Distance Learning Program for JEE Main, IIT-JEE Advanced - … · 2019-07-25 · 018 20182018 2016...

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SECTION-I

1. Ans. (A,B,D)

Sol. P(Ram gets six on third throw) ´ ´5 5 16 6 6 .

P(shyam gets six on or after third throw)

=25

211

Let 'P' be the probability for event in option(D).

{ {

æ ö= ´ + ´ + ´ Þ =ç ÷è ø14243

both obtain Noneobtainsixatleast oneofsix themobtainsix

1 1 1 5 5 5 8P .2 . P P6 6 6 6 6 6 33

2. Ans. (A,B,C,D)

Sol. Consider the interval [0, 1]

1g(x)g '(c )

x=

(By LMVT) where x Î (0, 1]

|g(x)| = |xg'(c1)|

|g(x)| £ |x||g(c1)| = |x||c1||g(c2)|

{Again using LMVT}

0 £ |g(x)| £ |x||c1||c2| – |cn||g(cn + 1)|

LEADER TEST SERIES / JOINT PACKAGE COURSETARGET : JEE (Main + Advanced)

DISTANCE LEARNING PROGRAMME

SOLUTION

PART-1 : MATHEMATICS ANSWER KEY

Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-Advanced

PAPER-1

where 0 < cn < cn–1 ..... < c2 < c1 < x < 1

Q 'n' can be increased indefinitely whichimplies cn®0 & as g(x) is continuous

|g(x)| = 0 " x Î [0, 1]

same argument can be applied in [1, 2]starting with |g(x)| £ |x – 1||g(c1)|

Þ g(x) = 0 for all x.

3. Ans. (A,B,C)

Sol. q =11cos14 q

b

a

= + +r ˆˆ ˆa i x j 3k

( )= + - +r ˆˆ ˆb i 4x 2 j 2k

=rr2 a b

4(1 + x2 + 9) = (16 + (4x – 2)2 + 4)3x(x – 2) + 2(x– 2) = 0

x = 2, = -2x3

=

rr

rr

a.b 1114a b

Q. 1 2 3 4 5 6 7 8 9 10A. A,B,D A,B,C,D A,B,C A A,C,D C,D A,C,D C A CQ. 11 12 13A. B D CQ. 1 2 3 4 5A. 8 5 2 7 6

SECTION-I

SECTION-IV

ALLEN

LTS-2/26

( )( )

+ - +=

+ 2

4 x 4x 2 6 11147 10 x

17x2 – 14x – 40 = 0x = 2

4. Ans. (A)

Sol.æ ö+ +

= + +ç ÷+è ø

201822018 2017

2018 2017x 2x 2 a x a x ..

x 1

( )+ + + + +

+ +1 2

1 0 2b b.. a x a ..

x 1 x 1 ( )+

+2018

2018b..

x 1

( ) ( )2æ ö+ + = + + +ç ÷+è ø

20182018 2016018 2018

0 11x 1 C x 1 C x 1

x 1

( ) ( )+ + + + + +2014 22018 2018 2008

2 1008 1009C x 1 ... C x 1 C

æ ö+ ç ÷+è ø

20182018

20181C

x 1Put x = 1

( )= =

æ ö æ ö+ = + = ®ç ÷ ç ÷è ø è ø

å å2018 20182018 2018

ii i

i 0 i 1

b 1 5a 2 A2 22

Put x = 0,

( )= + + + +2018 2018 2018 2018 200180 1 1008 10092 C C ... C C

( )

=

+ + + +

å14444444244444443

2018

ii 1

2008 2018 20181010 1011 2018

b

C C ... C

-S =

2018 20181009

i2 Cb

2

a0 = +2018 201810092 C

25. Ans. (A,C,D)Sol. 2,3,4 ...... n–1 n

1,3,4..... n

......... ... ..

1,2,3 n–1

In each sequence missing number can beaccomodated in (n – 1) ways. But in every

Þ consecutive pair one sequence isrecounted

Henceƒ(n) = n(n–1) – (n–1) + 1,ƒ(n) = n2 – 2n + 2

6. Ans. (C,D)

Sol. = - + +2y 4x 4x 2 x

= ( )- + +22x 1 1 xif x < 0

( )= - + -2y 2x 1 1 x

( )( )

-= -

- +2

2x 1 .2y ' 1

2x 1 1

<y ' 0 " x < 0if x > 0

( )= - + +2y 2x 1 1 x

( )( )

-= +

- +2

2 2x 1y ' 1

2x 1 1

-Þ > Þ < < ¥

3 1y ' 0 x2 3

-< Þ < <

3 1y ' 0 0 x2 3

Ö2

O Ö3–12 3Ö

12

minima at -

=3 1x2 3

7. Ans. (A,C,D)

Sol. ƒ(x) = loga(4ax – x2)3 , 22

é ùê úë û

a > 1 Þ ƒ(x) = 4ax – x2

must be " 3x , 22

é ùÎ ê úë û

O 3/2 2 2a 4aÞ 2a ³ 2 Þ a ³ 1Þ a > 10 < a < 1

All India Open Test/Leader Test Series/Joint Package Course/JEE (Advanced)/PAPER-1

ALLEN

LTS-3/26

Þ ƒ(x) = 4ax – x2

must be ¯ " 3x , 22

é ùÎ ê úë û

32a2

£ & 4a > 2

O 4a2a 32–

2a £

34 &

1a2

>

1 3a2 4

< £

( )1 3, 1,2 4

æ ù È ¥ç úè ûAnswer Q.8, Q.9 and Q.10(1) y2 = 4x (i) (x+1)2 + 2y2 = 1

( )++ =

2 2x 1 y 11 1/ 2

–2

(2) x2 + y2 – 2x = 0 (ii) ( ) ( )+ - = - - +2x y 3 4 2 x y 1

(x – 1)2 + y2 = 1

(1,2)x–y+1=0 x+y–3=0

(3) x2 – y2 + 2x = 0 (iii) (x + 3)2 + y2 = 1

(x + 1)2 – y2 = 1 –4 –3 2

–2

(4) (x + 2)2 + y2 = 4 (iv) y2 = –4(x + 4)

