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B.MAT PART TEST 1FOR OUR STUDENTS
TOWARDS
IIT----JOINT ENTRANCE EXAMINATION, 2012
SECTION I
1. (D) Work function = Energy required to just dislodge the electron in the ground state
= =
00
hch
=
34 8
96.63 10 3 10
330 10
=26
719.89 10
3.3 10
= 6 1019 J
2. (C) For H-bond formation, hydrogen must be covalently bonded to oxygen, nitrogenor fluorine which is more electronegative. In CH3F, hydrogen is attached to
carbon but not to fluorine. Hence there is least tendency for H-bond formationin CH3F.
3. (A) According to Boyles Law
1PV
at constant temperature
P =constant
V
IIT-JEE 2012PT1/CPM/P(I)/SOLNS
PAPER I SOLUTIONSCHEMISTRY PHYSICS MATHEMATICS
PART A: CHEMISTRY
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log P = log (constant) log V (represents graph C)
or log V = log constant log P (represents graph D)
P 1
V. Hence P vs V is a curve but not a straight line.
4. (A) Conc. of final solution (NaOH) = 10 mg per ml
= 10 g per litre
=1040
equivalent per litre
= 0.25 N
Conc. of initial solution of NaOH = 0.5 N
V1 N1 = V2 N2
or V2 =1 1
2
V NN
=500 0.5
0.25
= 1000 ml
V2 V1 = 1000 500 = 500 ml
Hence 500 ml of water should be added to 500 ml of 0.5 N NaOH to get theNaOH solution with conc of 10 mg/ml.
5. (C) The gas with higher value of van der Waals constant a undergoesliquefaction readily. Thus
NH3 (a = 4.17) undergoes liquefaction readily. It is very difficult to liquefy O 2
[a = 1.3]. Thus the order of readiness with which these gases can be liquefied is
NH3 > CH4 > N2 > O2
a = 4.17 2.253 1.39 1.3
6. (D) Oxidation number of the metal in amalgams and metal carbonyl is zero.
7. (B) Since the formula of the oxide is M2O3 the valency of the element M is 3
Weight of oxide = 0.559 g
Weight of the element M = 0.359 g weight of oxygen = 0.200 g
Eq. wt. of element M =0.359 8
14.360.2
=
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Atomic weight of M = Eq. wt valency
= 14.36 3
= 43.08
43
SECTION II
8. (A), (B)
Kinetic energy absolute temperature
Kinetic energy = 0 at zero kelvin temperature
At constant temperature, change in pressure of the gas does not affect kinetic
energy of the gas.
9. (A), (C)
Critical temperature, TC =8a
27Rb
Boyles temperature, TB =a
Rb
Inversion temperature, Ti =2aRb
Unit of gas constant, R = 0.082 lit. atm K
1
mol
1
= 8.31 joules mol1 K1
= 1.987 cal mol1 K1
= 8.31 kPa dm3 K1 mol1
10. (B), (C)
4SF One lone pair of electrons
4XeF Two lone pair of electrons
2Cl O Two lone pair of electrons
[ ]6IF One lone pair of electrons
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11. (A), (D)
SS
O
OO
ThiosulphateOxidation states of S are zero and +4
O
S
OOO S O
O
O
Pyrosulphate
Both sulphur atoms are in + 6 state
O
S
OOO S O
O
O
O
Perdisulphate
Both sulphur atoms are in + 6 state
OS
O
SO S O
O
OS
Tetrathionate
Oxidation states of sulphur are 0 and + 5
SECTION III
12. (B) orbital no. of nodal planes
(= )
2s 0
2p one
3d two
13. (D) The number of spherical node = n 1
For 3p orbital, n 1 = 3 1 1 = 1
For 3d orbital n 1 = 3 2 1 = 0
14. (B) 22 21 2
1 1 1RZ
n n
=
Z = 2 for He+ ion
For Lyman series limiting line n1 = 1 and n2 =
=
2 2 21 1 1R 2 1
=2
1R 4 1 4R
=
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15. (C) En =2
213.6 Z
n
eV/atom for H-like ions.
