JEE Advanced Physics Question Paper with Solution JEE Advanced 2021 Paper 1
A
3.07 cm
B
3.11 cm
C
3.15 cm
D
3.17 cm
Solution
So, we have given,
10VSD=9MSD⇒1VSD=910MSD
Least count =1MSD−1VSD=(1−910)MSD
⇒0.1MSD=0.1×0.1=0.01cm
As ' O ' of vernier scale lies before ' O ' of the main scale,
Zero error =−(10−6)×( Least Count )=−4×0.01=−0.04cm
Reading =3.1+1×( Least count )=3.1+1×0.01=3.11cm
Therefore, true diameter = Reading − Zero error
=3.11−(−0.04)=3.15cm
A
steps 1 and 2
B
steps 1 and 3
C
steps 1 and 4
D
steps 2 and 4
Solution
ΔQ1=nCpΔT>0
ΔQ2=nCVΔT<0
ΔQ3=nCpΔT<0
ΔQ4=nCVΔT>0
A
0.4
B
0.8
C
1.3
D
1.6
Solution
1f1=(1.5−1)(120+120)=120
1f2=(1.5−1)(−120−120)=−120
So, 1v−1−10=120
So, v=−20cm
and 1v′−1−30=1−20
So, v′=−12cm
So total magnification =(−20−10)(−12−30)=0.8
A
50
B
75
C
350
D
525
Solution
Total no. of decays in 60 minutes
=1000−1000(12)3=875
So, no. of α-decay =875×0.6=525
Answer: 0.50
Solution
After splitting 1st mass takes 0.5sec to reach ground.
Initial velocity is same for both mass at the highest point in vertical direction. Displacement and acceleration in vertical direction is also same
So, 2nd mass will also take 0.5sec to reach ground.
Answer: 7.50
Solution
Velocity of projectile at highest point 5m/sˆi
Since, there is no external force in horizontal direction so by conservation of momentum
m(5)=m2(0)+m2(v)
→v=10m/sˆi
Distance covered by second mass before landing = Range 2+10(t)=7.5m
Answer: 1.33
Solution
After long time we can replace the capacitor by open circuit.
i current in circuit =1/3 ampere
vAB=2−2(13)=43 volt
So, q1=CV=43μC
Answer: 0.67
Solution
After long time i circuit =23 ampere
vAB=2−2(23)=23 volt
So, q2=CV=(1)(23)=23μC
Answer: 1.73
Solution
Lets take two points (a,0) and (C,0) on equipotential circle.
Net potential at (C,0)=0
K(−q)C+Kq√3(C−2)=0
1C=1√3(C−2)
⇒√3C−2√3=C
⇒(√3−1)C=2√3
⇒C=2√3√3−1
Potential net at (a,0)=0
K(−q)a+Kq√3(2−a)=0
⇒1a=1√3(2−a)
⇒2√3−√3a=a
⇒a=2√31+√3
So, Radius =C−a2=2√3√3−1−2√3√3+12
=√3(1√3−1−1√3+1)=√3(√3+1−√3+13−1)
Radius =√3
Answer: 3.00
Solution
b=a+ radius
=2√3√3+1+√3=2√3+3+√31+√3
=3√3+31+√3=3
Center =(3,0)
A
For the same F, the value of a does not depend on whether the cylinder is solid or hollow
B
For a solid cylinder, the maximum possible value of a is 2μg
C
The magnitude of the frictional force on the object due to the ground is always μmg
D
For a thin-walled hollow cylinder, a=F2m
Solution
F−f=ma
fR=Iα (about center of mass)
a=Rα (pure rolling)
For hollow cylinder a=F2m,f=F2
For solid cylinder, a=2F3m,f=F3
Also for solid cylinder F2≤μmg
Therefore a ≤2μg
A
The light ray enters air if n2=n1
B
The light ray is finally reflected back into the medium of refractive index n1 if n2<n1
C
The light ray is finally reflected back into the medium of refractive index n1 if n2>n1
D
The light ray is reflected back into the medium of refractive index n1 if n2=1
Solution
The ray diagram for the following conditions are
The light finally must reflected back in medium of refractive index n1 for all values of n2.
A
The particle arrives at the point (x=ℓ,y=−h) at time t=2s.
B
→τ=2ˆk when the particle passes through the point (x=ℓ,y=−h)
C
→L=4ˆk when the particle passes through the point (x=ℓ,y=−h)
D
→τ=ˆk when the particle passes through the point (x=0,y=−h)
Solution
Time taken to reach from (−ℓ,−h) to (ℓ,−h) is given by
2ℓ=12at2
20=12(10)t2
t=2sec
→L=mvhˆk=mathˆk
→L=(0.2)(10)2(1)ˆk=4ˆk [ when particle passes through (ℓ,−h)]
→τ=d→Ldt=mahˆk=(0.2)(10)(1)ˆk=2ˆk (always)
A
The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9/5
B
There is an overlap between the wavelength ranges of Balmer and Paschen series
C
The wavelength of Lyman series are given by (1+1m2)λ0, where λ0 is the shortest wavelength of Lyman series and m is an integer
D
The wavelength ranges of Lyman and Balmer series do not overlap
Solution
For hydrogen atom, z=1
1λ=R(1n21−1n22),
where R=1.0973×107m−1
=1.1×107m−1= Rydberg constant.
