PLMGS(S) 4049 AM Prelim P1 Solutions 2021

Preliminary Examination 2021 Secondary 4 Express Additional Mathematics Qn No Working Total 1(i) 1 1 3Given tan 4 3 tan 4 Since principal values are 90 tan 90 , hence angle in 4th quad. 4cos 5 x x x x           ALTERNATIVE 22 2 sec 1 tan 3sec 1 4 25 16 5 4 4cos 5 xx x x        1 1(ii) 2cos 1 2sin 2 1 cossin 22 4Sub cos , 5 41 5sin 22 1sin 2 10 11sin or sin22 10 10 10 (reject) = 10 xx xx x x x xx          2 ̶ 3 O C x 4 α y

2            22 22 22 2 22 2 21 2 2 2 3 28 3 243 0 3 28 3 3 243 0 3 84 3 243 0 Let 3 , 84 243 0 ( 81)( 3) 0 ( 81) 0 or ( 3) 0 81 3 Hence, 3 81 or 3 3 3 xx xx xx x xx x y yy yy yy yy                     4 22 3 By comparison, 4 or 1 2 or 1 xx xx     4 3             2 2 2 22 2 22 2 1 ............ (1) 21 ............ (2)2 (1) (2), 21 1 = 2 1 2 2 1 2 2 2 1 1 3 2 3 0 4 3 2 4 1 3 9 12 4 12 y mx xxy x xxmx x mx x x x mx mx x x x m x m x b ac m m mm                                        2 2 2 22 12 4 3 Since curve doesn't meet line, there's no real roots. 4 0 Hence, 4 3 0 2 3 0 2 3 2 3 0 m m b ac m m mm           33 22 m   5 - m

4 (Alt seg theoreom) (Alt seg theoreom) (Alt angles, / / ) ( s in the same seg) i.e. , hence, bis ABF FEB y CBD BED x CBD FDB x FD AC FEB BED CBD yx FEB CBD BE                  ects . DEF 4 5(i) 2 2 dGiven 24 cm /s,d 4 d 8d d d d d d d d24 8 d d 24 d8 d3 cm/sd A t Ar A rr A A r t r t rr t r tr r tr               3 5(ii) 3 2 2 3 3 4 3 d 4d d d d d d d d3 4d d 12d When = 10 cm, d 120 cm /sd Rate of decrease in volume is 120 cm /s. Vr V rr V V r t r t V rtr V rt r V t             3 6(i) When t = 0, 2

0.6(0) 95 000 2 498 95 000 2 498(1) 190 N e N N      Initially, there are 190 people who were infected by the virus. 6(ii) When N = 0.08(95 000) = 7600,   0.6 0.6 0.6 0.6 0.6 0.6 95 000 76002 498 95 000 7600 2 498 95 000 2 4987600 10.5 498 7 332 7ln ln 332 70.6 ln 332 7ln 332 0.6 6.432041366 t t t t t t e e e e e e t t t                 The authority will issue home quarantine orders after 7 days. 3 6(ii) When t is very large, 0.6te  0, 95 000 4750020N   Hence, all 95 000 people will never be infected after a long period of time. 2 7(a)(i) (2 − 3𝑥)5 = 32 + 5(16)(−3𝑥) + 10(8)(9𝑥2) + 10(4)(−27𝑥3) + ⋯ 2

= 32 – 240x + 720x2 – 1080x3 + … 7(a)(ii)           23 3 3 2 3 2 2 3 5 1 – 32 – 240 720 – 1080 240 – 1080 480 1080 16 2 2 3 1 1 0 2 5 2 xx x x x x x x xx x x               coefficient of x3 = – 1560 2 7(b)           2 2 2 2 3 1 2 1General Term 2 1 2 1 2 1 2 Since the n r nr r rnr r n r r r nr x x n xr x n xxr n xr n xr                    expansion is dependent of , 3 0. Hence, 3 x n r nr     Therefore, n is any positive integer which is not a multiple of 3. 4 8(i) Maximum speed is 9.2 m/s 1

