Bmat FT4 Chemistry Solns

1

♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns

TEST-I

CHEMISTRY

PART-I SECTION–I

Single Correct Choice Type

1. (A) mass of solution = (100 + 1.04) g

volume of solution = ml01.1

04.101

molarity of silver acetate = 04.101

01.11000

167

04.1 ×× = 0.0622 M

For saturated solution ionic product of salt is equal to solubility product of the salt.

]Ag[ ]COOCH[K 3sp+−=

2)0622.0(=

23 M1087.3 −×=

2. (C) There is one C=C bond in C5H10 (decolourization of bromine). Its six possible isomers are

3222 CHCHCHCHCH −−−= , 323 CHCHCHCHCH −=− (Cis and Trans)

CH2 = C – CH2 – CH3

CH3

CH3 – C = CH – CH3 ,

CH3

, CH3 – CH – CH = CH2

CH3

3. (A) Br–Br: + AlBr3

. .

. . Br–Br–AlBr3 ⊕ –

O

O

C H

O

Br–Br–AlBr3 ⊕ –

O

O

C H

O

Br–AlBr3 –

⊕

Br H

O

O Br

C H

O

–AlBr4 –HBr –AlBr3

–

Given that ether groups are ortho and para directors and the aldehyde group is a meta director, you might predict the major product to have the new C-Br bond meta to aldehyde group. However the given product is correct. Consider the following resonance contributor.

O

O

C H

O

O

O

C H

O

⊕

. .

. .

. . . .

. . . .

–

Due to this resonance some of the electron lone pair density on the ether oxygen decreases, reducing its ability to stabilize an adjacent carbocation and thus weakening its influence as an ortho/para director. This resonance also adds electron density to the aldehyde group, decrease its electron withdrawing destabilization of an adjacent

2


carbocation, and thus weakening its influence as a meta director. This leaves the other oxygen as the dominant directing group. The net effect is EAS para to this strongest group.

4. (C)

5. (B) ∆G° for formation of one mole of kJ2

594Al

−=

J 10002

594nFEG ×−=°−=°∆

V965003

1000

2

594

nF

GE

××=°∆=°

V 02.1= 6. (D) [ ] [ ]yx BAkr =

yx3 )1.0()1.0(k104.2 =× − …(1)

yx3 )1.0()2.0(k108.4 =× − …(2)

yx2 )4.0()1.0(k108.3 =× − …(3) On solving 2y and 1x == Overall order of the reaction is 3. Rate constant depends on activation energy of the reaction. The stoichiometric

coefficient of reactants may not be equal to order of reaction.

7. (A) More is the conjugation, C–N bond acquires a double bond character and bond length decreases.

In compound (1), due to steric inhibition to conjugation – NO2 is pushed out of the plane, non conjugative to the ring.

In (4), two – OC2H5 groups at m-position are e– withdrawing through – I, do not allow the ring e– cloud to conjugate with the ring.

In (3), –OC2H5 is e– donating towards the ring hence the – NO2 group is the most conjugated to the benzene ring.

In compound (2), normal conjugation takes place.

3


8. (D)

CH2–CN CH2–CN

C2H5ONa C2H5OH NH

CN

H2O O

CN

H2O/H+

O

COOH

∆ O

CH–CN

CH2–C≡N

–

(4)

SECTION–II Multiple Correct Choice Type

9. (A, B, D) 10. (A, B) 11. (A, B, C, D)

CO2H Br2in CCl4

Br CO2H

Br

Na2CO3

–H+

Br

Br

C O

O

–

–Br –

Br C8H7Br

–CO2

Product shows geometrical isomerism and it exists in two stereoisomeric forms

having molecular formula, C8H7Br. 12. (A, B, C, D) For one electron system energy of electron depends only on principal quantum

number while for many electron system it depends on value of (n+l). 24th electron enters in 3d orbital. Hence, its principal quantum number is 3. 13. (A, B, C)

4


SECTION–III Paragraph Type

Paragraph for Question Nos. 14 to 16

14. (B) ZnCO3 →∆ ZnO + CO2 (P)

15. (A) ZnCO3 + 2HCl(dil) → ZnCl2 + H2O + CO2 (P) (Q)

−24)OH(Zn + 4H+ → Zn2+ + 4H2O

Zn2+ + H2S → ZnS↓ + 2H+ (R)

16. (A) ZnCl2 + 2NaOH → Zn(OH)2 + 2NaCl (white ppt.)