–4 –22

(–4,0)

8. Ans. (C)9. Ans. (A)10. Ans. (C)Answer Q.11,Q.12 and Q.13

(1) |z + 2| + |z + 4| = 4

Ö3

-Ö3–5 –3 –1

(2) |z+13| = –11 – Rez

–12

(3) |z + 2| = 1

–2–3 –1

(4) |z + 10| = 1

–11–10

–9

(i) |z + 10| + |z + 8| = 4

–11 –9 –7

(ii) + + + =22 2z 8z z 8z 2 z

y=2

–y=2

(iii) |z + 12| = |z + 11|

–12 –11x=–11.5


ALLEN

LTS-4/26

(iv) |z| = |z + 1|

–1 0

x= –1 2

11. Ans. (B)

12. Ans. (D)

13. Ans. (C)

SECTION-IV

1. Ans. 8

Sol. ˆˆ ˆr ai bj ck= + +r

r.(2i k) 2a b5 5+ +

a = =r

2a b , r c5

- +b = =

\ |a| = |b| = |g|

Þ A B

2a b 2a b c 5+ = - + =E5555555555555FE555F

5a3

b 0 42c3

ü= ± ï

ïï= ®ý

ïï= ±ïþ

1c65 4b6

a 0

üü= ± ïï

ïïïïïï ®= ± ýý

ïïïï=ïïïïþþ

a = 0

8 possible triplets.

2. Ans. 5

Sol. ( )ìì < < < <p < ïïï ïp = =í í

ï ïp ³ £ £ï ïî î

1 51 0, 0 x , x 10, sin x6 62ƒ sin x

1 1 51, sin x 1, x2 6 6

( )p = + + = - =ò ò ò ò1 1 / 6 5 / 6 1

0 0 1 / 6 5 / 6

5 1 2ƒ sin x dx 0 1 06 6 3

p = 2, q = 3 Þ p + q = 53. Ans. 2

Sol.

log(1 x)1/ y

0x 0

(1 tan2y)lim

sinx

+

®

-ò

( )1

log(1 x)x 0

L lim 1 tan(2log(1 x)) +®

= - +

(Using L-Hospital)L'e=

( )x 0

tan 2log(1 x)L' lim 2 22log(1 x)®

- += ´ = -

+

\ 21Le

=

4. Ans. 7Sol. A2 = I, l = 4

\ B2 = 81I

\ æ ö+

+ l =ç ÷è ø

2 2A BTr 782

5. Ans. 6

Sol.1E z 1 z z 1z

= + + + -

E = |(z +1)| + |2Re(z) – 1| Q |z| = 1

2 2(x 1) y 2x 1= + + + -

E 2 2x 2x 1= + + - let 2x – 1 = t

E t 3 t= + + t Î [–3,1]

Range of E is 133,4

é ùê úë û


ALLEN

LTS-5/26

SOLUTION

PART-2 : PHYSICS ANSWER KEY

SECTION-I1. Ans. (A,B,C)

Sol. = -2 22Zep (b a )5 ....(i)

3 30 0 0(R R ) R2Zep

5 bæ ö+ d -

= ç ÷è ø

0 0 0 03R R (R R2Zep5 b

d + dæ ö= ç ÷è ø

2 00

0

R6Zep R5 R

æ öd= ç ÷

è ø....(ii)

By camparing (i) and (ii)

2 2 2 00

0

Rb a 3RR

æ öd- @ ç ÷

è ø(R0 + dR0)2 – a2 = 3R0dR0

a2 = R02 + dR0

2 + 2R0dR0 – 3R0dR0

= R02 – R0dR0 {dR0

2 ® 0}

= 2 00

0

RR 1R

æ öd-ç ÷

è ø

2. Ans. (B,C)3. Ans. (A,B,C,D)

Sol.

x

y

M3R/8

3R/8O

COM2M

R/8

Where 32M f R3

æ ö= pç ÷è ø

eMM(3R / 8) 2M( 3R / 8)y

3M+ -

=

eMRy8

= -

For equillibrium, FB = weight of ball

3 3 34 2 2R R 2 R3 3 3

æ ö æ ö æ ör p = r p + r pç ÷ ç ÷ ç ÷è ø è ø è ø

4r0 = 6r

023r

r =

2 20

2M 2(2M)I R R5 5

= + where 32M R3

= p r

2

06MRI

5=

Icm = I0 – md2

= 226MR R3M

5 8æ ö- ç ÷è ø

= 2 2

26MR 3MR 369 MR5 64 320

- =

= 2 3369 2R R320 3

æ öp rç ÷è ø

5cm

123I R160

= rp

4. Ans. (A,C)

Sol. y1 = 0.02 xsin 400 t

330é ùæ öp -ç ÷ê ú

è øë û

2xy 0.02sin 404 t

330é ùæ ö= p -ç ÷ê ú

è øë ûy1+ y2 = 2(0.02)

x xsin 404 t cos 2 t330 330

é ù é ùæ ö æ öp - p -ç ÷ ç ÷ê ú ê úè ø è øë û ë û

= 402 2 x0.04sin x 402 t cos 2 t330 330

pæ ö æ öp - p - pç ÷ ç ÷è ø è ø

Q. 1 2 3 4 5 6 7 8 9 10A. A,B,C B,C A,B,C,D A,C A,C A,C A,B,D B B AQ. 11 12 13A. C A BQ. 1 2 3 4 5A. 8 2 3 3 3

SECTION-I

SECTION-IV


ALLEN

LTS-6/26

5. Ans. (A,C)Sol. If x be the initial length of the gas chambers

& A be the area of cross section of thecylinder then,

0VxA

= Also P0A = 2kx

200 0

P 2kx P V 2kxx

= Þ =

( )2 2 20 0

1 1k 2x K4x 2kx P V2 2

\ = = =

Similarly, ff

vP A 2kA

=

00

VP A 2kA

=

0f f

0 0 0

PP V P vP V V

Þ = Þ =

Now \ Q = dU + W

( )1 1 0 052 P V P V w2

= - +

= 12 P0V06. Ans. (A,C)7. Ans. (A,B,D)Sol. E = 24 + 20 cos120pt

Charge on C2

Q2 = EC2= (24 + 20cos120pt)1.5µC= 36 + 30cos120ptµC

C1 3µF C2 1.5µF

24V

e0=20V

22

dQidt

= = –30 × 120psin(120pt)