Ionization energy of He+ ion 2He e He+ +
, Z for He = 2
2
213.6 2
4 13.6 eV 1
= =
Ionization energy of Be3+ ion 3 4Be e Be+ +
, Z for Be = 4
2
213.6 4
1
=
= 16 13.6 eV
The ratio of ionization energies of He+ ion and Be3+ ion
= 4 13.6 : 16 13.6
= 1 : 4
16. (B) E =
hc=
27 10
86.62 10 3 10
4000 10
erg/photon
=19.86
4 1012 erg/photon.
= 4.965 1012 erg/photon
= 4.965 1019 Joules
No. of photons having 2 Joule of energy
=19
2
4.965 10
=1820 10
4.965
4 1018
SECTION IV
17. (2) An+5
3A O
+
(oxidation)
change in oxidation state = (5 n)
4MnO H
+
Mn2+
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change in oxidation no: 7 2 = 5
No. of equivalents of An+ = 0.268 (5 n)
No. of equivalents of 4MnO = 0.161 5
Both are same. Hence
0.268 [5 n] = 0.161 5
1.340 0.805 = 0.268 n
or n =0.535
20.268
18. (6) 25
Br 5e 5Br2
+
Add5
2 31
Br 5e B O2
+
2 33Br 5Br BO + ; no. of electrons involved is 5.
E = Eq. wt. of Br =6[A] 6[A]
5no. of e=
5E = 6[A]; compared with 5E = x[A]
or x = 6
19. (6) Fe (26) = 3d
6
4s
2
Fe2+ (24 e) = 3d6 4s0
No. of electrons in 3d orbital in Fe2+ ion = 6
20. (4) Density of gas = d =PMRT
For the gas A, dA = A AP M
RTor PA=
A
A
d RTM
For the gas B, dB = B BP M
RTor PB =
B
B
d RTM
Now A BB A
P MdA 3 2P dB M 1.5 1
= =
= 4
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21. (2) SO3 has sp2 hybridised sulphur central atom
S[16] 1s2 2s2 2p6 3s2 3p4 3s 3p
S* hybridisation 3s 3p 3d
3p2sp 3d
The three single electrons present in the three sp2 orbitals of S are involved information of bonds with the three oxygen atoms. Hence SO
3contains one
p p bond and two p d bonds.
22. (3) 2 2 24 3 42MnO 5SO 6H 2Mn 5SO + + + + + + 3H2O
or 2 2 24 3 43MnO 7.5SO 9H 3Mn 7.5SO + + + + + + 4.5H2O
no. of moles of 4MnO required to react with 7.5 moles of 23SO
completelyin the acid medium is 3.
23. (2) () =150 150
4V 37.5
= =
or = 2
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SECTION I
24. (A) v = a + bt +c
d t+
Hence [a] = [LT1
], [b] = [LT2
]
[d] = [T], [c] = [L]
[Q] =abd
c
=1 2LT LT T
L
= [LT2
]
unit of Q = unit of acceleration.
25. (B) F = Krn
, n > 0.
[K] = [M] [L]n + 1
[T]2
.
The coefficient of viscosity = kMx
Ey
Kz, k dimensionless quantity
Dim. = [M] [L1
][T1
] = [M]x
[M]y
[L]+2y
[T]2y
[M]z
[L](n + 1)z
[T]2z
= [Mx + y + z
] [L2y + (n + 1)z
] [T2y 2z
]
Hence x + y + z = 1, 2y + (n + 1)z = 1, 2y + 2z = 1
(i.e.,) z(2 n 1) = 2 or z =2
1 n.
And y =
1 1 4
[1 2z] 12 2 1 n
=
1 1 n 4
2 1 n
=
(n 3)
2(n 1)
+=
Hence
(n 3)
2(n 1)E
+
. But E , absolute temperature.
(n + 3)/2(n 1)
p
, p > 0
Hence p =(n 3)
02(1 n)
+>
n < 1 and further n > 0.
