www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
SOLUTIONS PAPER-1(PCM) CODE AA (UPSEE 2017)
PHYSICS Sol.1. (C) ray will be still totally internally reflected at interface. As n2 decreases , i
C
1
n2 also decreases, so condition i iC is still satisfied and there n1
sin
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 KEmax h max 3 1 2eV
Sol.3. (C) 100 cm 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. Sol.5. (B) along +y axis 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 KEi PEi KE f PE f 1 1 2 2 k OA R KE f k OB R 2 2 1 1 2 2 k 5 R R KE f k R R 2 2 2 KE f 8kR 0
Sol.7. (B) 9 J 1 1 (3)(6) 2 U Ceq V 2 3 9J 2 2 (3 6)
Sol.8. (A) experiences a force directed along the radial direction only. Circular motion is a special case in gravitational field. There may be straight line,elliptical paths, but force will be always directed toward the centre of the sphere. http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
Sol.9. (C) speed is maximum at t=4s. t 3
t 0, 2, 4
t 1
x A
x0
x A
v 0
vmax
v 0
mg k
Sol.10. (A) kx
F.B.D. of lower rod
kx
kx
kx
T2 2mg
T1 mg
kx kx 2mg mg x k
T2 2mg
T1 mg
Sol.11. (C) 9 We consider two possible cases: Case II
Case I nC
5 nB
nB
4
nA nC
4 5
nA
nC nA 4 5 9 nA nC 1
Sol.12. (A) 0.6 m / s By momentum conservation (10 103 )1000 (10 103 ) 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 3f0 60Hz Sol.15. (B) 4m / s 2 d2x x slope (t ) 2t 2 4 dt 2
2
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
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 from graph that current 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 field due to larger 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 1 3 2 5 4 3 5 0 2 2
Impulse during t=2 to t=4 is
Impulse =change in momentum mv f mvi 0 v f vi 3 speed 3
Sol.19. (D) 10 8
5
1
aQ ˆi ,aP 5ˆi 8 ˆj arel aP aQ 6ˆi 8 ˆj , arel 10
Sol.20. (A) pressure of 85cm of Hg Pgas Patm P 76cm of Hg 9cm of Hg 85cm of Hg
Sol.21. (C) work done by gas is negative Volume does not remain constant throughout the process AB . As T
PV nRT
,temperature
decreases initially as both P and V decreases .By area under curve ,net work done is negative. P
A
B
V
2 3
Sol.22. (A) 20 kgm s
dKE Pall F.v ma.v 100 103 20ˆi 10 ˆj .10ˆi 20 dt
Sol.23. (C) It remains stationary Maximum possible friction force F frictionMax. 0.5 2 g 10 N Maximum applied force Fapplied Max. 8 1 8 N http://www.myengg.com/engg/info/category/uptu/
www.myengg.com F
applied Max.
F
UPSEE 2017 Paper I - SET AA Solutions
frictionMax.
so the block will remain stationary
Sol.24. (A) 84kPa V p B V
Sol.25. (D)
0.004 6 84 kPa 2100 10 100 2
m1R
m2 R 2 3
I I ABC I AOC mi R 2 m2
AC
2
12
R 2 mi m2
2R 12
2
m1 R 2 m2
R2 3
Sol.26. (A) dark Path difference 13.5 13 12 n 12
So there will be minima. Sol.27. (A) It will be clockwise 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 qB
v ,q and B are same so
Rm
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 6F
6 i 6
3F
i
18 1 .5 A and V2 R2i 6 1.5 9V ; q2 3 9 27C 66 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. http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
Sol.33(C)
UPSEE 2017 Paper I - SET AA Solutions
20m 3
mg airVg 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 ,Hence power delivered to R will be maximum if R 5 Sol.35. (A) RA
1 4
l 2R
2 3 5
2
,RB
l R
2
RA RB
1 VA R A I 1 4 VB RB I 4
Sol.36. (A) Al Sol.37. (A) 7 kgm2s1 L 2 2 2 sin 90o 3 3 3 sin 0o 1 1 1sin 90o 8 1 7 2V 2 Sol.38. (A) 3R 2 2 V V V 2 2V 2 P Req R R 3R 3R 2 2 Sol.39. (C) A 0, B 1, C 0 A0 B 1 C 0
1
F 1
1
Sol.40. (D) remains same Intensity after passing through polaroid A is
I0
Intensity of unpolarised light 2 2 2 I B I A cos IA
Intensity after passing through polaroid B is Here is the angle between pass axes of A and B .Here their pass(transmission) axes always remain parallel to each other i.e.
