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β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
CHEMISTRY
TEST-II
PART-I
SECTIONβI
Single Correct Choice Type
1. (B) 2 4
4 2MnSO MnO+ +
β
The equivalent weight of 4
mol wtMnSO
2=
i.e., 0.2 M MnSO4 β‘ 0.4 N MnSO4
Normality of KMnO4 solution = 0.05 N
Therefore 20 Γ 0.4 = 0.05 V
20 0.4
V 160 mL0.05
Γ= =
2. (D) 2.303 56 2.303
K log log 460 14 60
= =
12.3032 0.3010 min
60
β= Γ Γ
0
dNKN
dtβ =
rate = decay constant Γ initial number of nucleides on substitution
0
2 2.303 0.301056 N
60
Γ Γ= Γ
0
56 60 1680N
2 2.303 0.3010 0.693
Γ= =
Γ Γ
= 2400 (approx)
3. (C) Molecular weight of phenol in solution f
f
1000 K w
W T=
β β
1000 5.12 20
1 1000 0.69
Γ Γ=
Γ Γ
= 148.4
Actual mol. Wt. of phenol ( )6 5C H OH 94=
Hence association of phenol in benzene takes place.
( )6 5 6 5 22 C H OH C H OH
1 β Ξ± 2
Ξ±
26
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
Number of particles before association = 1
Number of particles after association 12
Ξ±= β Ξ± +
12
Ξ± = β
1Normal mol.wt 942
obs.mol.wt 1 148.4
Ξ±β
= =
148.4 β 74.2 Ξ± = 94
148.4 94 54.4
0.73374.2 74.2
βΞ± = = =
% association = 73.3%
4. (A)
3CH
( )3
3 2
CrO
CH CO Oβ
( )3 2CH OCOCH
3H O+
ββ
CHO
5. (A) Ξ²-keto acids on heating easily undergo decarboxylation via six-membered cyclic
transition state
OC
2CH
C
O
H
O
2COββ 6 5C H Cβ
OH
2CH
6 5 3C H C CHβ β β
O6 5C H
6. (B) 2CH C
2CH O
O
3CH OHβ
2CH C
2CH O
Oβ
3O CHβ
H
+
Hβββ 2CH 3C O CHβ β
2CH O
O
β2 3CH C OCHβ β
2CH Oββ
O
H2 2 3HO CH CH COOCH
β
β β β β
7. (A) Trans ( ) 22Co en Cl
+ does not exhibit optical isomerism others exhibit optical
isomerism.
8. (D) As the size of the cation decreases its acidic strength increases. As the positive
charge decreases, the acidic strength increases. Hence the strength of acids follow
the order 3 < 1 < 2 or ( ) ( ) ( )3 2 3
2 2 26 6 6Al H O Fe H O Fe H O
+ + + < <
27
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
SECTIONβII
Multiple Correct Choice Type
9. (A, D)
22 5 2 5 2 2CO
C H CH COOH C H CH NOβ
ββ β β β β
2NO
(optically active)
(optically inactive)
O O does not exhibit tautomersim as the Ξ± β H atom is attached to sp2 carbon.
Cyclopenta dienyl cation (4n rule) is anti aromatic. In E1CB mechanism, the first step is the
ionization of the substrate giving H+ and carbanion.
10. (B, C)
2 2(g)2 2H O 2e H 2 OHβ Γ + β +
(cathode)
2 2(g)4 OH 2H O O 4eβ β + +
(anode)
Add
2 2(g) 2(g)2H O 2H Oβ +
is cell Redox reaction
11. (A, B, D)
2 6 0Fe 3d 4s+ 4 unpaired eβ
3 6 0Co 3d 4s+ 4 unpaired eβ
3 7 0Ni 3d 4s+ 3 unpaired eβ
3 4 0Mn 3d 4s+ 4 unpaired eβ
Fe+2
, Co+3
and Mn+3
have the same no. of unpaired electrons and hence have the same
magnetic moment n(n 2)+ i.e., 4 6 4.85 BMΓ =
12. (A, B, D)
HClO is the strongest oxidising agent and weakest oxy acid. HClO4 is the strongest oxy acid
and weakest oxidising agent. Solubility in water decreases from BeSO4 to BaSO4 due to
decrease in hydration energy while lattice energy of these salts is almost constant.
28
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
SECTIONββββIII
Integer Type
13. Cr(24) 1s2 2s
2 2p
6 3s
2 3p
6 4s
1 3d
5
Electron in the d-orbital has = 2 value.