0 < n < 1
PART B: PHYSICS
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26. (B) 0
= 1 m, 0
= 0.01 m, =v
0.6c
,
= v
0.02c
Now, = 2
0 2
v1
c.
Taking log, n = n 0
+1
2n
2
2
v1
c
Differentiating
= +
2
20
20
2
vd 1
cdd 1
2 v1
c
=
02
0
2
d v dv 1c c v
1c
d 0.6( 0.02)0.01
(1 0.36)
=
Max. fractional error in the measurement of is
d 0.6 0.02 1.20.01 0.01
0.64 64
= + + = +
= 0.01 + 0.01875 = 0.02875
Now, 1 1 0.36= = 0.80 m.
d = 0.023000 = 0.023
length of metre scale = [0.80 0.023]m
27. (A) OA 2i, AB 4 2 [cos i sin j]= =
4i 4 j=
= =
3BC r , OC 4i
OC OA AB BC= + +
= = +
3BC r 4i 2i 4i 4 j
2i 4 j= +
3r 4 16 2 5 km,= + = making an
angle = tan1
2
4
west of north
1r
2
r
A
B
3r
CE45
O
N
W
S
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28. (B) Initial velocity u = 144 kmph =3
1144 10 40 ms
60 60
=
Let m be the mass of the aircraft in kg
Resistive force F = (900 + V2)
mN
1000.
= + 2dV m
m (900 V )dt 1000
or2dV dV (900 V )
Vdt dx 1000
+= =
2
VdV dx,
1000900 V =
+
let V2
+ 900 = y2
2VdV = 2ydy
0x30
250 0
ydy 1dx
1000y= ,
500
30
x dy 5n
1000 y 3
= =
, V = 40 ms
1, y = 50
V = 0, y = 30
x0
=5
n3
km = 0.5 km
= =
25 5n 2 n 1.02
9 3
29. (B) With respect to the truck moving with acceleration a, the block is about to slide
down the incline (for minimum a).Let m be the mass of the block
and f the force of friction and =
30.
may
+ f mg sin = 0, f = N.
(along the incline)
+ ma
+ N mg cos = 0,
ay
= a cos .
a
= a sin
(normal to the incline)
N = m(g cos a sin )
ma cos + m (g cos a sin ) mg sin = 0
a(cos + sin ) = g(sin cos )
FBD of the block
mg =
30
ma
N f
ve+ve+
a
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a =g( cos sin )
( sin cos )
+
(i.e.,) a =
3 110 0.64
2 2
1 30.64
2 2
+
= 0.45 ms2
30. (B) Let 1 2 3C C i C j C k= + +
, F 2i j k= +
, F C j k = = +
1 2 3F C 2C C C 4 = + =
(1)
1 2 3
i j k
F C 2 1 1
C C C
=
= ( ) + + +
2 3 3 1 2 1
i C C j(2C C ) k(2C C ) (2)
Comparing it with
,
C2
+ C3
= 0, 2C3
+ C1
= 1, 2C2
C1
= 1
[Relation C2
+ C3
= 0 becomes redundant]
C3
= 1(1 C )
2
+ , C
2=
+ 1(1 C )
2 (3)
Substituting (3) in (1): 1 11(1 C ) (1 C )
2C 42 2
+ ++ + =
14 1
C 13
= =
Hence C1
= 1, C2
= 1, C3
= 1 C i j k= +
SECTION II
31. (A), (C)
Let u be the initial speed, making angle with the horizontal.
u
O
B
10 m
15 ms
5 mB
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Max. height h = 15 m =2 2u sin
2g
u2
sin2 = 2gh = 2 10 15 = 300 (1)
By the time, bird flies horizontally from B to B, the stone rises 5 m and thencomes down.
time taken t by the bird to fly from B to B =2h
2g
(i.e.,) t =2 5
2 2s10
= .
And hence u cos = speed of the bird = 5 ms1
.
u2 = 300 + 25 = 325 or u = 18 ms1
Distance covered by the bird = 5 2 = 10m
Now, u sin = 300 10 3=
u cos = 5
tan = 2 3 , = tan1
(2 3)
32. (A), (B), (C)
Let T be the tension in the string, T = 30 N.