0 . IB IA
I0 2
constant.
Hence during the rotation,
intensity of transmitted light through polaroid B remains same. Sol.41. (B) 4 days 8000 T1/2
4000 2000 1000
4days 4days
4days
4days
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
Sol.42. (B) 1035 A o
12420 12420 12420 1035 A o E 4 (16) E 2 E1
Sol.43. (D) 2V In reverse bias ,it is equivalent to open circuit condition. From figure given below 2V
VAB 2V
A B
100
Sol.44. (B) greater than P but less than 16P P T4 ,P T4 HereT 273 50 ,T 273 100 T T 2T 4
P 273 100 P 4 1 2 P 273 50 P
Sol.45. (D) It is not possible 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. Sol.46. (C)
k p r2 kpcos 900 r2
r
A
kpcos r2
kpcos 0 r2
p
k p cos k p 2 r2 r 2 2 b a
VA
Sol.47. (C)
R B1 .A1 B2 .A 2 Bb 2 cos 0o Ba 2 cos 180o B b 2 a 2 b 2 a 2 t
d 2 2 dt b a i R R R
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
Sol.48. (C)
UPSEE 2017 Paper I - SET AA Solutions
F MR F sin 30O O F cos 30O 30
120O
C
I F sin 300 R
MR 2 F 2 MR
Sol.49. (C) acceleration may be zero at t=2 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. 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 is moving on horizontal plane and it 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 2ˆj m / s v 1 v 0 e separation y vy 2 2 40 vapproach y vx remains same,Hence v final 4iˆ 2jˆ
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
CHEMISTRY Sol.51. (A) [Co(CO)4 ] and Ni(CO)4 [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 and become Pentagonal bipyramid with one lone pair. F
O
F
F
Xe
O
Cl
I
F
O
O
XeF6
O
Cl
Sb
F
F
Cl
O
Cl
Cl
IO 56
Cl
Orthoperiiodate 5
5 6 7 1 48 Octahedral
Octahedral 7 6 6 5 48 electrons Octahedral
SnCl62 4 6(7) 2 48 6 set of electrons octahedral
Sol.53. (B) H2S < NH3 < H2O < HF 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 NH 3 NH 3 Kb 14 9.45 log
NH 4Cl NH 3 Kb
log 3 0.47
log1014 (log109 log3) log
NH 4Cl NH3 Kb
log
1014 NH4Cl 1014 NH 4Cl log 109 3 NH3 Kb 109 3 NH 3 Kb
1014 1014 NH K NH Cl 0.011.85 105 NH 4Cl 0.35 3 b 4 9 9 10 3 10 3 2 Kp Sol.57. (D) 1 2 K p
2 HI ( g ) H 2 ( g ) I 2 ( g ) At eq.2(1 x ) x x Total moles at equilibrium =2-2x+x+x=2 ,here x= degree of dissociation
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
x x 2 P 2 Kp x2 x 2 2 K p 2 2 Kp x 2 1 x 1 2 K p 4 1 x P 2 4 1 x 4 Sol.58. (C) 171 6g 60 A 6% solution of sucrose C22H22O11 conc.= 0.06ml 1 molel 1 100 ml 342 30 For unknown solution conc.= 3 g per 100cc 30 gl 1 mole l 1 m 60 30 30 342 For isotropic solution m 171 342 m 60 Sol.59. (D) 491 kJ CH 3COOH 2O2 2CO2 2 H 2O 869 2 395 2 285 f H ( CH 3COOH )
Simplification gives 2 434.5 2 395 2 285 S H (CH 3COOH ) 1 or 434.5 395 285 f H 491 f H 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 a2 and b are negligible, finally reduced Vm