ZnCl2 + 4NH3 + H2O → [Zn(NH3)4]Cl2 + H2O


17. (C)

Se O

O + C

C

I

∆ A

Se O

O

I further heating I

NO2

I

NO2

,Cu, ∆, DMF

–SeO2

(B) is o, o´-disubstituted biphenyl – so B is optically active in nature.

18. (B) 1, 2-disubstituted alkenes prepared by selenoxide elimination are generally obtained largely as the E-isomers. Following syn elimination from a cylic transition state in which the two substituent groups are staggered.

C6H5CH2Br → C6H5CH2SeC6H5 → C6H5CHSeC6H5 C6H5SeNa C2H5OH

1. LiN(iso-C3H7)2

2. C2H5Br

D2O2, THF

C6H5CH–Se+

CH3CH–H O–

C6H5

H CH3 H

C6H5

C2H5

5


SECTION−−−−IV

Integer Type

19. Solubility of CaF2 = 1.7 × 10–3 g/100 c.c

= 17 × 10–3 g/1000 c.c

= 78

107.1 2−×

= 2.18 × 10–4 mol L–1

CaF2(s) +2)aq(Ca +

−)aq(F2

[Ca2+] = 2.18 × 10–4 mol L–1

[F–] = 2 × 2.18 × 10–4 mol L–1

Ksp = [Ca2+] [F–]2

= [2.18 × 10–4] [2 × 2.18 × 10–4]2 = 4.14 × 10–11

Hence, x ≈ 4.0

Ans. 4 20. Let there are ‘n’ palladium atoms per unit cell. All types of gaps of palladium metal are

occupied by hydrogen atoms. Number of hydrogen atoms occupying tetrahedron shaped gaps = 2n. Number of hydrogen atoms occupying octahedron shaped gaps = n. Total number of hydrogen atoms = 3n. Simplest ratio of Pd and H atoms is 1 : 3. So empirical formula of palladium hydride is PdH3.

Ans. 3 21. 4224 SOK )CN(Cu CuSO KCN2 +→+

2222 )CN( )CN(Cu )CN(Cu2 +→ (unstable) [ ]4322 )CN(CuK2 KCN6 )CN(Cu →+

Ans. 3

22. CH3–CH–CH=CH–CH–CH3

Br

* *

Br

There are three stereocentres in the given compound but the molecule has symmetrical structure. Therefore total six stereoisomers are possible.

CH3–C–CH=CH–C–CH3

Br Br

H H

6


1. d cis d 2. d trans d 3. l cis l 4. l trans l 5. d cis l or l cis d 6. d trans l or l trans d

Ans. 6

23.

C−OH

CH2–C6H5

SOCl2

O

–SO2

–HCl

C−Cl

CH2–C6H5

O

AlCl3

H3O+

C

O

Zn/Hg/HCl

(∆U = 9)

Ans. 9 24. A(x+n)+ + ne– → Ax+

(Oxidised form) (Reduced form)

Case-I: [Ax+] = 25, [A(x+n)+] = 75,

ERP = o

RPE + ]F.R[

]F.O[log

n

059.0

= o

RPE +

25

75log

n

059.0 …(1)

Case-II: ERP = o

RPE + ]40[

]60[log

n

059.0 …(2)

Now (2) – (1)

(0.115 – 0.1066) =

−2

3log3log

n

059.0

0.0084 = [ ])2log3(log3logn

059.0 −−

n = 2log0084.0

059.0 × = 0084.0

3010.0059.0 ×=

0084.0

0177.0 = 2.1 ≈ 2.