Horizontal force on leg and foot by the device
FH
= T + T cos 30 .
= (30 + 15 3 ) N.
Vertical force FV
= T + T sin 30 = 45 N.
Net force by the device = 72 N.
horizontal force on the device = (30 + 15 3 ) N = 56 N
Effective weight of leg and foot = 45 N = 4.5 kg wt.
33. (B), (C), (D)
x = 1 + t + t2
t3, 2
dxv 1 2t 3t
dt= = +
v = 0 at t = 2 4 126
+ = 1s (t = 13
s not admissible)
v > 0 for 0 < t < 1s
v < 0 for t > 1s.
T
30 N
3 kg
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a =dv
dt= 2 6t, a = 0 at t =
1s
3.
a > 0 for 0 < t 1
s3
Thus,
v and a
are parallel during 0 < t 1s.
v and a
are anti-
parallel only during1
s3
< t < 1s.
Now, at t = 0, x = 1 m; t = 1 s, x = 2 m, t = 2s, x = 1 m. Therefore, x = 1 m att0, 1s < t
0< 2s.
34. (A), (B), (C), (D)
2 kg
A f
F T
2 g
2a1N
fT
4a 1N 2N
4 kg
B
4 g
B
AF
a22 ms =
N1
+ 2a 2g = 0, N1
= 2(g a)
T + f F = 0, F = T + N1
(1)
But f = T = N1
max. force F = 2 N1
= 4 (g a) = 4 0.4 (10 2)
F = 12.8 N
SECTION III
35. (A) Let F = hx
cy
z,
MLT2
= Mx
L2x
Tx
Ly
Ty
Lz x = 1
2x + y + z = 1
F =
2
hc (x + y) = 2
y = 1 and z = 1 3 = 2
one unit of F =34 8
262
6.6 10 3 1019.8 10
1m
=
= 1.98 1025
N.
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36. (A) In new system, energy E =
hc
one unit = 1.98 1025
J
Now, 13.6 eV = 13.6 1.6 1019
J
ionization energy in above system
196
25
13.6 1.6 10 13.6 1.610
1.981.98 10
= =
j 1.1 107
units
37. (B) Let ux
= u cos , uy
= u sin
ax
= a, ay
= 0
x = u cos t 21
at2
; y = u sin t
x = cot y 2
2 2
1 ay
2 u sin = py qy
2
p = cot , q = 2 22 2 2 2
a a acosec (1 cot )
2u sin 2u 2u= = +
q = 2 2
2
a a(1 p ) or u (1 p )
2q2u+ = +
38. (C) = + = +
dx dyv i j ai bx jdt dt
(given)
dx
adt
= x = at + c1. At t = 0, x = 0 c
1= 0.
x = at.
dybx abt
dt= = y =
2abt
2( At t = 0, y = 0)
=2
22
ab x b x2 2aa
=
path of the particle is y = 2b
x2a
parabolic.
Ox
y
u
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39. (C) x = a sin t, y = a(1 cos t)
(i.e.,) x = 2a sin t2
cos t2
, y = 2a sin2 t2
sint
2
=
y
2a,
cost
2
=
y1
2a
x =y y y
2a 1 2ay 12a 2a 2a
=
SECTION IV
40. (2)
+ =
=
= =
1 A 1
2 1 21 2 3 4
3 2 3
B 3 4
2x x C
2x x CC , C , C , C
2x x C
x x C
are constants.
Bx
3x
2x Ax
1x
A
B 10 kg
1 kg
Differentiating twice w.r.t. time t, A 1 2 3 Bx 2x 4x 8x 8x= = = =
Let a be the acceleration of block B downward. Then, the acceleration of A is 8a
upward.
FBD of A:
T
g
8a T g = 8a (1)
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FBD of B:
8T
10 g
a
10 kg 10g 8T = 10a (2)
From (1) and (2), 10g 8(8a + g) = 10a
2g = 74a
a =2g xg
74 74=
x = 2
41. (2) Let a be the acceleration of wedge Balong horizontal direction.
