MODEL SOLUTIONS TO IIT JEE ADVANCED 2014

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1 MODEL SOLUTIONS TO IIT JEE ADVANCED Pper II Code PART I B A A C D B D C C B 6 C B D D C A C A B D. Rhc(Z ). Cu M. ZM Secon I K Z 8 Cu hc W mu hc 8 W + KE hc W + KE W + KE W + KE W + KE (KE KE ) W.7 ev. A(T ) R R (T ) 9R. T T T K R R cm 6 mm 6 8 d d dr R. Inlly mg > mm mm 6. mm 6 mm mv r mv s cenrfugl force cng rdlly ouwrd r G. R GM e e 6. g e Re Re g P G.R e g g R d(mg ) g R W R / R g g R R. 9R g N 7. Angle beween ST nd he vercl s

2 b.hg bscos h s cos bg KQ KQ. 8. E ; E E R R K.Q KQ E E R R E > E > E 9. B cos(9 ) B B sn B col r r / r / r B / h h h / h /. h. +.h h.6.. h... B sn n L L L B.77 B sn L.7 L.7L.6.7. sn. d. d d Q mole domc k. Secon II B Q moles Mono 7 k B 9 h B Q Q mc VT + work nc v T + Pv nc pt R 7R 7 T T T 7T 8 9 T T 9 B B 9 9 B h X. When prcle moves, boh process become sobrc. Q Q (nc pt) (nc pt) R 7R 7 T T T 7T T T Tkng boh gses s one sysem Q U + W PQ SR

3 W U n C T n C T V V R 6 R 6 7 R. v consn r v consn r v r v v v mm s m s 6. vv pson p v v r Volume v re of nozz;e P Q R S 8. P Q R S 9. f R R. For S he only mch cn be opon C becuse norml componen of he mss would be equl o (m + m cos). Hence frcon need o be (m + m )cos. r Secon III 7. R gh h' g h hh' h d

4 PART II D C B B C D A A C B 6 C D A C B D B C A C Secon I. Couplng recon of phenol s crred ou n lklne medum. C s, *s, s, *s,. Re [M ] z, p, p y p. Bolng pon decreses wh ncrese n brnchng. Presence of pr-mehoy phenyl group wll sblse he crbocon nermede nvolved n he recon 6. Cl CH Cl HO CH CH CH CH () CH MgBr O () HO C CH CH C CH CH O CH 7. H O H O + (O) NH OH + (O) NO + H O H O + NH OH NO + 6H O KIO + H O KIO + H O + O CH CH C CH CH 8. P + 8SOCl PCl + SO + S Cl complee OH / HO hydrolyss (P) 9. XeF 6 XeO 6HF OH / HO HXeO XeO6 Xe O HO. The process s equlbrum. H S sysem T.e., S sysem > nd S surroundngs <. d d d d d 6 cm My M Secon II. Incresed collson frequency of X wh he ner gs compred o h of Y decreses he speed of X NNH. HOCH CH CCH CH CH NOCH CH CCN CH NOCH CH CCCH CH H CH OCH CH CCCH CH Lndlr' CH O CH CH C C H O. Y conns CH C posve odoform es. HCl (Q) N [NCl] (M ) Terhedrl N KCN (R) [N(CN) ] Squre plnr clys s CH CH H group nd so gves

5 KOH ecess KOH (S) (S ) (M ) (Whe pp) 6. Zn Zn(OH) K [Zn(OH) ] (So luble) p-d bondng Secon III 7. [Cr(NH ) Cl ]Cl Cr + d. Hence prmgnec I ehbs cs rns somersm [T(H O) Cl](NO ) I ehbs onson somersm T + d Hence prmgnec [P(en)(NH )Cl]NO I ehbs onson somersm [P + d 8 prng kes plce. Hence dmgnec [Co(NH ) (NO ) ]NO I ehbs cs-rns somersm Co + d 6 prng kes plce. Hence dmgnec (). (c) - + p-d nbondng + - d-d nbondng d-d bondng

