( ) 32 2 2 22 f 1.5 1.5 7 f ' 3 3 1.5 3 0.5 0.75 1.5 3 0.5 0.75 0 since 0.5 0 xx x x xxx x xx = − ++ = −+ =− −+ = − +> − ≥ Since ( )f' 0x > , graph of f is a strictly increasing cubic graph which will only cut the x-axis once. Hence, there is only 1 real root for ( )f0 x = . [In fact, since f(1) = 8 > 0 and f(–2) = –10 < 0, it suggests that there is a root between –2 and 1.] “Show that the gradient of the curve is always positive. Hence, explain ...” Your explanation should make mention to the fact that ( )f' 0x >

2023 PROMO PRACTICE PAPER A Solutions 4ai 1 44 4 144 32 4 x h hhx hrx = − −= = − =−=+ This is a show question. Clear working has to be shown. A clearly labelled diagram will aid in our explanation. Extra Note: There are no similar trapeziums in the figure. Having the same corresponding angles does not mean that the figures are similar. If the above statement is true, then all rectangles are similar to each other. Alternative 23 84 xxx= ⇒=+ 2 88 8 244 r h hhr =+ += = + Alternative 2 32 4 2 4 r h hr −− = = + 4aii ( ) 21Volume of water, π4 23V r rh= ++ . 2 2 32 2 1 π4 2 223 44 1 π 44 43 16 2 13π 123 16 2 d 13 π 3 12d 3 16 hhVh hhhh hh h Vh hh   = ++ + +     = +++ ++ = ++  = ++ . When h = 1, d dd d dd 13 d9 π 3 123 16 d 81πd 16 d d 16 d9 π V Vh t ht h t h t h t = × = ++ × = × = The rate of increase of the depth of water is 116 m min9π − . We can use the given formula for volume of frustum of right circular cone, with base radius 2 m, and radius r m. You can only substitute constant values before differentiation. ( ) ( ) 2 2 1 π4 23 d1 π4 2d3 V r rh V rrh = ++ ≠ ++ r i s a variable too . We have to use implicit differentiation and product rule if we want to differentiate it directly. (Refer to tutorial 9.2 Q5: 11 1 uv f+= 22 1d 1d 0dd uv u tvt−−= ) Write your answers properly. 16 16 π9π9≠ 3 r x 2 x 8h+ 2 32− 2r− 4h 4 12 h 3 4 r

2023 PROMO PRACTICE PAPER A Solutions 5i ( )( ) ( ) ( ) ( ) 1 2 1 2 2 2 2 2 12 4 12 4 1 12 124 13 11 2212 12 2 4 2! 4 1312 12 8 128 1 15 2912 8 128 1 15 29 2 16 256 x x xx xx xxx xxx xx xx − − − − = −− = −−    −−      = − +− − + − +     = − ++ + =−− +  = −− +     When applying formula in binomial expansion, we can replace x with f(x), i.e. ( )( ) ( ) ( ) ( )( ) 2 1f 1f 1 f2! n x nx nn x + = + −++  5ii Let 3 4x= . 2 312 1 15 3 29 34 2 16 4 256 434 4 −    ≈− −   − 1 10932 409613 4 1 1093 409613 409613 1093 − ≈− − ≈− ≈ Alternative (if you ra tionalised the denominator) 13 1093 14209 1313 4096 4096− ≈− ⇒ ≈ When asked to substitute 3 4x= , we are required to do the substitution on both sides of the equation. 6i Since the narrowest part is at height 80m, the centre of the hyperbola is at ( )0,80 . k = 80. At ( )50,0 ,− ( ) ( ) 22 22 50 80 1ab − −= At ( )37.5,125− , ( ) ( ) 22 22 37.5 45 1ab − −= By GC, 22 1 11 1 ,900 3600ab= = a2 = 900, b2 = 3600 This question can be solved as a system of linear equations with unknowns 2 1 a and 2 1 b .