FBD of A:
a
mg
( )90 1N
ma
P
Q45 =
Perpendicular to PQ: N1
+ ma sin = mg cos
N1
= m(g cos a sin ) (1)
Parallel to PQ: mg sin + ma cos = ma (2)
FBD of B:
a
N
1N
Mg
1.6 kg
a
B
P
A
0.8 kg
Q
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N1
sin = Ma a =m
sin (g cos a sin )M
a (M + m sin2) = mg sin cos
a =2
mgsin cos 0.8 10
0.8M m sin 2 1.62
=
+ +
28 2ms2 2
= =
42. (3) Let a be the acceleration of each block.
For A,
Tmg
1N
60 60
O
A
B
a
T sin 60 + mg sin 30 = ma (1)
For B,
2N
T
amg
O
mg T sin 60 = ma (2)
T sin 60 + mg sin 30 = mg T sin 60
2T sin 60 = mg(1 sin 30)
T =
1mg 1
mg2
3 2 322
=
=5
N
3
.
=(n 2)
Nn
+, n = 3
A
B O
TT
( )1 kg
( )1 kg
60 =
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43. (5) T4
= 2T2
= 4T1
= 4T3, T
1= T
3
Let a1, a2, aA, aB, aC and aD be the accelerations of pulleys 1, 2 and blocks A, B,C, D respectively.
Bx
Ax
1x
2x
Cx
Dx
A
B
1T
2T
1T
1
2
2T
3T
3T
4T
C
D
Taking downward + ve,
xA
+ xB
2x1
= C1, a constant
aA
+ aB
= 2a1
(1)
xC
+ xD
2x2
= C2
aC
+ aD
= 2a2
(2)
And x1 + x2 = C3
a1
+ a2
= 0 (3)
For pulley 1,
1 g T1
= 1 aA
, 2 g T1
= 2 aB
4g 3T1
= 2(aA
+ aB
) = 4a1
(4)
Similarly for pulley 2,
1 C
1 D
3g T 3a
4g T 4a
=
= 24g 7T
1= 12(a
C+ a
D) = 24a
2 (5)
But a1
+ a2
= 0. Hence equation (4) and (5)
(4g 3T1
= 4a1) 6
24g 7T1
= 24a1
148g 25T 0 =
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148
T g25
=
Hence the tension in fixed string = 4T1
=1920
N25
=2
1920N
n
n = 5
44. (8) Let m be the mass per unit length. At
an instant t, let x be the length
hanging from the edge. Then FBD are:
N
( )m 1 x g
aT
a
T
mxg
T = m(1 x)a
mxg T = mxa
xg = a(x + 1 x) = a =dV dV
Vdt dx
=
By integration,
1.02 2
0.6
V gx
2 2
=
20V g(1 0.36)= , V0 = 0.8 10 =
8
10
n = 8
45. (3) Let P
be the compressive force (alongAO
)
A
DE
B C
O
0.4 m
0.6 m
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P = 4 1000 cos
= 4000 1824
= 3000 N
= 3 kN.
46. (3) Constraint equation:
2xB
+ xA
= constant
2aB
+ aA
= 0
For B
300 g 2T = 300 a2
3000 150 a = 2T
T = 1500 75a (1)
For A
200 g T F = 200 a
F = 2000 T 200 a
= 2000 1500 + 75a 200a
= 500 125a
= 375 N
power delivered
= FV = 375 2 =750
kW1000
=3
kW4
=n
n 1+kW,
n = 3
A2
Aa a 1 ms
= =
200 g
2T
300 g
BB
aa
2=
FT
Bx Ax
B A
M
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SECTION I
47. (C) Now1 1 1 1 4
1 .....14 16 64 5
14
+ + = =
1 1 1 1 21 ....
12 4 8 31
2
+ + = =
10 alog a log 104 2
20 35 3
=
( )1
t4 t2 2= where t = log10
a
4t =1
t t =
1
2
log10
a =1
2 a =
1
210 10=
48. (B) Sum to n terms
Sn
=n
[56 (n 1)( 2)]2
+
=n
[58 2n] n(29 n)2 =
= n2
+ 29n
= {n2
29n}
=
2 229 29
n2 2
=
2 229 29
n2 2
Sum is maximum when
2
29n2
is least.