Ans. 5
14. Number of chiral carbon atoms in the molecule 3 3CH CHCl CHBr CHCl CHβ β β β is 3.
This molecule has plane of symmetry. Hence meso isomer is possible. Two meso forms are
possible.
3 3CH C CH C CHβ β β β
H H H
Cl Br Cl
and
3 3CH C C C CHβ β β β
H
H
H
Cl
Br
Cl
are meso isomers.
Ans. 2
15. PM
dRT
= or dRT
PM
=
x
x
dx RTP
M= and y
y
dy RTP
M=
yx
y x
MP dx3 2 6
P dy M
= = Γ =
Ans. 6
16. In diamond carbon atoms are present in corners, face centres and 50% of tetrahedral sites of
a unit cell. Hence effective number of carbon atoms per unit cell is 1 1
8 6 4 88 2
Γ + Γ + =
Ans. 8
29
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
17.
P
O
OO
P
OP
O
P
O
O O
O
O
There are four shorter P β O bonds in P4O10 which have double bond character.
Ans. 4
18.
C C=Br
Cl
F
IC C=
Br
Cl F
I
C C=Br
ClF
I
C C=Br Cl
F I
C C=Br
Cl
F
IC C=
Br Cl
FI
Ans. 6
SECTIONββββIV
Matrix-Type
19. (A) β (r) ; (B) β (p), (q), (s) ; (C) β (p), (q), (s) ; (D) β (s), (t)
20. (A) β (s) ; (B) β (r), (t) ; (C) β (p) ; (D) β (q)
30
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
PHYSICS
PART-II
SECTIONβI
Single Correct Choice Type
21. Let us suppose that the rate of flow (volume flowing per second) V depends on (i) coefficient
of viscosity Ξ· (ii) radius of cross-section r and (iii) pressure gradient P
, where is the
length and P is the pressure difference.
zx y P
V K r
= Ξ·
, K-dimensionless constant.
3 1 x x x y z 2z 2zL T M L T L M L T
β β β β β =
x z x y 2z x 2zM L T+ β + β β β= β β
β΄ x + z = 0, β x + y β 2z = 3 and β x β 2z = β 1
i.e., x + 2z = 1 or x + z + z = 1
β΄ z = 1, x = β 1, y = x + 2z + 3 = 4
4
1 4 P KrV K r Pβ
β΄ = Ξ· = β Ξ·
Viscous resistance 1
4
PR K
V r
Ξ·= =
β΄β΄β΄β΄ (C)
22. In (a), Q is given by
Q = Q0 (1 β eβ Ξ±t
)
n Q = n Q0 + n (1 β eβ Ξ±t
)
β No log graph (neither b nor c)
In (d),
Q = Q0eΞ±t
, n Q = n Q0 + Ξ±t β (C)
β΄β΄β΄β΄ correct choice is (D)
23. Let at t = 0, train T be just crossing O and
car be at C.
1c
90 5v 90 kmph 25 ms
18
βΓ= = =
1 1t
5v 126 kmph 126 ms 35 ms
18
β β= = Γ =
Velocity of train w.r.t car : tc t cv v v= β
60Β°
tv
cvxC
O
y
T
31
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
1ctv (35 cos 60 , 35 sin 60 ) , v ( 25, 0) msβ= Β° β Β° = β
1tcx
85v 35 cos 60 25 42.5 ms
2
ββ΄ = Β° + = =
1tcy
3v 35 sin 60 35 30.3 ms
2
β= β Β° = β Γ = β
2 2 1tcv 42.5 30.3 52.2 msββ΄ = + =
At t = 0, position of the car C [It takes 3 min to cross O]
= (4.5 km, 0)
Position of train T = (0, 0)
(5 min. after crossing O)
At t = 8 min, position of train T w.r.t. car is
x = 42.5 Γ 8 Γ 60
= 20.4 km
y = β 30.3 Γ 8 Γ 60 = β 14.5 km
2 2d x y 25 kmβ΄ = +
β΄β΄β΄β΄ (D)
24. Max. acceleration
2 22v dva 2
r dt
= + =
22 2dv v4
dt r
= β
4
2
1 v4 1
4 r
= β β
4dv v
2 1dt 4r2
β΄ = β
Now, 4 4
2 6
v 256 10
4r 4 10
Γ=