Ans. 2

7


25. 2HI H2 + I2 1 0 0

(1–α)

α2

α2

Kc = 2

2

)1(4 α−α

= 2

2

)8.01(4

)8.0(

− = 4.

Now for the association equilibria

H2 + I2 2HI 2 2 0 (2–x) (2–x) 2x

cK ′ = cK

1 =

2

2

)2(

4

x

x

−

2

2

)2(

4

x

x

−=

4

1 ;

x

x

−2

2 =

2

1

x = 5

2.

I2 left = 2 – 5

2=

5

8 mole.

Volume of Na2S2O3 required can be calculated as follows.

I2 + 2Na2S2O3 → 2NaI + Na2S4O6.

Eq. of Na2S2O3 = Eq. of I2 left

1 × 1.6 × V = 5

8× 2 (n-factor I2 = 2 and that Na2S2O3 = 1)

V = 6.15

16

× = 2 L.

Ans. 2

26.

∆

COOH

(A)

OsO4

HIO4 (1,2 glycol splitting)

+ COOH

LiAlH 4

Diel’s Alder reaction

COOH

COOH

COOH

COOH

HO

HO

COOH

COOH

OHC

OHC

CH2OH CH2OH

CH2OH CH2OH

(B)

(C) (D)

Ans. 4

8


27. Number of moles of alcohol = 20

1

104

2.5 =

Number of moles of CH4 produced 203

4.2236.3 ==

lt

So number of –OH groups presents = 3

Ans. 3 28. 2NaOH + NiCl2 → Ni (OH)2 + 2NaCl 0.044 mole 0.025 mole 0 0.003 0.022 mole

At equilibrium, [Ni+2] = 0.012 M

Now, 1.6 × 10–14 = 22 ]OH[]Ni[ −+

⇒ [OH–]2 = 012.0

106.1 14−× = 1.33 × 10–12

⇒ [OH–] = 1.15 × 10–6

[ ] 86

14

1087.01015.1

10H −

−

−+ ×=

×=

( )81087.0logpH −×−= = 8

Ans. 8

9


PHYSICS

PART-II

SECTION – I


29. E = 02ε

σ =

x

V

∆∆−

E2 + E1 =

+−−−

12

010 = 10 V/m outside plates

E2 – E1 = 5.22

5 = V/m between the plates

As potential is decreasing from sheet with charge σ2 to sheet having charge σ1

∴∴∴∴ (A)

30. Time taken from B to C and then back from C to B are both equal

∴ Distance fallen is 4

H till C.

Hence answer is 4

3H

∴∴∴∴ (C)

H h

A C R/2 R/4 R/4 B

31. xA

R θ1

xB

R θ2

xB – xA

R

θ

R

xA=θ1tan R

xB=θ2tan R

xx AB −=θtan

⇒ 12 tantantan θ−θ=θ

∴∴∴∴ (B)

32. ∴∴∴∴ (B)

33. Restoring torque τ = ISα = Mg2

Lsin θ + mg L sinθ

⇒ α

+ 2

2

3mL

ML = θ

+ mM

gL2

(sin θ ≈ θ)

⇒ α = θ

+

+

LmM

mM

g

3

2

Mg

mg

Lθ

10


= θ+

+LmM

mMg

)3(2

)2(3

∴ ω2 =

++

mM

mM

L

g

3

2

2

3

T

π2 =

3(M 2m)g

2(M 3m)L

++

⇒ T =gmM

LmM

)2(3

)3(22

++π

∴∴∴∴ (B)

34. We shall write the condition of the equality to zero of the potential of the sphere, and hence of any point inside it (in particular, its centre), with the passage of time t. We shall single out three time intervals:

v

bt

v

bt

v

a

v

at ≥<≤< )3(,)2(,)1( .