11

dQ didt dt

= = (24 + 20cos(120pt)3µC

= –60 × 120ptsin(120pt)i = i1 + i2 = –90 × 120ptsin(120pt)i = 33.9 sin(120pt + p) mAMinimum energy stored

= 21 1CV2 2

= (3 + 1.5)(4)2µJ

= 36µJ

8. Ans. (B)9. Ans. (B)10. Ans. (A)11. Ans. (C)Sol. S1 closed & S2 & S3 open

G

R3

R2

R4R1

E

32

4 1

RRR R

= (wheat stone bridge)

12. Ans. (A)Sol. (S1 open, S2, S3 = Closed)

G

R3

R2

R4R1

E

E

Þ

R + R2 4E

R + R1 3

R

E

for current in galvanometer is zero.

E Þ 0 Þ ( ) ( )2 4 1 3

E E 0R R R R

- =+ +

R2 + R4 = R1 + R3

13. Ans. (B)Sol. S3 open, S1 & S2 closed

R

R3

R1

E

R4

E


ALLEN

LTS-7/26

No potential doop across this net chargeon capacitor

R + R3

R4x R1 2E E E

E

O

O

E

2R

O R E R 2E E

E

2R E

E

i = E/R

1 3 4

x 2E x E x 0 0R R R R- - -

+ + =+

x 2E x E x 0R 2R R- -

+ + =

2x – 4E + 2x – 2E + 2x = 06x = 6Ex = ER3 = R4 = R1 = RR2 = 2R

SECTION-IV1. Ans. 8Sol. Let the distance of the lens from the object

be lwhen a real image is formed on the screen.Then

l-1001

– l-1

= 231

On solving, we get l = (50 ± 10 2 ) cm.

Now, if the lens performs SHM and a realimage is formed after a fixed time gap, thenthis time gap must be one-fourth of the timeperiod.

P

Screen

50 cm100 cm

\ Phase difference between the two

positions of real image must be 2p

. As the

two positions are symmetrically locatedabout the origin, phase difference of any of

these positions from origin must be 4p

.

Þ 10 2 cm = A sin 4p

Þ A = 20 cm

To achieve this velocity at the meanposition,

u0 = Aw = AmK

\ Required impulse p = mu0 = A m/K =8 kg m/s.

2. Ans. 2Sol. The hand is a point mass which has a

moment of inertia with respect to thecenter of the wheel mr2 at the time of eachimpact. The angular momentumtransferred from the hand to the wheel inthe n-th hit (after the hit, the hand is atrest with respect to the point of impact) inmr (w – vn) where vn is the velocity of thepoint of impact after the n-th hit. Theangular momentum of the wheel after then-th hit is therefore

Ln = mr (w – vn) + Ln + 1That give a recurrent formula for thevelocities of the point of impact based onthe formula Ln = Iwn

Avn = w – vn + Avn – 1where we used the substitution A = I/(mr2)Now, we will try to compute the first fewterms of the progression. We get


ALLEN

LTS-8/26

1v1 A

w=

+, 2

Av 11 A 1 A

w æ ö= +ç ÷+ +è ø,

2

3A Av 1

1 A 1 A 1 Aæ öw æ ö= + +ç ÷ç ÷ç ÷+ + +è øè ø

, ...

The formula for vn will clearly be

n 1

nA Av 1 .....

1 A 1 A 1 A

-æ öw æ ö= + + +ç ÷ç ÷ç ÷+ + +è øè ø

Summing up a geometric series andsimplifying, we obtain

n

nAv 1

1 Aæ öæ ö- w -ç ÷ç ÷ç ÷+è øè ø

.

The velocity is v10 = 4.54 m·s–1.3. Ans. 3Sol. At depth x, the density is determined by

( ) ( )1 b 1xxh

r = r + r - r

For the rod to be at equilibrium, it isrequired that the total torque acting on therod is zero, i.e.

dM 0=òWe can express the elementary torqueacting on an infinitely small section of therod as dM = x(dFvz – dFg) where theelementary forces dFvz are given by

( )vz

r

mg xdF dx

hr

=r ,

gmgdF dxh

=

where m is the mass of the rod. Integratingfrom 0 to hcosj (where our x-axis isdirected vertically downwards and hcosjis the x-coordinate of the lower end of therod), we have

( )h cos

1 b 1r0

mg x mgx dx 0h

j æ öæ ör + r - r - =ç ÷ç ÷r è øè øò

l l,

( )1b 1

r r

h cos 1 02 3 h 2r j

+ r - r - =r rl l l

It remains to express rb and substitute thenumerical values. Eventually, we obtain

( ) 3b r 1 1

3 999kg m2cos

-r = r - r + r = ×j

4. Ans. 3

Sol. q h

3a/2

3atan2h

q =

2 × 4a T × cosq = 4alg

gcos2Tl

q =

2 2 2

3 gah2 4T g

lÞ =

- l

5. Ans. 3

Sol.D 1 rRTyd f M

æ ö= ç ÷ç ÷

è ø

y Tµ

y 1 Ty 2 T

D D=

y 1100 1y 2

D´ = ´

% charge in y = 12


ALLEN

LTS-9/26

SOLUTION

PART-3 : CHEMISTRY ANSWER KEY

SECTION I1. Ans.(A, C)

Sol.

2V0V0

3P0

P0

T1

T2RVolume

P

QPressure

(A) Consider two isotherms at T1 & T2 drawn like above. now T2 < T1. While moving from Q toR, we first encounter T2 & then T1, \ A is correct.