8(ii) 9.2sin 0.02vt sin 0.02𝑡 = 0 0.02𝑡 = 0, 𝜋, 2𝜋, … 𝑡 = 0, 50𝜋, 100𝜋 Since 0 ≤ 𝑡 ≤ 300, Megan made a turn at 50𝜋 = 157 seconds. (3.s.f) 2 8(iii) Megan only made 1 turn in her run at 50 s and completed her run at 300s.       9.2sin 0.02 d 460cos 0.02 when 0, 0. 0 460cos 0 460 Hence, 460cos 0.02 460 when 50 s, 460cos 0.02(50 ) 460 460( 1) 460 920 m when 300 50 s, 460cos 0.02 sv s t t s t c ts c c st t s s s t s                         (300 50 ) 460 441.67833 460 901.6783319 m Total 901.6783319 920 1821.678 m s s s         Total dist = 1820 m Alternative When t = 300, s = – 460cos[0.02(300)] + 460 = 18.321668 m Dist for 2nd phase = 920 – 18.321668 = 901.6783319 m Alternative 3 A t = 0 t = 157 s t = 300 s

𝑠 = ∫ 9.2 sin 0.02𝑡 𝑑𝑡 50𝜋 0 + |∫ 9.2 sin 0.02𝑡 𝑑𝑡 300 50𝜋 | = [− 9.2 cos 0.02𝑡 0.02 ] 0 50𝜋 + [− 9.2 cos 0.02𝑡 0.02 ] 50𝜋 300 = [−460 cos 0.02𝑡]0 50𝜋 + [−460 cos 0.02𝑡]50𝜋 300 = [−460 cos 0.02(50𝜋) + 460 cos 0] + |−460 cos 0.02(300) + 460 cos 0.02(50𝜋)| = (460 + 460) + |−441.67833 − 460| = 920 + 901.67833 = 1821.67833 = 1820 m Total dist = 1820 m 8(iv) d d d 9.2sin 0.02d 9.2(0.02)cos 0.02 230.184cos 0.02 or cos 125 50 va t at t at ta t a     When t = 200 s, a = – 0.1202704262 a = – 0.120 m/s2 (3 s.f) 2 9(i) xxy 4 1 4

      2 2 2 2 2 2 d1 11d4 d1 1d4 dFor to an increasing function, > 0.d 1Hence, 1 > 04 1 1 4 4 1 4 1 0 2 1 2 1 0 y xx y xx yy x x x x x xx            11 or 22xx   9(ii) 2 2 2 2 dFor any stationary points, 0d 1 1 04 1 1 4 4 1 1 4 1 2 y x x x x x x       2 9(iii) x 0.5- 0.5 0.5+ x –0.5- –0.5 –0.5+ Sign of d d y x – 0 + Sign of d d y x + 0 – Slope \ – / Slope / – \ d d y x changes from – to + as x increases through 0.5, hence 1 is a minimum point.2x d d y x changes from + to – as x increases through – 0.5, hence 1 is a maximum point.2x 3 10(a) 10 cos x  1 10(b)(i) period of g( )x =  or 180o 1 - x

10(b)(ii) period of f ( )x = 2 or 360o 1 10(b)(iii) 4 10(b)(iv) 4sin 2 2cos 1sin 2 cos2 xx xx   From the sketch, No. of solutions = 4 1 10(b)(v) When g(x) is shifted down by 0.5 units, it intersects f(x) 5 times. Hence, m = 0.5 1 11(i) Let the ht bet the of cylinder be h cm. 2 2 2 2 2 2 2 16 (2 ) 256 4 4(64 ) 2 64 cylinder's ht = 2 64 cm (shown) hr hr hr hr r      Alternative By Pythagoras’ Theorem, 2 1 –1 y 0 x 0.5 2̶ ̶ 3̶ 4 ̶ 2 3̶ 2 7̶ 4 – 0.5 5̶ 4  4 O 8 8 h 2r

22 2 1 ht 82 cylinder's ht = 2 64 cm (shown) r r   Alternative Let the ht bet the top of cylinder and sphere be x cm.           2 22 2 22 22 22 22 2 2 cylinder's ht = 2(8) 2 = 16 2 cm 88 88 88 88 cylinder's ht = 16 2 cm =16 2 8 8 16 16 2 64 2 64 cm (shown) x x xr xr xr xr x r r r                         11(ii)                            2 2 1 22 2 11 2 2 222 1 2 2 22 2 1 2 2 2 1 2 2 Curved surface area of cylinder, 2 2 2 64 4 64 d1 4 64 4 64 2d2 d 4 64 4 64d d 4 64 64d 4 64 2d d 64 4 (2) 32d d 64 8 32d d A rh A r r A r r A r r r rr A r r rr A r r rr rA r r rA r r A r                                        2 2 (shown) 64 r r 4 O x cm 8 r 8 – x