Denoting the charge of the sphere q(t), we obtain the following expression for an instant t

from the first time interval : 0)(21 =++

vt

tq

b

q

a

q,

hence q(t) = tb

q

a

qv

+− 21

I1(t) =

+−b

q

a

qv 21

For an instant t from the second time interval, we find that the fields inside and outside the sphere are independent, and hence

vt

qtq 1)( +=

b

q2− , b

qvtI 2

2 )( −=

Finally, as soon as the sphere absorbs the two point charges q1 and q2, the current will stop flowing through the “earthed” conductor, and we can write 0)(3 =tI . Thus,

I(t) =

≥

<≤−

<

+−

v

bt

v

bt

v

a

b

qv

v

at

b

q

a

qv

,0

,

,

2

21

∴∴∴∴ (D)

35. (A)

11


36. The magnetic field can be calculated as the superposition of fields of two cylindrical conductors, since the effects of the currents in the area of intersection cancel.

B+ =B– in magnitude

BP =B+ cos θ + B– cos θ

∴ BP = θ+ cos2B

= θπ

µcos.

2

'2

20

R

rI

∵ r cos θ = 2

R

BP = 2

'..2

0 RI

Rπµ

= R

I

πµ2

'0

• ⊗

A B

r r

B+ B–

θ

θ θ

P

'I = current over total area )( 2Rπ

∴ 'I = 2

22

23

3

RRR

I π×+π

⇒ 'I = 332

6

+ππI

∴ BP =332

6.

20

+ππ

πµ I

R =

R

I

)332(

3 0

+πµ

∴∴∴∴ (C) SECTION – II

Multiple Correct Choice Type 37. (A, B, C)

38. F

1 =

20

1)1( =−µR

⇒ 20 = )1( −µ

R … (1)

If equiconvex lens of f = 20

Then 20

1 =

R

2)1( −µ

From equation (1) and (2) we conclude

∴ R = 40(µ – 1) That option (A) is correct.

( ) ( )RRfL

1111

1 −µ=

∞−−µ=− for refraction at the convex surface

∴ power ( )

RfP

LL

11 −µ−==

For reflection at the silvered plane surface

α=MF

∴ power PM = 0

12


For refraction at the convex surface again

( )

RPL

1−µ−=

Hence power of the system

LML PPPP ++=

( )

RPP ML

122

−µ−=+=

∴ focal length of the system

( ) cm1012

1 −=−µ

−== R

PF

u = –15 cm

uv

11 + = eqF

1

⇒ 15

11 −v

= 10

1−

v

1 =

10

1

15

1 −

⇒ v = –30 cm so (B) is correct.

∴∴∴∴ (A, B) 39. Using work-energy theorem till string become taut Work done by qE = Change in K.E.

⇒ qEl = 021 2 −mv

⇒ v = m

qEl2

When string become taut an impulse is received and speed of the bob becomes

m

qElv

2

330cos 0 =

Further using work energy theorem speed of the bob becomes

m

qElv

25

'=

∴ time period ='

2v

lπ =

qE

ml

52

2π

After 1st half revolution in magnetic field. Work done by qE =Change in K.E., from lowermost to uppermost point.

qE ⋅ 2l =m

qElmmv

25

22

21 −

13


v1 = m

qEl

2

13

∴ Time period of 2nd revolution = 1

2

v

lπ =

2ml

13qEπ

∴ Angular speed of 1st revolution = ml

qE

2

5

∴ Angular speed of 2nd revolution = ml

qE

213

∴∴∴∴ (A, B, C) 40. (A, B, C)

41. (C, D)

SECTION −−−− III

Paragraph Type


From the plot the graph with +ve slope corresponds to E2 because it goes upto 10V. At E2 = 6 volt, E1 is in opposition to E2 and hence I of E1 is zero. Consequently E1 = 6V.

From graph , at E2 = 8V, I of E2 is 0.3 A and for E1 = 6V, I due to E1 is 0.1A.

