SOLUTIONS PAPER-1(PCM) CODE AA (UPSEE 2017) PHYSICS Sol.1. (C) ray wi will be stil ti ll total totally inte interrnal nally ref reflected ted at at inte i nterrface. As n2 decreases , iC
n also decreases, so condition i i n
sin 1
2
C
is still satisfied and there
1
will be still total internal reflection at interface. If angle of incidence is increased then ray will be still totally internally reflected at interface because i iC . Sol.2. (B) 2eV
h max 3 1 2eV Sol.3. (C) 100 cm KEmax
A
C
B
D
F
F
30
10
20
10
30
AD=30+10+20+30+10=100cm Sol.4. (B) conduction As atoms in the spoon vibrates about their equilibrium positions and transfer energy from one end to other end. This process is conduction.
along g +y axis axis Sol.5. (B) alon
dq
dE P
dq
Consider any two symmetric dipole elements, net contribution due to these elements at point P is along +y axis . Similarly by principle of superposition we can say that net electric field at P is directed along +y axis. Sol.6. (D) 8 kR 2 K Ei
PEi KE f PE f
0
1 2
1 2
2
k OA R
k 5 R R
2
KE f
KE f
1 2
1 2
k O B R
2
2
k R R
KE f 8kR Sol.7. (B) 9 J
2
U
1
Ceq V 2
1 (3)(6)
2
3 9J
2 2 (3 6) Sol.8. (A) experiences nces a force forcedir directed cted al along the theradi adial dir direction ction onl only. y. Circular motion is a special case in gravitational field. There may be straight line,elliptical paths, but force will be always directed directed toward the centre of the sphere.
Sol.9. (C) speed is is maxim axi mum at t=4s. =4s. t 3 t 0, 2, 4 t 1 x A x A x 0 v 0 v 0 vmax mg Sol.10. (A) k F.B.D. of lower rod kx kx kx
kx
T2 2mg
kx kx 2mg
x T1 mg
T2 2mg
T1 mg
mg k
Sol.11. (C) 9
We consider two possible cases: Case II
Case I nC
5 n B
n B
4
n A nC
4 5
n A
n A nC 1
nC nA
4 5 9
Sol.12. (A) 0 . 6 m / s
By momentum conservation
(10 103 )1000 (10 10 3 ) 400 10v
v 0.6m / s
Sol.13. (B) AP AQ Q P N O A
Q
P
For complimentary angles range ON is same for P and Q as points O and N are on same horizontal plane. From figure AP AQ Sol.14. (D) 60 Hz Second overtone n=3 f3
3f 0 60Hz
Sol.15. (B) 4 m / s 2
x slope ( t ) 2t 2
2
d2x dt 2
4
Sol.16. (C) 1s
In one time constant 63% change occurs in the value of current .The 63% of maximum current is 1.26 A .It is obvious obviou s from graph that current cu rrent 1.26A corresponds to time which is slightly greater than 0.9sec. Hence option having 1s is most appropriate. Sol.17. (C) zero a b
At the position of smaller loop, Magnetic Magnetic field due to larger loop loop is parallel to plane of smaller loop. Due to larger loop, magnetic flux linked with smaller loop is zero. Hence mutual induction is zero.
1 Mi2 0 Mi2 M 0 Sol.18. (D) 3 m / s
1
Impulse during t=2 to t=4 is
1
3 2 5 4 3 5 0
2 Impulse =change in momentum mv f
2
mvi 0 v f vi 3 speed 3
Sol.19. (D) 10 8
5
aQ
ˆi
1
,aP
5ˆi 8 ˆj arel aP aQ 6ˆi 8 ˆj , arel 10
Sol.20. (A) pre pressure ure of 85cm 85cmof Hg Pgas
Patm P 76cm of
Hg 9cm of Hg
85cm of
Hg
Sol.21. (C) work work done by by gasis negati negative ve
Volume does not remain constant throughout the process AB . As T
PV
,temperature nRT decreases initially as both P and V decreases .By area under curve ,net work done is negative. P
A
B 2
Sol.22. (A) 20 kgm s
dKE
V
3
Pall F .v ma .v 100 103 20ˆi 10 ˆj .10ˆ i 20
dt Sol.23. (C) I t re remains ains stationa tati onarry
Maximum possible friction force F
frictionMax.