For the Lyman series, n1=1 and n2=2,3,4,……∞
1λ=R(1−1n22)
λ=λmax when n _{2}=2
\frac{1}{\lambda_{\max }}=\frac{3 R }{4}
\Rightarrow \lambda_{\max }=\frac{4}{3 R }=121.5\, nm
\lambda=\lambda_{\min }, when n _{2}=\infty
\frac{1}{\lambda_{\min }}= R
\Rightarrow \lambda_{\min }=\frac{1}{ R }=91.1\, nm
Also, \lambda=\left(\frac{ n _{2}^{2}}{ n _{2}^{2}-1}\right) \frac{1}{ R }=\left[1+\frac{1}{\left( n _{2}^{2}-1\right)}\right] \lambda_{0}
=\left(1+\frac{1}{ m ^{2}}\right) \lambda_{0}
\lambda=\left(1+\frac{1}{ m ^{2}}\right) \lambda_{0}
where,
m ^{2}=\left( n _{2}^{2}-1\right)= an integer
m =\sqrt{ n _{2}^{2}-1}= not an integer
For the Balmer series, n _{1}=2 and n _{2}=3,4,5,6,........, \infty
\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{n_{2}^{2}}\right)
\lambda=\lambda_{\max }, when n _{2}=3
\frac{1}{\lambda_{\max }}=\frac{5 R }{36}
\lambda_{\max }=\frac{36}{5 R }=656.2\, nm
\lambda=\lambda_{\min }, when n _{2}=\infty
\Rightarrow \lambda_{\min }=\frac{4}{ R }=364.5\, nm
Hence, for the Balmer series,
\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{36 / 5 R }{4 / R }=\frac{9}{5}
For the Paschen series, n _{1}=3 and n _{2}=4,5,6, \ldots \ldots ., \infty
\frac{1}{\lambda}= R \left(\frac{1}{9}-\frac{1}{ n _{2}^{2}}\right)
\lambda=\lambda_{\max }, when n _{2}=4
\frac{1}{\lambda_{\max }}= R \left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7 R }{144}
\Rightarrow \lambda_{\max }=\frac{144}{7 R }=1874.7 \,nm
\frac{1}{\lambda_{\max }}= R \left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7 R }{144}
\Rightarrow \lambda_{\max }=\frac{144}{7 R }=1874.7\, nm
\lambda=\lambda_{\min }, when n _{2}=\infty
\frac{1}{\lambda_{\min }}=\frac{ R }{9}
\Rightarrow \lambda_{\min }=\frac{9}{ R }=820.2\, nm
A
Maximum current through R is 1.2 \times 10^{-6} ampere
B
Maximum current through R is 3.8 \times 10^{-6} ampere
C
Maximum charge on capacitor C_{0} is 8.4 \times 10^{-12} coulomb
D
Maximum charge on capacitor C_{0} is 2.4 \times 10^{-12} coulomb
Solution
Emf induced across the semi-circular conducting rod.
\varepsilon=\int\limits_{1}^{4} \frac{\mu_{0} I v d x}{2 \pi x}
=\frac{\mu_{0} I v}{2 \pi} \ln (4)=\frac{\mu_{0} I v}{\pi} \ln (2)
Since the semi-circular conducting rod is moving with a constant speed
v =3\, m / s,
then \varepsilon=\frac{\mu_{0} I v}{\pi} \ln (2)= constant
Maximum current through the resistor R is
i _{\max }=\frac{\varepsilon}{ R }=\frac{\mu_{0} Iv }{\pi R } \ln (2)
=\frac{4 \times 10^{-7} \times 2 \times 3 \times 0.7}{1.4}
=1.2 \times 10^{-6} ampere.
Maximum charge on the capacitor C _{0} is
q _{\max }= C _{0} \varepsilon= C _{0}\left(\frac{\mu_{0} Iv }{\pi} \ln (2)\right)
=5 \times 10^{-6} \times 4 \times 10^{-7} \times 2 \times 3 \times 0.7
=8.4 \times 10^{-12} coulomb.
