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