Maximum applied force F
applied applied Max.
0.5 2 g 10 N
8 1 8 N
F appl applie ied d Max. Max.
F fric fricti tionM onMax. ax.
so the block will remain stationary
Sol.24. (A) 84 kPa 0.004 V p B 2100 106 84 kPa 100 V
Sol.25. (D) m1R I
2
m2 R 2
3
I ABC I AOC mi R m2 2
AC
2
12
R
2
m m i
2 R 2
12
2
m1 R m2 2
R
2
3
Sol.26. (A) dark 1 1 Path difference 13.5 13 n 2 2 So there will be minima. Sol.27. (A) I t will beclockw clockwiise
As collision is elastic ,so after collision ball moves towards left with speed v .As walls and ground are smooth ,there is no tangential torque on the ball. Only normal forces and mg force pass through the centre of the ball ,so their torques about the centre are zero. Torques on the ball about its centre is zero . By I ,angular acceleration is zero hence angular velocity does not change. Sol.28. (A) electron R
mv
v ,q and B are same so R m
qB
Mass of electron is minimum for given options. Sol.29. (C) 8 Nm 1
iAB sin 90 o 2 2 2 2 8 Nm 2
2
Sol.30. (A) 84 m
Result should have only two significant numbers (same as 12m). Sol.31. (C) 27 C 18V i 6 F
6 i
6
3 F
i
18 66
1 .5 A
an d V2
R2 i 6 1 .5 9V ; q2 3 9 27 C
x
Sol.32. (B) 0
t
Initially velocity is constant ,so slope of x-t graph is constant and finite .Finally velocity becomes zero hence slope of x-t graph becomes zero.
Sol.33(C) 20 m 3
mg
air Vg 26 g 1.3Vg V 20m 3
Sol.34. (D) 5 3 0V
R
2 E
Since equivalent internal resistance of equivalent cell across the external resistor R is 2 3 5 ,Hence power delivered to R will be maximum if R 5
1 4
Sol.35. (A) R A
l
2 R
,RB
2
R A 1 VA RA I 1 l 2 R V R I 4 4 R B B B
Sol.36. (A) Al 2 1 Sol.37. (A) 7 kgm s
L 2 2 2 sin 90o
2V 2
Sol.38. (A) P
V2 Req
3R
V2
3 3 3 sin 0 o 1 1 1 sin90 o 8 1 7
R
R
V2 3R
2V 2 3R
2 2 Sol.39. (C) A 0 , B 1, C 0 A 0
1
B 1
1
C 0
Sol.40. (D)
F 1
remain mains same
Intensity after passing through polaroid A is IA
I0
Intensi Intensity ty of unpo unpola larise rised d light light
2 2 Intensity after passing through polaroid B is I B I A cos
2
Here is the angle between pass axes of A and B .Here their pass(transmission) axes always
constant. Hence during the rotation, 2 intensity of transmitted light through polaroid B remains same. same. days Sol.41 . (B) 4 days remain parallel to each other i.e. 0 .
8 0 00
4000 2000 1000
4days T1/2
4days
4days
4days
IB IA
I0
Sol.42. (B) 1 035 A o
1 2420
E
124 20
E2
E1
1242 0
4 ( 1 6 )
103 5 A o
2V
Sol.43. (D)
In reverse bias ,it is equivalent to open circuit condition. From figure given below
2V
VAB
2V A B
100 100
Sol.44. (B) gre greater than P but le l ess than 16P
P T4 , P T4 Here T 273 50 , T 273 100 T T 2T 4
P
273 100 P 1 24 P 273 50 P
I t is i s not not pos possible
Sol.45. (D)
It is not possible, because it will violate the second law of thermodynamics (Clausius statement) .If we consider imaginary case in which temperature of sample becomes more than 600K then it will radiated power is more than absorbed power. Hence it will correspond to decreasing temperature situation. So it is not possible to heat the sample to 900K.
