Bmat FT4 Chemistry Solns
Transcript of 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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
Paragraph for Question Nos. 17 to 18
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
PHYSICS
PART-II
SECTION – I
Single Correct Choice Type
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
= 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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
Paragraph for Question Nos. 42 to 44
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
MATHEMATICS
PART–III SECTION - I
Single Correct Choice Type
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
( )( ) ( )( )( )( ) ( )( )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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
Paragraph for Question Nos. 70 to 72
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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
♥ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT4/CPM/P(I)/Solns
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