ACJC 9758 2023 Prelim P2 Solution

1 2023 JC2 H2 Math Prelim P2 Marking Scheme Qn Solution 1 2 sin 2 dcos cos 2 x xxx++∫ Let cosux= d sind u xx =− 2 sin2 dcos cos 2 x xxx++∫ 2 2sin cos dcos cos 2 xx xxx= ++∫ 2 2cos ( sin )dcos cos 2 x xxxx=−− ++∫ 2 2cos d dcos cos 2 d xu xxx x =− ++ ∫ 2 2 d2 u uuu=− ++∫ 2 (2 1) 1 d2 u uuu +−=− ++∫ 22 21 1 d22 u uuu uu += −− ++ ++∫ 2 2 21 1 dd 172 () 24 u uuuu u +=−+ ++ ++ ∫∫ 21 21 2 21ln 2 tan 77 2 2cos 1ln cos cos 2 tan 77 uuu c xxx c − − += − +++ + += − + ++ + 2(i) ( )coty xa= + ( ) ( )( ) ( ) 2 22d cosec 1 cot 1d y xa xa yx = − + = −+ + = −+ 2d 1d y yx = −− ( ) 2 23 2 dd 2 21 2 2dd yy y y y yyxx = − = += + ( ) ( ) ( ) 3 2 3 222 42 dd d 26dd d 21 6 1 682 yy y yxx x yyy yy = + = −+− + = − ++ 2(ii) Since tan 0 0= , 1cot0 tan0= is undefined. Hence the Maclaurin series of ( )cot 0x+ cannot be found.

2 2(iii) πcot 2yx = +  When 0x= , πcos 02 0π 1sin 2 y= = = , 2d 10 1d y x = −− = − , ( ) ( ) 2 3 2 d 20 20 0d y x = += , ( ) ( )( ) 3 42 3 d 60 80 2 2d y x = − + += − 3 32 3! 3 xyx x x≈ −− = −− 2(iv) ( ) ( ) ( )( ) 1 1 2 3 32 323 23 22 1 122 1211 112 3 2 2! 2 1 1 11 12 3 24 1 111 2 324 11 7 2 4 24 y yxx xy xxxx xx xx xx x x xx x − − = ++ = +   −−  ≈ −− +− +        = −− − +     ≈ −− + −  = −+ − 3(i) d6 π d vR vgtm η+= Substituting all the values, 1.3806, 0.05, 0.2, 9.780R mgη = = = = gives 6.5059 6.5059 d 6.5059 9.780d d 9.780 6.5059d 1 d9.780 6.5059 ln 9.780 6.5059 6.5059 9.780 6.5059 e , e tc v vt v vt vtcv v tc vA A −− += ⇒= − ⇒= +− −⇒= +− ⇒− = = ± ⌠⌡ At 0, 0tv= = , 9.780 A= ( ) ( ) 6.5059 6.5059 6.51 9.780 6.5059 9.780e 1.5032 1 e 1.50 1 e t t t v v v − − − ∴− = ⇒= − = − 3(ii) As 6.5059t ,e 0 t−→∞ → , 1.5032v→ The ball approaches a velocity of 1.50 1ms− after a long time.

3 3(iii) d d xv t= ( ) 30 6.5059 0 1.5032 1 e d (using GC) 44.865 44.9 txt −= − = ≈ ∫ The distance covered by the ball after 30 seconds is 44.9 m. Alternative Method (not recommended as this is more tedious) ( ) 6.51d 1.50 1 ed tx t −= − ( ) 6.51 6.5059 1.50 1 e d e1.5032 6.5059 t t xt tc − − = − =++  ∫ When 1.50390, 0 6.5059tx c= =⇒= − When 30, t = 6.5059(30)e 1.50321.5032 30 6.5059 6.5059 44.9(3s.f.) x −=+−  = 4(a)(i ) 2 4 11 4 1 25 21 25rr r r = −++ + + 4(a) (ii) 2 22 4 11 4 1 25 21 25 11 59 11 7 11 11 ...9 13 11 2 32 1 11 2 12 3 11 2 12 5 11 1 1 5 7 23 25 12 1 1 3 5 23 25 nn rr rr r r nn nn nn nn nn = =    = −  ++ + +   = − +− +− + +− −+ +− −+ +− ++ =+− − ++ = −− ++ ∑∑ 4(a) (iii) 1 2 1 2 4 4 12 5 4 4 12 5 rr rr uu rr uu rr − − = + ++ −= ++

4 ( )1 2 22 2134 1 2 1 1 4 4 12 5 12 1 1... 3 5 23 25 12 1 1 3 5 23 25 117 1 1 3 5 23 25 nn rr rr n n nn n n uu rr uuuu u u uu nn uu nn u nn − = = −− − −=  ++ −+−++ − +− = − − ++ −= − − ++ = −− ++ ∑∑ 4(b) ( ) 33 33 3 22 3 33 23 lim 33lim lim57 57 33 lim lim7755 33 71 as , 0, 0.55 nn nn n n nn nn a n nx n nx nn xxn nn n nn xn nn →∞ →∞ →∞ →∞ →∞  −+ −+= =  ++     −+ −+      = =    ++       =− = < →∞ → → ∴By the root test, 3 3 0 3 57 r r r rx r ∞ = −+  +∑ converges for all values of x. 5(i) 22 11cos , coscos cos d sin d sinsin , sind cos d cos xy xy θθ θθ θθθθθ θθ θ = −= + = −− = −+ 22 22 sin 1 2 cos cos 2sin 1 cos cos sin 1d cos 1 d sin 1 cos 1 y x θ θθ θ θθ θ θ θθ − + −+ −= = =− − −− + 5(ii) At the point, 11cos , coscos cosxp yp pp= −= + 22 22 d cos 1 1 cosgradient of normald cos 1 1 cos yp p xp p −+= ⇒= +− Equation of normal: 2 2 2 22 2 2 22 2 22 2 1 1 cos 1cos coscos 1 cos cos cos 1 1 cos cos 1 cos 1 cos cos cos 1 1 cos 1 cos cos 1 cos cos 1 cos 1 cos 21 cos cos pyp xppp p p ppyx p pp p ppyx p pp ppyx pp   +−+= −−   −     ++ −−=−   −    ++ +−= + − ++= + − 