k p
Sol.46. (C)
r 2
0 kpcos cos90
r 2
r
A
p
cos kpcos
r 2
VA
kpcos cos0 r 2
k p cos k p 2 r2 r b2 a2
Sol.47. (C)
R B1 .A1 B2 .A 2 Bb2 cos 0 o Ba 2 cos180o
i
R
d dt R
b
2
a2 R
B b 2 a 2 b 2 a 2 t
Sol.48. (C)
F MR O
F sin30 O cos 30O 30 F cos
O
120 120
C
MR 2 F I F sin 30 R 2 MR acceleration ati on may may bezero at t=2 t=2 Sol.49. (C) acce 0
Let us consider an example in which particle is projected vertically upward from ground at t=0 and it reaches highest point at t=2. Then at that instant(t=2) velocity v=0 but acceleration=−g. acceleration=−g. Here displacement is non zero for the duration t=0 to t=2. For t>2 the ball again acquires velocity. In this example options(A),(B) and (D) are incorrect. Let us consider another example in which a particle part icle is moving on horizontal plane and i t comes to rest permanently at t=1 ,then this is the one of the special case in which acceleration of particle is zero at t=2.
Sol.50. (C) 4ˆi 2ˆ j m s
vseparation 1 v 0 e y vy 2 v 2 4 0 approach y
vx remains same,Hence v final
4ˆi 2ˆj
CHEMISTRY Sol.51. (A) [Co(CO) CO)4 ] and Ni(CO)4
[Co(CO (CO)4 ] has total electrons =3+2×4+1=12, Ni(CO)4 has total electrons =4+2×4=12 ;Thus isoelectronic Sol.52(B) (i), (ii) & (iv) XeF6 has distortion distortion and become Pentagonal bipyramid with with one lone pair. F
O
F
F
O
Xe
O
Cl
Cl
I
Sb
F
F
O
F
O
Xe F6
O
Cl
5
IO 6
Orthoperiiodate
Cl
Cl
Cl
5
5 6 7 1 48 Octahedral
Octahedral
7 6 6 5 4 8 electrons Octahedral 2
SnCl6
4 6(7) 2 48 electro trons ns octah octahedr edral al 6 set of elec
Sol.53. (B) H S< NH NH O < HF 2 3 < H 2 Sol.54. (D) BF3
BF3 can form π bond also in addition to σ bond. As F is more electronegative and octate in B is not complete. Sol.55. (D) H2O H2O will act as Brönsted acid as provided H + ion. Sol.56. (B) 0.354 gm NH 4Cl pOH log K log NH 4Cl pOH log NH 4Cl pOH pK b log b NH 3 NH3 NH3 K b
14 9.45 log
NH 4Cl NH3 K b
log10 (log10 14
1014 9
10
3
Sol.57. (D)
9
log 3 0.47
NH 4 Cl NH4Cl 1014 1014 NH 4Cl log 3) log log 9 log 9 10 3 NH3 K b NH3 K b 10 3 NH 3 K b
NH3 Kb NH 4Cl
1014 9
10
3
0.011. 1.85 105 NH4Cl 0.35
2 K p 1 2 K p
2 HI ( g ) H 2 ( g ) I 2 ( g ) At eq. 2(1 x )
x
x