But n is a natural number, n can be 15 or 14
S14
= S15
= 15 14 = 210.
PART C: MATHEMATICS
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49. (C) Given 2 sin cos sin = sin sin cos + sin cos sin
Dividing throughout by sin sin sin , we get2 cot = cot + cot
cot , cot , cot are in A.P.
tan , tan , tan are in H.P.
50. (A) Let S = 1 6 + 2 62
+ 3 63
+ + n 6n
6S = 1 62
+ 2 63
+ + (n 1) 6n
+ n 6n + 1
subtracting 5S = 6 + 62
+ . + 6n
n 6n + 1
=n
n 16(6 1)n 6
5
+
5S =n 1 n 16 6 5n 6
5
+ +
S =n 1(5n 1)6 6
25
+ +
a + b = n + 1 + 6
= n + 7
51. (C) Let E = 3 sec2 + 12 cosec
2
= 3 + 12 + 3 tan2 + 12 cot
2
= 15 + ( )2
3 tan 12 cot 2 3 12 +
= 15 + ( )2
3 tan 12 cot 12 +
= 27 + ( ) 2
3 tan 12 cot
E 27 Minimum value of E is 27.
It is attained when 3 tan 12 cot = .
52. (B)
xtan A tan B
x tan A y tan B y
xx y1
y
++
=+
+
cos A sin A sin B
cos B cos A cos BcosA
1cosB
+=
+
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sin A sin B
cos A cos B
+=
+
A B A B2sin cos
2 2A B A B
2cos cos2 2
+
=+
A Btan
2
+ =
53. (B) tan 2 =tan( ) tan( ) 4
1 tan( ) tan( ) 1 1
+ + =
+
2 =2
=
4
Now tan ( + ) + tan ( ) = 4
i.e.,1 tan 1 tan
41 tan 1 tan
+ + =
+
i.e.,2
2
2(1 tan )4
1 tan
+ =
i.e.,2
4cos 2
=
cos 2 =1
2
2 =3
=6
2 + 3 =2 2
+ =
SECTION II
54. (A), (D)
The first equation is sin + cos + 1 = x
x 1 = sin + cos
Similarly y + 1 = sin cos
(x 1)2
+ (y + 1)2
= 2
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x 1 cos sin 1 tan
y 1 cos sin 1 tan
+ + = = +
tan4
= +
cot2 4
= + +
3cot
4
= +
55. (A), (C)
The given equation is
2
2 2
2t 1 t
6 7 91 t 1 t
+ = + + where t = tan 2
16t2
12t + 2 = 0
8t2
6t + 1 = 0
1t
2= ,
1t
4=
But2
12tan 2
42 2tan1 3
1 tan 12 4
= = =
Also tan =
12 84
1 151
16
=
56. (A), (D)
Given2 2 2 2
22 2 2
2x z 2x zy
x z (x z) 2xz= =
+ +
2 2
2
2x z
(2y) 2xz=
{since x, y, z are in A.P.}
2 2
2
x z
2y zx
=
2y4
xy2z x
2z
2= 0
(2y2
+ xz) (y2
xz) = 0
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y2 zx since x, y, z are distinct terms in A.P.
2y2 + xz = 0 either x , y , z2
are in G.P.
orz
, y , x2
are in G.P.
57. (B), (C)
In an A.P. the sum of two terms which are equidistant from the first and fromthe last are the same.