Γ = 0.64
Hence maximum rate of decrease 22 1 0.64 1.2 msβ= β =
β΄β΄β΄β΄ (A)
25. FBD of B : 2TF
Ba0.5 kg
B
F β 2T = mBaB β¦ (1)
FBD of A:
T
Aa
2 kg
Agm
Am
T β mAg = mAaA β¦ (2)
32
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
By constraints, aA = 2aB
β΄ T β mAg = 2mAaA or 2T β 2mAg = 4mAaB β¦ (3)
Adding (1) and (3): F β 2mAg = (4mA + mB) aB
A A B BF 2m g (4m m ) aβ΄ = + + β¦ (4)
Given 2
2B
v 16a 16 ms
2s 2 0.5
β= = =Γ
β΄ F = 40 + (8.5) Γ 16 = 40 + 136.0 = 176 N
β΄β΄β΄β΄ (B)
26. Resistance of the wire AB : 2
Rr
Ο=
Ο
, Ο -resistivity of the material
dR d 2dr 0.05 0.5
2R r 0.5 5
β΄ = β = + Γ
= 0.3 β¦ (1)
For unstrained wire,
Voltage at D : o oD
v R vv
R R 2
Γ= =
+
Voltage at B : oB
vv
2=
β΄ P.D. between B and D = 0
For strained wire,
o o oD B
v v (R dR) v (R dR)v , v
2 R R dR (2R dR)
+ += = =
+ + +
A
B
C
D
RR
RR
ov
oB D
vv v v [2R 2dR 2R dR]
2(2R dR)β = β = + β β
+
o o
dR
v dR v R
dR dR22R 2 2
R R
= = β
+ +
β΄ % change o
v 0.3 100 15100 6.5%
v 2(2 0.3) 2.3
β Γ= Γ = = =
+
β΄β΄β΄β΄ (D)
27. 15 = 10I + 10 (I β I1)
= 20I β 10I1
β΄ 3 = 4I β 2I1
or 13 2II
4
+= β¦ (1)
10 Ξ©10 Ξ©
10 Ξ©
LR15 V
1(I I )β
I 1I
33
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
(10 β RL) I1 β 10(I β I1) = 0 β¦ (2)
11 L
(3 2I )I (10 R 10) 10I 10
4
++ + = =
I1 (40 + 2RL) = 15 + 10Il
O1
L L O L
E15 7.5I
30 2R 15 R R R= = =
+ + +
β΄ EO = 7.5 V
RO = 15 Ξ©
β΄β΄β΄β΄ (C) 7.5 V
15 Ξ©
LR
28. Let rm be the minimum distance of the
satellite (at A) in the orbit from the centre
of the earth and Vmax be the maximum
velocity. At the point P of launching, let
velocity Vo = 104 ms
β 1 make an angle Οo
with the radius vector joining the satellite
to the centre of earth at a distance ro .
or oΟ
maxv
E
mrP
ov
A
By the conservation of energy and angular momentum at point A and P,
2 2max 0
min o
1 GMm 1 GMmmV mV
2 r 2 rβ = β β¦ (1)
min max o o omr V mr v sin= Ο
omax o o
min
rV V sin
r= Γ Ο
ro = (6400 + 400) km = 6.8 Γ 106 m
rmin = (6400 + 200) km = 6.6 Γ 106 m
4 4max o o
6.8V 10 sin 1.03 10 sin
6.6β΄ = Γ Ο = Γ Ο
2 8 2 8 2
o
min o
1 1 1 11.03 10 sin 10 gR
2 2 r r
β΄ Γ Γ Γ Ο = Γ + β
Here, GM = gR2 = 10 Γ 6.4
2 Γ 10
12
2 12
2o 2 8 6
1 20 6.4 10 0.2sin
1.03 10 6.8 6.6 10
Γ Γβ΄ Ο = + Γ
Γ Γ
2
10.037 0.94 0.037
1.03= + = +
0.98
sin Οo = Β± 0.9899 β Οo = 82Β° or 180Β° = 90Β° Β± 18Β°
β΄β΄β΄β΄ (D)
34
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
SECTIONβII
Multiple Correct Choice Type
29. T + qE β mg = ma1
3 6 61
5010 10 20 10 500 10 10 ma
0.1
β β βΓ + Γ Γ β Γ Γ =
3 3 3 6110 10 10 10 5 10 500 10 aβ β β βΓ + Γ β Γ = Γ
qE
1a
mg
T
+
3
21 4
15 10a 30 ms 3g
5 10
ββ
β
Γβ΄ = = = β
Γ
OR : mg + qE β T = ma2 β
3 3 3 4 22 25 10 10 10 10 10 5 10 a , a 10 ms gβ β β β βΓ + Γ β Γ = Γ = = β
β΄β΄β΄β΄ (B, C)
30. Vde = 60 V
1I in 10 6Aβ΄ Ξ© =
2I in 20 3AΞ© =
