JEE Advanced Question Paper with Solution JE Advanced 2022 Paper 2

JEE Advanced Physics Question Paper with Solution JE Advanced 2022 Paper 2

Answer: 3

Solution

Torque about origin is zero So angular momentum about origin remains conserved.
|ijk1220222π|=|ijkxy0.5vxvy2π|ˆi[2×2π]ˆj[2π]+ˆk[1+2]=i[y×2π0.5vy]ˆj[x×2π0.5vx]+k[xvyyvvx]xvyyvx=3

Answer: 2

Solution

Th23090Po21484+n42+mβ01
230=214+4n
n=164=4
90=84+n×2m×1
90=84+4×2m×1
m=9290=2
Hence nm=42=2 Ans.

Answer: 5

Solution

For the balanced Wheatstone bridge
R1R2=0.50ρdxπr2x10.5ρdxπr2x
R1R2=+[1rx]0.50+[1rx]10.5

Answer: 4

Solution

B = e ^\alpha\left( m _{ e }\right)^\beta h ^\gamma k ^\delta
{[ B ]=\left[ e ^\alpha\right]\left[ m _{ e }\right]^\beta[ h ]^\gamma\left[ k ^\delta\right]}
{\left[ M ^1 T ^{-2} A ^{-1}\right]=[ AT ]^\alpha\left[ m ^\beta\right]^{[}\left[ ML ^2 T ^{-1}\right]^\gamma\left[ ML ^3 A ^{-2} T ^{-4}\right]^\delta}
M ^1 T ^{-2} A ^{-1}= m ^{\beta+\gamma+\delta} L ^{2 r +3\delta} T ^{\alpha-\gamma-4\delta} A ^{\alpha-2 \delta}
Compare : \beta+\gamma+\delta=1 ; 2 \gamma+3 \delta=0, \alpha-\gamma-4 \delta=-2, \alpha-2 \delta=-1
On solving \alpha=3, \beta=2, \gamma=-3, \delta=2
\alpha+\beta+\gamma+\delta=4

Answer: 4

Solution

image
1 \sin 60^{\circ}=\sqrt{\frac{3}{2}} \sin \theta
\Rightarrow \theta_1=45^{\circ}
\sqrt{\frac{3}{2}} \sin 45^{\circ}=\sqrt{3} \sin \theta_2
=\sqrt{\frac{3}{2}} \frac{1}{\sqrt{2}}=\sqrt{3} \sin \theta_2
=\theta_2=30^{\circ}
h \tan 60^{\circ}+d \tan 45^{\circ}+d \tan 30^{\circ}
\frac{1}{3} \sqrt{3}+\left(\frac{\sqrt{3}-1}{2}\right)+\left(\frac{\sqrt{3}-1}{2}\right) \frac{1}{\sqrt{3}}
\frac{2 \sqrt{3}+3 \sqrt{3}-3+3-\sqrt{3}}{6}
\frac{4 \sqrt{3}}{6}
\therefore n \frac{4 \sqrt{3}}{6}=\frac{8}{\sqrt{3}}
n = 4

Answer: 3

Solution

From Gauss law,
\phi_{\text {hemisphere }}+\phi_{\text {Cone }}=\frac{ q }{\varepsilon_0} \ldots . . (i)
Total flux produced from q in \alpha angle
\phi=\frac{ q }{2 \varepsilon_0}[1-\cos \alpha]
For hemisphere, \alpha=\frac{\pi}{2}
\phi_{\text {hemisphere }}=\frac{ q }{2 \varepsilon_0}
From equation (i)
=\frac{ q }{2 \varepsilon_0}+\phi_{\text {cone }}=\frac{ q }{\varepsilon_0}
\phi_{\text {cone }}=\frac{ q }{2 \varepsilon_0}
\frac{4 q }{6 \varepsilon_0}=\frac{ q }{2 \varepsilon_0}
n =3
Alternatively, \phi \propto no of electric field lines passing through surface q is point charge which has uniformly distributed electric field lines thus half of electric field lines will pass through hemisphere & other half will pass through conical surface.