eq. PVm=RT Sol.64. (D)
t 3/4
A 0 If t1/2 vs. A0 will be
is constant ,it is radioactive decay hence is not of zero order. Thus answer 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
Th232 t1/2Th 1 N0 / 232 1.4 1010 ; N 0 AvagardoNo W N0 / 238 7 Ra 238 t1/2 Ra W
238 5 1010 gm 9 232 2 10
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
Sol.67. (B)
UPSEE 2017 Paper I - SET AA Solutions
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. Hence Answer will be n=4,l=0, m=0,s=+
1 2
Sol.68. (A) 4, 6 Sol.69. (C)
COOH
COOH
Terlhalic acid. Sol.70. (B) cis-1, 3-dimethylcyclohexane
Two chiral centre of plane of symmetry so it is meso compound. Sol.71. (C) 2 ethyl-3 methyl pentanal OH
Sol.72. (A) CHCl3 aq. NaOH
In Reimer-Tieman ,phenol reacts with CHCl3 & 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)
20 Carbocation will be more stable than 10 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)
1,2 hydride shift
Br (ii)
H
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
Sol.80. (A) iii < i < iv < ii Acidic strength ∝ electron withdrawing group strength. R H I
1 1 1 I H R
Sol.81. (A) Only –I effect Due to SIR effect –NO2 group goes out of plane of the paper (-I). Sol.82. (C) AgNO3 Sol.83. (A) 6-ethyl-2-methyl-5-propyldecane CH3 (4)
CH2 CH2 CH CH3
H
C(6 )
CH3 CH2 CH2CH2
(10 )
(9 )
(8)
(3)
(7)
(5)
CH2 CH3
Sol.84. (B) D-fructose
(1)
(2)
CH2CH2 CH3
C H
CHO
CHO
OH
OH
OH OH
OH
CH2 OH
base
O
OH base OH
OH
OH
OH
OH
OH
OH
CH2OH
CH2 OH
CH2OH
D Manose
D Glucose
D Fructose
O Sol.85. (B) Benzoquionone
NH Cr O H SO 4 2 2
2
7
4
O
Benzoquinone
Conjugated diketones Sol.86. (C) 19 , H
H
H
C H H
H
19 and
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 Lewis acids but BCl3 is more electron deficient than AlCl3 so BCl3 s stronger Lewis acid than AlCl3 due to high electron 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
,NO2 NO2
NO2
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
Group has -M or –R effect so it cannot be used in Friedel Craft acylation ,this is deactivating & metadirecting group in ESR. O
O H3 C
Cl AlCl 3
C
C
Cl
NO2
NO2
Sol.90. (A) 4-butyl-1-ethyl-2-methylcycloheptane Naming according to closest set of locant rule
1 7 6 2 5 4 3
Sol.91. (D) 5-chloro-2-methylcyclohexanol CH3 3 4
1
2
5
OH
6
Cl
Sol.92. (A) -hydroxyaldehyde
R R CH 2
H or OH R CH 2 CHO Dilute
CH 2
C
OH
H
R condensation CHO
R CH 2
H 2O
CH
C
CHO
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
NO2
O 2N
O 2N
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
O 1mol
COOEt 2mol
total 4mol
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
Sol.95. (A) KO2 KO2 K O2
SiO2 Si4 2s2 2 p6