3I in 30 2AΞ© =
β΄ current incoming at a = current outgoing at b
β΄ I = I1 + I2 + I3 = 11 A
To find Vcd,
4(I4) = 8 (11 β I4) = Vcd
β΄ 12I4 = 88
4
88 22I A
12 3= =
4 Ξ©
10 Ξ©20 Ξ©30 Ξ©
8 Ξ©
6 Ξ©
e
d
c
1 2 3I I I I= + +a
b
1I2I3I
5I 4I
I
e
c
cd
22 88V 4 29 V
3 3β΄ = Γ =
ab ac cd de ebV V V V V= + + +
88 88 378
0 60 663 3
+= + + + =
466
3= = 155 V
β΄β΄β΄β΄ (B, C, D)
35
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
31. Process a β b : V β T : isobaric heating
(P = constant)
a
bc
TO
V
ab
c
TO
P
a b
c
VO
P
Process b β c : isochoric cooling (V = constant) P β T
Process c β a : isothermal (T = constant) compression
PV = constant : P is increasing with V decreases
β΄ correct graphs are (C) and (D)
β΄β΄β΄β΄ (C, D)
32. When S is in position 1 : 11
di10 0.5 50 i
dt= +
i.e., 11 1
di20 100 i 100 (i 0.2)
dt= β = β β
100 t11
1
di100 dt i (t) 0.2 Ke
i 0.2
ββ΄ = β β = +β
At t = 0, i1 = 0.
Hence, K = β 0.2 or i1(t) = 0.2 (1 β eβ 100 t
)
50 Ξ©
0.5 H2
10 V10 V
1
S
At t = 10 ms 1
s100
= , i1 = 0.2 (1 β eβ 1
) = 0.126 A = 126 mA
β΄ P.D. across resistor just before t = 10 ms = 50 Γ 0.126 = 5 Γ 1.26 = 6.30 V
When S is in position 2 (at tβ² = (t β 10) ms = 0)
22
di50 i 0.5 10
dt+ = β
β²
100 t2i (t ) 0.2 K e
β²ββ² β²β΄ = β +
At tβ² = 0, i2 = 0.126 A
β΄ Kβ² = 0.326 A
i2(tβ²) = 100 t0.326 e 0.2
β²β β
36
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
Now i2(tβ²) = 0
when 100 t 0.326e 1.63
0.2
β²= =
100tβ² = n 1.63 = 0.49
0.49
t s 4.9 ms (t 10) ms100
β²β΄ = = = β
β΄ t = 14.9 ms
Further, as tβ² β β, i2 = β 0.2 A = β 200 mA
After closing S in position 2, P.D. across resistor = 6.3 V
β΄ P.D. across inductor = β 10 β 6.3
= β 16.3 V
Alternate
100 t
2i 0.326 e 0.2β²β = β
2
t 0
L di100 0.326 0.5 16.3 V
dt β² =
= β Γ Γ = β
β΄β΄β΄β΄ (A, B, C, D)
SECTION ββββ III
Integer Type
33. Potential in central field V(r) = kr3
If the particle describes circular motion of radius a, then attractive force d dV
[mV(r)] mdr dr
β = β
should be equal to centrifugal force in magnitude. If v is the speed of the particle,
2
2mv3 mka
aβ΄ = or 3v 3 ka=
Angular velocity 3v 3ka
3 kaa a
Ο = = =
β΄ angular momentum 2 2L mr ma 3ka= Ο =
Let the particle be displaced slightly from its circular orbit, so that r = a + x, x << a
Total energy 2
2
2
1 LE mr mV(r)
2 2mr= + +
Let 2 4
3 3
2
m r 3kr 5U (r) mkr mkr
22mr
Γβ² = + =
Hence 2 31 5E mr mkr
2 2= +
37
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
Differentiating w.r.t. time,
215mr r mk.r r 0
2β + =
or 215r kr 0
2+ = r a x, r x= + =
2 2 2 2r (a x) a 2ax x= + = + +
= a2 + 2ax, x << a
215x 15 kax ka 0
2β΄ + + =
Let 1
X x ka2
= + X 15kaX 0β + =
β΄ angular frequency 15 ka 3 nkaΟ = =
β΄ n = 5
Ans.: 5
34. Charges on C1 and C2 are q1 = C1V, q2 = C2V
After the connection, Q = (C1 β C2) V.