(C) PRq = DH = nCp (TP – TR)

= (1) 0 0 0 05 P V 2P VR2 (1)R (1)R

æ öæ ö -ç ÷ç ÷è øè ø

= 0 05 P V2

-

\ C is correct(D) Q cyclic process ; DU = 0

qcycle = qPQ + qQR + qRP = –Wcycle

= 12

- (2P0.V0)

(1) 0 0 0 0QR 0 0

3 3P V P VR q –P V2 R R

æ öæ ö- + =ç ÷ç ÷è øè ø

+ 0 05 P V2

æ öç ÷è ø

Þ 3P0V0 0 0 QR 0 05 P V q P V2

- + =

Þ qQR = 0 03 P V2

-

2. Ans.(A, C, D)Sol. Total charge passed = å i.dt.

= 1 100 10010 102 1000 1000

æ ö æ ö´ + ´ç ÷ ç ÷è ø è ø

= 1 100 102 1000

æ ö+ ´ç ÷è ø

Amp. sec.

= 2 Amp. sec. = 2 coulomb

Q. 1 2 3 4 5 6 7 8 9 10A. A,C A,C,D A,B,D A,B,C A B,C,D B,D C C BQ. 11 12 13A. A C BQ. 1 2 3 4 5A. 6 2 1 3 5

SECTION-I

SECTION-IV


ALLEN

LTS-10/26

(A) w = z.Q , z = electrochemical equivalent

= 200gmsz 100gm/ coulumb2coulomb

= =

(C) Total charge needed = 2 coulumb

= 66.66 301000

æ ö´ç ÷è ø

coulumb = 2 coulumb.

(D) w = (100 gm / coulumb) × 1 100 10010 5 coulumb2 1000 1000

é ùæ ö æ ö´ ´ + ´ç ÷ ç ÷ê úè ø è øë û = 100 gm

3. Ans.(A, B, D)

Sol. NO

120°O

O–

: NO

134°O

:

4. Ans.(A, B, C)

Sol. (A) (B) n = 3

n = number of oxygen shared by pertetrahedra5. Ans.(A)

Sol.XeF + H O XeOF + 2HF 6 2 4¾®

sp d3 2

2 2

2x y

zd ,d-

XeF + 2H O XeO F + 4HF 6 2 2 2¾®sp d3

2zd

XeF + 3H O XeO + 6HF 6 2 3¾®sp3

None

6. Ans. (B, C, D)

Sol.O

Br(x)Br(y)

Br(z)

x x

xx

Br(x) most reactive for SN2 due to –I eff. of – C –

OGP

Br(z) most reactive for SN1 due to carbocation stablised by 5a–HTotal (4) chiral centreso total stereo isomer = 24 = 16


ALLEN

LTS-11/26

7. Ans. (B, D)

Sol. (A) 2 > 1 N – H

due to acidic hydrogen

(B) O 2Clhn

¾¾¾® O

Cl a–halogenation

(C) H Dimerised major product

(D) Only glucose give asazone8. Ans.(C)9. Ans.(C)10. Ans.(B)Sol. NaOH + HCl ® NaCl + H2O

t = 0 10 mmole 10 mmolet = 0 0 0 10 mole [NaCl] = 0.01M(neither ion hydrolysis to an appreciable extent)

V addedA

G

(I) , (i) , (b)

NH4OH + HCl ® NH4Cl + H2Ot = 0 10 mmol 10 mmolet = 0 0 0 10 mmole

Þ [NH4Cl] = 10 0.01M1000

=

Þ pH = w bpk pk logC2

- -

V added4

G

= 14 5 log(0.01)

2- -

5.5 (cation hydrolysis)(II), (iv), (c)


ALLEN

LTS-12/26

NaOH + CH3COOH ® CH3COONa + H2Ot= 0 10 mmole 10 m mole 0 0t= 0 0 0 10 m mole 0

[CH3COONa] = 10mmole 0.01M1000ml

=

pH = w bpk pk logC2

- - = 14 5 log(0.01)

2+ +

V addedA

G

= 8.5 (anion hydrolysis)

(II), (ii), (a)

NH4OH + CH3COOH ® CH3COON4 + H2Ot= 0 10 mmole 10 m mole 0 0t= 0 0 0 10 m mole 0

[CH3COONa] = 10mmole 0.01M1000ml

=

pH = w A bpk pk pk2

+ - =

14 5 5 72

+ -=

V addedA

G

= 8.5 (anion hydrolysis)(IV), (iii), (c)

11. Ans. (A)

Sol.

Me

Me – CH CH CH Cl2 2 2 – – – AlCl3

Me

CHCH3

[I] – [D] – [P]Me

+ Me – CH CH CH Cl2 2 2 – – + AlCl3

Me

CH

MeCH

+

Me – CH2 – CH2 – Cl + AlCl3 Me – CH CH CH2 2 2 – – H – shift Me – CH CH CH2 3 – – Carbocationintermediate

CHMe – CH CH CH2 3 – – AlCl3

Electrophilic Substituting

I – D – P


ALLEN

LTS-13/26

12. Ans. (C)

Sol. CH3 – CH – C – OH 2

O

Red PBr2

CH3 – CH – C – OH

O

Br

x

Optical activeHell vholard zelenski not follows free radical mechanism

13. Ans. (B)Sol. Product of IV is :

Me

COOH

H

NH2NaNO2

HClMe

COOH

H

OH

Diazotisationproduct of III is

CH3 – CH – C – OH

O

Br

Optical active

Aq.KOH CH3 – CH – COOH

OH

SECTION-IV1. Ans. (6)2. Ans.(2)Sol. (b) and (d) have atleast three isomers3. Ans.(1)

Sol.