Hence a2
+ a18
= a4
+ a16
= a8
+ a12
each =336
1123
=
a1 + a19 = 112
sum of all the terms = ( )1 1919
a a2
+
=19
1122
= 1064
Also a7
+ a13
= 112 and a5
+ a15
= 112
SECTION III
58. (A) tan7
tan cot16 2 16 16
= =
Similarly tan6
16
= cot
2
16
and tan5
16
= cot
3
16
given expression
= 2 2tan cot16 16
+
+ 2 2
2 2tan cot
16 16
+
+ 2 2
3 3tan cot
16 16
+
I bracket = = =
8 8 8 22 2 2
1 2 11 cos 14
2
III bracket =8 8 8 2
2 2 23 1 2 11 cos 14 2
= = + +
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II bracket =8
2 8 2 6
1 cos 2
= =
the whole expression = 2 + 8 2 2 2
= 34
59. (C) Given tan2 + cot
2 + 2 =
8
1 cos 4
i.e., (tan + cot )2
=8
1 cos4
2
8tan cot
12 121 cos
3
+ =
= 16
60. (B) Let A and R be the first term and common ratio of the G.P.
Now A Rp 1
= 64 (1)
A Rq 1
= 27 (2)
A Rr 1
= 36 (3)
(1)
(2)gives
3p q 4R
3
=
(3)
(2) givesr q 4
R 3
=
p q = 3r 3q
i.e., p + 2q = 3r
p 2q3
r
+=
61. (B) p = 8, q = 2
3r = 12
r = 4
This is possible.
A R7 = 64 and AR = 27
R =
1
24
3
R =2
3
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62. (A) f(0) = r = ve { r is + ve r cannot be 0}
f(1) = p + 2q r= 3r r = 2r = + ve
f(0) f(1) < 0
SECTION IV
63. (1) uk
= Sk
Sk 1
= { }1
k(k 1)(k 2)(k 3 k 1)16
+ + +
=1
k(k 1)(k 2)4
+ +
Let tk
=k
1 4
u k(k 1)(k 2)=
+ +
Let vk
= 4(k 1)(k 2)+ +
vk
vk 1
=4 4
(k 1)(k 2) k(k 1)
+ + +
= 2 tk
tk
= k k 11
(v v )2
[ ]n
k n 0k 1
1t v v
2== = [ ]0 n
1v v
2
=1 4 4
2 2 (n 1)(n 2)
+ +
kk 1
1t [2 0] 1
2
== =
64. (6) a + c = 2b
a3
+ c3
= 8b3
3ac 2b
a3
+ c3
+ 4b3
= 12b3
6abc
= 3b{4b2
2ac}
= 3b{(a + c)2
2ac}
= 3b(a2 + c2)
( )
3 3 3
2 2
2a 2c 8b6
b a c
+ +=
+
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65. (3) Given expression =
2 2
2 2
1 1
x w1 1
y z
=
+
+
1 1 1 1
x w x w
1 1 1 1
y z y z
=
1 1 1 13k 3k
x x x x
1 1 1 1k 2k k 2k
x x x x
+ +
+ + + +
where k is the common difference of the corresponding A.P.
=
23k ( 3k)
x3
23k ( k)
x
+
=
+
66. (9) Given2
cos x 3 3
4 41 sin x
3
=
2
3cosx 3 3
43 4sin x =
sin xcosx 3
sin 3x 4=
sin2x 3
2sin3x 4=
sin2x 3
sin3x 2=
2
2
sin 2x 3
4sin 3x
=
2
2
sin 2x12 9
sin 3x=
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67. (4) L.H.S. is (x 4)2
+ 5 which is 5
R.H.S. = 5 cos (y ) where tan =3
4
R.H.S. 5
Equality is possible.
L.H.S. = 5 when x = 4 (R.H.S. = 5 when y = )
68. (1) Now sin2
47 + sin2
13 + sin2
38 + sin2
22
= 1 cos2
47 + sin2
13 + 1 cos2
38 + sin2
22
= 2 {cos 60 cos 34 + cos 60 cos 16}
= 2 1
2{1 2 sin
217 + 1 2 sin
28}
= 1 + sin2
17 + sin2
8
given expression = 1 + sin2
17 + sin2
8 sin2
8 sin2
17
= 1
69. (6)sin(2 )
3sin
+ =
sin(2 ) sin 4
sin(2 ) sin 2
+ + =
+
2sin( ) cos2
2cos( )sin
+ =
+
tan( )2
tan
+ =
3tan( )6
tan
+ =