Hence the circuit takes the form

6V

+ – +

– 8V

0.3A 0.1A

0.2A

By Kirchoff’s voltage law R1 = 20 Ω and R2 = 40 Ω

42. (B)

43. (B)

44. (D)

14


Paragraph for Question Nos. 45 & 46

45. E = 2x

kQ

E2 = 2

3

0

34

4

1

x

x ρπ

πε, where x is the distance of any

point in the overlap region from the centre of the sphere with charge density ρ.

E2 = )(33

)(

00

xddxd −ε

ρ=ε−ρ+

Enet = 21 EE + = 00 33

)(

ερ+

ε−ρ xxd

+ρ –ρ

R R

1E

2E

d

E = 03ε

ρd

∴ (C)

46. V = ∫− Edx

∫2

1

V

VV = ∫ ε

ρ−R

dxR

003

V2 – V1 = 0

2

3ερ− R

|| V∆ = 0

2

3ερR

∴ (A) SECTION −−−− IV

Integer Type

47. 2

2

1

1

C

q

C

q= ; 021 2Qqq =+

vtd

AC

+ε

=0

01 ;

vtd

AC

−ε

=0

02

vtd

vtd

q

q

+−

=0

0

2

1

020

02 2Qq

vtd

vtdq =+

+−

; 00

02 2

2Q

vtd

dq =

+

( )vtdd

Qq += 0

0

02 2

2;

0

02

d

vQ

dt

dqI == = 20 amp

∴ n = 5 Ans.: 5

15


48. Let e– in hydrogen atom is excited to nth level. ∴ )1( =nKEE = ||8 ).(. nEPE

∴ 13.6 eV = eVn2

6.1328 × ⇒ n = 4

∴ E∆ =

−16

116.13 = eVeV 75.126.13

16

15 =×

Using conservation of linear momentum mv = 21 mvmv + … (1)

)( 12 vv − = v2

1 … (2)

∴ v1 = 4

3,

4 2

vv

v − … (3)

H-atom

After collision Before collision

m m neutron v1

v2 v

By energy conservation

∴ 2

2

1mv = Emvmv ∆++ 2

221 2

1

2

1

⇒

−−16

9

16

11

2

1 2mv = E∆ ⇒ 2

2

1mv = E∆

6

16 = eVeV 3475.12

3

8 =×

K.E. of neutron = 34eV Ans.: 4 49. Applying work energy theorem: Work done by gravity + Work done by buoyant force + Work done by external agent = Change in K.E. = 0

Hence, work done by external agent ⇒ 0

210

2

2

)(

ρρ−ρAgh

.

Ans.: 5 50.

VB B

30°

60°

VA = 1m/s

A

VA cos 60°

VB cos 60°

VA cos 60° = VB cos 60° VA = VB Ans.: 1

16


51. At the junction of L and C by Kirchoff’s current law, current through L at any instant is

dt

dqii RL −=

Voltage across L is dt

dqi

dt

dL

dt

diLV R

LL −==

−=dt

dq

dt

d

dt

diLV R

L

( )

−= CV

dt

d

dt

d

dt

diL R

+−=dt

dCV

dt

dVC

dt

d

dt

diL R

2

2

2

2

dt

CdV

dt

dC

dt

dV

dt

VdC

dt

dV

dt

dC

dt

diL R −−−−=

0as22

22

=−−=dt

Cd

dt

VdC

dt

dV

dt

dC

dt

diL R

= 4V Ans.: 4

52. I0 =4

1 of

12)4(

22 aaM

+

= 6

2aM = 168 kg – cm2.