34 bond : 34 lone pairs s

4. Ans.(3)

Sol. 1) PCl3 + 3H2O ¾® H3PO3 + 3HCl

2) PCl5 + 4H2O ¾® H3PO4 + 5HCl

3) IF7 + 4H2O ¾® HIO4 + 7HF

4) IF5 + 3H2O ¾® HIO3 + 5HF


ALLEN

LTS-14/26

5) P4O10 + 6H2O ¾® 4H3PO4

6) H2S2O7 + 2H2O ¾® 2H2SO4

7) H4P2O8 + 2H2O ¾® 2H3PO4 + H2O2

5. Ans. (5)

Sol. So ketone is CnH2nO = 100, n = 6

(1) CH3 – C – CH – CH – CH – CH2 2 2 3

O

(2) CH3 – CH – C – CH – CH – CH2 2 2 3

O CH3 – C – CH – CH – CH 2 3

O

CH3

(3) CH3 – C – C – CH 3

O CH3

CH3

(4) CH3 – C – CH – CH – CH 2 3

O CH3

(5) CH3 – CH – C – CH2

O CH3

CH3

(1) CH3 – C – CH – CH – CH – CH 2 2 2 3

O

SBH CH3 2 2 2 3– CH – CH – CH – CH – CH

OH

x

(2) CH3 – CH – C – CH – CH – CH2 2 2 3

O

SBH CH3 2 2 2 3– CH – CH – CH – CH – CH

OH

x

(3) CH3 – C – C – CH 3

O

SBH CH3 3– CH – C – CH

OH

x

CH3

CH3

CH3

CH3

(4) CH3 – C – CH – CH – CH 2 3

OCH3 2 3– CH – CH – CH – CH

OHCH3 CH3

(5) CH3 CH – C – CH2

O

CH3

CH3 SBH CH3 CH – CH – CH2

O

CH3

CH3

x

(6) CH3 – C – CH – CH – CH 2 3

OCH3 2 3– CH – CH – CH – CH

OHCH3 CH3

SBHx x x

Not consider due to 2–chiral


ALLEN

HS-15/26Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-2757575 [email protected] www.allen.ac.in

LEADER TEST SERIES / JOINT PACKAGE COURSETARGET : JEE (Main + Advanced)

DISTANCE LEARNING PROGRAMME

SOLUTION

PART-1 : MATHEMATICS ANSWER KEY

Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-Advanced

PAPER-2

SECTION-I1. Ans. (C)

Sol.22x 4x 4 x [ 2,2]ƒ(x)

4x 4 x (2, )ì + - Î -

= í+ Î ¥î

y

x2

–2 –1

–6

\ Range : [–6, ¥)2. Ans. (D)

Sol. Line x k y 2 z k

1 k 2- - -

= =

Plane 2x – 4y + z = 7\ 2k – 8 + k = 7 Þ k = 5& 2 – 4k + 2 = 0 Þ k = 1Þ no value of k exist.

3. Ans. (C)Sol. h2 + (k + 1)2 + h2 + (k – 2)2 = 3[h2 + (k – 1)2]

2h2 + 2k2 – 2k + 5 = 3h2 + 3k2 + 3 – 6kÞ h2 + k2 – 4k – 2 = 0Þ x2 + y2 – 4y – 2 = 0

4. Ans. (D)Sol.

aR(h,k)

PQ

Chord of contact of Rhx + 2ky = 6

compare with x cos ysin 12

q + q =

2h 2k 6cos sin

= =q q

h2 + k2 = 9which is director circle of x2 + 2y2 = 6.

5. Ans. (B)

Sol.

2/ a 1

10

a2

cos ax dxsec (ax 2)lim 1

a

1 -

-

®¥

+ò, let ax = t

/ a 1

10

a

cos t dtsec (t 2)lim 1

a

1 -

-

®¥

+=

ò

1

a 1

1cos 3a 2lim1 2sec 2 3a

-

®¥ -

p

= = =pæ ö+ç ÷

è ø

Q. 1 2 3 4 5 6 7 8 9 10A. C D C D B D C A,B,D B,C A,CQ. 11 12 13 14 15 16 17 18A. A,D A,D A,B,D A,B,D C C B C

SECTION-I

ALLEN

LTS-16/26

6. Ans. (D)x2 + y2 = 5 Þ (x – y)2 = 5 – 2xy

V = px2y – pyx2 = ( )xy 5 2xyp -

Let xy = t

( )V t 5 2t= p -

dv ( 2)t5 2t

dt 2 5 2t

é ù-Þ = p - +ê ú-ë û

= 5 – 2t –t = 0

Þ5

t3

= Þ5

xy3

=

7. Ans. (C)Sol. Let a point on y3 = x4 be (t3, t4)

3y2y' = 4x3

Þ3

2

4xy '

3y= Þ

4y ' t

3=

Equation of tangent is4 34t

y t (x t )3

- = -

Q it is a normal to x2 + y2 – 2x = 0\ it must pass through (1, 0)

Þ3 33

t 1 t4

- = - Þ3t

14

=

Þ t3 = 4

Now m = 4t3

Þ3

33mt 4

4æ ö = =ç ÷è ø

8. Ans. (A,B,D)Sol.

44

34

J E D

G FH

1 e3 e4

A B C x

y

A is area bounded by curve between x = 1,x = e4

x = e3 is point of inflectionNow, A < Ar.( ABGF) + Ar.(( BCDE)Þ A < (e3 –1)34 + (e4 – e3)44

Þ A < 256e4 – 175e3 – 81 (Ans. A)Also, A < Ar.( ACDJ)Þ A < (e4 – 1)44

Þ A < 256(e4 – 1) (Ans. B)

Again, A > Ar.(trapezium BFDC)

Þ ( ) ( )4 4 4 31A 3 4 e e2

> + -

Þ 3337A (e (e 1))2

> - (Ans. D)

9. Ans. (B,C)Sol. Put z = x + iy, (x,y Î R)

Þ x2 – y2 + 2ixy – 2iy + 2|y|i = 8iìíî

x2 – y2 = 0 Þ x = y or x = –y 2xy – 2y + 2|y| = 8Þ xy – y + |y| = 4 ....(1)Case-I : when x = y

Þ x2 – x + |x| = 4(i) if x > 0, then x = 2, y = 2

Þ z = 2(1 + i)(ii) if x < 0 Þ y < 0 (rejected)

Case-II : when x = –yÞ –x2 + x + |x| = 4

x2 – x – |x| = –4(i) if x > 0, then x2 – 2x + 4 = 0

Þ x Î f(ii) if x < 0 Þ x2 = –4

Þ x Î f10. Ans. (A,C)