Total moles at equilibrium equilibriu m =2-2x+x+x=2 ,here x= degree of dissociation dissoc iation
x x P 2 2 2 2 x 2 2 2 4 1 x P 2 4 1 x
K p
Kp
x
1 x
x
2 K p 1 2 K p
4 Sol.58. (C) 171
A 6% solution of sucrose C22H22O11 conc.=
6g 100 ml
0.06 ml 1
For unknown solution conc.= 3 g per 100cc 30 gl 1 For isotropic solution
60 34 2
30 m
m
30 34 3 42 60
30 m
60 342
mole l 1
mole l 1
171
Sol.59. (D) 491 kJ CH 3COOH 2O2
2CO2 2H 2O 869 2 395 2 285 f H (CH COOH ) 3
Simplification gives 2 434.5 2 395 2 285 S H (CH3COOH ) 1
434.5 395 285 f H 491 f H
or
2
Sol.60. (B) Sulphur Sol.61. (A) Cathode is Lead dioxide (PbO2) and anode is Lead (Pb) Sol.62. (D) ( ∆Ssystem+ ∆S surrounding ) > 0 Sol.63. (A) PVm=RT
At low pressure & high pressure V is very high ,thus
a 2
V m
and b are negligible, finally reduced
eq. PVm=RT Sol.64. (D) t 3/ 4
A 0 0
If t1/2 vs. A
is constant ,it is radioactive decay hence is not of zero order. Thus answer
will be t 3/4
A 0 Sol.65. (C) Hexane Sol.66. (A)
5 1010 gm
Let W be the weight of Ra 238 in equilibrium as Th
232
Ra W
238
t 1/2 Th t 1/2 Ra
1 N 0 / 232 1.4 1010 ; N 0 AvagardoNo 7 W N 0 / 238
238 232 2 10
9
5 1010 gm
Sol.67. (B)
n=4,
l=0,
m=0, s=+
1 2
For option (A) electron is in 3d For option (B) electron is in 4s For option (C) electron is in 4p For option (D) electron is in 5s According to Aufbau principle 4s<3d<4p<5s. 4 s<3d<4p<5s. Hence Answer will be n=4,l=0, m=0,s=+ Sol.68. (A) 4, 6 Sol.69. (C)
COOH
COOH
Terlhalic acid . Sol.70. (B) cis-1, 3-dimethylcyclohexane 3- dimethylcyclohexane
Two chiral centre of plane of symmetry so so it is meso compound. Sol.71. (C) 2 ethyl-3 methyl pentanal OH
Sol.72. (A) CHCl3
aq. Na NaOH
In Reimer-Tieman ,phenol reacts with CHCl 3 & aq. KOH/NaOH to give Selicil Aldehyde. Sol.73. (B) B < S < P < F I.P. in period increases left to right. I.P. in group up to down decreases. Sol.74. (B) 0
2 Carbocation will be more stable than 1
0
Substituted carbocation will more stable than simple. Sol.75. (A) Be3N2 Sol.76. (C) Cannizzaro reaction Sol.77. (C) BN Sol.78. (C) i > iv > ii > iii Acidity
∝
1 basicty
Sol.79. (C) (i),(ii)
Br
(i) (i )
1, 2 hyd ri rid e sh if ift
Br (ii)
H
1 2
Sol.80. (A) iii < i < iv < ii Acidic strength ∝ electron withdrawing group gro up strength.