Answer: 6

Solution

image
\ell>\ell_0 \rightarrow k = k _1
\ell<\ell_0 \rightarrow k = k _2
Time period of oscillation,
T =\pi \sqrt{\frac{ m }{ k _1}}+\pi \sqrt{\frac{ m }{ k _2}}
T =\pi \sqrt{\frac{0.1}{0.009}}+\pi \sqrt{\frac{0.1}{0.016}}
T =\frac{\pi}{0.3}+\frac{\pi}{0.4} \Rightarrow T =\frac{0.7}{0.12} \pi \Rightarrow T =5.83 \pi
T \approx 6 \pi
So, n =6

Answer: 3

Solution

image
u =-30 \,cm
f =-10 \,cm
v =\frac{ f _0}{ u - f }=-15 \, cm
\frac{1}{ v }+\frac{1}{ u }=\frac{1}{ f }
\frac{ du }{ dt }=-\frac{ v ^2}{ u ^2} \frac{ du }{ dt }
\overrightarrow{ v }_{ lm }=-\left(\frac{ v }{ u }\right)^2 \overrightarrow{ v }_{ om }
Given \vec{v}_1=\overrightarrow{0}
\overrightarrow{ v }_1-\overrightarrow{ v }_{ m }=-\left(\frac{-15}{-30}\right)^2\left(\overrightarrow{ v }_{ O / m }\right)
\overrightarrow{ v }_1-\overrightarrow{ v }_{ m }=-\frac{1}{4} \overrightarrow{ v }_0+\frac{1}{4} \overrightarrow{ v }_{ m }
\overrightarrow{ v }_0=15 \, cm / s \hat{ i }
\overrightarrow{ v }_{ I }=\overrightarrow{0 } cm / s
\frac{5}{4} \overrightarrow{ v }_{ m }=\frac{\overrightarrow{ v }_0}{4}
\overrightarrow{ v }_{ m }=\frac{\overrightarrow{ v }_0}{4}=\frac{15 cm / s \hat{ i }}{5}=3 m / s \hat{ i }
\left|\overrightarrow{ v }_{ m }\right|_{ m } cm / s =3

A

If r_B=\sqrt{\frac{3}{2}}, then the electric field is zero everywhere outside B.

B

If r_B=\frac{3}{2}, then the electric potential just outside B is \frac{k}{\epsilon_0}.

C

If r_B=2, then the total charge of the configuration is 15 \pi k.

D

If r_B=\frac{5}{2}, then the magnitude of the electric field just outside B is \frac{13 \pi k}{\epsilon_0}.

Solution

q _1=\int\limits_0^1 kr 4 \pi r ^2 dr =\frac{4 \pi k }{4}=\pi k
q _2=\int\limits_1^{ r } \frac{2 k }{ r } 4 \pi r ^2 dr =\frac{8 \pi k \left( r ^2-1^2\right)}{2}
q _2=4 \pi k \left[ r ^2-1\right]=4 \pi kr r ^2-4 \pi k
q _{ net }= q _1+ q _2
=4 \pi kr ^2-3 \pi k
q _{\text {net }}=\pi k \left[4 r ^2-3\right]
(A) E _{\text {net }}=0 \Rightarrow q _{\text {net }}=0 \Rightarrow r =\frac{\sqrt{3}}{2}
(B) V =\frac{ kQ _{ net }}{ r }=\frac{1}{4 \pi \varepsilon_0} \frac{\pi k \left(4 r ^2-3\right)}{ r }
V =\frac{ k }{4 \varepsilon_0}\left[4 r -\frac{3}{ r }\right]
=\frac{ k }{4 \varepsilon_0}\left[4 \times \frac{3}{2}-\frac{3 \times 2}{3}\right]=\frac{ k }{\varepsilon_0}
(C) q _{ net }=\pi k \left[4(2)^2-3\right]
=13 \pi k
(D) E _2 =\frac{ kQ }{ r ^2}
=\frac{1}{4 \pi \varepsilon_0} \frac{\pi k \left(4 r ^2-3\right)}{ r ^2}
=\frac{ k }{4 \varepsilon_0}\left[\frac{4\left(\frac{5}{2}\right)^2-3}{(5 / 2)^2}\right]
=\frac{ k }{25 \varepsilon_0}[25-3]=\frac{22}{25} \frac{ k }{\varepsilon_0}

A

When a voltage source of 6 V is connected across A and B in both circuits, P_1 < P_2.

B

When a constant current source of 2 Amp is connected across A and B in both circuits, P_1>P_2.

C

When a voltage source of 6 V is connected across A and B in Circuit-1, Q_1>P_1.