Let 1
qβ² and 2
qβ² be the final charges on C1 and C2 respectively, such that
1 21 1 22
1 2 1 2 1 2 1 2
q q q q (C C )QV
C C C C C C (C C )
β² β² β² β²+ β= = = =
+ + +
1 1 21
1 2
C (C C )q V
(C C )
ββ²β΄ =
+ and 2 1 2
21 2
C (C C )q V
(C C )
ββ² =
+
Now, electrostatic energy before connection
21 1 2
1E (C C ) V
2= +
After connection 22
21 22
1 2 1 2
(C C )1 Q 1E V
2 (C C ) 2 C C
β= =
+ +
Loss in energy ( ) ( )2
2 2
1 2 1 2 1 2
1 2
1 VE E C C C C
2 (C C ) β = + β β +
2
21 2 1 2
1 2 1 2
2C C C C VnV
(C C ) 2 (C C )= β = β
+ +
β΄ n = 4
Ans.: 4
38
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35. Let h be the height of the cliff. Parallel incident
ray AB and reflected ray (from sea) CDB from
distant star interfere with each other in the field
of view of telescope.
The geometrical path difference between the
rays
Ξ΄ = DB β EB = DB β DB sin 16Β°
= DB (1 β sin 16Β°)
But h
sin 37 0.6DB
= Β° =
D
cliff
B
E
A
C
37Β°
h
37Β°
16Β°
sea
h 7.2
(1 0.28) h 1.2 h0.6 6
β΄Ξ΄ = β = β =
Condition for destructive interference is (Taking reflection from denser medium)
Ξ΄ = nΞ», n = 0, 1, 2, β¦..
Now, n = 0 corresponds to h = 0. Hence for n = 1,
1.2 h = 240
240
h 200 m1.2
= =
= 0.2 km x
km10
=
β΄ x = 2
Ans.: 2
36. Maximum kinetic energy of photoelectrons
Kmax = hΞ½ β Ο
Photon energy 34 14
19
6.62 10 6 10h eV
1.6 10
β
β
Γ Γ ΓΞ½ =
Γ
2.5 eV
β΄ Kmax = 2.5 β 2 = 0.5 eV
Positive potential to be developed at the surface = 5 n
0.5 V V V10 10
= =
β΄ n = 5
[If V > 0.5 V, no electrons leave the surface]
Ans.: 5
39
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37. N0 = 82 β 10 = 72 cps
N1 = 19 β 10 = 9 cps
3
0
N 1 1
N 8 2β΄ = =
Let T be the half life. Then, t = 3T = 180
T = 60 s = 1 min
Ans.: 1
38.
2
2 1Z R L
C
= + Ο β
Ο
i
22
VI
1R L
C
=
+ Ο β
Ο
O2ViV , Ο
R LC
O1V
Output
2 2 2i
O12 2 20
V R L CV lim
RC ( LC 1)Οβ
+ Ο β Ο=
Ο + Ο β
= ViRC β Ο
Output iO2 i
2 2 20
V CV lim V
C RC ( LC 1)Οβ
β Ο= =
Ο Ο + Ο β
O1
O2
VRC
Vβ΄ = β Ο
β Ο = Οn β n = 1
Ans.: 1
SECTIONβIV
Matrix Type
39. (A) β (s), (B) β (p, q), (C) β (t), (D) β (r)
(p) F = Ξ±x, F β dx = dK 21K x
2β΄ = Ξ±
K versus x is a parabola, because
dK
F (B)dx
= O
K
x
(q) 2 UU 500 60x F 120 x
x
β β= + β = = β
β
120
a x 24 x5
= β = β
For SHM, 2 1a x 24 rad. sβ= β Ο β Ο = (B).