Ans.: 4

2cm

a O

a=12cm

M=7kg 10cm

2cm

A B

C

53. 22

221

220 gh

l ρ−

ωρ+ = P0

22

1

8l

ωρ = 220 ghP ρ+

ω = 2

1

220 )(8

l

ghP

ρρ+

Ans.: 8 54. The radar antenna receives signals directly from Venus and by reflection from the surface of

lake. Venus may be considered infinitely far away, so that disturbance at B and D have the same phase at every instant. The path difference between the reflected and direct ray is

BE – DE = BE – BE sin 20

17


E

20º

90ºD

B

A

C

y

C

Lake35º

= 20sin

35sin35sin

yy − = )20sin1(35sin

−y

Also there is a phase difference of π due to reflection at B. Thus the total phase difference at E is

d = π+−λπ

)20sin1(35sin

2 y

for a minimum at antenna. E, phase difference π−= )12( nd

π+−λπ

)20sin1(35sin

2 y = π−π )12( n

for minimum possible value of y, n = 2 (since for n = 1, y = 0)

)20sin1(35sin

,2 −λπ y

= 2π

y = 69301.348)20sin1(

35sin =λ−

° ≈ 349

Ans.: 9

55. Applying conservation of energy

mgl cos θ = 2

2

1mv

v = θcos2gl

BglqmgN θ−θ− cos2cos = l

glm 2)cos2( θ

⇒ N = θ+θ cos2cos3 glgBmg

N is maximum when cos θ = 1 ⇒ θ = 0

θ

⇒ Nmax = glqBmg 23 + = 5 N

Ans.: 5

18


56. Distance between two virtual images is h = mm321 =hh

(where h1 = 4.5 mm and h2 = 2mm)

Magnification in 1st case = m1 = 3

5.4 =

x

xl

u

v −= (where x is the object distance)

Magnification in 2nd case = m2 = 3

2 = ( )xl

x

−

⇒ 2

3 =

xl

x

− l = 150 cm

)(

11

xlx −−− =

f

1

Solving f = 36 cm

Ans. 6

19


MATHEMATICS

PART–III SECTION - I


57. (B) 012

51

2

5189

=+

++

+ba

( ) ( ) 0151522989 =−+−+ ba

( ) 53476152 −=−+ ba

⇒ 2134,762 =⇒−==− abba

58. (B) 5552 =+− baab

⇒ ( )2

555

2

512 −=

−+ ba

⇒ 7.5.3105)52)(12( ==−+ ba

Total number of positive solutions 8222 =××=

Total number of solutions 16= .

59. (C) nnnf 32 +=

⇒ nnn ff 321 =−+ and n

nn ff 3.32 12 =− ++

So option ‘C’ is correct.

60. (B) 32

1y

x

y

dy

dx

x=− take t

x=1

⇒ ∫−=⇒−=+ dyyeteytydy

dt yy 32/2/3 22

⇒ cey

ex

e yyy

+

−= 2/

22/

2/22

2

22

10)1( −=⇒= cy hence at exy −== ,2

61. (D) ( )

( ) ( ) ( )( )23133

6lim

)!3(!)1(

)!!.(33!.3 3

3

3

−−=

−−

=∞→ nn

n

nnn

nnP

nn

9

2=

62. (C) ba =+

+=+

+7log35log

7log5log3,

7log25log

7log5log2

20


( )( ) ( )( )( )( ) ( )( )7log5log37log25log7log35log7log5log2

7log5log37log5log27log35log7log25log1

++−++++−++

=−

−ba

ab

5=

63. (B) ( ) ( )cbcba

×β++α=

( ) ( )kjikjikjia ++−β+−+++−α= '3232

where β=β 7'

β=α⇒= 40.da

⇒ .124

7coscos14.627. 2 =θ⇒θ=α=ba

64. (B) 0722 =+−

−+bybx

ac

ca

⇒ 07222 =+−

− bybxb

⇒ 02

7 =

−−− xbyx

Line passes through

2

7,

2

7 maximum distance from origin

2

7

2

7

2

722

=

+

and

minimum tends to zero.