R H I
1
I
1
H
1
R
Sol.81. (A) Only –I effect Due to SIR effect effect –NO2 group goes out of plane of the paper (-I). (-I ). Sol.82. (C) AgNO3 Sol.83. (A) 6-ethyl-2-methyl-5-propyldecane CH3 ( 4)
C H 2C H 2 C H C H 3
H
C(6 )
C H3 CH2 C H2 C H2
(10)
(9 )
(8 )
(3)
(7)
(5)
C H2 C H 3
Sol.84. (B) D-fructose
(1)
(2)
CH2CH2 CH3
C H
CHO
CHO
OH OH
C H2 OH
OH
OH
OH
OH
O
OH
base
base
OH
OH
OH
OH
OH
OH
C H2 O H
C H2 OH
C H2OH
D Manos Manose e
D Gluco Glucose se
D Fructo Fructose se
O Sol.85. (B) Benzoquionone
NH 4 2 Cr2O7 H 2SO 4
O
Benzoquinone
Conjugated Conjugated diketones Sol.86. (C) 19 , H
H
H
H
C
H
and 19 and
H
H
H H
H
Sol.87. (B) (i) and (iii) As a plane of symmetry exist in compound(ii),there is no chirality in it. Hence (i) &(iii) will be
optically active. Sol.88. (A) BCl3 and AlCl3 are both Lewis acids and BCl3 is stronger than AlCl3 BCl3 and AlCl3 both are are Lewis Lewis acids but but BCl3 is more more electron electron deficient deficient than AlCl3 so BCl3 s stronger Lewis acid than AlCl3 due to high electron negativity.(E.N. negativity.(E.N. B-2.0, Al 1.5) Sol.89. (C) I, II, IV O
Compound
C
CH3
cannot be obtained by Friedel Craft acylation, In
,NO 2 NO2
NO2
Group has has -M or –R effect so it cannot be be used in in Friedel Craft acylation acylat ion ,this ,th is is deactivating deact ivating & metadirecting group in ESR. O
O
H3 C
C
Cl A lCl3
C
Cl
NO2
NO2
Sol.90. (A) 4-butyl-1-ethyl-2-methylcycloheptane Naming according according to closest set of locant locant rule
1
6 7 2 5 4 3
Sol.91. (D) 5-chloro-2-methylcyclohexanol CH3 3 4
1
2
OH
6
5
Cl
Sol.92. (A) -hydroxyaldehyde or OH R CH 2 CHO H
R R CH 2
Dilute
CH 2
C
OH
H
R
condensation CHO
H
R CH 2
O
2
CH
C
CHO
Aldoladdition Aldoladdition , unsaturated aldehyde
In aldol addition it is β hydroxyl aldehyde(aldol both alcohol and aldehyde group) But in aldol condensation the product is α,β unsaturated aldehyde Sol.93. (D)
O 2N
OH
O 2N NO2
O2 N
OH
NO 2
OH
OH
Highest dipole moment Sol.94. (A) 4 Grignard reagent will react on CO bond there are four C-O bond .These will be addition of Grignard reagent at four positions. HO
1mol
COOEt O 1mol
2mol
total 4mol
Sol.95. (A) KO2 KO2 K O2
SiO2 Si4
2O2
3d 0 BaO BaO2
Diamagnetic
2s 2 2 p6 2
Diamagnetic
2s 2 2 p 6
Ti4
2O2
2s2 2 p6 TiO2
17e in vala valanc ncee cellso cellso unpair unpaired ed elec electro tron n prese present nt &show &show para parama magm gmeti etism sm
2O2
Ba
XeCon eConfig figurat uratio ion n
Diamagnetic
NeConf eConfig igur urat atio ion n
Sol.96. (C) CH 3 Nucleophilicity Nucleophilicity order CH 3 NH 2 OH F Sol.97. (D) For lead +2, for tin +4
PbO PbO 2 + Pb
2PbO 2PbO,,
ΔG 0 < 0
i.e. ΔG 0 is negative so reaction is fisiable i.e. for Pb,+2 Oxidation state is more stable.
2SnO,, 2SnO
SnO SnO 2 +Sn
ΔG 0 > 0
i.e. ΔG 0 is positive so reaction is nonfisiable i.e. for Sn,+4 Oxidation state is more stable. Correct answer is (D) For lead +2, for tin +4 For lead +2, for tin t in +4 Oxidation State are more characteristics. Sol.98. (A) 1-Phenyl-2-butane C H 2 CH CH CH3 Geometrical isomerism. (i) Ph CH CH 2 CH CH CH3 (ii)
No Geometrical isomerism
Ph CH 2
C CH CH 2 CH3
(iii)
No Geometrical Geometrical isomerism
Ph
P
(iv)
C CH CH3 Ph
No Geometrical Geometrical isomerism
Sol.99. (C) associate At CMC they associate. Sol.100. (B) CH3
CH3
O
CHO
O
CHO
will be most reactive.