A
\beta=0 when a=g / \sqrt{2}
B
\beta > 0 when a=g / \sqrt{2}
C
\beta=\frac{\sqrt{2}-1}{\sqrt{2}} when a = g / 2
D
\beta=\frac{1}{\sqrt{2}} when a = g / 2
Solution
\left(P_{1}-P_{2}\right) d s=\rho d s d \sqrt{2}(g \sin 45-a)
\left(P_{1}-P_{2}\right)=\rho d(g-a \sqrt{2})
\beta=\frac{\left(P_{1}-P_{2}\right)}{\rho g d}=\left(1-\frac{a \sqrt{2}}{g}\right)
When a=g / \sqrt{2}, \beta=0
When a=g / 2, \beta=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)
Answer: 4
Solution
r _{a} =\frac{\sqrt{2 m _{a} q _{a} V }}{ q _{a} B }
r _{ S } =\frac{\sqrt{2 m _{ S } q _{ s } V }}{ q _{ s } B }
\frac{ r _{ S }}{ r _{a}} =\sqrt{\frac{ m _{ s }}{ q _{ S }} \frac{ q _{a}}{ m _{ a }}}=\sqrt{\left(\frac{32}{1}\right)\left(\frac{2}{4}\right)}
\frac{ r _{ s }}{ r _{a}} =4
Answer: 49
Solution
Angular momentum of disc about O is
L _{ DO }= M \left(\frac{3 a }{4}\right)\left(\frac{3 a }{4}\right) \Omega+\frac{ M }{2}\left(\frac{ a }{4}\right)^{2}(4\, \Omega)
Angular momentum of rod about O is
L _{ RO }=\frac{ Ma ^{2}}{3} \Omega
So, L _{0}= L _{ DO }+ L _{ Ro }=\frac{49}{48}\left( Ma ^{2} \Omega\right)
So, n =49
Answer: 9
Solution
- C \frac{ dT }{ dt }=\left( T ^{4}- T _{ s }^{4}\right)
\int\limits_{200}^{100} \frac{ dT }{ T ^{4}- T _{ s }^{4}}=\int\limits_{0}^{ t }-\frac{1}{ c } dt
-\frac{1}{3}\left[\frac{1}{ T ^{3}}\right]_{200}^{100}=-\frac{1}{ c }\left( t _{1}\right)
\Rightarrow \left[\frac{1}{(100)^{3}}-\frac{1}{(200)^{3}}\right]=\frac{3}{ c } t _{1}
Similarly, \left[\frac{1}{(50)^{3}}-\frac{1}{(200)^{3}}\right]=\frac{3}{ c } t _{2}
\frac{ t _{2}}{ t _{1}}=\frac{\left[\frac{1}{(50)^{3}}-\frac{1}{(200)^{3}}\right]}{\left[\frac{1}{(100)^{3}}-\frac{1}{(200)^{3}}\right]}=9
JEE Advanced Chemistry Question Paper with Solution JEE Advanced 2021 Paper 1
A
B
C
D
Solution
Na in liquid NH_3 reduces non terminal alkyne into trans alkene.
A
B
C
D
Solution
In option (B), given configuration represents meso - butane - 2, 3-diol and due to intramolecular hydrogen
bonding, the gauche form is more stable.
Option (C) and (D) does not represent meso - isomer.
A
0.74
B
0.63
C
0.52
D
0.48
Solution
Packing fraction (f)=\frac{3 \times \frac{4}{3} \pi r_{+}^{3}+1 \times \frac{4}{3} \pi r_{-}^{3}}{a^{3}}
=\frac{1 \times \frac{4}{3} \pi\left[3\left(\frac{r_{+}}{r_{-}}\right)^{3}+1\right]}{\left(\frac{a}{r}\right)^{3}}
Now 2 r _{-}= a
\therefore \frac{a}{r_{-}}=2
Also, \frac{r_{\text {土 }}}{r_{-}}=0.414
So, f=\frac{1 \times \frac{4}{3} \times 3.14\left[3 \times(0.414)^{3}+1\right]}{(2)^{3}}
=0.634 \approx 0.63
A
3.87 and 2.84
B
4.90 and 1.73
C
3.87 and 1.73
D
4.90 and 2.84
Solution
\left[ Cr \left( NH _{3}\right)_{6}\right]^{3+} has d ^{3} configuration, so as per CFT,
N =3 and \mu=\sqrt{3(3+2)}=3.87 \,BM
\left[ CuF _{6}\right]^{3-}, has d ^{8} configuration and weak field ligand.
So N =2 and \mu=\sqrt{2(2+2)}=2.84\, BM
Answer: 1.62
Solution
Molar mass of P = 40
\underset{(0.1 \text{mole})}{ P} \xrightarrow{75\%} \underset{(0.1 \times \frac{3}{4})}{Q}
\underset{(0.1 \times \frac{3}{4})}{3Q} \xrightarrow{40\%} \underset{(\frac{1}{3} \times 0.1 \times \frac{3}{4} \times 0.4)}{R}
So, moles of R = 0.01 mole
Molar mass of (R) = 162
So, x = 0.01 \times 162 = 1.62\,g
Answer:
Solution
Molar mass of ' U ' =122 \,g or 100 \,g
\underset{(0.1 \text { mole })}{ P } \xrightarrow{100 \%} \underset{(0.1 \text { mole })}{ S }
\underset{(0.1\text{mole})}{2S} \xrightarrow[(80\%)]{\text{Aldol condensation}} \underset{(\frac{0.1}{2} \times 0.8)}{T} \xrightarrow{80\%} \underset{(\frac{0.1}{2} \times 0.8\times .8)}{U}
So, mass of 'U' =\frac{0.1}{2} \times 0.8 \times 0.8 \times 100=3.20\, gm
Or
Mass of ' U ' =\frac{0.1}{2} \times 0.8 \times 0.8 \times 122=3.90\, gm
Answer: 166.28
Solution
X ( s ) \rightleftharpoons Y ( s )+ Z ( g )
