Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh;...

32
1 CODE 6 Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472 DATE : 21/05/2017 PAPER - 1 (Code - 6) Answers & Solutions for JEE (Advanced)-2017 Time : 3 hrs. Max. Marks: 183 INSTRUCTIONS QUESTION PAPER FORMAT AND MARKING SCHEME : 1. The question paper has three parts : Physics, Chemistry and Mathematics. 2. Each part has three sections as detailed in the following table : Section Question Type Number of Questions Full Marks Category-wise Marks for Each Question Partial Marks Zero Marks Negative Marks Maximum Marks of the Section Single Correct Option 6 +3 If only the bubble corresponding to the correct option is darkened 0 If none of the bubbles is darkened –1 In all other cases 18 3 One or more correct option(s) 7 +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened 0 If none of the bubbles is darkened –2 In all other cases 28 1 Single digit Integer (0-9) 5 +3 If only the bubble corresponding to the correct answer is darkened 0 In all other cases 15 2 fgUnhekè; e

Transcript of Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh;...

Page 1: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

1

CODE

6

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005Ph.: 011-47623456 Fax : 011-47623472

DATE : 21/05/2017

PAPER - 1 (Code - 6)

Answers & Solutionsfor

JEE (Advanced)-2017

Time : 3 hrs. Max. Marks: 183

INSTRUCTIONS

QUESTION PAPER FORMAT AND MARKING SCHEME :

1. The question paper has three parts : Physics, Chemistry and Mathematics.

2. Each part has three sections as detailed in the following table :

Section QuestionType

Number of Questions Full Marks

Category-wise Marks for Each Question

Partial Marks Zero Marks Negative Marks

MaximumMarksof the

Section

SingleCorrectOption

6 +3If only the bubble corresponding to the correct option

is darkened

— 0If none of the

bubbles is darkened

–1In all other

cases

183

One ormore

correctoption(s)

7 +4If only the bubble(s)

corresponding toall the correct

option(s) is(are)darkened

+1For darkening a bubblecorresponding to each

correct option, provided NO incorrect option is

darkened

0If none of the

bubbles is darkened

–2In all other

cases

281

Singledigit

Integer(0-9)

5 +3If only the bubblecorresponding to

the correct answeris darkened

— 0In all other

cases

— 152

fgUnh ekè;e

Page 2: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

2

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

PHYSICS

[ kaM-1 (vf/ dre vad : 28)

bl [ kaM esa l kr i z' u gSaA

çR;sd i z'u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ft uesa ,d ;k ,d l s vf/ d fodYi l gh gSaA

i zR; sd i z' u ds fy, vks-vkj-, l - i j l kjs l gh mÙkj (mÙkjksa) ds vuq: i cqycqys (cqycqyksa) dks dkyk djsaA

i zR; sd i z' u ds fy, vad fuEufyf[ kr i fjfLFkfr ; ksa esa l s fdl h , d ds vuql kj fn;s t k;saxs %

i w.kZ vad : +4 ; fn fl i QZ l kjs fodYi (fodYi ksa) ds vuq: i cqycqys (cqycqyksa) dks dkyk fd; k gSA

vkaf' kd vad : +1 i zR; sd l gh fodYi ds vuq: i cqycqys dks dkyk djus i j] ; fn dksbZ xyr fodYi dkykugha fd; k gSA

' kwU; vad : 0 ; fn fdl h cqycqys dks dkyk ugha fd; k gSA

½.k vad : –2 vU; l Hkh i fjfLFkfr ; ksa esaA

mnkgj.k % ; fn ,d i z' u ds l kjs l gh mÙkj fodYi (A), (C) vkSj (D) gSa] r c bu rhuksa ds vuq: i cqycqyksa dks dkys djus i j+4 vad feysaxs_ fl i QZ (A), (D) ds vuq: i cqycqyksa dks dkyk djus i j +2 vad feysaxsa_ rFkk (A) vkSj (B) ds vuq: i cqycqyksadks dkyk djus i j –2 vad feysaxs D;ksafd ,d xyr fodYi ds vuq: i cqycqys dks Hkh dkyk fd; k x; k gSA

1. fp=k esa fn[ kk; s x, i fj i Fk esa L = 1H, C = 1F, R = 1kgSA ,d i fjorhZ oksYVrk (V = V0sint) l zksr l s Jzs.kh l acaèk gSAfuEu esa l s dkSu l k (l s) dFku l gh gS@gSa\

V t0 sin

L = 1 H C = 1 F R = 1 k

(A) t c fo| qr èkkjk oksYVrk dh l edyk esa gksxh rks og vkofÙkZ R i j fuHkZj ugha djsxh

(B) t c >>106 rad. s–1, i fji Fk l aèkkfj=k (capacitor) dh r jg O;ogkj djrk gS

(C) t c ~ 0 gksxh rc i fji Fk esa cgrh èkkjk ' kwU; ds fudV gksxh

(D) t c = 104 rad.s–1 gksxh rc fo| qr èkkjk (electric current) oksYVrk dh l edyk esa gksxh

mÙkj (A, C)

gy >> 106 i jXL = L

= = 106 ds fy, 106 × 10–6

= 1 >> 106 ds fy, XL >> 1

XC = 6 –61 1 1

10 10C

= 106 i j >> 106 ds fy, Xc << 1

R = 1 k

R LC –6 –6

1 1

10 10 = 106 rad.s–1

0 Xc i j / kjk 0

Page 3: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

3

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

2. ekuoh; i "Bh; {ks=ki Qy yxHkx 1 m2 gksrk gSA ekuo ' kjhj dk r ki eku i fjos' k ds r ki eku l s 10 K vfèkd gksrk gSA i fjos' k

rki eku T0 = 300 K gS] bl i fjos' k r ki eku ds fy, 4 20 460 WmT gSA t gk¡ LVhi Qku&cksYV~t eku fu; rkad (Stefan-

Boltzmann constant) gSA fuEu esa l s dkSu l k (l s) dFku l gh gS@gSa\

(A) ekuoh; ' kjhj l s 1 l sadM esa fudVre fofdfjr Åt kZ 60 t wy (60 Joules) gS

(B) i "Bh; {ks=ki Qy ?kVkus (t Sl s% fl dqM+us l s) l s ekuo vi us ' kjhj l s fofdfjr Åt kZ ?kVkrs gSa ,oa vi us ' kjhj dk r ki eku vuqjf{krdjr s gSa

(C) i fjos' k r ki eku vxj T0 l s ?kVr k gS (T0 << T0) r c ekuo ds ' kjhj dks r ki eku dk vuqj{k.k djus ds fy,30 04W T T vfèkd Åt kZ fofdfjr djuh i M+rh gS

(D) ekuoh; ' kjhj ds r ki eku esa vxj l kFkZd of¼ gks rc i zdk' k pqEcdh; fodj.k Li SDVªe dh f' k[ kj r jax&nSè; Z (peak in the

electromagnetic spectrum) nh?kZ r jax&nSè;Z dh vksj foLFkkfi r gksrh gS

mÙkj (A, B, C)

gy ' kjhj dk mi ki p; ra=k r ki dks cuk;s j[ kus ds fy, vfr fjDr Åt kZ mRi Uu djsxk

ekuk vkUr fjd mi ki p; ds dkj.k] mRi Uu ' kfDr I gS]

t c ' kjhj dk r ki fu; r gS] Q usV

= 0

I = A [T4 – 40T ]

dI = 4A 30T (dT0) = 4A 3

0T .T0 (fodYi C)

= 4× 460 × 10300 60 J/s (fodYi A)

3. oÙkkdkj pki okys ,d xqVds dk nzO;eku M gSA ;s xqVdk ,d ?k"kZ.k jfgr est i j fLFkr gSA est ds l ki s{; (in a coordinatesystem fixed to the table) xqVds dk nkfguk dksj (right edge) x = 0 i j fLFkr gSA nzO; eku m okys ,d fcanq d.k (point mass)dks oÙkkdkj pki ds mPpre fcanq l s fojkekoLFkk l s NksM+k t krk (released from rest) gSA ; s fcanq d.k oÙkkdkj i Fk i j uhps dh vkSjl jdrk gSA t c fcanq d.k xqVds l s l ai dZ foghu gks t krk gS] r c ml dh r kR{kf.kd fLFkfr x vkSj xfr gSA fuEu esa l s dkSul k (l s)dFku l gh gS@gSa\

R

Rm

M

y

x

x = 0

(A) fcanq d.k (m) dk LFkku 2 mRxM m

gS

(B) xqVds (M) ds l agfr dsUnz ds foLFkki u dk X ?kVd (X co-ordinate) mR

M m

gS

(C) xqVds (M) dk osx 2mV gRM

gS

(D) fcanq d.k (m) dk osx 2

1

gRmM

gS

Page 4: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

4

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

mÙkj (B, D)

gy v = M dk osx

u = m dk osx

mu = –MV ...(i)

2 21 12 2

mgR mu MV ...(ii)

2 – 2

1 1

gR m gRu vm mMM M

rFkkMx = m(R – x) t gk¡ x = CykWd M dk foLFkki u

x(M + m) = mR mRx

m M

ck;ha vksj

;k – mRxm M

4. , d l eku jSf[ kd ?kurkokys (uniform mass per unit length) mèokZèkj Mksj ds fupys fl js i j , d xqVdk M yVdk gqvk gSA

Mksj dk nwl jk fl jk n<+ vkèkkj (fcanq O) l s l ayXu gSA rajx&nSè; Z 0 dh vuqi zLFk r jax Li an (Li an 1, Pulse 1) fcanq O i j

mRi Uu dh xbZ gSA ; s r jax Li Un fcanq O l s fcanq A rd TOA l e; esa i gq¡prh gSA xqVds M dks fcuk fo{kksfHkr fd; s gq, fcanq A

i j fuekZ.k dh xbZ r jax&nSè; Z 0 dh vuqi zLFk r jax Li an (Li an 2, pulse 2), fcanq A l s fcanq O rd TAO l e; esa i gq¡prh gSA fuEu

esa l s dkSul k (l s) dFku l gh gS@gSa\

MA

O Pulse 1

Pulse 2

(A) Li an 1(pulse 1) dh r jax&nSè;Z fcanq A rd i gqapus esa yEch gks t k,xh

(B) l e; TAO = TOA

(C) Mksj ds vuqfn' k i zsf"kr fdl h Hkh Li an dk osx ml dh vkofÙk ,oa r jax&nSè; Z i j fuHkZj ugha gS