Hence most appropriate option is CH3
O
CHO
MATHEMATICS Sol.101. (A) x
log 2
y
1 y
2 x y y which gives x log 2 2 log2 x 1 y 1 2 Sol.102. (D) 2 x 0 y is well defined when log10 (1 x ) 0 & x 2 0, Hence 2 x 0 Sol.103. (B) A =
1 , B = 1
For continuity of f(x) at x
2
&
lim f( f ( x ) 2 lim f (x ) A B f x
2
x
2
, we have
2
2 2
2 2
& lim f (x ) 2 lim f ( x ) 0 f x
2
x
2
A B 2 & A B 0 A 1, B 1 2 Sol.104. (B) x
continuity of f(x) at x=0.
lim f (x (x ) f (0 (0) for x 0
Sol.105. (D)
y a
2 x
a
cot y is of the form 0 . 0 2 2a y a y a Using L’Hospital rule lim sin . tan 2a y a 2 lim sin
y a
lim n does not exist but lim Ln exists n 0 when n is odd lim n so lim n does not exist 2 w h e n n i s e v e n n n lim Ln 0 exists n Sol.107. (B) x 0
Sol.106. (C)
n
x2 2 tan 1 x is valid for 0 x ,so negative times shows the answer x 0 Since cos 2 1 x Sol.108. (A) ( 1 , 0 ) , ( - 1 ,- 4 ) 1 1
Slope of tangent to curve at point ( , ) is
dy dx
2
i.e. 3
1 which is parallel to line having
( , )
So 3 2 1 4 which gives 1 .The point ( , ) lies on the curve so (1, 0) 0) & (1, 4)
. Sol.109. (C) 25
a b
, bc
, c a a b . c b a b . b c . c a
2
a b c b. b . c a a b c 25
0, 4 .Points are
slope 4.
Sol.110. (C)
2x - y + 1 = 0
Equation of chord joining the points P(1,4) & (3,8) on the parabola is 2 x y 2 0 . Tangent parallel to this chord will have the slope i .e.
dy dx
2 ∴Equation of tangent at
( , )
on the curve with
slope 2 is 2x - y + 1 = 0
3 3 , 2 4
Sol.111. (B) x
Given y
Differentiating w.r.to x, we have t 3t 2 dt . Differentiating 2
0
d 2y dx 2
point of inflection
3 , 3 2 4 Sol.112. (C)
Clearly P
dy dx
x2 3x 2 &
d2y dx 2
2 x 3 . At the
0 & second derivative changes sign whi le passing through the point of i nflection.
. -2
2 x sin x 0 is c o s x 0 2
lim 2 x tan x lim c o s x x x 2
form.
Use L’ Hospital Rule, we get result -2. Sol.113. (D) 2x - y - 1 = 0 The point of intersection of the curve y x 2 with the bisector of the first quadrant i.e. y=x is (1,1) {Neglect the point (4,0) as it does not satisfy y=x}. Equation of normal to the curve at (1,1) is 2 x y 1 .
2 1 Sol.114. (D)
y
2
y5
dy Write y tan( x y) as x y tan y ,then differentiating directly implicitly ,we get dx Sol.115. (D) 600 or 1200 1
1 y 2 2
y5
Let sides AB, BC & AC be c,a,b respectively in ABC . 1
1 2
Area of triangle bcsin bcs in A 10 3 5.8 sin A sin A 2
Sol.116. (C)
tan 1
3 a 2
600 or 120 0
41 2
Equation of curves c1 : y x 2 & c 2 : 9 x 2 16 y 2 25 .Let m1 & m2 be the slope of the tangents to these curve at the point of intersection (1,1) m1 2 & m 2
Similarly at the point of intersection (-1,1) 2 tan 1 Sol.117.
4
16
,So 1 tan 1
9 16 18 1 16
2
ta n 1
m1 m2 1 m1m 2
41 2
(C) 1
For maxima –minima So x
9
dy dx
0 2 sec 2 x 1 tan x 0 x
4
&
d2 y dx 2 0 x 4
is the point point where function y 2 tan x tan 2 x has maximum value.
1 tan 1
41 2
y at x 1
∴Maximum value
4
Sol.118.