Sol. 2

y dy xdx1 y-

=-

2

2y dy xdx2 1 y

-=

-

2

2y dy xdx2 1 y

-æ ö =ç ÷-è ø

( )2d 1 y xdx- =2

2 x1 y C2

- = +

put x = 0, y = 1 Þ C = 0

Þ 4

2 x1 y4

- = Þ4

2x y 14

+ =

11. Ans. (A,D)

Sol. 1 1 11tan x tan tan 3y

- - -+ =

1xy 3x1y

+=

-


ALLEN

LTS-17/26

Þ xy + 1 = 3y – 3x

Þ3y 1 10x 3y 3 y 3

-= = -

+ +y + 3 = 5, 10y = 2, 7 Þ x = 1, 2

12. Ans. (A,D)Sol. Locus of P is a circle X2 + Y2 = 4

and locus of Q is |X| + |Y| = 2

(0,2)

(2,0)X

Y

O

13. Ans. (A,B,D)

Sol.aì

- + = íbî

2x 3x b 0 or 3

ab

x2 – ax + 6 = 0 3

aìí bî

or ab

Case-I : 3

.3 6

a + b =ìía b =î

Þ a = 1, b = 2 or a = 2, b = 1Þ a = 7, b = 2 and a = 5, b = 2

Case-II : 3 3

6

a + b =ìíab =î

Þ (3 – 3b)b = 6Þ b2 – b + 2 = 0Þ no real value of b.

14. Ans. (A,B,D)Sol. adj(M) = 2N, adj(N) = M

Þ |adjM|= 8|N|and |adjN|=|M|Þ |M|2 = 8|N|and |N|2 =|M|Þ |N| = 2Now, MN = adj(N).M = |N|IÞ MN = 2I Þ (B)Now adj(M2N) + adj(MN2)

= adj(M.2I) + adj(2IM)= 4(2N) + 4(M)Now, adj(MN–1) = adj(2N–2) Þ (A)= 4(adjN)–2 = 4M–2 Þ (C)

Paragraph for Question 15 and 1615. Ans. (C)

Sol. Case : I nW

n+1B

n–1W

nB

urn A urn B2w

1w

+ + + +

+ + + +´ + ´ =n n 2 n 1 n 1 n 1

1 2 1 1 12n 1 2n 2 2n 1 2n 2

1 2 1 2

C C C C . C 1325C C C C

Þ - - =229n 46n 24 0( ) ( )Þ - + =n 2 29n 12 0

n = 216. Ans. (C)

Sol.´

=´ + ´

2 645 15

2 6 3 9 135 15 5 15

Paragraph for Question 17 and 1817. Ans. (B)

Sol. ( ) ( ) ( )- - =22ƒ x .ƒ ' x 2ƒ ' x 0

2 x

( )( ) ( )- =12 ƒ x 1 ƒ ' x

2 xintegrating both the sides

( )( )- - +2ƒ x 1 x C

Q ƒ(0) = 2 c = 1

( )( )Þ - = +2ƒ x 1 x 1

( )Þ = + +ƒ x 1 x 1

Now, ( )

+ +® ®

- + -=

x 0 x 0

ƒ x 2 x 1 1lim limx x

( )+®

+ -= =

+ +x 0

x 1 1 1lim2x x 1 1

18. Ans. (C)

Sol. Q ( ) ( ) ( ) ( )= Þ =

1 1g ' y g ' 4ƒ ' x ƒ ' 64

( )( )=Qƒ 64 4

= =1 96196


ALLEN

LTS-18/26

SOLUTION

PART-2 : PHYSICS ANSWER KEY

SECTION-I1. Ans. (C)

Sol. Force on m due to M(F) = 2 2GMm

( x )+l

Net downward force on m = 2Fcosq

MM

qx

m

l l

= 2 2 2 2

GMm x2 .x x+ +l l

= 2 2 3 / 22GMm x

( x )+l

if x << l

Fnet = 32GMmx ma=

l

a = 3

P32GM x T 2

2GMÞ = p

l

l

For TS

MMl l

F

w

F = 2MGM M

(2 )= w l

l

2

Þ2

2 3GM GM4 4

= w Þ w =ll l

Þ3

S4T 2GM

= pl

P

S

T 1T 2 2

=

2. Ans. (C)Sol. When identical resistor is inserted in BD

then it forms a wheatstone bridge due towhich current in the bulb becomes zero andit becomes dark.

3. Ans. (A)Sol. For mean position : qE = kx

x = qEk = Amplitude A

–veextreme

meanposition

+veextreme

Kx qEqE

v=0A A

x

ÞqEAk

=

Net force on block at any instantFnet = qE – kx

Fnet = –kqExk

æ ö-ç ÷è ø

= – k(x – x0)Fnet = –kXma = –kX

a = – k .Xm

ÞmT 2k

= p

4. Ans. (B)Sol. Let area of needle be A and length be l.

For needle to remain in equilibriumForce due to surface tension ³ weight ofneedle2S (A )g³ rl l

2SAg

£r

Q. 1 2 3 4 5 6 7 8 9 10A. C C A B C A A B,C B,C,D B,DQ. 11 12 13 14 15 16 17 18A. A,D D A,C B,C,D B A B D

SECTION-I


ALLEN

LTS-19/26

22D 2S 8SD

4 g gp

£ Þ £r pr

on solving D 1.53mm£ & independent oflength.

