Instructions for Section 1
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1 Instructions for Sction 1 Choos th rspons tht is corrct for th qustion. A corrct nswr scors 1, n incorrct nswr scors 0. Mrks will not b dductd for incorrct nswrs. You should ttmpt vry qustion. No mrks will b givn if mor thn on nswr is compltd for ny qustion. QUESTION 1 3 If 66 6 is divisibl by D. -1 E. 1 3, thn is qul to QUESTION Lt f ( ) 1 3 nd g( ). Th vlus of for which th grphs of y f () nd y g() hv two uniqu intrsction points r D. E. 0 nd 0 Th School For Ecllnc 017 Mthmticl Mthods Emintion Qustions Pg 1
2 QUESTION 3 Th linr fctors of r D. E QUESTION Th fctors of y y 6 r ( y )( y 3) ( y )( y 3) ( y )( y 3) D. ( y 3)( y ) E. ( y )( y 3) QUESTION P( ) ( )( b)( c), whr Th qution P ( ) 0 hs ctly, b nd c r thr diffrnt positiv rl numbrs. 1 rl solution distinct rl solutions 3 distinct rl solutions D. distinct rl solutions E. distinct rl solutions QUESTION 6 If 0 whr is positiv rl constnt, thn 0 only 0 nd 1 1 nd D. 0 nd E. 0 nd log Th School For Ecllnc 017 Mthmticl Mthods Emintion Qustions Pg
3 QUESTION 7 Th simultnous linr qutions 3y 0 ( 1) y 0 whr is rl constnt, hv infinitly mny solutions for R 3, R\ 3, D., 3 E. R\, 3 QUESTION 8 Th systm of simultnous qutions my 1 nd y 18 whr m R will hv no uniqu solution for m m Thr r no vlus of m D. m R E. m R\ Th School For Ecllnc 017 Mthmticl Mthods Emintion Qustions Pg 3
4 QUESTION Two functions f nd g r dfind s follows: Th function f ( g( )) B : f : A R with rul f ( ) 3 g : B R with rul will b dfind if g ( ) 3 A : B : 3 D. A : E. B : QUESTION 10 Considr th two rltions h: R /{ 1) R whr Th composit function g h( ) is dfind s log 1 for R log 1 for (1, ) log 1 for R D. 1 log 1 for /{ 1} R E. log 1 for R/{ 1} 1 h ( ) 1 1 g:(1, ) R whr g( ) log ( 1) nd Th School For Ecllnc 017 Mthmticl Mthods Emintion Qustions Pg
5 QUESTION 1 QUESTION -coordint of intrsction points: f ( ) g( ) Discriminnt: Two distinct solutions: Howvr, if 0 Two distinct solutions: ( 3) (1) ( )(1) qution (1) bcoms which hs only on solution., 0. QUESTION 3 QUESTION Answr is B Th School For Ecllnc 017 Mthmticl Mthods Emintion Qustions Pg
6 QUESTION Answr is B QUESTION 6 Lt m in th qution 0. Thn m m 0 nd so ( m )( m 1) 0. Thrfor Also m m 1 nd so nd so 1 which mns tht log which mns tht log 1 0. QUESTION 7 Answr is B For infinitly mny solutions, th qutions must b idnticl. Equting cofficints givs. Howvr, th solution lon is not providd. Substitut in you will gt: 3 nd 3 (th othr nswrs) into both qutions nd whn 3, 3 3y 0 y 0 Multiply th first qution by - nd th scond by 3. This will rsult in two idnticl qutions, mning tht 3 will lso produc n infinit numbr of solutions. Ths typs of qustions r hrd to gt corrct unlss th nswrs r givn to you s in this qustion, or unlss you us th discriminnt or mtrics, which will b covrd ltr. QUESTION 8 Answr is A No uniqu solution mns no solution or infinitly mny solutions. For no solution to ist: Lins must b prlll (grdints must b th sm) nd th Y intrcpts must b diffrnt. my 1 y 18 my 60 10y 36 my y m Th School For Ecllnc 017 Mthmticl Mthods Emintion Qustions Pg 6
7 For infinitly mny solutions to ist: Lins must b prlll (grdints must b th sm) nd th Y intrcpts must b th sm. Thrfor: Th rtio of th cofficints of th nd trms must b qul. Th constnts indpndnt of nd must b th sm. y y Thr r no vlus of m for which infinitly mny solutions ist. QUESTION Rquir rn g dom f. 3 dom f : 3 0. It is thrfor rquird tht: 3 QUESTION 10 g h( ) 1 log log 1 log 1 log ( 1) log 1 Th domin of g (h()) is th sm s th domin of h() which is R / { 1}. Th School For Ecllnc 017 Mthmticl Mthods Emintion Qustions Pg 7
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