K _{ p }=\frac{ p _{ z }}{ p ^{0}}, also \Delta G ^{0}=- RT \ln k _{ p }
=- RT \ln \left(\frac{ p _{ z }}{ p ^{0}}\right)
Now, \Delta G ^{0}=\Delta H ^{0}- T \Delta S ^{0}
- RT kn \left(\frac{ p _{z}}{ p ^{0}}\right)=\Delta H ^{0}- T \Delta S ^{0}
\ln \left(\frac{p_{z}}{p^{0}}\right)=-\left(\frac{\Delta H^{0}}{R}\right) \frac{1}{T}+\frac{\Delta S^{0}}{R} ....(1)
(1) \Rightarrow \ln \left(\frac{p_{z}}{p^{0}}\right)=-\left(\frac{\Delta H^{0}}{10^{4} R}\right) \times \frac{10^{4}}{T}+\frac{\Delta S^{0}}{T} \ldots(2)
Slope of the line =-\frac{\Delta H ^{0}}{10^{4} R }
=\frac{[-7-(-3)]}{12-10}=-2
\therefore \Delta H ^{0}=2 R \times 10^{4}
=2 \times 8.314 \times 10^{-3} \times 10^{4}
=1.66 .28\, kJ \,mol ^{-1} K ^{-1}
Answer: 141.34
Solution
X ( s ) \rightleftharpoons Y ( s )+ Z ( g )
K _{ p }=\frac{ p _{ z }}{ p ^{0}}, also \Delta G ^{0}=- RT \ln k _{ p }
=- RT \ln \left(\frac{ p _{ z }}{ p ^{0}}\right)
Now, \Delta G ^{0}=\Delta H ^{0}- T \Delta S ^{0}
- RT kn \left(\frac{ p _{z}}{ p ^{0}}\right)=\Delta H ^{0}- T \Delta S ^{0}
\ln \left(\frac{p_{z}}{p^{0}}\right)=-\left(\frac{\Delta H^{0}}{R}\right) \frac{1}{T}+\frac{\Delta S^{0}}{R} ....(1)
(1) \Rightarrow \ln \left(\frac{p_{z}}{p^{0}}\right)=-\left(\frac{\Delta H^{0}}{10^{4} R}\right) \times \frac{10^{4}}{T}+\frac{\Delta S^{0}}{T} \ldots(2)
Putting the value of \Delta H ^{0} in equation (2), we get
-3=-\left(\frac{2 R \times 10^{4}}{10^{4} R }\right) \times \frac{10^{4}}{7}+\frac{\Delta S ^{0}}{ R }
-3=-2 R \times \frac{10^{4}}{ T }+\frac{\Delta S ^{0}}{ R }
-3=-2 \times \frac{10^{4}}{1000}+\frac{\Delta S ^{0}}{ R }
-3=-20+\frac{\Delta S ^{0}}{ R }
\therefore \frac{\Delta S ^{0}}{ R }=17
\therefore \Delta S ^{0}=17 \times 8.314
=141.34 \,J\,K ^{-1} mol ^{-1}
Answer: 100.10
Solution
0.1 molal AgNO _{3} (aq) solution
AgNO _{3} \longrightarrow Ag ^{+}( aq )+ NO _{3}{ }^{-}( aq )
i =1+(2-1) \times 1=2(\alpha=1, given )
\Delta T _{ b }= i \times k _{ b } \times m
\Delta T _{ b }=2 \times 0.5 \times 0.1=0.1
So, boiling point of solution ' A ' is =100.10^{\circ} C = x
Answer: 2.5
Solution
Let solution ' B ' is prepared by mixing 1 \,L ( =1000\, g ) of solution 'A' with 1 \,L (= 1000\, g ) of solution of BaCl _{2}.
So, molality of new solution =\left(\frac{ i _{1} \times m _{1}+ i _{2} \times m _{2}}{2}\right)
=\left(\frac{3 \times 0.05+3 \times 0.05}{2}\right)
=0.15
Now, Elevation of boiling point of solution ' B ' be \left(\Delta T_{b}{ }^{1}\right)
\Delta T _{ b }^{1}=0.15 \times k _{ b }
=0.15 \times \frac{1}{2}
=0.075
Now, T _{ b }{ }^{1}=100.075{ }^{\circ} C
So, difference of boiling point of ' A ' and ' B '
=100.10-100.075=0.025= y \times 10^{-2} (given)
So, y =2.5
A
B
C
D
Solution
Since, we have to get the product (x) of (\alpha)_{0}=-52.7^{0}, i.e. the enantiomer of above product. Which is only possible from (C) \& (D).
A
B
C
D
Solution
In option (B) and (C), NO _{2} group (an EWG) is not present ortho or para position wrt the leaving group, so ArSN ^{2} reaction will not be possible.
A
The process of precipitating colloidal sol by an electrolyte is called peptization.
B
Colloidal solution freezes at higher temperature than the true solution at the same concentration.
C
Surfactants form micelle above critical micelle concentration (CMC). CMC depends on temperature.
D
Micelles are macromolecular colloids.
Solution
(A) Process of precipitating colloidal solution by using an electrolyte is called "COAGULATION" and not peptisation.
(B) Since, molar mass of sol is much higher than true solutions, so magnitude of any colligative properties is smaller than true solutions.
\left(\Delta T _{ f }\right)_{\text {sol }}<\left(\Delta T _{ f }\right)_{\text {true solution }}
So, freezing point of sols > freezing point of true solution.
So, option (B) is correct.