5. Ans. (C)6. Ans. (A)7. Ans. (A)Sol. We have

2T – 20 = 2b (i)20 – T = 2a (ii)

m1

m2

P1

P2

b2kg

b

2kg

T

a

a – 2b = 0 Þ a = 2b (iii)2T – 20 = 2b40 – 2T = 8b20 = 10b Þ b = 2m/s2

\ a = 4 m/s2

Also T = 12 N

\ vwave = 212 20m / s

3 10- =´

apulley = 2 m/s2

\ 0.2 = 20t + 21 2t2

Þ t2 + 20t – 0.2 = 0t = 0.01 s

8. Ans. (B, C)Sol. Since no heat transferred Þ process is

adiabatic

\ Watm + Wgas = DKEbullet

Wgas = 4

0 0 fi i f f 8P 5 10 P VP V P V1 0.5

-´ ´ --=

g -

Also 8P0 × (5 × 10–4)g = P0 Vfg

Þ Vf = 2 /3 4 3 4 38 5 10 m 20 10 m- -´ ´ = ´

\ Wgas = 4 4

0 08P 5 10 P 20 100.5

- -´ ´ - ´ ´

= 4

0P 10 20(2 1) 400J0.5

´ ´ -=

Given 10–2 × x = 5 × 10–4 Þ x = 5 × 10–2 mAlso 10–2 L = 20 × 10–4 Þ L = 20 × 10–2 m\ Watm =

( ) 40 f i 0P V V P (15 10 ) 150 J-- = - ´ = -

L

x

\ DKEbullet = 250 J9. Ans. (B, C, D)

Sol. r1

r1

r1

r1

f

f

1 sin f = 17 sin r

2

since rays enters from air to glass Þ 2 cr £ q

Thus no TIR from back surface10. Ans. (B, D)Sol. Since inductor is ideal Þ EMF inductor =

Potential difference inductor and EMFacross resistor = 0


ALLEN

LTS-20/26

11. Ans. (A, D)Sol. flux through the strip,

( )2 20 0i id 2 a x dx2 (a x) 2 (a x)m mé ùf = + -ê úp - p +ë û

= 2 20 1 1i 2 a x dx2 o a x a xm é ù+ ´ -ê úp - +ë û

i

i

x

a

dx

02 2

2 ia dxd .a x

mf =

p -

Total flux through the ring,a

02 2

0

2 ia dx2a x

é ùmf = ê ú

pê ú-ë ûò

a10

0

4 ia xsina

-é ùm æ ö= ç ÷ê úp è øë û

= 04 ia2

m pé ù= ê úp ë û

02 iaf = m

Mutual inductance, M = 020

2 iaQ 2 ai i^

m= = m

Also, 22

MdiMdi / dt dt

^

^

e= Þ e =

0

2 t

2 00

22 ai.e .æ öç ÷

mè øe = mm

e2 = 4ai

Induced current in the ring, iR = 2

Re

(R resistance)

R4ai 2ii2 a

= =p p

Magnetic field at the centre,

Bnet = 0 0 0i i (2i / ). . .2 a 2 a 2 am m m p

+ -p p

= 0 0ii.a a

m m-

p p

= 0

12. Ans. (D)Sol. Current in the element,

i = –KA dQdx

Q2Q1 dx i

x K= Qa

2

1

Q

0 Q

dQi dx AQ

= - µò òl

1

2

Qi A lnQ

æ ö= µ ç ÷

è øl ...(i)

Also, 1

Qx

0 Q

dQi dx AQ

= - µò ò

1Qi x A lnQ

æ ö= µ ç ÷

è ø

Using (i), x /

1 1

2

Q QQ Q

æ ö=ç ÷

è ø

l

x /2

11

QQ QQ

æ ö= ç ÷

è ø

l

13. Ans. (A, C)

Sol. Velocity gradient

dv r cos 0dy r sin

w q -=

q

w

rq dy

= tanw w

=q q

(as q is small)


ALLEN

LTS-21/26

Considering elemental ring of radius x &thickness dx torque on this elemental ring,

d (2 xdx) xwé ùt = p hê úqë û

w

x dx

Þ Net torque, 32 R3

wt = p h

q

14. Ans. (B, C, D)

Sol. f = f0 0

s

v vv v

æ ö-ç ÷-è ø

If source is receding away, then

f = f0 s

vv v

æ öç ÷+è ø

so f decreases as vs increasesIf observer receding away, then

f = f0 0v v

v-æ ö

ç ÷è ø

so, f decreases as v0 increases but at highspeeds f will increase and at very highspeeds no frequency will be detected

15. Ans. (B)

Sol.BdV Bdgdy

Vr

r = - =r

Bdr = r2g dy

r = r + 2ghB

r

= gh1B

ræ ör +ç ÷è ø

2dy dy

B=

rò ò

0

1 1 ghB

é ù- - =ê úr rë û

0

1 1 ghB

- =r r

0

1 gh 1B

- =r r

0

0

BB gh

rr =

- r = 1

00

gh1B

-ræ ör -ç ÷è ø

= 00

gh1B

ræ ör +ç ÷è ø

16. Ans. (A)

Sol.dp gdy

= -r = –g(r0 +by)

dp = r0gh + 2ghb

2

17. Ans. (B)Sol. Least energetic photon corresponds to

transition from n = 2 to n = 1,

so 2 2

2 21500P 1500 2P 1 2 1

´l = =

- -

= 2000 Å 200nm=

18. Ans. (D)

Sol. 21hc 1240eV nmE 6.2eV

200-

D = = =l

only in option (D), there is difference of6.2 V in energy of first & second level.i.e, E2 – E1 = 6.2 eV


ALLEN

LTS-22/26

SOLUTION

PART-3 : CHEMISTRY ANSWER KEY

SECTION-I1. Ans. (D)

Sol. [Ag+] for pptn of I– = sp –k (AgI)

M.[I ]

171610 10

0 1

-

- = =

[Ag+] for pptn of Br– = sp –k (AgBr)

M.[Br ]

131210 10

0 1

-

- = =

Þ [I–] where AgBr starts precipitating = sp –k (AgI)

M(Ag )

175

1210 1010

-

+ -= =

Þ % [I–] remaining in solution = –

. %.

510 100 0 010 1

´ =

Þ % [I–] precipitated =99.99%

2. Ans. (B)

Sol. ph = 3 Þ [H+] = 10–3 M

let the degree of dissociation be a

Þ–

..