40
β£ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns
(r) For conservative electric field, C
E d 0β =β«
and for non-conservative field, C
E d 0β β β« (D)
(s) 1 2
1 1 1 1( 1) , f
f R R ( 1)
= Β΅ β β β
Β΅ β
V R
R V
f ( 1)1
f ( 1)
Β΅ β= <
Β΅ β, because R VΒ΅ < Β΅ (A)
(t) F qV B F V= Γ β β₯
No work done, βK = 0
But P mV=
changes (in direction) (C)
due to F
on charged particle.
40. (A) β (q, r, s); (B)β(r); (C)β(p, r, s, t); (D)β(p, q)
1S
2S
10 Ξ© 5 Ξ©
5 mH60 VI
A
B
200 FΒ΅
(p) S1 closed, S2 open: AB
60t 0 , I 6A, V 0
10
+= = = =
6 3RC 10 200 10 2 10 sβ βΟ = = Γ Γ = Γ = 2 ms
(q) S1 closed, S2 open: ABt , I 0, V 60 Vβ β = =
Ο = 2 ms
(r) S1 open, S2 closed: 60
t , I 4 A15
β β = = , VAB = 4 Γ 5 = 20 A
3L 5 10
sR 5
βΓΟ = =
= 1 ms
(s) S1 and S2 closed: at steady state : AB
60I 4A, V 20 V
15= = =
(t) At t = 0+ s (S1 and S2 closed): AB
60I 6A, V 0
10= = =
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MATHEMATICS
PART-III
SECTION β I
Single Correct Choice Type
41. (B) Ξ±, Ξ² are the roots of x2 + x + 1 = 0 β Ξ±, Ξ² are the complex cube roots of unity so
that Ξ± + Ξ² = β 1, Ξ±Ξ² = 1.
β΄ sum of the roots of the required equation = 2(Ξ±4 + Ξ²4
) + Ξ² + Ξ±
= 2(Ξ± + Ξ²) + (Ξ² + Ξ±) 3 3( 1)Ξ± = Ξ² =β΅
= β 3
Product of the roots = (2 ) ( 2 )Ξ± + Ξ² Ξ± + Ξ²
2 22( ) 5= Ξ± + Ξ² + Ξ±Ξ²
( ) 2
2 2 5 2 1 2 5= Ξ± + Ξ² β Ξ±Ξ² + Ξ±Ξ² = β + = 3
Hence the required equation is x2 β ( β 3) x + 3 = 0
β x2 + 3x + 3 = 0
42. (D) Let x[x]
f (x)sin | x |
=
Then left hand limit of f(x) as x β β 1
h 0
( 1 h) [ 1 h] 1 ( 2) 2Lt (h 0)
sin | 1 h | sin 1 sin 1β
β β β β β Γ β= > = =
β β
Right hand limit h 0
( 1 h) [ 1 h] 1 1 1Lt (h 0)
sin | 1 h | sin 1 sin 1β
β + β + β Γ β= > = =
β +
β left hand limit
β΄ limit does not exist.
43. (B) Clearly f(x) does not exist at x = 0 and x = β 3
β΄ the domain of 2
2
x 2xf (x)
x 3x
β=
+ is R β β 3, 0
and x 2 5
f (x) 1x 3 x 3
β= = β
+ +
Now 1f (f (x)) xβ =
1
51 x
f (x) 3ββ β =
+
1 5f (x) 3
1 x
β + =β
( )1
2
d 5f (x)
dx (1 x)
β =β
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44. (A)
( )
12 9 3 6
3 3 35 3
2 5
2 5dx
2x 5x dtx xdx
t1 1x x 1 1x x
+ + = = β
+ + + +
β« β« β« where 2 5
1 1t 1
x x= + +
( )
10
2 25 3
1 1 xc
22t x x 1
= = ++ +
( )
( )
12 91
35 30
2x 5x 1 1 1dx 0
2 9 18x x 1
+ β΄ = β =
+ +β«
45. (D) Let 1 2 3a a i a j a k= + +
Then 1a i aβ = , 2 3a j a , a k aβ = β =
( ) ( ) ( ) ( ) ( )1 2 3a i a i a a i a a j a a kβ΄ β Γ = Γ + Γ + Γβ
( )1 2 3a a i a j a k a a 0= Γ + + = Γ =
46. (C) 2 2 2 2a b c 8R+ + =
2 2 2sin A sin B sin C 2β + + = ( )a 2R sin A etc.,=β΅
1 cos 2A 1 cos 2B 1 cos 2C
22 2 2
β β ββ + + =
β cos 2A + cos 2B + cos 2C = β 1
β 2 cos (A + B) cos (A β B) + 2 cos2 C β 1 = β 1
β β 2 cos C cos (A β B) + cos (A + B) = 0 ( )cos C cos (A B)= β +β΅
β cos A cos B cos C = 0 β one of A, B, C is a right angle.