2

7,0

SECTION – II

Multiple Correct Choice Type

65. (B), (C), (D)

Case possible 0)( >xf and )(xf decreasing

or )(,0)( xfxf < increasing

and )(xf and )(' xf have to change sign simultaneously to maintain the condition

which is not possible also ( ) 0)(').(2')(2 ≤= xfxfxf

66. (A), (B), (C), (D)

0.2)1(2)1( 816

71616

6

0

1616

0

=+−−=− ∑∑==

CCCC rr

rr

r

r

9158

16

71616

6

0 2)1( C

CCCr

r

r

=−=−∑=

= 5.7.11.13

67. (A), (D)

02322

=+

+

+

+

+

q

z

y

z

x

z

y

z

x

z

y

z

xp

21


To factorise make 0=∆

02

3

2

111

2

31

2

121

222 =

−

×−×−×××+×× qpqp

and also for perpendicularity 01=++ qp

⇒ 0144 2 =−− pp 0)2()74( =−+ pp

4

3,

4

7,3,2 =−=−== qpqp

68. (A), (C), (D)

Let ,pdx

dy = we have

xpyp =+2

⇒ ''2 xppppp +=+

⇒ xp =2 or 0'=p

which gives yx 42 = or )12( ++= ccxy

or )2(1 +=− xcy

69. (A), (B), (C), (D)

Point of contact will lie on xy = (let say (α, α)) and slope of tangent can be 1± and

24

12 +α+α=α aa eliminating α we get

24

1

2

1

2

122

++

−±=

−±a

a

aa

a

a

On solving we get 12

60113,

2

3,

3

2 ±=a

SECTION – III Paragraph Type


70. (D) ∫ π=a

o

dxxnnh |sin|2)(

Let α+=n

ka ,

n

10 <α≤ , Ik ∈

∫ ∫α

π+π=n

dxxndxxnk/1

0 0

|sin|2|sin|2

22


π

πα−+π

=n

n

n

k )cos1(24

π

πα−+π

α−=n

n

a

a

a

nh )cos1(2)(4)(

πα−

ππα−+

π=

aan

n 4)cos1(24

a

nh )( is independent of n if 0=α .

i.e. kan = ⇒ kNa =

a should be divisible by 2, 3, 4 i.e. minimum (a) = 12.

71. (A) ∫=a

dxxxh0

2 |cos|.sin)1(

⇒ ∫∞→∞→=

a

aadxxx

aa

h

0

2 |cos|.sin1

lim)1(

lim

Let α+π= na π<α≤0

⇒ ∫ ∫ ∫α+π π α

∞→∞→

+

α+π=

n

nndxxxdxxxn

ndxxx

0 0 0

222 |cos|.sin|cos|sin1

.lim|cos|.sinlim

∫π

π=

π=

0

2

3

2|cos|sin

1dxxx

72. (D) =

∞→)(lim nh

n

∫ ∑ ∫−

=

+

∞→∞→π=π

1

0

1

0

1

/

|sin|lim|sin|limn

r

n

r

nrnn

dxxnxdxxnx

Let ∑ ∫−

=∞→

+π

+⇒

=+=1

0

/1

0

sinlimn

r

n

ndtt

n

rnt

n

rxt

n

r

dttntn

rn

r

n

n∑ ∫

−

=∞→π

+=1

0

/1

0

|sin|lim

∫ ∫ π=1

0

1

0

|sin|. dttdxx

π

=π

= 12.

2

1

23


Paragraph for Question Nos. 73 and 74 Sol.: 73. (C) α= cos3AP

In APC∆

)sin(

5

sin

cos3

A−π=

αα

⇒ 12

25cot =α

α

α α

P

5 4

3

A B

C

π–A

74. (A) α=α−

α=

∆∆

tan.3

4

)º90sin(.3..2

1

sin.4..2

1

PB

PB

PAB

PBC

25

12.

3

4=

25

16=

SECTION – IV Integer Type

75. Let

+ bjLjb

aiMbjSaiQ3

4,

2),(),( , ( )ajaiR +

be required points: T is intersection of LM and SR

i.e. )(3

4

22aibjbjj

baij

bai µ+=

−+λ++

S T

R

M

Q P

L

5

3−=λ and 5

2=µ one comparing the coefficient, we get

+≡ ia

bjT5

2

Area of abba

MRT20

3

25

.3

2

1 =××=∆ .