310 0 10 01

a = =

Þ i = 1 + (y – 1)a = 1.1

Þ p(osmotic pressure) = (1.1) (0.01) RT = 0.011 RT

3. Ans.(D)

Sol. (A) N2H4 + 2I2 ¾® 4HI + N2

(B) N2H4 + 2O2 ¾® 2H2O2 + N2

(C) N2H4 + 2CuSO4 ¾® Cu + N2 + 2H2SO4

4. Ans.(C)

Sol. (A)

H

N+ Cr CrH H H

2 O O

O O

O–

OO–

(B) (NH4)2 Cr2O7 D¾¾® N2 + 2 3(green)

Cr O + 4H2O (this section is used for making artificial volcano)

(C) Metal nitrate D¾¾® metal oxide + NO2 + O2

(D) Metal nitrates are generally soluble in water

Q. 1 2 3 4 5 6 7 8 9 10A. D B D C D A C A,B,C,D A,B,C A,B,C,DQ. 11 12 13 14 15 16 17 18A. B,C A,D A,D A,B,C,D D B B B

SECTION-I


ALLEN

LTS-23/26

5. Ans. (D)

Sol. CH – CH – CH – CH2 33

CH3Cl2 CH – CH – CH – – Cl23 2 CH

CH3

CH – CH – CH – 3 3CH

CH3

Cl

CH – C – CH – 3 2 3CH

CH3

Cl

Cl – CH – CH – CH – 2 2 3CH

CH3

6. Ans. (A)

Sol.SO3

Low temperature

High temperature

SO3H

(X)

SO3H

[KCP]

[TCP]

7. Ans. (C)

Sol. + CH Cl2 2 (excess)AlCl3

CH Cl2 – CH2

X

NBS

Br

Y

Hydrolysis PCC

Br OH O

Z8. (A, B, C, D)

Refer theory


ALLEN

LTS-24/26

9. Ans. (A, B, C)Sol. (A) Carbon atoms occupy all lattice points and also are present in alternate tetrahedral voids.

(B) Two atoms in the closest neighbourhood are the ones present at any lattice point and intothe tetrahedral void.

Þ nearesta ad r 3 32 2

8 4æ ö

= = ´ =ç ÷ç ÷è ø

(C) Contribution from atoms at corner = 18 18

´ =

contribution from atoms at face centres = 16 32

´ = contribution from atoms in half tetrahedral

voids = 1 8 42

´ =

Þ Total atoms (effective) = 810. Ans.(A, B, C, D)Sol. NO + I NO + I 2 2

– – ¾®

FeSO4

[Fe(H O) NO]2 5

+ISO4

Brown coloured complex

[Fe(H O) NO]2 5

+I

I

SO4

d

d

d

d d

x –y2 2

xy

z2

yz xz

d dx –y2 2 z2d d dxy yz xz

Fe [Ar]3d 7Þ

µ = 15 B.M., brown coloured complex due to charge transfer, hybridisation Þ sp3d2

11. Ans.(B, C)Sol. (A) Na+ & KÅ ions can form alums.

(B) K SO .Al (SO ) .24H O 2 4 2 4 3 2BaCl2 dil. HNO3BaSO4 Not Soluble

white ppt.(C) Alums are used as a mordants in dyeing industry.(D) All types of alums are not coloured.

12. Ans.(A, D)

Sol. (A) N Cº+ – LAH NH – CH3

(D) O + N – MeH2 N – Me H /Pd2 CH – NH – CH3–H O2


ALLEN

LTS-25/26

13. Ans. (A, D)Sol. (A)

HO – C

O

C – OH + CH – CH2 2

O OH OH

H

O

C

O

C – O – CH – CH – O2 2 + nH O2

N Molen

(D) COOH

COOH+ CH – 2 CH2

OH OH

HC – O – CH – CH – O 2 2 —

C – O – CH – CH – O 2 2 —

O

On

Pthallic Acid Glyptal14. Ans. (A, B, C, D )

Sol.H H

2

(i) RMg X(ii) FeCl¾¾¾¾®

H HMgX

+ R – H

FeCl2

MgX

MgX

Fe

15. Ans. (D)

Sol. m mol of compound = mg

g / mol300 2

150=

m mole of H2O produced = mg

g / mol324 18

18=

Þ m mole of H atom = (18 × 2)

Þ m mole of H atom in one m mole of compound = 18 2 18

2´

=

m mole of C in 2 m mole compound = m mole of H2CO3

= m moleof NaOH

2 = . mmole0 3 80 12

2´

=

Þ m mole of C in 1 m mole compound = 6

m mole of N2 = .5 394 1 2

760 0 0821 3009æ öæ ö´ç ÷ç ÷

è øè ø;

Þ m mole of N present in 1 m mole of compound = 2Þ C6H18Ox.N2 º mol.wt = (6 × 12 + 18 × 1 + 16 × x + 14 × 2) = 150Þ x = 2Þ formula = C6H18N2O2


ALLEN

LTS-26/26

16. Ans. (B)

Sol. CO H COn n2 2 312

1000æ ö= =ç ÷è ø

Nn 22

1000=

Þ CO Nn n .2 2 0 014+ =

17. Ans.(B)

Sol. (A) No change observed in test tubes.

(B) + ¾¾¾¾¾¾® ¯2H S2Basic medium Brown ppt

Sn SnS

+ ¾¾¾¾¾¾® ¯2H S42Basic medium

pptYellowSn SnS

(C) + ¾¾¾¾¾® ¯dil. NaOH22

white pptSn Sn(OH)

+ ¾¾¾¾¾® ¯dil. NaOH24

white pptSn Sn(OH)

(D) + -¾¾¾¾® ¯ ¾¾¾¾®NaOH NaOH2 22 4excesswhite ppt so lub le

Sn Sn(OH) [Sn(OH) ]

+ -¾¾¾¾® ¯ ¾¾¾¾®NaOH NaOH2 24 6excesswhite ppt so luble

Sn Sn(OH) [Sn(OH) ]

18. Ans.(B)

Sol. when tin(II) & tin (IV) compound dissolve in excess NaOH then [Sn(OH)4]–2 & [Sn(OH)6]–2 ionsare formed respectively.


ALLEN

Distance Learning Program for JEE Main, IIT-JEE Advanced - … · 2019-07-25 · 018 20182018 2016...

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Transcript of Distance Learning Program for JEE Main, IIT-JEE Advanced - … · 2019-07-25 · 018 20182018 2016...