D

When a constant current source of 2 Amp is connected across A and B in both circuits, Q_2 < Q_1

Solution

Case (i)
When both switches are open equivalent resistance in circuit 1
R _{ C _1}=\frac{16}{11} \Omega
Equivalent resistance in circuit 2
R _{ C _2}=\frac{6}{11} \Omega
For voltage source
P =\frac{ V ^2}{ R }
P \propto \frac{1}{ R }
R _{ C _1}> R _{ C _2}
\Rightarrow P _2> P _1 \text { (Option (A) correct) }
For constant current source
P=i^2 R
P \propto R
\Rightarrow P_1>P_2 \text { (Option (B) correct) }
Case-II
When switch is closed
R _{ C _1}^{\prime}=\frac{5}{11} \Omega
R _{ C _2}^{\prime}=\frac{1}{2} \Omega
R _{ C _1}^{\prime} < R _{ C _1}
For voltage source
P \propto \frac{1}{ R } \Rightarrow Q _1> P _1(\text { Option }( C ) \text { correct) }
\& R _{ C _1}^{\prime}> R _{ C _2}^{\prime}
For current source P \propto R
Q _1> Q _2

A

If the surface of the bubble is a perfect heat insulator, then \left(\frac{r_1}{r_2}\right)^5=\frac{P_{a 2}+\frac{2 S}{r_2}}{P_{a 1}+\frac{2 S}{r_1}}

B

If the surface of the bubble is a perfect heat insulator, then the total internal energy of the bubble including its surface energy does not change with the external atmospheric pressure.

C

If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible, then \left(\frac{r_1}{r_2}\right)^3=\frac{P_{a 2}+\frac{4 S}{r_2}}{P_{a 1}+\frac{4 S}{r_1}}.

D

If the surface of the bubble is a perfect heat insulator, then \left(\frac{T_2}{T_1}\right)^{\frac{5}{2}}=\frac{P_{a 2}+\frac{4 S}{r_2}}{P_{a 1}+\frac{4 S}{r_1}}.

Solution

image
P_{ gas }=P_{ a }+\frac{4 S}{r}
PV ^{\gamma}= constant [adiabatic process]
\left( Pa _1+\frac{4 S }{ r _1}\right)\left(\frac{4}{3} \pi r _1^3\right)^{5 / 3}=\left( P _{ a _2}+\frac{4 S }{ r _2}\right)\left(\frac{4}{3} \pi r _2^3\right)^{5 / 3}
\frac{r_1^3}{r_2^3}=\left(\frac{P_{a_2}+\frac{4 S}{r_2}}{P_{a_1}+\frac{4 S}{r_1}}\right)
P^{1-y} T^y= constant
\left(P_{a_2}+\frac{4 S}{r_2}\right)^{1-5 / 3} T_2^{5 / 3}=\left(P_{a_1}+\frac{4 S}{r_1}\right)^{1-5 / 3} T_1^{5 / 3}
\left(\frac{T_2}{T_1}\right)^{5 / 3}=\left(\frac{P_{a_1}+\frac{4 S}{r_1}}{P_{a_2}+\frac{4 S}{r_2}}\right)^{-2 / 3}
\left(\frac{T_2}{T_1}\right)^{5 / 2}=\left(\frac{P_{a_2}+\frac{4 S}{r_2}}{P_{a_1}+\frac{4 S}{r_1}}\right)

A

For \beta=\frac{1}{4} and z_0=\frac{25}{7} R, the particle reaches the origin.

B

For \beta=\frac{1}{4} and z_0=\frac{3}{7} R, the particle reaches the origin.

C

For \beta=\frac{1}{4} and z_0=\frac{R}{\sqrt{3}}, the particle returns back to z=z_0.

D

For \beta>1 and z_0>0, the particle always reaches the origin.

Solution

W _{ el }+ W _{ ext }= k _{ f }- k _{ i }
qv _{ i }- qv _{ f }+ W _{ ext }= k _{ f }- k _{ i }
\frac{ q \sigma}{2 \epsilon_0}\left[\sqrt{ R ^2+ Z ^2}- Z \right]-\frac{ q \sigma R }{2 \epsilon_0}+ CZ = k _{ f }-0
C =\frac{ q \sigma B }{2 \epsilon_0}
Substitute \beta \& Z, calculate kinetic energy at Z =0
If kinetic energy is positive, then particle will reach at origin
If kinetic energy is negative, then particle will not reach at origin.

A

The phase difference between the two rays is independent of d.

B

The two rays interfere constructively at the detector.

C

The phase difference between the two rays depends on n _1 but is independent of n_2.

D

The phase difference between the two rays vanishes only for certain values of d and the angle of incidence of the beam, with \theta being the corresponding angle of refraction.

Solution

image
Optical path difference \rightarrow
\Delta x = n _1( d \tan \theta) \sin \alpha- n _2( d \tan \theta) \sin \theta
=\left( n _1 \sin \alpha- n _2 \sin \theta\right) d \tan \theta
=0
\Rightarrow \Delta \phi=0

A

The magnitude of the total work done in the process A \rightarrow B \rightarrow C is 144 \,kJ.