6 PART I C B D A D D A C B B 6 B C D B A D D A D C Secon I. Crds 6 Envelops 6 If we ssume C s correspondng o C No: of rerrngemens!!!!!! No of rerrngemens 9 Tol +9. ( + b) b y + c + b +c s () y c y c b ( + b) c + b + b c c + b b + b b cos6 Therefore, c 6 b sn c b sn 6 r R b b 6 sr s bc s bc b 6 c bc b 8c c y c c. S Le P cos, sn nd Q P cos, sn Tngen o he crcle P s P cs y sn cos + ysn y co + () sn () s ngen o y 8 P Q Q sn cos cos Pons P nd Q re (, ) nd (, ) Respecve. A pon on y 8 s (, ) Here, n Pons R nd S re (, ) nd (, ) R Are of PQRS Are of PQR + Are of QRS + S. Tol no of wys we cn rrnge G nd B s! We need o enumere for ech poson. Possble posons re B B B B B G B B G B R

7 B G B B G G B G G B B G G B G In ech cse Boys nd Grls my be permened mongs hem!! Tol no. of wys 6 6 Probbly. Le p() + p(p() gves ( + ) ( ) + ( ) + Neher rel nor comple 6. sn + sn sn Sn + sncos (sn sn ) sn + sn cos + sn cossn + sn (sn) [cos cos] (sn) (cos cos) m. (sn) (, ) m(cos cos) n (, ) Ther produc cnno be. Therefore, no soluon 7. du cos ec.. d co d sn cos cosec d e e du d e e Pu u log co du d e u co e u n co n e e sn cos sn cos sn cosec e + e cosec (e + e ) 7 (cosec) 7 7 cos ec d e e log du e e log e e 6 du coeffcen of n ( ) f() F() f. d F () f(). dy y y d dy d f() y logf() + c f() f() F() F() e. c e d e c e d e dy y d d. F e Soluon s y y c Gven f() c y d 7

8 f() y f d d d sn d cos sn.. d. Secon II.. P( + + s odd) For o be odd cses Cse : Only one odd, res even Probbly Cse : All hree re odd. Probbly 7 Requred probbly. P(,, re n AP) Le d opons P 7 d Le d opons P 7 Le d opons P Le d opon P Le d opon P Requred Probbly. Snce PQ s focl chord nd P( ) Q s, R(r, r) nd k(, ) QR PK r r r r r r r r r. s s r y + () s + y s + s () 8 + y s + s + y + s () y + ( + s ) + y ( + ) y. g() lm d h h h h g lm d h h sn A sn cos d Inegrn sn cos d sn cos d sn h d sn h sn h sn h lm h h 6. g() lm e h d h g ' log d

9 log d sn g' log co d Clerly g 7. (P) f() + b + c Secon III f() c f() + b b b + b 6 b or b wo polynoml (P) () (Q) f() sn( ) + cos( ) f () (n( ) sn( ) n( ), soluon Q () (R) d f() e d (S) Now S () e 8 s n odd funcon fd 8. (P) y cos(cos 7 ) Cos y cos y y (Q) ( )y 9( y ) ( ) y y + y ( ) 9 y, ( ) y + y 9y y y 9 y (P) (9) () n k k k k k ( ). sm. cos k n 8 (Q) 8 () (R) Equon of norml he (, ) Whch s perpendculr o + y 8 s y h () Equon of norml 6 cos, sn 6 y.e. cos sn.e. 6 sn, cos y sn cos () 6 sn cos sn cos 6 cos sn (n ) (n ) n n or Thus (R) () (S) n n > soluons. Then only one s posve (S) () 9. f () f () f () f f () f ()

10 From he grphs we cn see h f s connuous nd one- one S () F s ono bu no one-one P () f f s neher connuous nor one- one R () lm sum lm f s dfferenble. From he grph f s connuous Opon (D) kt kt 6. z k cos s n k e 8 (P) z e, e...e Clerly z k. z j for somke k nd j (P) (Q) z. z j z k e, z e k z e hs se for se of comple no. Q () (R) (z ) (z ) (+z +.+ z 9 ) (z z ) (z z )..(z z 9) + z + z +..+z 9 Le z ( z ) ( z )..( z 9) z z... z 9 (R) () 9 k (S) cos f k, p k k cos (S) ()