Mid Year Examination F.4 Mathematics Module 1 (Calculus & Statistics) Suggested Solutions

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Christina Stevenson
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1 Mid Ya Eamination 3 F. Matmatics Modul (Calculus & Statistics) Suggstd Solutions Ma pp-: 3 maks - Ma pp- fo ac qustion: mak. - Sam typ of pp- would not b countd twic fom wol pap. - In any cas, no pp maks will b dductd in tos stps studnts could not sco any maks. Sction A (30 maks) No. Suggstd Solutions Maks Rmaks a. lim lim ( )( ) lim ( ) ( )( ) lim Dnominato coct ( ) lim ( )( ) b. ( ) lim lim lim 3( ) ( ) 3( ) ( ) [No backt fo. pp-] (t igst pow) 3 lim Som studnts dividd bot nominatos and dnominatos by. [Sould b.] Pag

2 . ( ) 3 Epand ( ) t tm C ( ) Fo constant tm, 0 0 3( 0 ) C C wic is not an intg. T is no constant tm. 3( 0 ) 0 0 ( ) ( ) [Sould b ( ). mak is dductd.]. 3 3 ( 3) ( 3)......!!!! Epand on of t pssion Epand coctly 3, not Altnativ Solution 3 3 ( ) ( 3) ( 3)......!!!! Epand coctly Epand 3, not 3 Pag

3 9 [ 3...] [ ( 3) ] ( 3) Som studnts pand! ( )! 3... in dnominato instad of [Btt wit 3.! ( ). Lt $P b oiginal amount of mony, % b quid intst at: % P ( 0) P 0. ln 0. 0 ln Rquid intst at 6.93% ( 0) 6.93 (co. to d.p.)!... No mak dductd.] P P, 0.07 (co. to d.p.) [Sould b Tn intst at 6.93% (co. to d.p.)] Lt % b intst at, Lt b intst at, 0 P P P P ( 0) % 0. [Sould b % 0 P P.] 0. [If unit includd in t vaiabl, sould b P P ] P P % [Wong compound intst fomula fo intst compoundd continuously] % 0 [ dductd bcaus P is not dclad and usd.] 000 % [ dductd as ason abov. dductd bcaus studnts just spcifid on on cas (oiginal 000, nw 000) only.] 6a. Lt y 6 y y 6 y y 6 0 Quadatic ( 6 )( y ) y 0 y 6 o y 6 o ln 6 (j.) ln 6 Accpt.33 Pag 3

4 Som studnts did not cogniz t quadatic pattn and tak ln on bot sids of t quation. 6 0, 6 0 [ 6b. ln( ) ln( 3) ln ln( ) ln 6 ln ( ) 6 ] ln A ln B ln AB o (j.) Altnativ Solution ln ln 3 ( ) ln ln( ) ln( 3) ln 3 Som studnts did not jct t answ. A lot of studnts did not cogniz t altnativ solution abov, wic is muc mo fast. 3, 3 [Sould b ( )( ) 0, (j.) o. mak dductd fo just simplify out t facto.] 7a. ( 3 ) ( a) 3 3 [ ( 3 ) ( 3) 6( 3)...] 3a 3 [ 0...] [ 3a 3a...] [...] a Epand ( 3a 0) ( 3a 3a )... 3a 0 3a 3a 3a 3a 37 3a 3a 3 0 3a a 0 ( 3 )( a ) a a 0 3 ( 3 ) 0... o a [Sould b 0...] Pag

5 7b. Fo a, Us (a) 3 Cofficint of Fo a, 70 Cofficint of ,37 Pag

6 Sction B (30 maks) No. Suggstd Solutions Maks Rmaks a. ( ) C b. Fo, ( ) ( 0) ( 9900)... ( ) Daw as common facto M, M is an intg. Rmaind 00 6 is not divisibl by. Altnativ Solution Fo, ( ) ( 0) ( 9900)... Fom scond tm onwads, all tms a divisibl by. T fist tm is not divisibl by. T is a maind of is not divisibl by. 00 ( ) , 90 is not divisibl by. 9a. T numb nds wit. [No poof is givn on wy t numb nds wit.] 9b. f ( ) f!! 3! Daw as common facto Us (a) Pag 6

7 9c. f ' ( ) f f lim st pincipls 0 lim ' f f ( ) f lim d. Fo, y Us (a). [No limit.] Fo, slop f Equation of tangnt is: ' Equation is f maks awadd.] ' y y y 0a ( )!! ( )!!!( ) 0b. [Studnts misundstood tat f ' ( )(!) Put into sult of (c) is quation instad of slop. No!!! as fist stp bfo poof. [pp-. Not yt povd. Cannot wit.] Wit down n (!) [ ( )!!] n ( ) n Us (a) (!! ) ( 3!! )... ( n )! n!!!! [pp-. No backt.] kt a P A Epand n ln P kt ln A ln P kt ln A Pag 7

ln P kt ln A [Studnts a not familia wit t opation of ln.] b (i) T 6 0 ln P...0 0.9 0.6 ln P Coct staigt lin. b (ii) Pass toug y-int. Accpt y-int btwn. to.7. ln P kt ln A O 0 t Fig.

8 ln P kt ln A [Studnts a not familia wit t opation of ln.] b (i) T 6 0 ln P ln P Coct staigt lin. b (ii) Pass toug y-int. Accpt y-int btwn. to.7. ln P kt ln A O 0 t Fig. b (iii) By gap, a) ln A.6 ln A y-int o k slop. 6 A (co. to sig. fig.) Accpt A fo y-int. k Accpt ot answs accoding to gap. ln P kt ln A, but us ln A y-int o k slop [Only mak dductd. No answ mak will b givn in public am.].. k [Studnts usd points fom t tabl. Howv, (.,) is not on t staigt lin. Studnts sould us points fom t staigt lin and t distanc btwn two points is as fa as possibl. mak dductd.] Pag

9 c By (b), P Fo t 0. 0.t t, 0 0 P 0 T numb of bactia is 0 aft a vy long tim. f.t. 0.t Fo t, 0.t. [Sould b 0 ] d Q P 9 0.t 0.t 9 Witold mak fo inquality sign. 0.t 0.t 9 0 Lt y 0.t,. t y 9y 0 ( )( y ) y 0 y o y 0 t o. 0.t ln T 6.. t 0 ln 0 6. (co. to 3 sig. fig.) (j.) Som studnts cannot daw a suitabl staigt lin sults in not gtting good valus fo vaiabl A and k. Tfo ty cannot fom a quadatic quation. Q P 9,000,000 [Sould b Q P 9] 0.t 0.t t t 0.t 0.t ( ); ( ) 0.t 0.t 0.t 0.t ln[ ] ln ln [Studnts a wak in ponntial and logaitmic opation] ln Q ln P 9 [Studnts did not undstand t qustions] ; -- END OF SUGGESTED SOLUTION -- Pag 9

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