B

The magnitude of the work done in the process B \rightarrow C is 84\, kJ.

C

The magnitude of the work done in the process A \rightarrow B is 60 \,kJ.

D

The magnitude of the work done in the process C \rightarrow A is zero.

Solution

For adiabatic process ( A \rightarrow B )
P _{ A } V _{ A }^\gamma= P _{ B } V _{ B }^\gamma
10^5 \times(0.8)^{\frac{5}{3}}=3 \times 10^5\left( V _{ B }\right)^{\frac{5}{3}}
\Rightarrow V _{ B }=0.8 \times\left(\frac{1}{3}\right)^{0.6}=0.4
Work done in process A \rightarrow B
W _{ AB }=\frac{ P _{ A } V _{ A }- P _{ B } V _{ B }}{\gamma-1}
\Rightarrow W _{ AB }=\frac{10^5 \times 0.8-3 \times 10^5 \times 0.4}{\frac{5}{3}-1}
\Rightarrow W _{ AB }=-60 kJ =\Rightarrow\left| W _{ AB }\right|=60 kJ
Work done in process B \rightarrow C (Isothermal process)
W _{ BC }= nRT \ell n \frac{ V _{ C }}{ V _{ B }}= P _{ B } V _{ B } \ell n \frac{ V _{ C }}{ V _{ B }}
\Rightarrow W _{ BC }=3 \times 10^5 \times 0.4 \ell n \frac{0.8}{0.4}
\Rightarrow W _{ BC }=84 kJ
Work done in process C \rightarrow A
W _{ CA }= P \Delta V =0 (\because \Delta V =0)
So total work done in the process A \rightarrow B \rightarrow C
W _{ ABC }= W _{ AB }+ W _{ BC }+ W _{ CA }=-60+84+0
W _{ ABC }=24 kJ
So correct options are (B,C,D)

A

2

B

5

C

\frac{7}{2}

D

\frac{9}{2}

Solution

v=\omega(2 R)
v=\omega_0 R: no slipping
\therefore \omega_0=2 \omega
image
\overrightarrow{ L }= m \overrightarrow{ r } \times \overrightarrow{ v }_{ c }+ I _{ c } \omega_0
= M 2 Rv +\frac{1}{2} MR ^2 \omega_0
=4 MR ^2 \omega+\frac{1}{2} MR ^2(2 \omega)=5 MR ^2 \omega
\therefore n =5

A

1.72 \times 10^{-7} m , 1.20 \, eV

B

1.72 \times 10^{-7} m , 5.60 \,eV

C

3.78 \times 10^{-7} m , 5.60 \,eV

D

3.78 \times 10^{-7} m , 1.20 \, eV

Solution

\frac{ hc }{\lambda}=\phi+6 \ldots \text { (i) }
\frac{ hc }{4 \lambda}=\phi+0.6 \ldots \text { (ii) }
\frac{3 hc }{4 \lambda}=5.4 eV \therefore \phi=1.2 eV
\Rightarrow \frac{3}{4} \times \frac{6.63 \times 10^{-24} \times 3 \times 10^8}{5.4 \times 1.6 \times 10^{-19}}=\lambda=1.72 \times 10^{-7} m

A

2.22 \pm 0.02 mm , \pi(1.23 \pm 0.02) mm ^2

B

2.22 \pm 0.01 mm , \pi(1.23 \pm 0.01) mm ^2

C

2.14 \pm 0.02 mm , \pi(1.14 \pm 0.02) mm ^2

D

2.14 \pm 0.01 mm , \pi(1.14 \pm 0.01) mm ^2

Solution

LC =\frac{0.1}{100}=0.001 \,mm
Zero error =4 \times 0.001=0.004 \,mm
Reading 1=0.5 \times 4+20 \times 0.001-0.004=2.16 \,mm
Reading 2=0.5 \times 4+16 \times 0.001-0.004=2.12\, mm
Mean value =2.14 mm
Mean absolute error =\frac{0.02+0.02}{2}=0.02
Diameter =2.14 \pm 0.02
Area =\frac{\pi}{4} d ^2

A

\vec{B}=\frac{-\mu_0 I}{L}\left(\frac{3}{2}+\frac{1}{4 \sqrt{2} \pi}\right) \hat{k}

B

\vec{B}=-\frac{\mu_0 I}{L}\left(\frac{3}{2}+\frac{1}{2 \sqrt{2} \pi}\right) \hat{k}