(C) Micells are formed greater than or equal to CMC and above KRAFT temperature.
So option (C) is also correct.
(D) Micelles are ASSOCIATED colloids and not Macromolecular colloids.
A
B
C
D
Solution
(I) \xrightarrow[\text{expansion}]{\text{Reversible isotherm}} (II) \xrightarrow[\text{expansion}]{\text{Reversible adiabatic}} (III)
So, the correct option is/are:
Reversible isothermal process is isoenthalpic while reversible adiabatic process is isoentropic.
A
A mixture of PbS and PbO undergoes self-reduction to produce Pb and SO_2.
B
In the extraction process of copper from copper pyrites, silica is added to produce copper silicate.
C
Partial oxidation of sulphide ore of copper by roasting, followed by self-reduction produces blister copper.
D
In cyanide process, zine powder is utilized to precipitate gold from Na \left[ Au ( CN )_{2}\right].
Solution
(A) PbS +2 PbO \xrightarrow{\text { self reduction }}3 Pb + SO _{2} \uparrow
So option (A) is correct.
(B) In the extraction Cu from copper pyrite CuFeS _{2}
SiO _{2} is added to remove FeO as slag FeSiO _{3}.
So option (B) is wrong.
(C) \underset{\text { concentrated ore }}{ CuF e S _{2}} \xrightarrow{\text { roasting }} \underbrace{ Cu _{2} S + FeS }_{\text {(matte) }}+ SO _{2}
FeS + O _{2} \longrightarrow FeO + SO _{2}
Cu _{2} S + O _{2} \longrightarrow Cu _{2} O + SO _{2}
Cu _{2} S +2 Cu _{2} O \longrightarrow \underset{\text { (Bister copper) }}{6 Cu }+ SO _{2}
So, option (C) is correct.
(D) Zn +2 Na \left[ Au ( CN )_{2}\right] \rightarrow Na _{2}\left[ Zn ( CN )_{4}\right]+2 Au \downarrow
So, option (D) is correct.
A
Pb \left( NO _{3}\right)_{2} and Zn \left( NO _{3}\right)_{2}
B
Pb \left( NO _{3}\right)_{2} and Bi \left( NO _{3}\right)_{3}
C
AgNO _{3} and Bi \left( NO _{3}\right)_{3}
D
Pb \left( NO _{3}\right)_{2} and Hg \left( NO _{3}\right)_{2}
Solution
Pb ( OH )_{2}, Zn ( OH )_{2} and Bi ( OH )_{3} are white precipitates but Hg ( OH )_{2} (unstable) is not. PbCl _{2} is white ppt.
So, option (A), (B) and (C) are correct.
Answer: 6
Solution
Possible structural isomers are:
JEE Advanced Mathematics Question Paper with Solution JEE Advanced 2021 Paper 1
A
x^{2}+y^{2}-3 x+y=0
B
x^{2}+y^{2}+x+3 y=0
C
x^{2}+y^{2}+2 y-1=0
D
x^{2}+y^{2}+x+y=0
Solution
Image of orthocentre about any side of the triangle lies on its circumcircle.
We can observe that
x^{2}+y^{2}+x+3 y=0 is satisfied by (-1,0) and (1,-1) both
So, option (B) is the correct choice
A
\frac{11}{32}
B
\frac{35}{96}
C
\frac{37}{96}
D
\frac{13}{32}
Solution
Area of shaded region
=\frac{1}{2}\left(\frac{2}{3}+\frac{3}{4}\right) \frac{1}{4}+\frac{1}{2} \times \frac{2}{3} \times \frac{1}{2}
=\frac{11}{32}
A
\frac{1}{5}
B
\frac{3}{5}
C
\frac{1}{2}
D
\frac{2}{5}
Solution
S_1
F_2 = F_1 \cup S_1
S_2
G_1 \cup S_2 = G_2
S_3
(i)
\{1, 2\}
\{1, 2, 3, 4\}
\{1, x\}
\{1,2,3,4,5\}
\{1,2\}
(ii)
\{2, 3\}
\{1,2,3,4\}
\{1,x\}
\{1,2,3,4,5\}
\{2,3\}
\{x, y\} (where x and y
or \{2,3,4,5\}
(iii)
\{1,3\}
\{1,3,4\}
\{1, x\} are other than 1)
\{1,2,3,4,5\}
\{1,3\}
(i) P_{1}=\frac{{ }^{2} c_{2}}{{ }^{3} c_{2}} \cdot \frac{{ }^{3} c_{1}}{{ }^{4} c_{2}} \cdot \frac{{ }^{2} c_{1}}{{ }^{5} c_{2}}=\frac{1}{60}