47. (C) 21z i z iz 1 0
zβ = β β β =
2z iz 1 0β β + + =
2(iz) (iz) 1 0+ + =
β΄ iz = Ο or Ο2 where Ο is a complex cube root of unity.
β΄ iz = Ο β z = β iΟ
( )
2005 2005
2005 2005
1 1z ( i )
z iβ΄ β = β Ο β
β Ο
21i i i
i= β Ο + = β Ο β Ο
Ο
( )2i i 1 i= β Ο + Ο = β Γ β =
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48. (B) y divides z β z = ky where k is a positive integer.
2xz 2 1008 ky
yx z 1008 ky
Γ Γβ΄ = =
+ +
1008
1008 ky 2016 k k2016 y
β + = β =β
β k is any divisor of 1008 except 1
Now 4 21008 2 3 7= β β
β the number of divisors of
1008 = (4 + 1) (2 + 1) (1 + 1) = 30
But this includes 1 also.
β΄ the required number of sequences = 30 β 1 = 29
SECTION β II
Multiple Correct Choice Type
49. (B, D)
[sin ΞΈ + 1] + [cos ΞΈ ] = 0 β [sin ΞΈ] + [cos ΞΈ] = β 1
β if both [sin ΞΈ] and [cos ΞΈ] are negative: then RHS = β 2 which is impossible.
β΄ one of them is negative and the other is zero.
3
, or , 22 2
Ο Ο β ΞΈβ Ο Ο
50. (A, C)
Consider the numbers a a b b b c c c c
, , , , , , , ,2 2 3 3 3 4 4 4 4
Their
12 3 4 9a b c a b c
A.M G.M9 4 27 256
+ +β₯ β β₯
Γ Γ
2 3 4 9a b c 2 4 27 256β β€ Γ Γ Γ
and the greatest value is obtained when
a b c a b c
22 3 4 2 3 4
+ += = = =
+ +
i.e., when a = 4, b = 6, c = 8
51. (B, D)
n n n n 2 n n0 1 2 n(1 x) C C x C x ..... C x+ = + + + +
Also n n n n n 1 n0 1 n(x 1) C x C x ..... Cβ+ = + + +
β΄ coefficient of xn β 1
in n n n n n n n n0 1 1 2 n 1 n(1 x) (x 1) C C C C ..... C Cβ+ + = β + β + + β
i.e., coefficient of xn β 1
in (1 + x)2n
= n n n n n n
0 1 1 2 n 1 nC C C C ..... C Cββ + β + + β
i.e., 2n
n 1RHS C β= 2n
n 1 n 1=
β +
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n 1
n
2n 2 n 1 n 1s
s n n 2 2n
+ + β +β΄ = β
+
(2n 1) (2n 2) 15
n (n 2) 4
+ += =
+
β n2 β 6n + 8 = 0
β (n β 2) (n β 4) = 0
β n = 2 or 4
52. (A, B, C)
P(E1) = probability that the first digit is 4 = 2 1
4 2=
similarly 2 3
2 1P(E ) P(E )
4 2= = =
P(E1 β© E2) = probability that the first and second digits are 4 1
4=
similarly 1 3 2 3
1P(E E ) P(E E )
4β© = β© =
1 2 1 2
1P(E E ) P(E ) P(E )
4β΄ β© = = β© etc.
Hence (A), (B), (C) are true.
1 2 3
1P(E E E )
4β© β© = β P(E1) β P(E2) β P(E3)
SECTION β III
Integer Type
53. 2
2
cot 2x tan 2x 1 tan 2x2
cot 4x (1 tan 2x)tan 2x
2 tan 2x
β β= =
ββ
Ans. 2
54. 2
xlog (x 5)x 16
β = β x > 0, x β 1 and (x β 5)2 = 16
β x β 5 = Β± 4 β x = 1 or 9.
But x β 1 β x = 9.