17 e in valance cellso unpaired electron present &show paramagmetism
2O2 Diamagnetic 2s 2 2 p6
TiO2 Ti4 3d 0
2O2 Diamagnetic 2s 2 2 p6
Ba2
BaO2
2O2
XeConfiguration
Diamagnetic
NeConfiguration
CH 3
Sol.96. (C) Nucleophilicity order CH 3 NH 2 OH F
Sol.97. (D) For lead +2, for tin +4
PbO 2 + Pb 2PbO, ΔG 0 < 0
i.e. ΔG0 is negative so reaction is fisiable i.e. for Pb,+2 Oxidation state is more stable. SnO 2 + Sn 2SnO, ΔG 0 > 0
i.e. ΔG0 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 +4 Oxidation State are more characteristics. Sol.98. (A) 1-Phenyl-2-butane (i) Ph CH 2 CH CH CH 3 Geometrical isomerism. CH 2 CH CH CH3
(ii)
No Geometrical isomerism
Ph CH 2 C CH 2 CH3
(iii)
No Geometrical isomerism
Ph
Ph
(iv)
Ph
C CH CH3
No 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
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
MATHEMATICS y 1 y
x log 2
Sol.101. (A)
2x y y which gives x log 2 2 log 2 x 1 y 1 2 Sol.102. (D) 2 x 0
y is well defined when log (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 & 2 , we have 10
lim f (x) 2 lim f (x) A B f 2 x x 2 2 2 & lim f (x) 2 lim f (x) 0 f 2 x x 2 2 2 A B 2 & A B 0 A 1, B 1
Sol.104. (B) lim f (x) f (0) x 0
2 x
for continuity of f(x) at x=0. 2x
a ya lim sin y a 2
Sol.105. (D)
y 0 cot 2a is of the form 0 . ya y a . tan Using L’Hospital rule lim sin y a 2 2a
Sol.106. (C)
lim n
n
does not exist but
0 when n is odd lim n 2 when n is even n lim Ln 0 exists
so
lim n
n
lim Ln
n
exists
does not exist
n
Sol.107. (B)
x 0 2
Since cos 11 xx 2 tan x is valid for 0 x ,so negative times shows the answer x 0 Sol.108. (A) ( 1 , 0 ) , ( - 1 ,- 4 ) dy Slope of tangent to curve at point ( , ) is dx i.e. 3 1 which is parallel to line having slope 4. 1
1
2
2
( , )
So 3
which gives 1 .The point ( , ) lies on the curve so (1,0) & (1, 4) . Sol.109. (C) 25 a b , b c , c a a b . c b a b . b c . c a 2
1 4
2 a b c b . c a a b c 25
0, 4 .Points
are
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
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 dy parallel to this chord will have the slope i.e. 2 ∴Equation of tangent at ( , ) on the curve with dx
slope 2 is 2x - y + 1 = 0 Sol.111. (B)
3 3 , 2 4
x
Given
y t 2 3t 2 dt
dy d2y x 2 3x 2 & 2 x 3 . At the dx dx2
. Differentiating w.r.to x, we have
0
d2y
point of inflection
dx2
0 & second derivative changes sign while passing through the point of inflection.
3 3 , . 2 4
Clearly P
Sol.112. (C) - 2 lim 2 x tan x cos x 2
2 x sin x 0 lim is form. x x cos x 0 2
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 2x y 1 .
2 1 y2
Sol.114. (D) Write
y5
y tan(x y) as x y tan 1 y
,then differentiating directly implicitly ,we get
Sol.115. (D) 600 or 1200 Let sides AB, BC & AC be c,a,b respectively in
2
.