(ii) P_{2}=\frac{1}{3}\left(\frac{{ }^{3} c _{1}}{{ }^{4} c _{2}} \times \frac{{ }^{2} c _{2}}{{ }^{5} c _{2}}+\frac{{ }^{3} c _{2}}{{ }^{4} c _{2}} \times \frac{{ }^{2} c _{2}}{{ }^{4} c _{2}}\right)=\frac{2}{45}
(iii) P_{3}=\frac{1}{{ }^{3} c_{1}} \times \frac{{ }^{2} c_{1}}{{ }^{3} c_{2}} \times \frac{{ }^{2} c_{2}}{{ }^{5} c_{2}}=\frac{1}{45}
Conditional probability =\frac{P_{1}}{P_{1}+P_{2}+P_{3}}=\frac{1}{5}
S_1 | F_2 = F_1 \cup S_1 | S_2 | G_1 \cup S_2 = G_2 | S_3 | |
---|---|---|---|---|---|
(i) | \{1, 2\} | \{1, 2, 3, 4\} | \{1, x\} | \{1,2,3,4,5\} | \{1,2\} |
(ii) | \{2, 3\} | \{1,2,3,4\} | \{1,x\} | \{1,2,3,4,5\} | \{2,3\} |
\{x, y\} (where x and y | or \{2,3,4,5\} | ||||
(iii) | \{1,3\} | \{1,3,4\} | \{1, x\} are other than 1) | \{1,2,3,4,5\} | \{1,3\} |
A
P is TRUE and Q is FALSE
B
Q is TRUE and P is FALSE
C
both P and Q are TRUE
D
both P and Q are FALSE
Solution
\because z_{1}=e^{i \theta_{1}}
So, z_{2}=e^{i\left(\theta_{1}+\theta_{2}\right)}
z_{3}=e^{i\left(\theta_{1}+\theta_{2}+\theta_{3}\right)}
:
z _{10}= e ^{ i \left(\theta_{1}+\theta_{2}+\ldots \ldots+\theta_{10}\right)}= e ^{ i (2 \pi)}
Sum of all the chord length < Circumference
So, \sum\left|z_{2}-z_{1}\right| \leq 2 \pi
Also, 2\left|z_{2}-z_{1}\right| \geq\left|z_{2}^{2}-z_{1}^{2}\right|
Hence, 2\left(\left|z_{2}-z_{1}\right|+\ldots . .+\left|z_{10}-z_{1}\right|\right) \leq 2(2 \pi)=4 \pi
So for, we have P \leq 2 \pi and Q \leq 4 \pi
Answer: 76.25
Solution
P_{1}=1 - (Probability that 3 chosen numbers are less than 81 )
=1-\left(\frac{80}{100}\right)^{3}=1-\frac{64}{125}=\frac{61}{125}
So, \frac{625}{4}
P_{1}=\frac{625}{4} \times \frac{61}{125}=76.25
Answer: 24.5
Solution
P_{2}=1-(Probability that 3 chosen numbers are greater than 40 )
=1-\left(\frac{60}{100}\right)^{3}=1-\left(\frac{3}{5}\right)^{3}=\frac{98}{125}
So, \frac{125}{4}
P_{2}=\frac{125}{4} \times \frac{98}{125}=24.5
Answer: 1
Solution
x+2 y+3 z=\alpha
4 x+5 y+6 z=\beta
7 x+8 y+9 z=\gamma-1
Equation (1)+(3)-(2)=0. Equation (2) provides
\alpha+\gamma-1-2 \beta=0
|M|=\begin{vmatrix}\alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1\end{vmatrix}\left(C_{1} \rightarrow C_{1}+C_{3}\right)
=\begin{vmatrix}\alpha+\gamma & 2 & \gamma \\ \beta & 1 & 0 \\ 0 & 0 & 1\end{vmatrix}\left(R_{1} \rightarrow R_{1}-2 R_{2}\right)
\begin{vmatrix}\alpha+\gamma-2 \beta & 0 & \gamma \\ \beta & 1 & 0 \\ 0 & 0 & 1\end{vmatrix}=\begin{vmatrix}1 & 0 & \gamma \\ \beta & 1 & 0 \\ 0 & 0 & 1\end{vmatrix}=1
Answer: 1.5
Solution
x+2 y+3 z=\alpha
4 x+5 y+6 z=\beta
7 x+8 y+9 z=\gamma-1
Equation (1)+(3)-(2)=0. Equation (2) provides
\alpha+\gamma-1-2 \beta=0
Plane P is x-2 y+z-1=0
D=\left(\frac{|-2-1|}{\sqrt{1+4+1}}\right)^{2}=1.5
Answer: 9
Solution
Locus C=\left|\frac{(x \sqrt{2}+y-1)(x \sqrt{2}-y+1)}{\sqrt{3}}\right|=\lambda^{2}
2 x^{2}-(y-1)^{2}=\pm 3 \lambda^{2}
for intersection with y=2 x+1
2 x^{2}-(2 x)^{2}=\pm 3 \lambda^{2}
-2 x^{2}=-3 \lambda^{2} (taking - ve sign )
x=\pm \sqrt{\frac{3}{2}} \lambda
Distance between R and S =2\left|\sqrt{\frac{3}{2}} \lambda\right| \sec \theta (\tan \theta is slope of line)
=\sqrt{6}|\lambda| \sqrt{5}
So, \sqrt{30}|\lambda|=\sqrt{270}(\lambda=\pm 3)
\lambda^{2}=9
Answer: 77.14
Solution
Equation of perpendicular bisector y=-\frac{1}{2} x+1
