DHS Quiz 1 Sequences and Series Maclaurin Suggested Solutions

Name: Index Number: Class: DUNMAN HIGH SCHOOL Quiz 1 Year 5 MATHEMATICS (Higher 2) 9758 22 August 2024 Additional Material: Printed Answer Booklet 40 minutes READ THESE INSTRUCTIONS FIRST Answer all the questions. Write your answers on the Printed Answer Booklet. Follow the instructions on the front cover of the answer booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you must present the mathematical steps using mathematical notations and not calculator commands. You must show all necessary working clearly. The number of marks is given in brackets [ ] at the end of each question or part question.

1 The first three terms of an infinite series are 16, x, 9. (i) Find the value(s) of x, if the series is (a) geometric, (b) arithmetic. [3] (ii) If all the terms in part (i)(a) are positive, calculate its sum to infinity. [2] (iii) Given that the sum of the first n terms in part (i)(b) is less than 64,− find the least value of n. [3] Qn Suggested Solution Comments / Common Mistakes 1(i) (a) 9 16 x x= 2 144x = 12x= (b) 16 9xx− = − 12.5x= • Formulate equation using common ratio or consecutive terms of GP • Formulate equation using common difference of AP • Students should recognise that information in part (ii) may not necessarily apply to part (i). For part (i)(a), students may make the wrong assumption that x or r (common ratio) must be positive hence missed out on one solution. • Note that the question asked for values of x and not d or r. (ii) 16 6431 4 S == − • Students should apply first term 1 ratioS = − formula for GP sum • Students may mistook 1 2 39, , 12,U U x U= = = hence taking the common ratio 4 3r = instead (then sum to infinity won’t exist) 2 16 9 16 9 12 a ar x ar x = = = =  = OR

(iii) Let the number of terms be n.  2(16) ( 1)( 3.5) 642 n n+ − − − [32 3.5 3.5] 128nn − + − 23.5 35.5 128 0nn− −  2.82 or 12.96nn−  OR From GC, least n = 13 • Students may apply wrong AP sum formula or wrong common difference • Note exact answers are not required hence students can use GC to solve the inequality graphically or by table • Students are reminded that the least value of n = 13 is obtained from the solution to the inequality n > 12.96, and not just derived by rounding up n = 12.96 to the nearest integer. Total marks: 8 n nS + 64 12 25 13 1−

2 (i) Find 1 1 2 rn r=   in terms of n. [2] A sequence is such that 0 3u = and 1 1 2 r rruu − −=  for 1.r (ii) Show that 0 1 1 .2 rn n r uu =  =− Hence find nu in terms of n. [2] (iii) Show that 14 2 2 , 2 n Sn = − +  where S denotes the sum of the first n terms of the sequence. [2] (iv) Determine, with reason, whether (a) nu converges, [1] (b) S converges. [1] Qn Suggested Solution Comments / Common Mistakes 2(i) 1 2 3 1 1 1 1 1 1 ...2 2 2 2 2 11122 11 2 11 2 rnn r n n =          = + + + +                    −  = − =−   Concepts used: 1. Sum of GP = ( )1 1 nar r − −

(ii) ( ) ( ) ( ) 1 11 1 11 1 2 0 1 1 0 1 2 ... ... rnn rr rr nn rr rr nn n uu uu u u u u u u uu − == − == −  =− =− = + + + − + + + =−   From (i) and 0 3u = 113 2 14 2 n n n n u u − = −  = −  (ii) Alternative: Using 1 1 2 r rruu − −=  When 1,r= 1 10 1 2uu −=  When 2,r = 2 21 1 2uu −=  When 3,r = 3 32 1 2uu −=  … When 1,rn=− 1 12 1 2 n nnuu − −− −=  When ,rn= 1 1 2 n nnuu − −=  Sum up these equations, 0 1 1 2 rn n r uu = −=  (Shown) 113 1 4 22 nn nu     = + − = −       from (i) and 0 3u = Concepts used: 1. Observe pattern to perform method of differences.

(iii) 11 00 1 0 14 2 14 2 111 2 4 11 2 14 2 2 (shown)2 rnn r rr rn r n n Su n n n −− == − =  = = −    =−   −  =− − = − +    Concepts used: 1. Sum of GP = ( )1 1 nar r − − 2. Understand that the first n terms of this sequence starts from 0u to 1nu − . (iv) When ,n→ 1 02 n  → . (a) 4nu → (constant), nu converges. (b) ,S → S does not converge. Total marks : 8 11 00 0 1 1 ... ... nn rn rr n n n n uu u u u u u u −− == −  + + +  + + + 

3 A famous entrepreneur, Elon Tusk, has designed a new rocket booster for his company SpaceY. A rocket booster consists of the following parts as shown in the following diagram. The design of the Aft Skirt can be viewed using a vertical cross sectional view with the stated dimensions (see dotted box above). Due to the amount of heat and pressure generated from the thrust of the rocket during lift off, the sides of the Aft Skirt will tilt outward by a very small angle,  , while keeping the slant length at 3 m (see diagram below). Before the lift off, the diameter of the circular base o f the Aft S kirt is given by π4 6sin 3  +   m. In order for the booster to function properly during the lift off, the diameter at the bottom of the Aft Skirt must be less than 9.197 m. 4 m 3 m Vertical cross section of Aft Skirt 4 m 3 m m π4 6sin 3 +  m Magnified View of Aft Skirt Rocket Booster 3 m 3 m 

Given that  is a sufficiently small angle, show that  satisfies the inequality 2 0a b c+ +  where a, b and c are constants to be determined. Hence find the range of values for , giving your answers correct to 6 decimal places. [4] Qn Suggested Solution Comments / Common Mistakes 3 Diameter at the bottom of the Aft Skirt π4 2 3sin 3 = +  +  π4 6sin 9.1973 πsin 0.866173 31cos sin 0.8661722    + +   + + For small  , 2 2 31 1 0.866172 2 2 3 1 3 0.86617 04 2 2 Using GC, 0.000289 or 1.154411 (6dp) Since 0 and it is small, 0 0.000289 (6d.p)       − +  − + + −     • Common for students to misinterpret the meaning of “diameter” in this question. Diameter refers to the whole length of the base of the skirt. • Note that only the slant length remains unchanged at 3 m when the skirt opens. • Apply the compound angle formula from MF27 and considered small  to apply the formulas: 2 sin and cos 1 . 2     − • Since the skirt opens 1. by a small angle, rej. 1.554411  2. outward, 0  . Total marks : 4 3