2023 PROMO PRACTICE PAPER A Solutions 6ii Recall the main features that we expect to see in a hyperbola. 6iii r = 30 A diagram would be useful to visualize this question. Observe that the centres of the hyperbola and circle are the same. We should capitalise on the graph drawn to deduce the value of r. 7i 2 2 dd 6 2, 3 2dd d32 d 62 xy t tttt yt t xt = += + += + Since tangent is parallel to the y-axis, d d y x is undefined. 16 20 3tt+=⇒= − 2 1 1532 23 33x  = − + − −= −   ∴ Equation of tangent is 2x=− . When a line is parallel to x- axis, d 0d y x = . When the line is parallel to y-axis, d d y x is undefined. A line that is parallel to y- axis has equation of the form xk= . 7ii 2 2 5132 3 22 33xt t t = + − = + − ≥− Least value of x is –2. When we are considering d 0d y x = , we are finding stationary values of y. For those who considered d 0d x t = , you would need to check that x is a minimum. r

2023 PROMO PRACTICE PAPER A Solutions 7iii When y = 0, 32 0tt += t = 0 or 1t =− 52 or 33xx= −= − Read the question: There is no mention about the range of values of t that can be used, so on the GC, we have to adjust the T min. We need to zoom in appropriately to see the shape of the graph. Although there is no mention of 2x=− in this part of the question, it is implied that we need this value from the earlier parts in the question. 8 ( ) 2 2 2 2 1tan 2 d1 1 secd2 2 d1 111 12 sec sec tand2 222 2 11 1 dsec tan22 2 d yx y xx y x xxx yx xy x =  =    =    = =  8i 232 32 dd d dd d yy y yxx x = + When x = 0, y = 0, d1 d2 y x = , 2 2 d 0,d y x = 3 3 d1 d4 y x = 33 1 1 11 4 2 3! 2 24yxx x x≈+ =+ We have to differentiate the given expression in the question. The formula for Maclaurin expansion is ( ) ( ) ( ) ( ) 2 f f 0 '0 '' 02! x xf x f = + ++  There is no need to replace x with 1 2 x here. (compare with Q5i) 8ii 11tan d 2ln sec22xx x C  = +   ⌠⌡ Since , 222 xx ππππ−<< − << . 11cos 0 sec 022xx  >⇒ >   Hence 11tan d 2ln sec22xx x C   = +     ⌠⌡ Do not forget to divide by d1 1 d2 2 xx  = , a constant. We need to remove the modulus notation if it is possible to do so in the question. We do not integrate the result from (i) here, question did not ask for a maclurin series, or an approximation. 5,03 − 2 ,03 − 22, 27 − y O x

2023 PROMO PRACTICE PAPER A Solutions 8iii 3 24 24 1 11 1 12ln sec d2 2 24 4 96 1 11ln sec 2 8 192 2 x x x x x xC Cxx x     ≈ + = ++        ≈+ +  ⌠⌡ When x = 0, 002 C C=⇒= 24111ln sec 2 8 192xx x  = ++   There are many methods to solve this question. However, if it is a hence, or deduce question, this will be the only acceptable method. We will need to find the value of the arbitrary constant as well. Note for those who used series from MF26, You have to check range of values of x for valid expansion. 9ai Consider ( ) 2 i 3 4iab+= − 22 2 i 3 4ia b ab−+ = − Comparing the real and imaginary parts 22 3ab−= and 24ab=− 2b a −⇒= 2 2 4 3a a−= 42 3 40aa− −= ( )( ) 22 4 10aa− += ( )2 Since is a real numberaa=± When 2,a= 1,b=− When 2,a=− 1,b= Hence the 2 roots are 2i− or 2i−+ 9aii 2 4i 4i 7 0zz− +−= ( ) ( ) 2 4i 4i 4 4i 7 2z ±− − −= 4i 16 28 16i 2z ±− + −= 2i 3 4iz= ±− 2iz= + or 2 3iz= −+ 9b ( ) ( ) 22