For point of intersection 2 x^{2}-\frac{1}{4} x^{2}=\pm 3 \lambda^{2}
x =\pm \sqrt{\frac{12}{7}} \lambda (taking + ve sign)
Distance =\left|2 \cdot \sqrt{\frac{12}{7}} \cdot 3 \cdot \sec \theta\right|
=2 \cdot \sqrt{\frac{12}{7}} \cdot 3 \cdot \sqrt{\frac{5}{2}}
=3 \cdot \sqrt{\frac{60}{7}}
D =\frac{9 \times 60}{7}=77.14
A
F = PEP and P ^{2}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
B
\left| EQ + PFQ ^{-1}\right|=| EQ |+\left| PFQ ^{-1}\right|
C
\left|( EF )^{3}\right|>| EF |^{2}
D
Sum of the diagonal entries of P^{-1} E P+F is equal to the sum of diagonal entries of E+P^{-1} F P
Solution
Let A =\begin{bmatrix} C _{1} & C _{2} & C _{3}\end{bmatrix}, B =\begin{bmatrix} R _{1} \\ R _{2} \\ R _{3}\end{bmatrix}
AP =\begin{bmatrix} C _{1} & C _{3} & C _{2}\end{bmatrix} and PB =\begin{bmatrix} R _{1} \\ R _{3} \\ R _{2}\end{bmatrix}
and P ^{2}= I
P ( EP )= P \begin{bmatrix}1 & 3 & 2 \\ 2 & 4 & 3 \\ 8 & 18 & 13\end{bmatrix}=\begin{bmatrix}1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3\end{bmatrix}= F
| E |=\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18\end{bmatrix}\left( R _{3} \rightarrow R _{3}-3 R _{2}-2 R _{1}\right)=\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 0 & 0 & 0\end{bmatrix}=0
\Rightarrow | F |=0
\left| EQ + PFQ ^{-1}\right|=\left| EQ + PPEPQ ^{-1}\right|=\left| EQ + EPQ ^{-1}\right|=| E |\left| Q + PQ ^{-1}\right|=0
| EQ |=| E || Q |=0,\left| PFQ ^{-1}\right|=| P || F |\left| Q ^{-1}\right|=0
(D) P ^{-1} EP + F = PEP + F =2 F \left(\right. as \left.P ^{-1}= P \right)
E + P ^{-1} FP = E + P ^{-1} PEPP =2 E ( trace ( E )= trace ( F ))
A
f is decreasing in the interval (-2,-1)
B
f is increasing in the interval (1,2)
C
f is onto
D
Range of f is \left[-\frac{3}{2}, 2\right]
Solution
f^{\prime}(x)=\frac{5 x(x+4)}{\left(x^{2}+2 x+4\right)^{2}}
f(0)=-\frac{3}{2} (point of local minima)
f(-4)=\frac{11}{6} (point of local maxima)
A
P\left(E \cap F \cap G^{C}\right) \leq \frac{1}{40}
B
P\left(E^{C} \cap F \cap G\right) \leq \frac{1}{15}
C
P(E \cup F \cup G) \leq \frac{13}{24}
D
P\left(E^{C} \cap F^{C} \cap G^{C}\right) \leq \frac{5}{12}
Solution
P(E)=\frac{1}{8}, P(F)=\frac{1}{6}, P(G)=\frac{1}{4}, P(E \cap F \cap G)=\frac{1}{10}
(A) P\left(E \cap F \cap G^{C}\right)=P(E \cap F)-P(E \cap F \cap G)
\leq P ( E )- P ( E \cap F \cap G )
\leq \frac{1}{8}-\frac{1}{10}
\leq \frac{5-4}{40}
\leq \frac{1}{40}
(B) P\left(E^{C} \cap F \cap G\right)=P(F \cap G)-P(E \cap F \cap G)
\leq P(F)-P(E \cap F \cap G)
\leq \frac{1}{6}-\frac{1}{10}
\leq \frac{10-6}{60}
\leq \frac{4}{60}
\leq \frac{1}{15}
(C) P ( E \cup F \cup G ) \leq P ( E )+ P ( E )+ P ( G )
\leq \frac{1}{8}+\frac{1}{6}+\frac{1}{4}
\leq \frac{13}{24}
(D) P\left(E^{C} \cap F^{C} \cap G^{C}\right)=1-P(E \cup F \cup G)
\geq 1-\frac{13}{24}
\geq \frac{11}{24}>\frac{10}{24}>\frac{5}{12}
A
| FE |=| I - FE || FGE |
B
( I - FE )( I + FGE )= I
C
EFG = GEF
D
( I - FE )( I - FGE )= I
Solution
G ( I - EF )=( I - EF ) G = I
\Rightarrow G - GEF = G - EFG = I ... (1)
(A) | FE |=| I - FE || FGE |=| FGE - FE FGE |
=| FGE - F ( G - I ) E |=| FGE - FGE + FE |=| FE |
(B) ( I - FE )( I + FGE )= I + FGE - FE - FEFGH
= I + FGE - FE - F ( G - I ) E = I + FGE - FE - FGE + FE = I
(C) From (I) it is true