(B) ( a
b ) 4 ab 138
2
2
Let M be the middle point of the segment AB. So
Length OM (a2
a b , 24 . Since O M MA and 2
M
a b 3 length MA MA , but MA , 13 . So using
2
OM
3
MA
.We get
b 2 ) 4 a b 1 38
Sol.119.
(A) 3 sin x
4 cos x
0 So x 0.Hence f ( x ) 3 sin( x ) 4 cos( x ) given ,But f is odd, so f ( x) f ( x ) where x 0 f ( x ) 3 sin x 4 cos x
Let x
Sol.120. (A) continuous at x = 0 but not differentiable differentiab le at x = 0 1
Since x x tan 1 lim x 0
2 f ( x ) f (0)
x ,So
x
2
lim f ( x ) 0 f (0 (0) x 0
f is continuous, but
1
lxim0 tan 1 does not exit, so not differentiable at x=0.
x0
x
Sol.121. (B) 48
According to question 2 &
Sol.122. (C) a
2
,b
3
2
1
3
6
2
24 . Solving 6, 54 48
1 6
At the point of Maxima or Minima &2 . a , b
dy dx
0 i.e. at x 1 & x 2 ,we have
dx
a
2bx 1 which is 0 at x=1 x
d y d y 0 ,so maximum. dx2 0 , so minimum & dx2 at x1 at x 2 2
, Clearly
dy
2
1
Sol.123. (C)
4 Let P be a point inside the circle z z0 r. Probability of the point P which lies within the circle of radius
r 2 4 r r 1 is z z0 is 2
2
r
2
4
1
Sol.124. (A)
3
Required chance Sol.125. (A) 30
5! 6! 2!
1 3
0
Given L : 3 sin A 4 cos B 6 & M : 4sinB 3 cos A 1 in ABC ,So L2 M2 implies sin(A B)
sin C sin(1800 A B)
1
C 30 0 or 150 0 . Discard
2
0
C 150 because
3
0
less than 30 . Hence 3 sin A 4 cos B 4 6 a contradiction.∴ C 30 c
1 a
sin
0
1 a
1
dx
t
2
Put x so I
x x
x
a
2
reduces to
1 a
dt 2
1 t2 a
.Hence I c
2
for this value of C, A will be
2
Sol.126. (B)
1
1 a
sin
1 a x
Sol.127.
2 y d y e dx 2
(A)
dy dx
2
sin x
Differentiating the given relation w.r.to x, we get e y 2 y d y e dx 2
e
2
2 1 2
& M : co cos a cos b
6
,So L2 M2 implies cos(a b) 0
2
While LM (using cos(a b) 0 ) gives sin(a b)
I
e
(C)
eb
3 2
3 ea
dy tan & dy tan So, f (a) 3 , f (b) 1 dx dx 3 4 x a x b
Given x
cos x 0 ,Again d.w.r.to x
dy sin x 0 dx
y
Given L : sin a sin b
b
dx
3
Sol.128. (C)
Sol.129.