1 1 3 bcsin A 10 3 5.8sin A sin A a 600 or 1200 2 2 2
Area of triangle Sol.116. (C)
ABC
1 y dy 2 dx y5
tan 1
Equation of curves
41 2
c1 : y x 2 & c 2 : 9x 2 16y 2 25
curve at the point of intersection (1,1) m
1
.Let
m1 & m 2
2 & m2
9 16
be the slope of the tangents to these
,So
1
tan 1
m1 m2 41 1 tan 1 1 m1m 2 2
9 16 tan 1 41 18 2 1 16
2
Similarly at the point of intersection (-1,1)
2
tan
1
Sol.117. (C) 1 For maxima –minima
d2 y dy 0 2sec 2 x 1 tan x 0 x & 2 0 dx 4 dx x 4
So
x is the point where function y 2 tan x tan 2 x has maximum value. 4
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
∴Maximum value yat x 1 4
2
2
Sol.118. (B) (a b ) 4 ab 138 Let M be the middle point of the segment AB. So M a 2 b , 24 . Since OM MA and
Length 2
OM 3 length MA , but MA a b , 13 . So using OM 3 MA .We get 2
2
( a b ) 4 ab 138
Sol.119. (A) 3 sin x − 4 cos x Let x 0 So x 0.Hence f ( x ) 3sin( x ) 4 cos( x ) given
,But f is odd, so
f ( x) f ( x ) where x 0 f ( x ) 3sin x 4cos x
Sol.120. (A) continuous at x = 0 but not differentiable at x = 0 Since 2 x x tan x1 2 x ,So lim f (x) 0 f (0) f is continuous, but 1
x 0
f (x) f (0) 1 lim lim tan 1 does not exit, so not differentiable at x=0. x 0 x 0 x 0 x
Sol.121. (B) 48 According to question Sol.122. (C) a
2 &
24 . Solving 6, 54 48 2
2 1 ,b 3 6
dy dy a 0 i.e. at x 1 & x 2 ,we have 2bx 1 which is 0 at x=1 dx dx x d2 y d2y 2 1 &2 . a , b , Clearly 2 0 , so minimum & 2 0 ,so maximum. 3 6 dx at x 1 dx at x 2
At the point of Maxima or Minima
1 4
Sol.123. (C)
Let P be a point inside the circle
z z 0 r. Probability of the point P which lies within the circle of radius
2
r 4 r r 1 is z z0 is 2 2 2 r 4
Sol.124. (A)
1 3
1 Required chance 5! 6! 3 2!
0
Sol.125. (A) 30 Given L : 3sin A 4cos B 6 & M : 4sinB 3cos A 1 in ABC ,So L M implies sin(A B) 12 2
2
0 1 C 300 or 1500 . Discard C 150 because for 2 0 0 3 30 . Hence 3sin A 4cos B 4 6 a contradiction.∴ C 30 2 1 1 a (B) c a sin x
sin C sin(1800 A B)
less than Sol.126. Put
x
1 t
so I
dx x x
2
a
2
reduces to
1 dt a 1 2 2 t a
.Hence I c
this value of C, A will be
1 1 a sin a x
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
d 2 y dy 2 sin x ey dx 2 dx
Sol.127. (A)
Differentiating the given relation w.r.to x, we get ey
dy cos x 0 ,Again dx
dy e y sin x 0 2 dx dx
3 2
L : sin a sin b
1 6 & M : cos a cos b 2 2
,So L M implies 2
While LM (using cos(a b) 0 ) gives sin(a b) 23 Sol.129. (C) e 3 e dy tan & tan So, f (a) 3 , f (b) 1 Given dy 3 4 dx dx b
d.w.r.to x
2
d2y
Sol.128. (C) Given
ey
2
cos(a b) 0
a
x a
b
x b
∴
b
b d x e f x dx e x f x eb f b ea f a eb 3e a a dx a
I e x f x f x dx 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 0 8 9 :b9 0 0 a 5:5 b
2:3 9 : b 9 R3 R2 a 4 : 4
For no solution rankA≠rank A b , So a 5 , b 5 Sol.131. (A) 5 x 2 3x 1 0 Required equation
1 1 1 1 x2 x 0
Where 3 & 5 as , are roots of
x 2 3x 5 0
5x 2 3x 1 0
h2 3a h 3
Sol.132. (D)
ah
Volume of the solid of revolution
V
y 2 dx
(The figure is bounded by x=a,x=a+h,y=0)
a ah
V
x
2
a 2 dx
a
Sol.133. (A)
h2 3a h 3
y 2 1 x log
c 1 1 x
Given diff. Eq. can be written as dy 1 x y y ,Let y t dx 2(x 1) 2(x 1) 2
dt 1 x t dx (x 1) (x 1)
2
where
I.F. e
so 2y 1
1 x dx
dy dt .Hence eq. reduces to dx dx
1 (x 1)
http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
UPSEE 2017 Paper I - SET AA Solutions
2
t.IF. Q.IF.dx c y 1 x log
Hence solution
c 1 1 x
Sol.134. (D) 10m / sec 2 x t 5t 2 7t 3 , v
dx 12 10t 7 5 10t 7 t dt 10
5 x 2 7 x 439
Sol.135. (C)
2x 10m / s 2 2 dt 12
; a d
t
10
=0
Obviously p,q satisfy the equation 5x 2 7 x 3 0 .Hence Given 5 p 4q & 5q 4 p. The required equation x 2 x 0
pq
5 x 2 7 x 439
7 3 , pq 5 5
=0
Sol.136. (B) 1 x 1 Let
2 2 1 sin x , given 3 sin 1 x sin 1 x 3 4 x 2 3 sin 1 sin 3 4 sin 2
3 ,Hence 2 2 6 6
Sol.137. (B) x
Sol.138.