(D) Mksj ds eè; fcanq i j Li an 1(pulse 1) ,oa Li an 2(pulse 2) dk osx l eku gS

mÙkj (B, C, D)

gy fdl h fcUnq i j Tv ,

pw¡fd osx ,d fcUnq i j r uko rFkk i zfr bdkbZ yEckbZ nzO;eku i j fuHkZj djr k gS] bl fy, O l s A rd

rFkk A l s O rd l e; l eku gksxkA

TOA = TAO

pw¡fd vkofÙk l Hkh fcUnq i j fu; r jgrh gSA bl s l eku cuk; s j[ kus ds fy, v

l eku gksxk rFkk v, O(T mPpre) i j mPpre

gS bl fy, , O i j mPpre gksxkA

Page 5: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

5

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

5. ,d xksykdkj fo| qr&jks/ h r kez r kj (insulated copper wire) dks A ,oa 2A okys nks {ks=ki Qyksa ds oy; ksa esa O; kofrZr

fd; k x; k gSA rkjksa ds vfrØe.k fcanq fo| qr jks/ h jgr s gS (t Sl k fp=k esa n' kkZ; k x; k gS)A l ai w.kZ oy; dkxt + ds r y

esa fLFkr gSA dkxt ds r y ds vfHkyEcor fLFkj rFkk ,dl eku pqEcdh; {ks=k B l oZ=k mi fLFkr gSA oy; vi us

l keqnkf; d O; kl ksa l s cus v{k ds i fjr% l e; t = 0 l s dks.kh; osx (angular velocity) l s ?kweuk ' kq: djr k gSA

fuEu esa l s dkSul k(l s) dFku l gh gS@gSa\

× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×

area 2A

area A

B

× × × × × × × ×

(A) çsfjr fo| qr okgd cy (emf induced) oy; ksa ds {ks=ki Qyksa ds ; ksx ds l ekuqi kfrd gS

(B) nksuksa oy; ksa l s mRi Uu vf/ dr e dqy çsfjr fo| qr okgd cy (net emf) dk vk; ke] NksVs oy; esa mRi Uuvfèkdre çsf"kr fo| qr okgd cy ds vk; ke ds cjkcj gksxk

(C) t c oy; ksa dk r y dkxt + ds r y l s vfHkyac fn' kk esa gksrk gS r c vfHkokg ds i fjor Zu ds nj vf/ dregksrh gS

(D) nksuksa oy; ksa l s mRi Uu dqy çsfjr fo| qr okgd cy (emf induced) cos t ds l ekuqi krh gS

mÙkj (B, C)

gy = BA cost

e = B sint

ywi 1 ds fy,

1 = BA cost : |e1| = AB sint

2 = 2BA sint |e2| = 2BA sint

e1 o e2 , d nwl js ds foi jhr gSa] bl fy, usV çsfjr fo-ok-c- dk vk; ke = 2BA – BA = BA

e1 rFkk e2, t = /2 ; k = 90° i j f' k[ kj gksaxsA

6. , d l i kV IysV (flat plate) vYi ncko ds xSl (gas at low pressure) esa] vi us r y dh vfHkyac fn' kk esa] ckã cy F dsi zHkko esa vxzl fjr gSA IysV dh xfr ,xSl v.kqvksa ds vkSl r xfr u l s cgqr de gSA fuEu esa l s dkSul k (l s) dFku l gh gS@gSa?

(A) IysV l oZnk ' kqU;sr j fLFkj Roj.k (constant non-zero acceleration) l s pyrh jgsxh

(B) i zfr xkeh ,oa vuqxkeh i "B ds ncko dk var j uvds l ekuqi krh gS

(C) dqN l e; ds ckn ckã cy F vkSj i zfr jksèkd cy l arqfyr gks t k,axs

(D) IysV }kjk vuqHko gqvk i zfr jksèkd cy ds l ekuqi krh gS

Page 6: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

6

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

mÙkj (B, C, D)

gy n = i zfr bdkbZ vk; ru v.kqvksa dh l a[ ; k

u = xSl v.kqvksa dh vkSl r pky

t c IysV v pky l s xfr ' khy gS] r c eq[ ; cy ds l ki s{k v.kqvksa dh l ki sf{kd pky = v + u

l Eeq[ k VDdj esa i zfr VDdj IysV dks LFkkukUr fjr l aosx = 2m(u + v)

l e; t esa VDdj dh l a[ ; k 1 ( )2

u v n tA

t gk¡ A = i "Bh; {ks=ki Qy

bl fy, vkxs dh l rg l s l e; t esa LFkkukUr fjr l aosx = m(u + v)2nAt

i hNs dh l rg l s l e; t esa LFkkukUrfjr l aosx = m(u – v)2nAt

usV cy = mnA[(v + u)2 – (u – v)2]

= mnA[4vu]F v

Pvxz – P

i ' p = mn[u + v]2 – mn[u – v]2

= mn[4uv] = 4mnuvP uv

7. , d l ef}ckgq fi zTe dk fi zTe dks.k A gS (isosceles prism of angle A)A bl fi zTe dk vi orZukad gSA bl fi zTe dkU;wure fopyu dks.k (angle of minimum deviation) m = A gSA fuEu esa l s dkSu l k (l s) dFku l gh gS@gSa\

(A) t c i gys r y i j vki ru dks.k 21 sin 1 sin 4cos 1 cos

2Ai A A

gS] r c bl fi zTe ds fy, f}rh; ry l s fuxZr

fdj.k fi zTe ds i "B l s Li ' khZ; gksxh (tangential to the emergent surface)

(B) U;wure fopyu esa vki fr r dks.k i1 , oa i zFke vi orZd ry ds vi orZd dks.k r1 = (i1/2) }kjk l acafèkr gS

(C) Tkc fi zTe dk vki ru dks.k i1 = A gS rc fi zTe ds Hkhr j i zdk' k fdj.k fi zTe ds vkèkkj ds l ekukUr j gksxhA

(D) fi zTe dk vi orZukad ,oa fi zTe dks.k (A), 11cos2 2

A

}kjk l acafèkr gS

mÙkj (A, B, C)

gy m = (2i) – A

2A = 2i i = A ds fy, r dh x.kuk

i = A rFkk r = A/2 (nk; ha vksj ds gy dks nsf[ k, )

sin2

sin2

A A

A

A

i1c

A

r A1 = /2

A Ar A2 = /2

2sin cos2 2

sin2

A A

A 1sinA = sinr

2cos2A

sinA = 2cos .sin2A r

Page 7: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

7

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

1sini1 = × sin(A – C)

2sin cos2 2sin

2cos2

A A

rA

= sin2A

= 2cos sin cos – cos sin2 C CA A A

2Ar

= 2 12cos sin 1– sin – cos2 CA A A

= 21 cos2cos sin 1– –

2 2cos2

A AAA

= 2

1 cos2cos sin 1– –2 4cos 2cos

2 2

A AAA A

i1 = –1 2sin sin 4cos – 1 – cos2AA A

U;wure fopyu ds fy, ] 1 2Ar

–1cos2 2A

–12cos2

A

[ kaM - 2 (vf/ dre vad % 15)bl [ kaM esa ikap ç'u gSaA

çR; sd ç' u dk mÙkj 0 l s 9 rd (nksuksa ' kkfey) ds chp dk ,d ,dy vadh; i w.kk±d gSA

çR; sd ç'u ds fy, vks- vkj- , l - i j l gh i w.kk±d ds vuq: i cqycqys dks dkyk djsaA

çR; sd ç' u ds fy, vad fuEufyf[ kr i fjfLFkfr ; ksa esa l s fdl ,d ds vuql kj fn; s t k; saxs%

i w.kZ vad : +3 ; fn fl i QZ l gh mÙkj ds vuq: i cqycqys dks dkyk fd; k gSA

' kwU; vad : 0 vU; l Hkh i fjfLFkfr ; ksa esa

8. ,d gkbMªkst u i jek.kq dk ,d bysDVªkWu ni DokaVe l a[ ; k (quantum number) okys d{k l s nf DokaVe l a[ ; k (quantum

number) ds d{k esa ços' k djr k gSA Vi rFkk Vf çkFkfed ,oa vafre fLFkfr t Åt kZ,a gSaA ; fn 6.25,i

f

vv

r c nf

dh U; wure l EHkkoh l a[ ; k (smallest possible nf) gSmÙkj (5)

gy2

2PE 27.2 Zn

H ds fy, Z = 1

227.2PEn

1 227.2PEi

i

Vn

2 2

27.2PE ff

vn

Page 8: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

8

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

22

2 6.25i f f

f ii

V n nV nn

nf = 2.5 ni

52

f

i

nn

ni feuV = 2

nf feuV = 5nf = 5

9. ,do.khZ çdk' k (monochromatic light) vi orZukad n = 1.6 okys ekè; e esa çxkeh gSA ; g çdk' k dk¡p dh phrh(stack of glass layers) i j fupys l r g l s = 30° dks.k i j vki fr r gksrk gS (t Sl k fd fp=k esa n' kkZ; k x; k gS)Adk¡pksa ds Lr j i jLi j l ekar j gSA dk¡p ds phrh ds vi orZukad ,dfn"V nm = n – mn, Øe l s ?kV jgs gSaA ; gk¡m Lr j dk vi orZukad nm gS vkSj n = 0.1 gSA çdk' k fdj.k (m – 1) ,oa m Lr j ds i "Bry l s l ekar j fn' kk esankbZa vksj l s ckgj fudyrk gSA rc m dk eku gksxk

m n m n – n m n – ( – 1)m – 1

321

n n – 3n n – 2n n – n

mÙkj (8)

gy i zFke i jr rFkk m oha i jr ds eè; Lusy fu; e yxkus i j

sin sin90º n n m n

11.6 1.6 0.1 12

m

0.8 80.1

m

10. vk; ksMhu dk l eLFkkfud (isotope)1311, ft l dh v/ Z&vk; q 8 fnu gS] -{k; ds dkj.k t suksau (xenon) ds l eLFkkfudesa {kf; r gksrk gSA vYi ek=k dk 1311 fpfëòr (labelled) l hje (serum) ekuo ' kjhj esa vUr%f{kIr (inject) fd; k x; k]ft l ek=kk dh vWfDVork (activity) 2.4 × 105 csdsjsy (Becquerel) gSA ; g l hje : f/ j / kjk esa vk/ s ?kaVs esa ,dl ekufor fjr gksr k gSA vxj 11.5 ?kaVs ckn 2.5 ml jDr 115 csdsjsy dh vWfDVork n' kkZrk gS] r c ekuo ' kjhj esa jDrv k; r u ( y hVj esa) gS ( v ki ex 1 + x for |x| << 1 ,oa ln 2 0.7 dk mi ; ksx dj l dr s gSaA)

mÙkj (5)

gy ekuk jDr dk dqy vk; ru V ml gS

2.5 ml jDr l fØ; rk = 115 Bq

1 ml jDr l fØ; rk = 1152.5

Page 9: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

9

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

l Ei w.kZ jDr dh l fØ; rk = 115 V Bq2.5

5115 2.4 102.5

tV e

ln2 11.55 8 242.4 10 e

0.7 11.55 8 242.4 10 e

15 242.4 10 e

5 12.4 10 124

5 23 2.5V 2.4 1024 115

5000 ml

jDr dk dqy vk; ru = 5 L

11. i "B&ruko (surface tension) –10.1Nm4

S

ds æo ds ,d cwan dh f=kT; k R = 10–2 m gS] ft l s K l e: i cwanksa esa

foHkkft r fd; k x; k gSA i "B&Åt kZ dk cnyko U = 10–3 Joules gSA ; fn K = 10 gS r c dk eku gksxk

mÙkj (6)

gy ekuk NksVh cw¡n dh f=kT; k = r

3 34 43 3

R K r

13R K r ...(i)

S(K4r2 – 4R2) = 10–3

22 3

23

0.14 104

Rk R

K

12 23 1 10R K

13 1 100K

13 101K

310 101

a 6

Page 10: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

10

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

12. ,d fLFkj L=kksr vkofÙk f0 = 492 Hz dh èofu mRl ft Zr djr k gSA 2 ms–1 ds xfr ds vi xeuh dkj l s ; g èofui jkofrZr gksrh gSA èofu L=kksr i jkofrZr l adsr dks çkIr dj ds ewy l adsr i j vè; kjksfi r (superpose) djr k gSArc i fj.kkeh fl Xuy dh foLi an&vkofÙk (beat frequency) gS

(èofu dh xfr 330 ms–1 gSA dkj èofu dks ml dh çkIr gqbZ vkofÙk i j i jkofrZr djrh gSA)

mÙkj (6)

gy fLFkj L=kksr }kjk mRl ft Zr èofu dh vkofÙk = f0 = 492 Hz

xfr ' khy dkj }kjk i jkofrZr èofu dh vkofÙk

0'

c

c

c vf f

c v

t gk¡ c = ok;q esa èofu dh pky = 330 m/s

vc = dkj dh pky = 2 m/s

gy djus i j

332' 492 498 Hz328

f

i fj.kkeh fl Xuy dh foLi Un vkofÙk = f ' – f0

= (498 – 492) Hz

= 6 Hz

[ kaM-3 (vf/ dre vad : 18)

bl [ kaM esa l qesy i zdkj ds Ng i z'u gSaA

bl [ k.M esa nks Vscy gSa (i zR; sd Vscy esa 3 dkWye vkSj 4 i afDr ; ka gSa)

i zR;sd Vscy i j vk/ kfjr rhu i z'u gSaA

çR;sd i z' u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ft uesa fl iQZ ,d fodYi l gh gSA

i zR; sd i z' u ds fy, vks-vkj-, l - i j l gh mÙkj ds vuq: i cqycqys dks dkyk djsaA

i zR; sd i z' u ds fy, vad fuEufyf[ kr i fjfLFkfr ; ksa esa l s fdl h , d ds vuql kj fn;s t k;saxs %

i w.kZ vad : +3 ; fn fl i QZ l gh fodYi ds vuq: i cqycqys dks dkyk fd; k gSA

' kwU; vad : 0 ; fn fdl h cqycqys dks dkyk ugha fd; k gSA

½.k vad : –1 vU; l Hkh i fjfLFkfr ; ksa esaA

uhps nh x;h Vscy ds rhu dkyeksa esa mi yC/ l wpuk dk mi ;qDr <ax l s l qesy dj i z'uksa Q.13, Q.14 vkSj Q.15 dsmÙkj nhft ;sA

, d pkt Z; qDr d.k (bysDVªkWu ; k i zksVksu) vkjafHkd xfr l s ewy fcUnq (x = 0, y = 0, z = 0) i j i zLrqr (introduced) gksrk

gSA fLFkj rFkk ,dl eku fo| qr~ {ks=k E , oa pqEcdh; {ks=k

B l oZ=k mi fLFkr gSA d.k dh xfr

, fo| qr {ks=k

E rFkk pqEcdh;

{ks=k B fuEu dkWyeksa 1, 2 , oa 3 esa Øe' k% n'kkZ;s x; s gSaA E0, B0 ds eku / ukRed gSaA

Page 11: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

11

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

(I) bysDVªkWu (i) (P)

(II) bysDVªkWu (ii) (Q)

(IV) i zksVksu (iv) (S)

(III) i zksVksu (iii) (R)

0

0

2E

xB

0

0

Ey

B

0

0

0

2E

xB

0E E z

0 E E y

0 E E x

0E E x

0 B B x

0B B x

0B B y

0B B z

l s

l s

l s

l s

13. fdl fLFkfr esa d.k +z–v{k vuqfn' k dqaMfyuh i Fk (helical path along positive z-axis) dk vuql j.k djsxk\

(A) (III) (iii) (P)

(B) (IV) (ii) (R)

(C) (IV) (i) (S)

(D) (II) (ii) (R)

mÙkj (C)

gy i zksVkWu dq.Myhnkj i Fk esa xfr djsxk rFkk v{k / ukRed z-fn' kk ds vuqfn'k gSA

; fn 0

0

2 Ev x

B,

0E E z rFkk

0B B z

14. fdl fLFkfr esa d.k l h/ h js[ kk esa ½.kkRed y-v{k (negative y-axis) dh fn' kk esa pysxk\

(A) (III) (ii) (P) (B) (III) (ii) (R)

(C) (II) (iii) (Q) (D) (IV) (ii) (S)

mÙkj (B)

gy i zksVkWu ½.kkRed y-fn' kk ds vuqfn' k l jy js[ kk esa xfr djsxk t c

0v ,

0 E E y rFkk

0B B y

15. fdl fLFkfr esa d.k vpy xfr l s l h/ h js[ kk esa pyu djr k gS\

(A) (III) (ii) (R) (B) (II) (iii) (S)

(C) (IV) (i) (S) (D) (III) (iii) (P)

mÙkj (B)

gy bysDVªkWu fu; r osx l s l jy js[ kk esa xfr djsxk ; fn

0

0

Ev y

B , 0

E E x ,

0B B z

Page 12: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

12

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

uhps nh x;h Vscy ds rhu dkyeksa esa mi yC/ l wpuk dk mi ;qDr <ax l s l qesy dj i z'uksa Q.16, Q.17 vkSj Q.18

ds mÙkj nhft ;sA

P

V

1 2

P

V

1

2

P

V

1 2

P

V

1

2

(i) l erki h;

l evk; r fud (Isochoric)

l enkch; (Isobaric)

: ¼ks"e (adiabatic)

1 2 2 2 1 11 ––1

W P V PV

1 2 2 1–W PV PV

1 2 0W

21 2

1–

VW nRT ln

V

(I)

,d vkn'kZ xSl fofHkUu pØh; Å"eki kfrd i zØeksa l s xqt jrk gSA ; g fuEu dkWye esa vkjs[ k }kjk n' kkZ; k x; k gSA dsoy fLFkfr l s fLFkfr t kusokys i Fk dh vksj è; ku nsaA bl i Fk i j fudk; i j gqvk dk; Z gS A ; gk¡ fu; r nkc ,oa fu; r vk; ru Å"ek&/ kfjrkvksa dk vuqi kr gS A xSl ds eksyksa dh l a[ ; k gSA

(ideal gas) 3 — 1 2 (work done on the system)

g (ratio of the heat capacities) (moles)

P V W

n

(P)

(ii)(II) (Q)

(iii)(III) (R)

(iv)(IV) (S)

16. fuEu fodYi ksa esa dkSu l k l a; kst u l gh gS\

(A) (II) (iv) (P)

(B) (IV) (ii) (S)

(C) (III) (ii) (S)

(D) (II) (iv) (R)

mÙkj (C)

gy fodYi (C) l gh l a; kst u gSA

W1 2 = 0 l evk; r fud i zØe

Page 13: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

13

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

17. fuEu fodYi ksa esa l s dkSu l k l a; kst u vkn' kZ xSl esa èofu dh xfr dh eki ds l a' kks/ u esa i z; qDr Å"ekxfrd i zfØ; k dks l ghn' kkZrk gS\

(A) (I) (iv) (Q)

(B) (III) (iv) (R)

(C) (I) (ii) (Q)

(D) (IV) (ii) (R)

mÙkj (A)

gy : ¼ks"e i zØe dks ,d vkn' kZ xSl esa èofu dh pky ds fu/ kZj.k esa l a' kks/ u ds : i esa i z; qDr fd; k t krk gSA

l gh fodYi (A) gS

:¼ks"e i zØe esa fudk; i j fd; k x; k dk; Z = 2 2 1 1– –

– 1P V PV

18. fuEu fn, fodYi ksa esa dkSul k l a; kst u U = Q – P V i zfØ; k dk vdsys l gh i zfr fuf/ Ro djr k gS\

(A) (II) (iii) (P)

(B) (III) (iii) (P)

(C) (II) (iii) (S)

(D) (II) (iv) (R)

mÙkj (A)

gy fodYi (A) l gh i zn' kZu gSA

W1 2 = –PV2 + PV1 l enkch;

END OF PHYSICS

Page 14: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

14

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

CHEMISTRY

[kaM[kaM[kaM[kaM[kaM-1 (vf/dre vad (vf/dre vad (vf/dre vad (vf/dre vad (vf/dre vad : 28)))))

• bl [kaM esa lkrlkrlkrlkrlkr iz'u gSaA

• çR;sd iz'u ds pkjpkjpkjpkjpkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ,d ;k ,d ls vf/d,d ;k ,d ls vf/d,d ;k ,d ls vf/d,d ;k ,d ls vf/d,d ;k ,d ls vf/d fodYi lgh gSaA

• izR;sd iz'u ds fy, vks-vkj-,l- ij lkjs lgh mÙkj (mÙkjksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk djsaA

• izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %fdlh ,d ds vuqlkj fn;s tk,axs %fdlh ,d ds vuqlkj fn;s tk,axs %fdlh ,d ds vuqlkj fn;s tk,axs %fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +4 ;fn fliQZ lkjs lgh fodYi (fodYiksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk fd;k gSA

vkaf'kd vad : +1 izR;sd lgh fodYilgh fodYilgh fodYilgh fodYilgh fodYi ds vuq:i cqycqys dks dkyk djus ij] ;fn dksbZ xyr fodYi dkykugha fd;k gSA

'kwU; vad : 0 ;fn fdlh cqycqys dks dkyk ughaughaughaughaugha fd;k gSA

½.k vad : –2 vU; lHkh ifjfLFkfr;ksa esaA

• mnkgj.k % ;fn ,d iz'u ds lkjs lgh mÙkj fodYi (A), (C) vkSj (D) gSa] rc bu rhuksa ds vuq:i cqycqyksa dks dkyk djus ij+4 vad feysaxs_ fliQZ (A), (D) ds vuq:i cqycqyksa dks dkyk djus ij +2 vad feysaxsa_ rFkk (A) vkSj (B) ds vuq:i cqycqyksadks dkyk djus ij –2 vad feysaxs D;ksafd ,d xyr fodYi ds vuq:i cqycqys dks Hkh dkyk fd;k x;k gSA

19. lewg 17 ds rRoksa ds X2 v.kqvksa dk jax buds oxZ esa uhps tkus ij ihys jax ls /hjs&/hjs cSaxuh jax esa cnyrk gSA ;g fuEu esa

ls fdlds iQyLo:i gS\

(A) lkekU; rki ij oxZ esa uhps tkus ij X2 dh HkkSfrd voLFkk xSl ls Bksl esa cnyrh gS

(B) oxZ esa uhps tkus ij π*-o* dk varj ?kVrk gS

(C) oxZ esa uhps tkus ij HOMO-LUMO dk varj ?kVrk gS

(D) oxZ esa uhps tkus ij vk;uu mQtkZ ?kVrh gS

mÙkjmÙkjmÙkjmÙkjmÙkj (B, C)

gygygygygy jax çdk'k ds vo'kks"k.k ls bysDVªkWu ds vkè;koLFkk ls mPp voLFkk esa mÙksftr gksus ds dkj.k mRiUu gksrk gSA oxZ esa uhps

pyus ij ÅtkZ Lrj lehi vk tkrs gSa rFkk HOMO-LUMO ds eè; vUrjky de gks tkrk gS

HOMO, π* gS

LUMO, σ* gS

20. HClO4 vkSj HClO ds ckjs esa lgh dFku gS(gSa)

(A) HClO4 vkSj HClO nksuksa esa dsUnzh; ijek.kq sp3 ladfjr gS

(B) ½.kk;u ds vuqukn fLFkjhdj.k (resonance stabilization) ds iQyLo:i HClO4, HClO ls vf/d vEyh; gS

(C) HClO4 dk la;qXeh {kkj (conjugate base) H

2O ls nqcZy {kkj gS

(D) Cl2 dh H

2O ds lkFk vfHkfØ;k gksus ij HClO

4 curk gS

mÙkjmÙkjmÙkjmÙkjmÙkj (A, B, C)

gygygygygy (A) HCIO4 rFkk HClO nksuksa esa dsfUnz; ijek.kq sp3 ladfjr gS

(B) HClO4 > HClO vEyh; lkeF;Z

+

4 4HClO H + ClO

−⎯⎯→+

HClO H + ClO−⎯⎯→

Page 15: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

15

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

CI

O–

O

OO

CI

O

O

O–

O

CI

O

OO

O–

CI

O

O–

OO

4ClO

− esa ½.kkos'k pkj vkWDlhtu ij iSQyk gqvk gSA vr% mÙke vuquknh LFkkf;Ro gS

(C) HClO4 > H

2O vEyh; y{k.k

ClO4

– < OH– la;qXeh {kkj

çcy vEyksa ds la;qXeh {kkj nqcZy gksrs gSaA

21. fuEufyf[kr ;kSfxd dk(ds) vkbZ- ;w- ih- ,s- lh- (IUPAC) uke gS(gaS)

ClH C3

(A) 1-Dyksjks-4-eSfFky csathu (B) 1-eSfFky-4-Dyksjkscsathu(C) 4-Dyksjks Vksyqbu (D) 4-eSfFkyDyksjks csathu

mÙkjmÙkjmÙkjmÙkjmÙkj (A, C)

gygygygygy IUPAC uke

(A)

Cl

CH3

1- -4-Dyksjks esfFkycsathu

1

2

3

4

(C)

CH3

Cl

4-DyksjksVksyqbu

1

2

3

4

22. ,d xqykch jax okys MCl2·6H

2O(X) vkSj NH

4Cl ds tyh; foy;u esa vf/D; tyh; veksfu;k ds feykus ij] ok;q dh mifLFkfr

esa ,d v"ViQydh; ladj (octahedral complex) Y nsrk gSA tyh; foy;u esa ladj Y 1:3 fo|qr vi?kV~; (electrolyte) dhrjg O;ogkj djrk gSA lkekU; rki ij vf/D; HCl ds lkFk X dh vfHkfØ;k ds ifj.kkeLo:i ,d uhys jax dk ladj Z curkgSA X vkSj Z dk ifjdfyr izpdj.k ek=k pqEcdh; vk?kw.kZ (spin only magnetic moment) 3.87 B.M. gS] tcfd ;g ladjY ds fy, 'kwU; gSA fuEu esa ls dkSulk (ls) fodYi lgh gS (gSa)\

(A) tc 0°C ij X vkSj Z lkE;koLFkk esa gSa rks foy;u dk jax xqykch gS

(B) Y esa dsUnzh; /krq vk;u dk ladj.k (hybridization) d2sp3 gS

(C) Z ,d prq'iQYdh; (tetrahedral) ladj gS

(D) Y esa flYoj ukbVªsV feykus ij flYoj DyksjkbM ds dsoy nks lerqY; feyrs gSa

mÙkjmÙkjmÙkjmÙkjmÙkj (A, B, C)

gygygygygy (X) = CoCl2⋅6H

2O, ;k, [Co(H

2O)

6]Cl

2 (xqykch)

Co =2+

[Ar]

3d

H2O nqcZy {ks=k yhxs.M gSA vr% bysDVªkWuksa dk ;qXeu ugha gksxk

∴ v;qfXer bysDVªkWuksa dh la[;k n = 3

⇒s

n(n 2)B.M.μ = +

= 15 B.M.= 3.87 B.M.

Page 16: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

16

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

2 22 6 3(aq) 3 6 2[Co(H O) ] 6NH [Co(NH ) ] 6H O+ ++ +���⇀

↽���

O2 ; [Co(NH

3)6]2+ dks [Co(NH

3)6]3+, esa vkWDlhÑr djsxhA vr% vxz fn'kk esa LFkkukUrfjr gksxk

∴ Y = [Co(NH3)6]3+Cl

3[1 : 3 ladqy]

Co(III) = [Ar]

3d

NH3 çcy {ks=k yhxs.M gS

∴ bysDVªkWuksa dk ;qXeu gksxk

∴ n = 0

μ = 0 B.M.

[Co(NH ) ] = [Ar]3 6

3+

3d 4s 4p

d sp2 3

vkSj, [Co(H O) ] + 4Cl2 6

2+ –[CoCl ] + 6H O; H = +ve4 (aq) 2

2– Δ( )

( )

X

xqykch jax( )Z

uhyk

0°C ij lkE; i'p fn'kk esa LFkkukUrfjr gksxkA vr% xqykch jax

23. fuEufyf[kr ladyu vfHkfØ;kvksa (addition reactions) ds fy, lgh dFku gS (gSa)

(i)

H C3

H

H

CH3

Br /CHCl2 3

M N vkSj

(ii)H C

3

H H

CH3 Br /CHCl

2 3

O P vkSj

(A) O vkSj P le:i v.kq gSa

(B) (M vkSj O) vkSj (N vkSj P) ,uUVhvksesjks (enantiomers) ds nks ;qxy gSa

(C) (M vkSj O) vkSj (N vkSj P) MkbZLVhfjvksesjksa (diastereomers) ds nks ;qxy gSa

(D) nksuksa vfHkfØ;kvksa esa czksfefudj.k Vªkal ladyu }kjk c<+rk gS

mÙkjmÙkjmÙkjmÙkjmÙkj (C, D)

gygygygygy C = CH

CH3

H C3

H

foi{k

Br /CHCl2 3

CH3

Br

Br

CH3

H

H+

CH3

H

H

CH3

Br

Br

eslks ,oa (M N)

CH3

Br

H

CH3

H

Br+

CH3

H

Br

CH3

Br

H

çfrfcEc leko;oh ;qXeO P

(i)

C = CCH

3

H

CH3

H

Br2

(ii)

Page 17: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

17

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

lei{k izfr jslhfedlei{k izfr jslhfedlei{k izfr jslhfedlei{k izfr jslhfedlei{k izfr jslhfed

M rFkk N eslks gS (le:i)

O rFkk P çfrfcEc leko;oh ;qXe gS

(C) (M rFkk O) rFkk (N rFkk P) vçfrfcEc f=kfoe leko;fo;ksa ds nks ;qXe gSA

(D) czksehuhdj.k çfr&;ksx }kjk lEiUu gksrh gS

24. ,d vkn'kZ xSl dks (p1, v

1, T

1) ls (p

2, v

2, T

2) rd fofHk voLFkkvksa ds v/hu iQSyk;k x;k gSA fuEufyf[kr fodYiksa esa lgh

dFku gS (gSa)

(A) tc bls vuqRØe.kh; rjhds ls (irreversibly) (p2, v

2) ls (p

1, v

1) rd fLFkj nkc p

1 ds fo:¼ nck;k tkrk gS rks xSl ds

mQij fd;k x;k dk;Z vf/dre gksrk gS

(B) xSl dh vkarfjd mQtkZ esa cnyko (i) 'kwU; gS ;fn bls T1 = T

2 ds lkFk iQSyko mRØe.kh; (reversible) rjhds ls fd;k

tk,] vkSj (ii) /ukRed gS ;fn bls T1 ≠ T

2 ds lkFk :¼ks"e (adiabatic) ifjfLFkfr;ksa ds v/hu mRØe.kh; (reversible)

iQSyko fd;k tk;

(C) ;fn iQSyko eqDr :i ls fd;k tk; rks ;g lkFk&lkFk nksuksa lerkih (isothermal) ,oa :¼ks"e (adiabatic) gS

(D) tc v1 ls v

2 rd :¼ks"e voLFkk ds v/hu bldk mRØe.kh; (reversible) iQSyko fd;k tk; rks xSl }kjk fd;k x;k dk;Z

v1 ls v

2 rd lerkih (isothermal) voLFkkvksa ds v/hu mRØe.kh; iQSyko esa fd;s x, dk;Z dh rqyuk esa de gS

mÙkjmÙkjmÙkjmÙkjmÙkj (A, C, D)

gygygygygy (A) vuqRØe.kh; lEihM+u esa P o V vkjs[k ds uhps dk {ks=kiQy mRØe.kh; lEihM+u esa P o V vkjs[k ds uhps ds {ks=kiQy ls

vfèkd gS

P1

P2

P

(P , V )1 2

(P , V )2 2

V1

V2

V

(P , V )1 1

vuqRØe.kh;mRØe.kh;

(C) eqDr çlkj esa Pex

= 0 ⇒ w = 0

,d vkn'kZ xSl ds lerkih; eqDr çlkj ds fy,,

ΔU = 0 ⇒ q = 0 ∴ :¼ks"e Hkh gS

(D)

V1

V2

V

P

W > Wlerkih; :¼ks"elerkih;

:¼ks"e

25. L vkSj M nzoksa ds feJ.k }kjk cuk;s ,d foy;u esa nzo M ds xzke&v.kqd fHk (mole fraction) ds fo:¼ nzo L ds okLinkc dks fp=k esa fn[kk;k x;k gSA ;gk¡ x

L vkSj x

M, L vkSj M ds Øe'k% xzke&v.kqd fHkÂksa dks fu:fir djrs gSaA bl fudk; dk

(ds) mi;qDr lgh dFku gS (gSa)

Page 18: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

18

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

PL

Z

xM1 0

(A) fcanq Z 'kq¼ nzo M ds ok"i nkc dks fu:fir djrk gS vkSj xL = 0 ls x

L = 1 rd jkmYV dk fu;e (Raoult's law) dk

ikyu gksrk gS

(B) fcanq Z 'kq¼ nzo M ds ok"i nkc dks fu:fir djrk gS vkSj tc xL → 0 rks jkmYV dk fu;e (Raoult's law) dk ikyu

gksrk gS

(C) 'kq¼ nzo L esa L-L ds chp esa vkSj 'kq¼ nzo M esa M-M ds chp esa varjk&v.kqd fØ;k,a L-M ds chp esa varjk&v.kqdfØ;kvksa ls izcy gSa tc mUgsa foy;u esa fefJr fd;k tkrk gS

(D) fcanq Z 'kq¼ nzo L ds ok"i nkc dks fu:fir djrk gS vkSj tc xL → 1 rks jkmYV dk fu;e (Raoult's law) dk ikyu

gksrk gS

mÙkjmÙkjmÙkjmÙkjmÙkj (C, D)

gygygygygy Z

x = 0M

x = 1L

'kq¼ L

xM

→x = 1M

M'kq¼

pL

fcUnq Z 'kq¼ nzo L ds ok"i nkc dks iznf'kZr djrk gS

xL

→ 1 ij foy;u cgqr ruq gS] L foyk;d cu tkrk gSA L esa m dk cgqr ruq foy;u yxHkx vkn'kZ gS rFkk jkÅV fu;e

(pL = x

Lp

L

°) dk ikyu djrk gSA

vkSj fcUnqfdr js[kk (vkn'kZ foy;u ds fy, visf{kr) ds Åij xzkiQ }kjk fufnZ"V gS fd blesa /ukRed fopyu gS

∴ L-M ikjLifjd fØ;k < L - L ,oa M - M ikjLifjd fØ;k

[kaM[kaM[kaM[kaM[kaM 2 (vf/dre vad(vf/dre vad(vf/dre vad(vf/dre vad(vf/dre vad % % % % % 15)))))

• bl [kaM esa ik¡p iz'u gSaA

• izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,d ,dy vadh; iw.kkZ ad,dy vadh; iw.kkZ ad,dy vadh; iw.kkZ ad,dy vadh; iw.kkZ ad,dy vadh; iw.kkZ ad gSA

• izR;sd iz'u ds fy, vks-vkj-,l- ij lgh iw.kkZad ds vuq:i cqycqys dks dkyk djsaA

• izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfRk;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs%fdlh ,d ds vuqlkj fn;s tk,axs%fdlh ,d ds vuqlkj fn;s tk,axs%fdlh ,d ds vuqlkj fn;s tk,axs%fdlh ,d ds vuqlkj fn;s tk,axs%

iw.kZ vad % +3 ;fn fliQZ lgh mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

26. fuEufyf[kr oxZ (species) esa çR;sd dsUnzh; ijek.kq ij ,dkdh bysDVªku ;qXeksa dh la[;k dk ;ksx gS

[TeBr6]2–, [BrF

2]+, SnF

3 rFkk [XeF

3]–

(ijek.kq la[;k % N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54)

Page 19: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

19

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

mÙkjmÙkjmÙkjmÙkjmÙkj (6)

gygygygygy

Te

Br

Br

Br

F

Br

Br

Br

Br

F

esa ,d ,dkdh ;qXe gS

esa nks ,dkdh ;qXe gS

Xe

S N≡

F

F

F

F

F

esa rhu ,dkdh ;qXe gS

esa ,dkdh ;qXe ugha gSF

27. ,d 'kq¼ inkFkZ ds ,d fØLVyh; Bksl dh iQyd&dsfUnzr ?ku (face-centred cubic) lajpuk ds lkFk dksfLBdk dksj (cell

edge) dh yEckbZ 400 pm gSA ;fn fØLVy ds inkFkZ dk ?kuRo 8 g cm–3 gS] rks fØLVy ds 256 g esa mifLFkr ijek.kqvksa dh

dqy la[;k N × 1024 gSA N dk eku gS

mÙkjmÙkjmÙkjmÙkjmÙkj (2)

gygygygygy a = 4 × 10–8 cm (a = dksj yEckbZ)

d = 8 g cm–3 (?kuRo)

3

A

ZMd

N a= M = vk.kfod nzO;eku (g/mol)

Z → 1 ,dd dksf"Bdk esa ijek.kqvksa dh la[;k

3 23 –24

AdN a 8 6 10 64 10

M 76.8 g/molZ 4

× × × ×= = =

256 g esa Bksl ds eksy = 3.33 eksy

ijek.kqvksa dh la[;k = 3.33 × NA = 20 × 1023 = 2 × 1024

28. fuEufyf[kr esa ls ,jksesfVd ;ksfxd (;ksfxdksa) dh la[;k gS

++

+

Page 20: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

20

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

mÙkjmÙkjmÙkjmÙkjmÙkj (5)

gygygygygy +

+

,,

, rFkk ,jksesfVd gS

rFkk vu,jksesfVd gSa tcfd rFkk + çfr,jksesfVd ;kSfxd gaS

29. H2, He

2, Li

2, Be

2, B

2, C

2, N

2, vkSj F

2 esa çfrpqEcdh; Lih'kht (diamagnetic species) dh la[;k gS

(ijek.kq la[;k % H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, F = 9)

mÙkjmÙkjmÙkjmÙkjmÙkj (6)

gygygygygy H2 2

sçfrpqEcdh;

2

2 * 1

1 1He

s s

+ σ σ vuqpqEcdh;

2 * 2 2

2 1 1 2Li

s s sσ σ σ çfrpqEcdh;

2 * 2 2 * 2

2 1 1 2 2Be

s s s sσ σ σ σ çfrpqEcdh;

2 * 2 2 * 2 1 1

2 1 1 2 2 2 2B

x ys s s s p p

σ σ σ σ π = π vuqpqEcdh;

2 * 2 2 * 2 2 2

2 1 1 2 2 2 2C

x ys s s s p p

σ σ σ σ π = π çfrpqEcdh;

2 * 2 2 * 2 2 2 2

2 1 1 2 2 2 2 2N

x y zs s s s p p p

σ σ σ σ π = π σ çfrpqEcdh;

2 * 2 2 * 2 2 2 2 * 1 * 1

2 1 1 2 2 2 2 2 2 2O

z x y x ys s s s p p p p p

σ σ σ σ σ π = π π = π vuqpqEcdh;

2 * 2 2 * 2 2 2 2 * 2 * 2

2 1 2 2 2 21 2 2 2F ,

z x y x ys s p p ps s p p

σ σ σ σ σ π = π π = π çfrpqEcdh;

30. ,d nqcZy ,d{kkjdh; vEy ds 0.0015 M tyh; foy;u dh pkydRo (conductance) ,d IykfVfuÑr Pt (platinized Pt)

bysDVªksM okys pkydrk lSy dk mi;ksx dj ds fu/kZfjr dh x;hA 1 cm2 vuqçLFk dkV ds {ks=kiQy okys bysDVªksM+ks ds chp dhnwjh 120 cm gSA bl foy;u dh pkydRo dk eku 5 × 10–7 S ik;k x;kA foy;u dk pH 4 gSA bl nqcZy ,d{kkjdh; vEy

dh tyh; foy;u esa lhekUr eksyj pkydrk (limiting molar conductivity) ( )o

mΛ dk eku Z × 102 S cm–1 mol–1 gSA Z

dk eku gS

mÙkjmÙkjmÙkjmÙkjmÙkj (6)

gygygygygyl

CA

⎛ ⎞κ = ⎜ ⎟⎝ ⎠

7 120

5 101

−= × ×

= 6 × 10–5 S cm–1

M

1000

M

κ ×λ =

5

4

6 10 1000

15 10

−× ×=

×

⇒ λM

= 40 S cm2 mol–1.

[H+] = Cα = 10–4

Page 21: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

21

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

4

4

10 1

1515 10

−α = =×

⇒ M

º

M

λα =

λ

⇒M

°λ = 40 × 15 = 600 = 6 × 102 S cm–1 mol–1

[kaM[kaM[kaM[kaM[kaM-3 (vf/dre vad (vf/dre vad (vf/dre vad (vf/dre vad (vf/dre vad : 18)))))

• bl [kaM esa lqesy izdkj ds NgNgNgNgNg iz'u gSaA

• bl [kaM esa nks Vscy gSa (izR;sd Vscy esa 3 dkye vkSj 4 iafDr;k¡ gSa)A

• izR;sd Vscy ij vkèkkfjr rhurhurhurhurhu iz'u gSaA

• çR;sd iz'u ds pkjpkjpkjpkjpkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa fliZQ ,d ,d ,d ,d ,d fodYi lgh gSA

• izR;sd iz'u ds fy, vks-vkj-,l- ij lgh mÙkj ds vuq:i cqycqys dks dkyk djsaA

• izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,saxs %fdlh ,d ds vuqlkj fn;s tk,saxs %fdlh ,d ds vuqlkj fn;s tk,saxs %fdlh ,d ds vuqlkj fn;s tk,saxs %fdlh ,d ds vuqlkj fn;s tk,saxs %

iw.kZ vad : +3 ;fn lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

uhps nh x;h Vscy ds rhu dk¡yeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa vkSj ds mÙkj nhft;sA

rjax iQyu] ,d xf.krh; iQyu gS ftldk eku bysDVªkWu ds xksyh; /zqoh; funsZ'kkad ij fuHkZj djrk gS vkSj DokaVe la[;k

vkSj ls vfHkyf{kr gksrk gSA ;gk¡ uwfDyvl ls nwjh gS] dksfV'kj gS] vkSj fnUx'k gSA Vscy esa fn, x;s xf.krh; iQyuksa esa ijek.kq Øekad gS vkSj cksj f=kT;k gSA

vkfcZVy

Q.31, Q32

ψl m r (colatitude) (azimuth)

z a (Bohr radius)

(I) 1s (orbital) (i) (P)

(II) 2s (orbital) (ii) (radial) (Q)

(Probability density)

(III) 2p (orbital) (iii) (R)

(Probability density)

(IV) 3d (orbital) (iv) xy- (S) n = 2 n = 4

n = 2

n = 6

1

0

z

z

θ φ

dkye dkye dkye 3I 2

vkfcZVy ,d f=kT;kRed uksM uwfDyvl ij izkf;drk ?kuRo

vkfcZVy uwfDyvl ij izkf;drk ?kuRo

vf/dre gS

vkfcZVy lery ,d uksMh; ry gS bysDVªksu dks voLFkk ls voLFkk rd

mÙksftr djus dh mQtkZ] bysDVªksu dks

voLFkk ls voLFkk rd mÙksftr djus ds

fy, vko';d mQtkZ ls xquk gS

2

Q.33

(r, , ) n, n,j,m θ φ

0

3

2

, ,0

zr

an j m

Ze

a

⎛ ⎞− ⎜ ⎟⎝ ⎠⎛ ⎞ψ ∝ ⎜ ⎟

⎝ ⎠

ψn,l,m

1(r)

a r/a0

0

30

1

a∝

0

5 zr

2 an,j,m

0

Ze cos

a

⎛ ⎞− ⎜ ⎟⎝ ⎠⎛ ⎞ψ ∝ θ⎜ ⎟⎝ ⎠

27

32

Page 22: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

22

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

31. He+ vk;u ds fy, fuEufyf[kr fodYiksa esa ls dsoy xyrxyrxyrxyrxyr (INCORRECT) la;kstu gS

(A) (I) (iii) (R) (B) (II) (ii) (Q)

(C) (I) (i) (R) (D) (I) (i) (S)

mÙkjmÙkjmÙkjmÙkjmÙkj (A)

gygygygygy 1s d{kd vfn'kkRed gS vr ψ cosθ ij fuHkZj ugha djsxh

vr% (A) xyr gS

32. gkbMªkstu ijek.kq ds fy, fuEufyf[kr fodYiksa esa ls dsoy lghlghlghlghlgh la;kstu gS

(A) (I) (iv) (R) (B) (II) (i) (Q)

(C) (I) (i) (S) (D) (I) (i) (P)

mÙkjmÙkjmÙkjmÙkjmÙkj (C)

gygygygygy H-ijek.kq ds fy,

1s-d{kd 0

3 Zr

2 a

0

Ze

a

⎛ ⎞−⎜ ⎟⎝ ⎠

⎛ ⎞ψ ∝⎜ ⎟

⎝ ⎠

vkSj, 4 2

3E E

16− =

6 2

2E E

9− =

vr% 6 2 4 2

27(E E ) E E

32− × = −

33. dkye&1 esa fn, x;s vkfcZVy (Orbital) ds fy, fuEufyf[kr fodYiksa esa ls fdlh Hkh gkbMªkstu&leku Lih'kht (species) ds

fy, dsoy lghlghlghlghlgh la;kstu gS

(A) (IV) (iv) (R) (B) (III) (iii) (P)

(C) (I) (ii) (S) (D) (II) (ii) (P)

mÙkjmÙkjmÙkjmÙkjmÙkj (D)

gygygygygy H- ds leku Lih'kht ds fy, dsoy D lgh gS

D;kasfd

(A) esa] 2

z3d , ds fy, xy ry uksMh; ry ugha gS

(B) esa] 2pz d{kd esa f=kT;h; uksM ugha gS

(C) esa] 1s d{kd esa f=kT;h; uksM ugha gS

uhps nh x;h Vscy ds rhu dk¡yeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dk¡yeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dk¡yeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dk¡yeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dk¡yeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa 34, 35 ,oa ,oa ,oa ,oa ,oa 36 ds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sA

dkye 1, 2 3 vkSj esa Øe'k% vkjfEHkd inkFkZ] vfHkfØ;k voLFkk,a] vkSj vfHkfØ;kvksa ds izdkj gSaA

(I) (Toluene)VkyqbZu

(II) (Acetophenone)vflVksiQsuksau

(III) (Benzaldehyde)csfUtYMgkbM

(IV) (Phenol)isQuksy

(i) NaOH/Br2

(ii) Br /hv2

(iii) (CH CO) O/CH COOK3 2 3

(iv) NaOH/CO2

(P) (Condensation)la?kuu

(Q) (Carboxylation)dkcksZfDLydj.k

(R) (Substitution)izfrLFkkiu

(S) (Haloform)gkyksiQeZ

Page 23: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

23

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

34. fuEufyf[kr fodYiksa esa ls dsoy lghlghlghlghlgh la;kstu ftlesa vfHkfØ;k ewyd (Radical) izfØ;k }kjk c<+rh gS] gS

(A) (II) (iii) (R) (B) (I) (ii) (R)

(C) (III) (ii) (P) (D) (IV) (i) (Q)

mÙkjmÙkjmÙkjmÙkjmÙkj (B)

gygygygygy

CH3

CH2

CH2

CH – Br2

+ Br

+ Br2

+ HBr

+ Br

Br2

Br + Brhν

35. fuEufyf[kr fodYiksa esa ls dsoy lghlghlghlghlgh la;kstu tks fd nks fHkUu dkcksZfDlfyd vEy nsrk gS] gS

(A) (II) (iv) (R) (B) (I) (i) (S)

(C) (III) (iii) (P) (D) (IV) (iii) (Q)

mÙkjmÙkjmÙkjmÙkjmÙkj (C)

gygygygygy

CH = O

CH – C3

CH – C3

O

O

O

CH COOK3

CH = CH – COOH

flusfed vEy

;g nks T;kferh; :iksa esa ik;k tkrk gS

C –H

H – C – COOH

C –H

HOC – C – H

O

lei{k leko;ofoi{k leko;o

;g i£du vfHkfØ;k dk ewy mnkgj.k gS

36. csUtksbZd vEy ds la'ys"k.k (synthesis) ds fy, fuEufyf[kr fodYiksa esa ls dsoy lghlghlghlghlgh la;kstu gS

(A) (II) (i) (S) (B) (IV) (ii) (P)

(C) (I) (iv) (Q) (D) (III) (iv) (R)

mÙkjmÙkjmÙkjmÙkjmÙkj (A)

gygygygygy C CH3

O

NaOH

Br2

COO Na– +

+ CHBr3

H O/H2

+

COOH

;g gSyksiQkWeZ vfHkfØ;k gS

Page 24: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

24

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

MATHEMATICS

[kaM[kaM[kaM[kaM[kaM-1 (vf/dre vad (vf/dre vad (vf/dre vad (vf/dre vad (vf/dre vad : 28)))))

bl [kaM esa lkrlkrlkrlkrlkr iz'u gSaA

çR;sd iz'u ds pkjpkjpkjpkjpkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ,d ;k ,d ls vf/d,d ;k ,d ls vf/d,d ;k ,d ls vf/d,d ;k ,d ls vf/d,d ;k ,d ls vf/d fodYi lgh gSaA

izR;sd iz'u ds fy, vks-vkj-,l- ij lkjs lgh mÙkj (mÙkjksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk djsaA

izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %fdlh ,d ds vuqlkj fn;s tk,axs %fdlh ,d ds vuqlkj fn;s tk,axs %fdlh ,d ds vuqlkj fn;s tk,axs %fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +4 ;fn fliQZ lkjs lgh fodYi (fodYiksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk fd;k gSA

vkaf'kd vad : +1 izR;sd lgh fodYilgh fodYilgh fodYilgh fodYilgh fodYi ds vuq:i cqycqys dks dkyk djus ij] ;fn dksbZ xyr fodYi dkykugha fd;k gSA

'kwU; vad : 0 ;fn fdlh cqycqys dks dkyk ughaughaughaughaugha fd;k gSA

½.k vad : –2 vU; lHkh ifjfLFkfr;ksa esaA

mnkgj.k % ;fn ,d iz'u ds lkjs lgh mÙkj fodYi (A), (C) vkSj (D) gSa] rc bu rhuksa ds vuq:i cqycqyksa dks dkyk djus ij+4 vad feysaxs_ fliQZ (A), (D) ds vuq:i cqycqyksa dks dkyk djus ij +2 vad feysaxsa_ rFkk (A) vkSj (B) ds vuq:i cqycqyksadks dkyk djus ij –2 vad feysaxs D;ksafd ,d xyr fodYi ds vuq:i cqycqys dks Hkh dkyk fd;k x;k gSA

37. ekuk fd X vkSj Y bl izdkj dh nks ?kVuk;sa (events) gSa fd P(X) = 1 1, |

3 2P X Y vkSj 2

|5

P Y X gSA rc

(A)4

( )15

P Y (B)2

( )5

P X Y ∪

(C) 1|

2P X Y (D)

1( )

5P X Y ∩

mÙkjmÙkjmÙkjmÙkjmÙkj (A, C)

gygygygygy1 1 2

( ) , ,3 2 5

X YP X P P

Y X

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) 1

( ) 2

P X YXP

P YY

⎛ ⎞ ⎜ ⎟⎝ ⎠

( ) 2

( ) 5

P X YYP

P XX

⎛ ⎞ ⎜ ⎟⎝ ⎠

4 1 2( ) , ( ) , ( )

15 3 15P Y P X P X Y⇒

( ) ( ) ( )

( ) ( )

P X Y P Y P X YXP

P Y P YY

⎛ ⎞ ⎜ ⎟⎝ ⎠

4 2

115 15

4 2

15

P(X Y) = 1 4 2

3 15 15

9 2 7

15 15

Page 25: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

25

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

38. ekuk fd x ls NksVk ;k x ds leku lcls cM+k iw.kk±d (integer) [x] gSA rc f(x) = xcos((x + [x])) fuEu esa ls fdu fcUnq(vksa)ij vlrr (discontinuous) gS\

(A) x = 0 (B) x = 2

(C) x = –1 (D) x = 1

mÙkjmÙkjmÙkjmÙkjmÙkj (B, C, D)

gygygygygy f(x) = xcos((x + [x]))

cos ,[ ]

cos , [ ]

x x x

x x x

⎧ ⎪ ⎨ ⎪⎩

le gS

fo"ke gS

Li"Vr% f(1+) f(1), f(2+) f(2), f(–1+) f(–1–)

ijUrq f(0) = f(0+) = f(0) = 0 vr% f, x = 1, –1, 2 ij vlrr~ gS ijUrq x = 0 ij lrr~ gS

39. ekuk fd : (0,1)f � ,d lrr iQyu (continuous function) gSA rc fuEu iQyuksa esa ls dkSuls iQyu(uksa) dk(ds) eku

vUrjky (interval) (0, 1) ds fdlh fcUnq ij 'kwU; gksxk\

(A)0

( )sinx

x

e f t t dt ∫ (B) x9 – f(x)

(C) 2

0

( ) ( )sinf x f t t dt

∫ (D) 2

0

( )cosx

x f t t dt

mÙkjmÙkjmÙkjmÙkjmÙkj (B, D)

gygygygygy 2

0

( ) ( )cosx

g x x f t t dt

2

0

(0) 0 ( )cos 0g f t t dt

∫ pw¡fd 0 ( ) cos < 1f t t

12

0

(1) 1 ( )cos 0g f t t dt

9( ) ( ) (0) 0 (0) 0g x x f x g f

(1) 1 (1) 0g f

rFkk (0, 1) ds fy,

0( )sin 0

xx

e f t t dt ∫

rFkk

/2

0( ) ( )sin 0f x f t t dt

∫ (0,1)x

40. ekuk fd a, b, x vkSj y bl izdkj dh okLrfod la[;k;sa (real numbers) gSa fd a – b = 1 vkSj y 0 gSaA ;fn lfEeJ la[;k

(complex number) z = x + iy, Im1

az by

z

⎛ ⎞ ⎜ ⎟⎝ ⎠ dks larq"V djrh gS] rc fuEu esa ls dkSulk(ls) x dk(ds) lEHkkfor eku

gS(gSa)?

(A) 21 1 y (B) 2

1 1 y

(C) 21 1 y (D) 2

1 1 y

mÙkjmÙkjmÙkjmÙkjmÙkj (C, D)

Page 26: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

26

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

gygygygygy

a – b 1, y 0, z = x + iy

Im1

az by

z

⎛ ⎞ ⎜ ⎟⎝ ⎠

( ) 1lm

( 1) 1

ax b ayi x iyy

x iy x iy

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

2

2 2

( )( 1) ( 1) ( )lm

( 1)

ax b x ay ay x i iy ax by

x y

⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠

ay(x + 1) – y(ax + b) = y(x + 1)2 + y3

ax + a – ax – b = x2 + 2x + 1 + y2

a – b = x2 + y2 + 2x + 1

1 = x2 + y2 + 2x + 1

(x + 1)2 = 1 – y2

21 1x y

21 1x y

41. ;fn 2x – y + 1 = 0 vfrijyo; (hyperbola) 2 2

21

16

x y

a

dh Li'kZjs[kk (tangent) gS rks fuEu esa ls dkSu lh ledks.kh; f=kHkqt

(right angled triangle) dh Hkqtk;sa ugha gks ldrh gS(gSa)\

(A) 2a, 8, 1 (B) a, 4, 1

(C) a, 4, 2 (D) 2a, 4, 1

mÙkjmÙkjmÙkjmÙkjmÙkj (A, B, C)

gygygygygy y = 2x + 1, 2 2

21

16

x y

a

dh Li'kZ js[kk gS

y = 2x + 1 dh 2 216y mx a m ls rqyuk djus ij

m = 2 rFkk a2m2 – 16 = 1 4a2 = 17

2a, 4, 1 ledks.kh; f=kHkqt dh Hkqtk,¡ gSa

42. fuEu esa ls dkSu lk(ls) okLrfod la[;kvksa ds 3 × 3 vkO;wg (matrix) dk oxZ (square) ugha gS(gSa)?

(A)

1 0 0

0 1 0

0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(B)

1 0 0

0 1 0

0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(C)

1 0 0

0 1 0

0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(D)

1 0 0

0 1 0

0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

mÙkjmÙkjmÙkjmÙkjmÙkj (A, B)

Page 27: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

27

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

gygygygygy pw¡fd fodYi (A) o (C) esa vkO;wg ds lkjf.kd /ukRed gSa

vr% ;g vkO;wg ds oxZ ds :i esa fu:fir fd, tk ldrs gSa ijUrq fodYi (B) o (D) esa vkO;wg ds lkjf.kd eku ½.kkRed gSa]vr% bls vkO;wg ds oxZ ds :i esa fu:fir ugha fd;k tk ldrkA

43. ;fn ijoy; (parabola) y2 = 16x dh ,d thok (chord), tks Li'kZjs[kk (tangent) ugha gS] dk lehdj.k 2x + y = p rFkk eè;fcUnq(midpoint) (h, k) gS] rks fuEu esa ls p, h ,oe~ k ds lEHkkfor eku gS(gSa)\

(A) p = 2, h = 3, k = –4 (B) p = –1, h = 1, k = –3

(C) p = –2, h = 2, k = –4 (D) p = 5, h = 4, k = –3

mÙkjmÙkjmÙkjmÙkjmÙkj (A)

gygygygygy thok dh lehdj.k 2x + y = p gS

ijoy; ds lkFk izfrPNsnh fcanq ds fy,

2( 2 ) 16p x x 2 2

4 (4 16) 0x p x p⇒

p = 128p + 256

vr% p = –2

eè; fcanq (h, k) ds fy, thok dk lehdj.k

28( ) 16yk x h k h

28 8yk x k h⇒

–8x + ky = k2 – 8h

rqyuk djus ij]

2

2

8 8

2 1

x y p

k k h

p

28 4

16 8 4

2 4

k h p

h p

p h

k = –4, p = 2, h = 3

[kaM[kaM[kaM[kaM[kaM 2 (vf/dre vad(vf/dre vad(vf/dre vad(vf/dre vad(vf/dre vad % % % % % 15)))))

• bl [kaM esa ik¡p iz'u gSaA

• izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,d ,dy vadh; iw.kkZ ad,dy vadh; iw.kkZ ad,dy vadh; iw.kkZ ad,dy vadh; iw.kkZ ad,dy vadh; iw.kkZ ad gSA

• izR;sd iz'u ds fy, vks-vkj-,l- ij lgh iw.kkZad ds vuq:i cqycqys dks dkyk djsaA

• izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfRk;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs%fdlh ,d ds vuqlkj fn;s tk,axs%fdlh ,d ds vuqlkj fn;s tk,axs%fdlh ,d ds vuqlkj fn;s tk,axs%fdlh ,d ds vuqlkj fn;s tk,axs%

iw.kZ vad % +3 ;fn fliQZ lgh mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

44. ekuk fd f : � � bl izdkj dk vodyuh; iQyu (differentiable function) gS fd f(0) = 0, f 32

⎛ ⎞ ⎜ ⎟⎝ ⎠

,oe~ f ' (0) = 1

gSaA ;fn x 0,2

⎛ ⎤ ⎜ ⎥⎝ ⎦

ds fy;s x

g x f t t t t f t dt

2

( ) [ '( )cosec – cot cosec ( )]

∫ gS] rc x

g x0

lim ( )

Page 28: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

28

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

mÙkjmÙkjmÙkjmÙkjmÙkj (2)

gygygygygy/2

( ) ((cosec )( ( )))

x

g x d t f t

∫ /2{(cosec ). ( )}x

t f t

blfy,] 0 0

lim ( ) lim ( ).cosec2x x

g x f f x x

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ 0

( )3 lim

sinx

f x

x

0

( )3 lim 2

cosx

f x

x

45. ,d ledks.kh; f=kHkqt (right angled triangle) dh Hkqtk;sa lekUrj Js<h (arithmetic progression) esa gSaA ;fn bldk {ks=kiQy24 gS rc bldh lcls NksVh Hkqtk dh yEckbZ D;k gS\

mÙkjmÙkjmÙkjmÙkjmÙkj (6)

gygygygygy ()2 = 2 + (–)2 dk iz;ksx djus ij

+

4

rFkk] 1

( ) 242

8

2

⎫⎬ ⎭

lcls NksVh Hkqtk = 6

46. okLrfod la[;k (real number) α ds fy;s] ;fn jSf[kd lehdj.k fudk; (system of linear equations)

x

y

z

2

2

1 1

1 –1

11

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

ds vuUr gy (infinitely many solutions) gSa] rc 1 + α + α2 =

mÙkjmÙkjmÙkjmÙkjmÙkj (1)

lehdj.k dks AX = B ds :i esa iqu% fy[kk tkrk gS

tgk¡

2

2

1 1

, ,1 1

11

x

A X By

z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

adj( ).

det( )

A BX

A

vifjfer gy ds fy,] det(A) = 0 rFkk adj(A).B = 0

= 1, –1

ijUrq , 1 ds cjkcj ugha gS D;ksafd bl fLFkfr esa lehdj.k vlaxr gSA

blfy,] = –1

i.e., 21 1

47. p ds fdrus ekuksa ds fy;s o`Ùk (circle) x2 + y2 + 2x + 4y – p = 0 ,oe~ funsZ'kkad v{kksa (coordinate axes) esa dsoy rhufcUnq mHk;fu"B (common) gSa\

mÙkjmÙkjmÙkjmÙkjmÙkj (2)

gygygygygy laHko fLFkfr;k¡

p = –1 p = 0

Page 29: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

29

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

48. v{kjksa A, B, C, D, E, F, G, H, I, J ls 10 yEckbZ ds 'kCn cuk;s tkrs gSaA ekuk fd x bl rjg ds mu 'kCnksa dh la[;k gS

ftuesa fdlh Hkh v{kj dh iqujko`fr ugha gksrh gS] rFkk y bl rjg ds mu 'kCnksa dh la[;k gS ftu esa dsoy ,d v{kj dh

iqujkofr nks ckj gksrh gS o fdlh vU; v{kj dh iqujkofr ugha gksrh gSA rc y

x9

mÙkjmÙkjmÙkjmÙkjmÙkj (5)

gygygygygy Li"Vr% x = 10!

y dh x.kuk djus ds fy, ,d v{kj lfEefyr ugha djuk gS] ,slk 10 rjhdksa ls gks ldrk gSA

'ks"k 9 v{kjksa esa ls ,d dh iqujkofÙk gksxh

10 9

1 1

10! 5 9 10!

2!y C C

5 9 10!

59 9 10!

y

x

[kaM[kaM[kaM[kaM[kaM-3 (vf/dre vad (vf/dre vad (vf/dre vad (vf/dre vad (vf/dre vad : 18)))))

bl [kaM esa lqesy izdkj ds NgNgNgNgNg iz'u gSaA

bl [kaM esa nks Vscy gSa (izR;sd Vscy esa 3 dkye vkSj 4 iafDr;k¡ gSa)A

izR;sd Vscy ij vkèkkfjr rhurhurhurhurhu iz'u gSaA

çR;sd iz'u ds pkjpkjpkjpkjpkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa fliZQ ,d ,d ,d ,d ,d fodYi lgh gSA

izR;sd iz'u ds fy, vks-vkj-,l- ij lgh mÙkj ds vuq:i cqycqys dks dkyk djsaA

izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,saxs %fdlh ,d ds vuqlkj fn;s tk,saxs %fdlh ,d ds vuqlkj fn;s tk,saxs %fdlh ,d ds vuqlkj fn;s tk,saxs %fdlh ,d ds vuqlkj fn;s tk,saxs %

iw.kZ vad : +3 ;fn lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa 49, 50 ,oa ,oa ,oa ,oa ,oa 51 ds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sA

dkWye 1, 2 3 (conic), (tangent)

(point of contact)

rFkk esa Øe'k% dkWfud dkWfud ij Li'kZjs[kk dk lehdj.k rFkk Li'kZfcUnq fn;s x;s gSa

(I) + = x y a2 2 2

(II) + = x y a2 2 2

a2

(III) = 4y ax2

(IV) x a y a2 2 2 2

– =

(i)

(ii)

(iii)

(iv)

2my m x a

2 1y mx a m

2 2 – 1y mx a m

2 2 1y mx a m

(P)

(Q)

(R)

(S)

2

2,

a a

mm

⎛ ⎞⎜ ⎟⎝ ⎠

2 2,

1 1

ma a

m m

⎛ ⎞⎜ ⎟

⎝ ⎠

2

2 2 2 2

1,

1 1

a m

a m a m

⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠

2

2 2 2 2

–1,

– 1 – 1

a m

a m a m

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

49. ;fn mi;qDr dkWfud (dkWye 1) ds fcUnq1

3,2

⎛ ⎞⎜ ⎟⎝ ⎠

ij Li'kZjs[kk 3 2 4,x y gS] rc fuEu esa ls dkSulk fodYi dsoy lgh

la;kstu gS\

Page 30: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

30

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

(A) (IV) (iii) (S) (B) (II) (iv) (R)

(C) (IV) (iv) (S) (D) (II) (iii) (R)

mÙkjmÙkjmÙkjmÙkjmÙkj (B)

50. 2a ds fy, mi;qDr dkWfud (dkWye 1) ij ,d Li'kZjs[kk [khph tkrh gS ftldk Li'kZfcUnq (–1, 1), rc fuEu esa ls dkSulk

fodYi bl Li'kZ js[kk dk lehdj.k izkIr djus dk dsoy lgh la;kstu gS\

(A) (III) (i) (P) (B) (I) (i) (P)

(C) (II) (ii) (Q) (D) (I) (ii) (Q)

mÙkjmÙkjmÙkjmÙkjmÙkj (D)

51. ;fn mi;qDr dkWfud (dkWye 1) ds Li'kZfcUnq (8, 16) ij Li'kZjs[kk 8y x gS] rc fuEu esa ls dkSulk fodYi dsoy lgh

la;kstu gS\

(A) (III) (i) (P) (B) (III) (ii) (Q)

(C) (II) (iv) (R) (D) (I) (ii) (Q)

mÙkjmÙkjmÙkjmÙkjmÙkj (A)

iz- la-iz- la-iz- la-iz- la-iz- la- 49 lslslslsls 51 ds fy, gyds fy, gyds fy, gyds fy, gyds fy, gy

lgh la;kstu fuEu gSI, ii, Q

II, iv, R

III, i, P

IV, iii, S

49. fn, x, fodYiksa esa ls A, B lgh la;kstu gSa ftudh tk¡p vko';d gSA blesa ls (ftudh tk¡p de djuh gS)x2 + a2y2 = a2,

2

2 2 2 2

31 14, 12 , 21 12 2

a m

a m a m

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

tks fd fn;k x;k gS] blfy, (B) lgh gSA

50. (–1, 1), 2a ds fy, x2 + y2 = a2 ij fLFkr gS] blfy, lgh mÙkj dsoy (D) gSA

51. pw¡fd y = x + 8 rFkk bldk Li'kZ fcanq (8, 16) dsoy y2 = 4ax dks larq"V djrk gS

uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/ lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa 52, 53 ,oa ,oa ,oa ,oa ,oa 54 ds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sAds mÙkj nhft;sA

(P) (0, 1) f o/Zeku gS(i) lim ( ) 0x

f x

(ii) lim ( )x

f x

(iii) lim ( )x

f x

(iv) lim ( ) 0x

f x

(Q) ( , ) f e e2esa ßkleku gS

(R) (0, 1) f esa o/Zeku gS

(S) f esa ßkleku gS( , ) e e2

(I) ( ) = 0 (1, ) f x x e 2 ds fy, fdlh

(II) ( ) = 0 (1, ) f x x e ds fy, fdlh

(III) = 0 (0, 1) x ds fy, f x( ) fdlh

(IV) = 0 (1, e) x ds fy, f x( ) fdlh

1

2 (limiting behaviourat infinity)

3 (increasing/decreasing) (nature)

ekuk fd gSaA

dkWye esa ,oe~ ds 'kwU;ksa dh lwpuk nh xbZ gSaA

dkWye esa ,oe~ ds vuUr dh rjiQ lhek ij O;ogkj dh lwpuk nh xbZ gSA

dkWye esa ,oe~ ds o/Zeku@ßkleku gksus dh izÑfr dh lwpuk nh xbZ gSA

f x x x x( ) = + log , (0, ) e

f x f x f x

f x f x f x

f x f x

( ), ( ) ( )

( ), ( ) ( )

( ) ( )

Page 31: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

31

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

52. fuEu esa ls dkSulk fodYi dsoy lgh la;kstu gS\

(A) (III) (iii) (R) (B) (I) (i) (P)

(C) (II) (ii) (Q) (D) (IV) (iv) (S)

mÙkjmÙkjmÙkjmÙkjmÙkj (C)

53. fuEu esa ls dkSulk fodYi dsoy lgh la;kstu gS\

(A) (II) (iii) (S) (B) (I) (ii) (R)

(C) (IV) (i) (S) (D) (III) (iv) (P)

mÙkjmÙkjmÙkjmÙkjmÙkj (A)

54. fuEu esa ls dkSulk fodYi dsoy xyr xyr xyr xyr xyr la;kstu (only INCORRECT combination) gS\

(A) (I) (iii) (P) (B) (III) (i) (R)

(C) (II) (iv) (Q) (D) (II) (iii) (P)

mÙkjmÙkjmÙkjmÙkjmÙkj (B)

iz- la-iz- la-iz- la-iz- la-iz- la- 52 lslslslsls 54 ds fy, gyds fy, gyds fy, gyds fy, gyds fy, gy

( ) log loge e

f x x x x x

(I) (1) 1 log1 1log1 1 0f

2 2 2 2 2 2 2 2( ) log log 2 2 2 0e e

f e e e e e e e e

(I) lR; gSA

(II)1

( ) 1 log 1e

f x xx

(1) 1f

1( ) 1 0f e

e

(II) lR; gSA

(III)2

1 1( ) 0f x

xx

lHkh (0, )x ds fy,

blfy,] ( )f x ßkleku gS

blfy, (0, 1) esa ( )f x dk U;wure eku (1) 1f gS

blfy, (III) vlR; gSA

(IV) (1) 2 0f

2

1 1( ) 0f e

ee

blfy, (IV) vlR; gSA

Page 32: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...3 JEE (ADVANCED)-2017 (PAPER-1) CODE-6 2. ekuoh; i"Bh; {k=kiQy yxHkx 1 m2 gkrk gAS ekuo 'kjhj dk rkieku ifjo'k d rkieku l 10 K vfèkd

32

JEE (ADVANCED)-2017 (PAPER-1) CODE-6

dkWyedkWyedkWyedkWyedkWye 2

0

log1lim ( ) lim log log lim log e

e e ex x t

t

f x x x x x tt t

0

1 log log

lim0

e e

t

t t t

t

(i) vlR; gS

(ii) lR; gS

(iii) 0

1lim ( ) lim log lim log

e ex x t

f x x t tx

(lR;)

(iv) 2 2

1 1 1lim ( ) lim 0x x

xf x

xx x

(lR;)

dkWyedkWyedkWyedkWyedkWye 3

(P)1

( ) log 0e

f x xx

lR; gS

(0, 1)x ds fy,] 1

(1, )x

log ( , 0)e

x

(Q)1

( ) loge

f x xx

ßkleku iQyu gS

1( ) 1 0f e

e

2

2

1( ) 2 0f e

e

lR; gSA

(R)2

1 1( ) 0f x

x x

blfy, ( )f x ßkleku gS

blfy,] (R) vlR; gSA

(S) lR; gSA

iz'u i=k lekIriz'u i=k lekIriz'u i=k lekIriz'u i=k lekIriz'u i=k lekIr