(D) ( I - FE )( I - FGE )= I - FGE - FE + FEFGE
= I - FGE - FE + F ( G - I ) E = I - FGE - FE + FGE - FE = I -2 FE
A
S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^{2}}{10 x }\right), for all x >0
B
\displaystyle\lim _{n \rightarrow \infty} \cot \left(S_{n}(x)\right)=x, for all x>0
C
The equation S _{3}( x )=\frac{\pi}{4} has a root in (0, \infty)
D
\tan \left(S_{n}(x)\right) \leq \frac{1}{2}, for all n \geq 1 and x>0
Solution
S_{n}=\displaystyle\sum_{k=1}^{n} \cot ^{-1}\left\{\frac{1+k(k+1) x^{2}}{x}\right\} ;
\displaystyle\sum_{k=1}^{n} \cot ^{-1}\left\{\frac{k(k+1) x^{2}+1}{(k+1) x-k x}\right\}
S_{n}=\displaystyle\sum_{k=1}^{n} \cot ^{-1}(k x)-\cot ^{-1}(k+1) x
t_{1}=\cot ^{-1}(x)-\cot ^{-1}(2 x)
t_{2}=\cot ^{-1}(2 x)-\cot ^{-1}(3 x)
t_{3}=\cot ^{-1}(3 x)-\cot ^{-1}(4 x)
:
t_{n}=\cot ^{-1}(n x)-\cot ^{-1}((n+1) x)
S_{n}=\cot ^{-1}(x)-\cot ^{-1}((n+1) x)
\Rightarrow S_{n}=\cot ^{-1}\left(\frac{(n+1) x^{2}+1}{n x}\right)
S_{10}=\cot ^{-1}\left(\frac{11 x^{2}+1}{10 x}\right)=\frac{\pi}{2}-\tan ^{-1}\left(\frac{11 x^{2}+1}{10 x}\right)
(B) \displaystyle\lim _{n \rightarrow \infty} \cot \left(S_{n}(x)\right)=\displaystyle\lim _{n \rightarrow \infty} \cot \left(\cot ^{-1}\left(\frac{(n+1) x^{2}+1}{n x}\right)\right)
=\displaystyle\lim _{n \rightarrow \infty} \frac{n x^{2}+x^{2}+1}{n x}=x
(C) S_{3}(x)=\cot ^{-1}\left(\frac{4 x^{2}+1}{3 x}\right)=\frac{\pi}{4} \Rightarrow \frac{1+4 x^{2}}{3 x}=1
\Rightarrow 4 x^{2}-3 x+1=0 have imaginary roots
(D) \tan \left(S_{n}(x)\right)=\tan \left(\cot ^{-1}\left(\frac{1+(n+1) x^{2}}{n x}\right)\right)
=\frac{n x}{1+(n+1) x^{2}}=\frac{1}{\frac{1}{n x}+\frac{(n+1) x}{n}}
A
\alpha=-1
B
\alpha \beta=4
C
\alpha \beta=-4
D
\beta=4
Solution
w=c+i d, \arg (w) \in(-\pi, \pi]
\arg \left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4} ; z_{0}=-\frac{5}{2}+\frac{3}{2} i
z_{0}+\alpha=\left(z_{0}+\beta\right) i ; z_{0}+\alpha=z_{0} i+\beta i ; z_{0}(1-i)=\beta i-\alpha (by Rotation)
-\frac{5}{2}+\frac{3}{2} i+\alpha=\left(-\frac{5}{2}+\frac{3}{2} i\right) i+\beta i
-\frac{5}{2}+\frac{3}{2} i+\alpha=-\frac{5}{2} i-\frac{3}{2}+\beta i
-\frac{5}{2}+\frac{3}{2}+\frac{3}{2} i+\frac{5}{2} i+\alpha=\beta i ;-1+4 i=\beta i-\alpha( As \alpha, \beta are real number )
\Rightarrow -\alpha=-1, \alpha=1 ; \beta=4
Answer: 4
Solution
3 x^{2}-4\left|x^{2}-1\right|+x-1=0
Case-I: |x| \geq 1
3 x^{2}-4 x^{2}+4+x-1=0
-x^{2}+x+3=0
x^{2}-x-3=0
x=\frac{1 \pm \sqrt{1+12}}{2}=\frac{1 \pm \sqrt{13}}{2}=\frac{1+\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2}
|x| < 1
3 x^{2}+4 x^{2}-4+x-1=0
7 x^{2}+x-5=0
x=\frac{-1 \pm \sqrt{1+20 \times 7}}{14}=\frac{-1 \pm \sqrt{141}}{14}
So, number of real roots =4
Answer: 2
Solution
A B=\sqrt{23}=C
B C=3=a
C A=4=b
\frac{\cot A+\cot C}{\cot B} ; \frac{\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}}
=\frac{\cos A \sin C+\cos C \sin A}{\sin A \cdot \sin C \cdot \frac{\cos B}{\sin B}}=\frac{\sin (A+C) \cdot \sin B}{\sin A \cdot \sin C \cdot \cos B}
=\frac{\sin B \cdot \sin B}{\sin A \cdot \sin C \cdot \cos B}
=\frac{b^{2}}{a c \cdot \frac{\left(a^{2}+c^{2}-b^{2}\right)}{2 a c}}=\frac{2 b^{2}}{a^{2}+c^{2}-b^{2}}
=\frac{2 \times 16}{9+23-16}=\frac{2 \times 16}{32-16}
=\frac{2 \times 16}{16}=2
Answer: 7