Ans. 1
55. Let a and r be the first term and the common ratio of the G.P.
Then p 1 q 1 r 1x aR , y aR , z aRβ β β= = =
q r r p p q (q r) (r p) (p q) (p 1) (q r) (q 1)(r p) (r 1)(p q)x y z a Rβ β β β + β + β β β + β β + β ββ΄ β β =
= a0R
0 = 1
Ans. 1
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56. The equation of the circle with centre at ( )0, 5 is
( )2
2 2x y 5 r+ β =
i.e., 2 2 2x y 2 5 y r 0+ β β =
For x, y to be rational y = 0 and r2 should be rational.
When y = 0, x = Β± r
β΄ the points on the circle with rational coordinates are (r, 0) and ( β r, 0)
Ans. 2
57. Let y = mx + c be a tangent to 2 2x y
16 2
+ =
β΄ c2 = 6m
2 + 2 β¦ (1)
Given 3 m
tan 451 3m
β= Β°
+ = 1
3 m
11 3m
ββ =
+ or β 1 β
1m
2= or β 2
For each of these values of m there are two values of c from (1)
β΄ the number of tangents with the given condition = 4
Ans. 4
58.
n
r 1
n n2 2
rr 1 r 1
n2 2
r 1
1 n n
2 r n n 1 n n 72
(2r 1) n n n 1
=
= =
=
β = + + + =
β + +
β
β β
β
2 2
2 2 2
n n n
n(n 1) n n 1 n n 72
n n n n 1
β + + + + =
+ +
2 2 2
2 2 2
1 1 1
n n n n n 1 n n 72
n n n n 1
β + + + + =
+ +
2
2
1 0 0
n n n 1 0 72
n 0 n 1
β + =
+
2 2 1
3 3 1
C C C
C C C
β β β β
β n(n + 1) β 72 = 0 β (n + 9) (n β 8) = 0
β n = 8
Ans. 8
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SECTION β IV
Matrix Type
59. (A) ββββ s; (B) ββββ s; (C) ββββ p; (D) ββββ r
Sol.: (A) 1 14 2 4 2cos cos sin sin 2
3 3 3 3
β βΟ Ο Ο Ο + = Ο β + Ο β
= Ο
(B) 1 1 1 1 2 3
tan 1 tan 2 tan 3 tan4 1 2 3
β β β β Ο ++ + = + Ο +
β Γ
1 1 1 a btan a tan b tan if ab 1
1 ab
β β β ++ = Ο + >
β β΅
1tan ( 1)4
βΟ= + Ο + β
4 4
Ο Ο= + Ο β = Ο
(C) 1 11 1 2cos 2 sin 2
2 2 3 6 3
β β Ο Ο Ο+ = + Γ =
(D) Let 1 1
2 2
1 1tan tan .... to n terms
2 1 2 2
β β+ + =β β
Then n n
1 1
2 2r 1 r 1
1 2tan tan
2r 4r
β β
= =
= =
β β
n
1
r 1
(2r 1) (2r 1)tan
1 (2r 1) (2r 1)
β
=
+ β β=
+ + β β
n
1 1
r 1
tan (2r 1) tan (2r 1)β β
=
= + β ββ
1 1tan (2n 1) tan 1β β= + β
1
2r 1
1tan
2 4 42n
ββ
=
Ο Ο Ο β΄ = β =
β
60. (A) ββββ s; (B) ββββ p; (C) ββββ r; (D) ββββ q
Sol.: (A) 2
2
(ydx xdy)xdy ydx xy dx xdx
y
β ββ = β =
2x x x
d xdx cy y 2
β β = β β + =
2x y 2 (cy x)β = β
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(B) 2
3 2 2
2
dy y xyx dy x ydx xy dx
dx x
β+ = β =
put y = vx
2dvv x v v
dxβ΄ + = β
2
dv dx 1 1 dxdv 2
x v 2 v xv 2v
β = β β =
ββ
2 2v 2log log cx y 2x cx y
v
β β = β β =
(C) ( )3 dy2x 10y y 0
dxβ + =
2
3
dy y dx 2x10y
dx dy y10y 2xβ = β + =
β
which is a linear equation in y.
β΄ the solution is
2 2dy dy
2y yxe e 10y dyβ« β«
= β β«
i.e., 2 4 5xy 10 y dy 2y c= = +β«
(D) The equation of the tangent at (x, y) on y = f(x) is
dy
Y y (X x)dx
β = β
The Y-intercept of this tangent is obtained by putting X = 0
2dy dyY y x i.e., 2xy y x
dx dxβ΄ = β = β
i.e., 2
ydx xdy2xdx
y
β=
2x
d x cy
β = +
β«
2x y x cyβ = β