Ans. 3 76. Let the first equation have the roots 1,, γβα and the second equation have the roots

2,, γβα .

Then ,51 −=γ+β+α p=β+αγ+αβ )(1

,72 −=γ+β+α p=β+αγ+αβ )(2

⇒ 0))(( 21 =β+αγ−γ as 0)( 21 =β+α⇒γ≠γ

7,5 21 −=γ−=γ .

Ans. 2

24


77. 2353)1530)(23()23( 1123 −λ=−=

Hence remainder is 30. Ans. 3 78. )sin(sinsin yxyx +=+

⇒ 02

sin2

sin2

sin =

+ yxyx

⇒ π=+ 12hyx , π= 22hx or π= 32hy , Zhhh ∈321 ,,

Only six solutions are possible Ans. 6

X

Y

79.

−++=

− 14

114

41

14

42

22

4

nn

n

n

⇒

−∑= 14

42

416

1 n

n

n

+−

−+=∑

= 121

121

81

4116

1 nnn

−+=33

11

8

14

33

44+=

Hence, 433

4433 =

+ .

Ans. 4 80. Required plane contains z-axis and is perpendicular to 0143 =++− zyx . Hence its

equation is 034 =+ yx its distance from (1, 2, 3) is 2916

64 =+

+ units.

Ans. 2

81. The abscissas form the series 1...8

1

4

1

2

1 =+++

Ans. 1

82. On adding and subtracting the curves we get

05920),1(4)3( 22 =+−−=+ yyyx respectively

Let point of intersection be ),)(,)(,)(,( 24231211 yxyxyxyx

⇒ 2021 =+ yy and 0, 21 >yy

The required sum of distances = 211

21

21 )2()1(4)2()3( −+−∑=−++∑ yyyx

)(2 211 yyy +=∑= 40=

Ans. 4

25


83. 113

14

13

12

13

11

13

10)( −

−

+

+

=xxxx

xf

01314

ln1314

1312

ln1312

1311

ln1311

1310

ln1310

)(' <

−

+

+

=xxxx

xf

Since 0)0( >f and )(xf is decreasing, 0)( =xf has exactly one root at x = 2.

Ans. 1 84.

A

=

−−−−

0

00

2

0

00

2

0

00

2

0

00

2

0

0

0

0

00

0

0

00

0

32

1 321 n

n ⋯

⋯⋯

⋯

⋮

⋯

⋯⋯

⋯

⋮

A2 =

−−−−−−−−

0

00

2

0

00

2

0

00

2

0

00

2

0

0

0

0

00

0

0

00

0

32

1

0

00

2

0

00

2

0

00

2

0

00

2

0

0

0

0

00

0

0

00

0

32

1 321321 nn

nn ⋯

⋯⋯

⋯

⋮

⋯

⋯⋯

⋯

⋮

⋯

⋯⋯

⋯

⋮

⋯

⋯⋯

⋯

⋮

A2

∑

=

−−−−−

0

00

2

0

00

2

0

00

2

0

00

2

0

0

0

0

0

0

0

0

0

0

0

0

0

00

2

0

0

0

0

00

0

0

00

0

32

1 321 nii

n ⋯

⋯⋯

⋯

⋮

⋯

⋯

⋯

⋮

⋯

⋯⋯

⋯

⋮

⇒ ( )AiA i−∑= 2.2

⇒ ( ) AiAnin 1

2.−−

∑=

⇒ ( ) ( )ii

nin aiATrace ∑∑= −− 12. ( )nii −

∑= 2.

⇒ ( ) ( ) .....8

3

4

2

2

12limlim

/1 +++=∑= −

∞→∞→

i

n

nn

niATrace = 2

Ans. 2

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Bmat FT4 Chemistry Solns