5 5(iii) 22 2 1 cos 1 cos: 00 21 cos cos ppPy xpp ++=⇒= + −  22 2 2 1 cos 1 cos cos 122 cos 1 cos cos pp px p pp   +− −=−=   +   21 cos:0 2 cos pQx y p  +=⇒=  2 2 1 coscos 1 0, 2 cos pp OP p  −−< ∴ =   Area OPQ = 221 1 cos 1 cos222 cos cos pp pp    −+      ( ) 4 22 2 1 cos2 2 sec coscos p ppp  −= = − 5(iv) Let the area of OPQ be A. ( ) ( ) 22 2 2 sec cos d dd d2 2sec tan 2cos sinddd d A pp A Ap ppp ppt pt t = − =×= + When π 3p= , ( ) ( ) 2d ππ ππ4 cos sin sec tan 0.1d 33 33 134 4 3 0.122 1.7 3 2.94 (3 s.f.) A t = + = + = = 6(i) Let Y be random variable “number of yellow chips in a box”. B(36,0.3)Y  P( 9) 0.32544Y ≤= P(5 9)P( 4 9) P( 9) P( 9) P( 4) P( 9) 0.978152 0.978 (3 sf.) YYY Y YY Y ≤≤> ≤= ≤ ≤− ≤= ≤ = ≈ 6(ii) Let W be random variable “number of boxes with at most 9 yellow chips”. B(59,0.32544)W  Required probability = P( 13) 0.32544 0.00825W = ×= 6(iii) Let X be random variable “number of boxes containing at most 9 yellow chips in a carton”. B(75,0.32544)X  E( ) 75 0.32544 24.408 Var( ) 75 0.32544 (1 0.32544) 16.46466 X X = ×= = × ×− =

6 Since n = 30 is large, 16.46466N(24.408, ) 30X  approximately by Central Limit Theorem. P( 25) 0.21211 0.212 (3 sf.)X >= ≈ 7(i) The monthly sales profits is expected to increase by m thousands of dollars for every additional thousand dollars spent on the monthly advertisement expenditure. 7(ii) Substituting 7.208333x = and 242 12 ky += into the least squares regression line of y on x: 242 0.3604 3.0194 7.20833312 22.12524066 12 242 23.503 23.5 k k k + =+× = ×− = ≈ 7(iii) 7(iv) The product moment correlation coefficient for Model A = 0.968 The product moment correlation coefficient for Model B = 0.982 Since the product moment correlation coefficient between x and ln y is closer to 1 as compared with the product moment correlation coefficient between x and y, it suggests that there is a stronger positive liner correlation between x and ln y. Furthermore, from the scatter diagram, y is increasing at an increasing rate as x increases, hence Model B will be the more appropriate model. 7(v) Using GC, the least squares regression line of ln y on x is ln 2.0867 0.13579 ln 2.09 0.136 yx yx = + = + When 11x= , ln 2.0867 0.13579 11 35.88754326 y y = +× = Hence the estimate for the sales profit is $35,900 (or $35,888). 7(vi) Since x = 11 does not fall within the given data range 4 10x≤≤ , the estimate obtained by extrapolation will not be reliable. 0 Advertisement expenditure, x 4 15. 10 32. Sales profits, y

7 8(a) (i) Number of ways = 26 3 61 6! 497296800CC× ×= Alternative: Number of ways = 3 126 25 24 23 22 21 497296800C×××××× = 8(a)(i i) Case 1: 3 letters with two different colours each Number of ways = ( ) 326 3 32 6! 50544000CC× ×= Case 2: 1 letter with 3 different colours and 1 letter with 2 colours and 1 letter with 1 colour Number of ways = 26 3 3 321 3! 6! 101088000CCC× × ××= OR 26 25 3 24 3 1 12 11 6! 101088000C CC C C× × × × ×= OR 26 25 2 3 3 1 21 12 6! 101088000C CC CC× × × × ×= Total number of ways = 50544000 101088000 151632000+= 8(b)(i ) 2 21 1 (1)41 40 66 66(2 )(1 ) (41 )(40 ) 65 117 1508 0 294 or (Rejected)5 cc cc cc c c cc c ++ × = −−−−−++ + += + + +−= = − Alternative: 12 1 2 1 1 241 40 41 40 41 40 66 cc c cccc cc −×+×× +×=++++ ++ 265 117 1508 0 294 or (Rejected)5 cc c +−= = − 8(b)(i i) Method 1: Case 1: red flower and non-red bead 6 28 28 245 44 165× ×= Case 2: red non-flower and non-red flower 11 5 1 245 44 18× ×= Required probability = 28 1 223 165 18 990+= Method 2: Case 1: red bead and non-red flower 17 5 17 245 44 198× ×= Case 2: red flower and non-red non-flower

8 6 23 23 245 44 165× ×= Required probability = 17 23 223 198 165 990+= Method 3: Case 1: