32
1 CODE 4 Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472 DATE : 21/05/2017 PAPER - 1 (Code - 4) 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

Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

  • Upload
    others

  • View
    0

  • Download
    0

Embed Size (px)

Citation preview

Page 1: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

1

CODE

4

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

DATE : 21/05/2017

PAPER - 1 (Code - 4)

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...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

2

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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

(B) 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

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

(D) l e; TAO = TOA

mÙkj (A, B, 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

Page 3: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

3

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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

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

(B) 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

(C) 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

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

Page 4: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

4

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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 coordinate

system 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 juhps dh vkSj l jdrk gSA t c fcanq d.k xqVds l s l ai dZ foghu gks t krk gS] rc ml dh r kR{kf.kd fLFkfr x vkSj xfr gSA fuEuesa l s dkSul k (l s) dFku l gh gS@gSa\

R

Rm

M

y

x

x = 0

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

1

gRmM

gS

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

M m

gS

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

gS

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

gSS

mÙkj (A, B)

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 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 v,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?

Page 5: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

5

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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

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

(C) IysV }kjk vuqHko gqvk i zfr jksèkd cy vds l ekuqi krh gS

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

mÙkj (A, B, C)

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] = 4mnuv

P uv

5. 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 = 104 rad.s–1 gksxh rc fo| qr èkkjk (electric current) oksYVrk dh l edyk esa gksxh

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

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

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

mÙkj (C, D)

gy >> 106 i j

XL = L

= = 106 ds fy, 106 × 10–6

Page 6: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

6

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

= 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

6. , 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 dk

U;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) 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

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

A

}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) 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)

mÙkj (A, C, D)

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

1sini1 = × sin(A – C)

2sin cos2 2sin

2cos2

A A

rA

= sin2A

= 2cos sin cos – cos sin2 C CA A A

2Ar

Page 7: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

7

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

= 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

7. 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) 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

(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) 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

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

mÙkj (A, B, D)

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 A)

= 4× 460 × 10300 60 J/s (fodYi D)

Page 8: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

8

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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

bl [ kaM esa i kap 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. 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 gksxkmÙkj (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

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)Ad k¡pksa d s Lr j i j Li j l ekar j gSA d k¡p d s phr h d s v i or Zukad , d fn"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

Page 9: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

9

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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 jDrvk; ru (yhVj esa) gS (vki 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

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. ,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) gS

Page 10: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

10

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

mÙkj (5)

gy2

2PE 27.2 Zn

H ds fy, Z = 1

227.2PEn

1 227.2PEi

i

Vn

2 2

27.2PE ff

vn

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 = 5

nf = 5

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

Page 11: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

11

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

[ 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 gSA½.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

(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 vpy xfr l s l h/ h js[ kk esa pyu djr k gS\

(A) (IV) (i) (S) (B) (III) (ii) (R)(C) (II) (iii) (S) (D) (III) (iii) (P)

mÙkj (C)

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

14. 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) (IV) (i) (S) (B) (II) (ii) (R)

(C) (III) (iii) (P) (D) (IV) (ii) (R)

mÙkj (A)

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

Page 12: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

12

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

15. 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) (R) (B) (IV) (ii) (S)

(C) (III) (ii) (P) (D) (II) (iii) (Q)

mÙkj (A)

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

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)

Page 13: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

13

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

16. 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) (III) (iii) (P) (B) (II) (iii) (S)

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

mÙkj (A)

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

W1 2 = –PV2 + PV1 l enkch;

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

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

(C) (II) (iv) (R) (D) (III) (ii) (S)

mÙkj (D)

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

W1 2 = 0 l evk; r fud i zØe

18. 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) (III) (iv) (R) (B) (I) (ii) (Q)

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

mÙkj (D)

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 (D) gS

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

– 1P V PV

END OF PHYSICS

Page 14: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

14

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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. ,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 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

(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 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

mÙkjmÙkjmÙkjmÙkjmÙkj (A, C, D)

gygygygygy (A)

V1

V2

V

P

W > Wlerkih; :¼ks"elerkih;

:¼ks"e

(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) 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 lsvfèkd gS

Page 15: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

15

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

P1

P2

P

(P , V )1 2

(P , V )2 2

V1

V2

V

(P , V )1 1

vuqRØe.kh;mRØe.kh;

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

ClH C3

(A) 4-eSfFkyDyksjks csathu (B) 4-Dyksjks Vksyqbu

(C) 1-Dyksjks-4-eSfFky csathu (D) 1-eSfFky-4-Dyksjkscsathu

mÙkjmÙkjmÙkjmÙkjmÙkj (B, C)

gygygygygy IUPAC uke

(B)

CH3

Cl

4-DyksjksVksyqbu

1

2

3

4

(C)

Cl

CH3

1- -4-Dyksjks esfFkycsathu

1

2

3

4

21. 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)

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) '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

(C) 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

(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 (B, D)

gygygygygy Z

x = 0M

x = 1L

'kq¼ L

xM

→x = 1M

M'kq¼

pL

Page 16: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

16

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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

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

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

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

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

mÙkjmÙkjmÙkjmÙkjmÙkj (B, 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)

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

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

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

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

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

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

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

Page 17: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

17

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

mÙkjmÙkjmÙkjmÙkjmÙkj (A, C, D)

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.

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

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

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

(B) Cl2 dh H

2O ds lkFk vfHkfØ;k gksus ij HClO

4 curk gS

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

2O ls nqcZy {kkj gS

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

mÙkjmÙkjmÙkjmÙkjmÙkj (A, C, D)

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

(C) HClO4 > H

2O vEyh; y{k.k

ClO4

– < OH– la;qXeh {kkj

çcy vEyksa ds la;qXeh {kkj nqcZy gksrs gSaA

(D) HClO4 > HClO vEyh; lkeF;Z

+

4 4HClO H + ClO

−⎯⎯→

Page 18: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

18

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

+

HClO H + ClO−⎯⎯→

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

25. 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) oxZ esa uhps tkus ij π*-o* dk varj ?kVrk gS

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

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

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

mÙkjmÙkjmÙkjmÙkjmÙkj (A, D)

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

[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)

mÙkjmÙkjmÙkjmÙkjmÙkj (6)

gygygygygy

Te

Br

Br

Br

F

Br

Br

Br

Br

F

esa ,d ,dkdh ;qXe gS

esa nks ,dkdh ;qXe gS

Page 19: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

19

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

Xe

S N≡

F

F

F

F

F

esa rhu ,dkdh ;qXe gS

esa ,dkdh ;qXe ugha gSF

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

++

+

mÙkjmÙkjmÙkjmÙkjmÙkj (5)

gygygygygy +

+

, ,, rFkk ,jksesfVd gS

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

28. H2, He

2, Li

2, Be

2, B

2, C

2, N

2, –

2O 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;

Page 20: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

20

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

29. ,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 dhdqy 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[;k3 23 –24

AdN a 8 6 10 64 10

M 76.8 g/molZ 4

× × × ×= = =

256 g esa Bksl ds eksy = 3.33 eksyijek.kqvksa dh la[;k = 3.33 × N

A = 20 × 1023 = 2 × 1024

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 dh

nwjh 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 1205 10

1

−= × ×

= 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

4

4

10 1

1515 10

−α = =×

⇒ M

º

M

λα =

λ

⇒M

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

Page 21: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

21

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

[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

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

(A) (I) (i) (S) (B) (II) (ii) (Q)

(C) (I) (iii) (R) (D) (I) (i) (R)

mÙkjmÙkjmÙkjmÙkjmÙkj (C)

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

vr% C xyr gS

Page 22: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

22

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

32. 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) (II) (ii) (P) (B) (I) (ii) (S)

(C) (IV) (iv) (R) (D) (III) (iii) (P)

mÙkjmÙkjmÙkjmÙkjmÙkj (A)

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

D;kasfd

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

(C) esa] 2

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

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

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

(A) (I) (i) (P) (B) (I) (iv) (R)

(C) (II) (i) (Q) (D) (I) (i) (S)

mÙkjmÙkjmÙkjmÙkjmÙkj (D)

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− × = −

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...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

23

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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

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

(C) (II) (iii) (R) (D) (I) (ii) (R)

mÙkjmÙkjmÙkjmÙkjmÙkj (D)

gygygygygy

CH3

CH2

CH2

CH – Br2

+ Br

+ Br2

+ HBr

+ Br

Br2

Br + Brhν

35. 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) (I) (iv) (Q)

(C) (IV) (ii) (P) (D) (III) (iv) (R)

mÙkjmÙkjmÙkjmÙkjmÙkj (A)

gygygygygy C CH3

O

NaOH

Br2

COO Na– +

+ CHBr3

H O/H2

+

COOH

;g gSyksiQkWeZ vfHkfØ;k gS

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

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

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

mÙkjmÙkjmÙkjmÙkjmÙkj (D)

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

Page 24: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

24

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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) 1|

2P X Y

(C)2

( )5

P X Y ∪ (D)1

( )5

P X Y ∩

mÙkjmÙkjmÙkjmÙkjmÙkj (A, B)

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...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

25

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

38. 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) 2

0

( ) ( )sinf x f t t dt

(C) 2

0

( )cosx

x f t t dt

∫ (D) x9 – f(x)

mÙkjmÙkjmÙkjmÙkjmÙkj (C, 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

39. 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 (B, D)

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

⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠

Page 26: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

26

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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

40. ;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) a, 4, 1 (B) 2a, 4, 1

(C) a, 4, 2 (D) 2a, 8, 1

mÙkjmÙkjmÙkjmÙkjmÙkj (A, C, D)

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

41. 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 = –1 (B) x = 1

(C) x = 0 (D) x = 2

mÙkjmÙkjmÙkjmÙkjmÙkj (A, B, 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

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, C)

Page 27: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

27

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

gygygygygy pw¡fd fodYi (B) o (D) 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 (A) o (C) 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 = –1, h = 1, k = –3

(B) p = 2, h = 3, k = –4

(C) p = –2, h = 2, k = –4

(D) p = 5, h = 4, k = –3

mÙkjmÙkjmÙkjmÙkjmÙkj (B)

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

ijoy; ds lkFk izfrPNsnh fcanq ds fy,

2( 2 ) 16p x x

2 24 (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

Page 28: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

28

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

44. 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

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. 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 ( )

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

Page 29: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

29

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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 rhu

fcUnq mHk;fu"B (common) gSa\

mÙkjmÙkjmÙkjmÙkjmÙkj (2)

gygygygygy laHko fLFkfr;k¡

p = –1 p = 0

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

Page 30: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

30

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

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. 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) (I) (ii) (Q) (B) (I) (i) (P)

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

mÙkjmÙkjmÙkjmÙkjmÙkj (A)

50. ;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\

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

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

mÙkjmÙkjmÙkjmÙkjmÙkj (B)

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\

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

(C) (II) (iv) (R) (D) (III) (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 gS

I, ii, Q

II, iv, R

III, i, P

IV, iii, S

49. (–1, 1), 2a ds fy, x2 + y2 = a2 ij fLFkr gS] blfy, lgh mÙkj dsoy (A) gSA

Page 31: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

31

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

50. 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

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 e2 esa 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

( ), ( ) ( )

( ), ( ) ( )

( ) ( )

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

(A) (I) (iii) (P) (B) (II) (iv) (Q)

(C) (II) (iii) (P) (D) (III) (i) (R)

mÙkjmÙkjmÙkjmÙkjmÙkj (D)

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

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

(C) (II) (iii) (S) (D) (IV) (i) (S)

mÙkjmÙkjmÙkjmÙkjmÙkj (C)

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

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

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

mÙkjmÙkjmÙkjmÙkjmÙkj (C)

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

Page 32: Time : 3 hrs. Max. Marks: 183 Answers & Solutions...(B) dNq l e; d sckn ckã cy F vkjS ifzrjksèkd cy l rafqyr gk t k,xas (C) IysV }kjk vuHqko gvk ifzrjkèkd cy v d sl ekuikrh gS (D)

32

JEE (ADVANCED)-2017 (PAPER-1) CODE-4

(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

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