∴ 12 sin 12
2
2
1 y 1
2
i.e
1 1 x 2 2
1 x y 32 8
2
1 y 1
Required ellipse x
x y 3 e 2
where
e
1 2
2 2 og e 2 e 1
(C)
2
2
ex dx Put 1 e x t x 1 e 0
Mean value M 2 1 0 1 2e dx
M
x
0
Sol.139. (A) oge
tan
,We get
2 M 2 og e 2 e 1
y x 2 sin c 4 2
dy x y x y x y sin sin 2 cos sin dx 2 2 2 2
.
So separating the variables and integrating oge
tan
y x 2 sin c 4 2
Sol.140. (C) - 9/2 2 x 9 2x 9 2x 9 R1 R2 R3
2
2x
2
7
6
2x
7 9 0 x 1, , 2 2
Sol.141. (D) a =1 , b = 0 a ib cos log i 4i cos 4i log i i 1 a 1,b 0
2
Sol.142. (B) increases in ( 0 , 1 ) but decreases in ( 1 , 2 ) y 2 x x2
so
0 for 0 x 1 dy 1 x dx 1 ( x 1) 2 0 for x 1, 2
So f increases in (0,1) & decreases in (1,2). http://www.myengg.com/engg/info/category/uptu/
www.myengg.com
Sol.143. (B)
5 3
5 3
as the line y=x intersect lines
Sol.144. (A) log
UPSEE 2017 Paper I - SET AA Solutions
5 5 5 5 2x 5 5 at points , & , 3 3 3 3
.
z 2 7
z 2 3
1 3 z 2 1 sin 6
,since
1 sin 1 6 2
So
z 2 3 1 z2 7 3 z 2 1 2
Sol.145. (C) 1 3 n 1 2
Let S Rewrite
n
1 4 13 40 121 364 ....... Tn 1 Tn Sn 1 4 13 40 121 364 ....... Tn 2 Tn 1 Tn
& Sn Sn 1 3 32 33 ....... Tn Tn 1 Tn Tn 1.
3n 1 3n 1 & Tn 3 1 2
Alternative: put options directly. Sol.146. (D) y x 2 sin x
5 , 3 3
has tangent parallel to x axis at the points
5 , 3 3
and
dy 5 0 for x 0 , U , 2 dx 3 3 dy 5 0 for x , dx 3 3
Sol.147. (D) 9 xC 21/ 3 x 6 31/ 3 6 6
1 x9 6 x 6 xC 1/ 3 1/ 3 x 6 2 3 6 Sol.148. (B) 256 If cardinality of A=m & Cardinality of B is n ,then total no. of relations from A to B is 2mn . Here m=4,n=2 ∴ 28 256
Sol.149. (A) 0 0 .00025 Using x 2 ,we get y 4 4 which is less than , So 4 2 For 0.001, the 0.00025
Sol.150. (D)
f 0
f (x) f (0) where f (0) 0(given),So L f 0 x 0 x 0
f 0 lim
http://www.myengg.com/engg/info/category/uptu/