dy
b
∴
d f x f x dx dx e f x dx e f x e f b e f a e x
b
x
b
a
a
a
b
3 ea
a
Sol.130. (B) a 5 , b 5
1 1 2 : 3 A b 3 5 3 : b 2 6 a : 2 1 1 R 2 3R1 , R 3 2R1 0 8 0 8
2:3 1 1 9 : b 9 R 3 R 2 0 8 9 : b 9 0 0 a 5 : 5 b a 4 : 4 For no solution rankA≠rank A b , So a 5 , b 5 Sol.131. (A) 5 x 2 3x 1 0 2:3
1 1 1 1 Required equation x 2 x 0 Where 3 & 5 as , are roots of x 2 3x 5 0 5x 2 3x 1 0 Sol.132. (D)
h 2
3a h
3
a h
Volume of the solid of revolution V y2 dx (The figure is bounded by x=a,x=a+h,y=0) a
ah
V
x
2
a 2 dx
h
a
Sol.133. (A) y 2
3
2
3a h
1 x log
c
1 x
1
Given diff. Eq. can be written as y
dy 1 y2 dx 2( x 1)
dt dx
1 ( x 1)
t
x 2( x 1)
x ( x 1)
,Let y 2 t so 2y 1
dx where I.F. e 1 x
dy dx
1 ( x 1)
dt dx
.Hence eq. reduces to
Hence solution t. IF. Q.IF.dx c
y 2 1 x log
c
1 x
1
Sol.134. (D) 10m / sec 2 x t 5t 2 7t 3 , v
dx dt
10t 7 5 10t 7 t
d 2x ;a 10m / s 2 2 dt t 12
12 10
10
Sol.135. (C) 5 x2 7 x 439 = 0
Obviously p,q satisfy the equation 5 x 2 7 x 3 0 .Hence p q
7 5
, pq
3 5
Given 5 p 4q & 5q 4 p. 5 x2 7 x 439 = 0
0 The required equation x 2 x Sol.136. (B)
1
x
2
1 2
Let sin 1 x , given 3 sin 1 x sin 1 x 3 4 x 2
3 ,Hence 2
2
6
Sol.137. (B) x
Sol.139. dy dx
6
2
2
1 8
x
1
2
2
2 0 0 1 e
(A) og e tan
1
1
2
2
x
i .e
2 2 x x 1 y 1 e
32
y
y 3
1
where e 2
2
2 e2 1
(C) 2 og e
Mean value M
1
12 y 1 2
Required ellipse Sol.138.
1
∴ sin
3 sin1 sin 3 4 sin2
e x
2
dx x
y
4
M 0
2 sin
1e
x
2
x dx Put 1 e
x
2 2 e 1
t ,We get M 2 og e
c
x y xy x y sin sin 2 cos sin . 2 2 2 2
So separating the variables and integrating og e tan Sol.140.
y
4
2 sin
(C) - 9/2 2 x 9 2 x 9 2 x 9
R1 R2 R3
Sol.141. (D)
2
2x
2
7
6
2 x
7
9
2
2
0 x 1, ,
a =1 , b = 0
Sol.142. (B) increases increase s in ( 0 , 1 ) but decreases in ( 1 , 2 ) a ib cos log i 4i cos 4i log i i 1 a 1,b 0 2
y 2 x x2
so
dy dx
0 0 1 ( x 1) 2 1 x
for 0 x 1 for x 1, 2
So f increases in (0,1) & decreases in (1,2).
x
2
c
Sol.143. (B)
5
3
5
5 5 5 5 as the line y=x intersect lines 2x 5 5 at points , & , . 3 3 3 3 3 Sol.144. (A) z 2 7
log
z 2 3 1 ,since 3 z 2 1 sin
1 1 So 6 2
sin
6
1
Sol.145. (C)
Let
Sn
2
3
n
z 2 3 1 z 2 7 3 z 2 1 2
1
1 4 13 40 121 364 ....... Tn 1 Tn
Sn 1 4 13 40 121 364 ....... Tn 2 Tn 1 Tn Rewrite & Sn Sn 1 3 32 33 ....... Tn Tn 1 Tn 3n 1 3n 1 Tn 1. & Tn 3 1 2 Alternative: put options directly.
Sol.146. (D) y
3
5
,
3
x 2 sin x has tangent parallel to x axis at the points
3
,
5 3
and
5 0 for x 0 , U ,2 dx 3 3 dy 5 0 for x , dx 3 3 dy
Sol.147. (D) x 6
9 31/ 3
6
1 x 9 x6 6 6 x C x6 21/ 3 31/ 3 x 1/ 3 C 6 2
Sol.148. (B) 256 mn If cardinality of A=m & Cardinality of B is n ,then total no. of relations from A to B is 2 .
8
256 Sol.149. (A) 0 0 .00025
Here m=4,n=2 ∴ 2
Using
x2
,we get y 4 4 which is less than ,
So 4 2 For 0.001, the 0.00025
Sol.150. (D) f 0
f 0 lim
x 0
f (x (x ) f (0 (0) x 0
where f (0) 0 (given ) , So L f 0