C

\vec{B}=\frac{-\mu_0 I}{L}\left(1+\frac{1}{4 \sqrt{2} \pi}\right) \hat{k}

D

\vec{B}=\frac{-\mu_0 I}{L}\left(1+\frac{1}{4 \pi}\right) \hat{k}

Solution

\overrightarrow{ B }=\frac{\mu_0 I }{4 \pi L } \sin 45^{\circ}(-\hat{ k })+\frac{\mu_0 I \pi}{4 \pi \frac{ L }{2}}(-\hat{ k })+\frac{\mu_0 I }{4 \pi \frac{ L }{4}} \times \frac{\pi}{2}(-\hat{ k })

JEE Advanced Chemistry Question Paper with Solution JE Advanced 2022 Paper 2

Answer: 6

Solution

image
image
K _{ a _2}=1.2 \times 10^{-2}=\frac{(1- x )\left(1.8 \times 10^{-2}- x \right)}{(1+ x )}
Since x is very small (1+x) \simeq 1 and (1-x) \simeq 1
x =\left(1.8 \times 10^{-2}-1.2 \times 10^{-2}\right) M
\left[ SO _4^{2-}\right]=\left(1.8 \times 10^{-2}-0.6 \times 10^{-2}\right) M
=1.2 \times 10^{-2} M
image
K _{ sp }= s \left( s +1.2 \times 10^{-2}\right)=1.6 \times 10^{-8}
\left( PbSO _4\right)
Here, \left( s +1.2 \times 10^{-2}\right) \simeq 1.2 \times 10^{-2}( since ' s ' is very small)
s\left(1.2 \times 10^{-2}\right)=1.6 \times 10^{-8}
\Rightarrow s=\frac{1.6}{1.2} \times 10^{-6} M = X \times 10^{- Y } M
\Rightarrow Y =6

Answer: 5

Solution

0.1 mole ionic salt in 1.8 \,kg water at 35^{\circ} C
Vapour pressure of solution =59.724\, mm of Hg
Vapour pressure of pure H _2 O =60.000\, mm of Hg
Let the number of ions present per formula unit of the ionic salt be ' x
'
image
Moles of water =\frac{1.8 \times 10^3}{18}=100 moles
Relative lowering of vapour pressure \frac{ P ^{\circ}- P _s}{ P ^{\circ}}= Mole fraction of non - volatile particles
\frac{ P ^{\circ}- P _s}{ P _{ s }}=\frac{\text { moles of non }-\text { volatileparticles }}{\text { moles of water }}
\frac{60.000-59.724}{59.724}=\frac{0.01+0.09 x }{100}
(0.276) \times 100=0.59274+(0.59274 \times 9) x
27.6-0.59274=(0.59274 \times 9) x
\Rightarrow x \simeq \frac{27}{0.6 \times 9}=5

Answer: 7

Solution

\Lambda^{\circ}\left(U_{ m } Y _{ p }\right) = m \times \lambda_{ u ^p}^0+ p \times \lambda_{ Y ^{+-}}^0=250
25 m +100 p =250
m +4 p =10....(1)
\Lambda^{\circ}\left(V_{ m } X _{ n }\right)= m \times \lambda_{ v ^{2+}}+ n \times \lambda_{ x ^{m-}}^0=440
100 m +80 n =440
5 m +4 n =22....(2)
image
From the extrapolation of curve
\Lambda^{\circ}\left(Z_{ m } X _{ n }\right)=340
m \times \lambda_{ z ^{n+}}^{\circ}+ n \lambda_{ x ^{m-}}^{\circ}=340
50 m +80 n =340
5 m +8 n =34.....(3)
(3) - (2) \Rightarrow 4 n =12 \Rightarrow n =3
Putting in (2) we get m =2
Putting in (1) we get p =2
m + n + p =2+3+2=7

Answer: 4

Solution

Xe +2 O _2 F _2 \rightarrow XeF _4+2 O _2
3 XeF _4+6 H _2 O \rightarrow 2 Xe + XeO _3+\frac{3}{2} O _2+12 HF
\therefore One mole of XeF _4 gives 4 moles of HF on hydrolysis.

Answer: 6

Solution

AgNO _3 \rightarrow 2 Ag +2 NO _2+1 / 2 O _2
- Both NO _2 \& O _2 are paramagnetic
- NO _2 is odd electron molecule with one unpaired electron
- O _2 has two unpaired electrons
image
Total number of antibonding electrons =6

A

If empirical formula of compound 3 is P _3 Q _4, then the empirical formula of compound 2 is P _3 Q _5.

B

If empirical formula of compound 3 is P _3 Q _2 and atomic weight of element P is 20 , then the atomic weight of Q is 45 .

C

If empirical formula of compound 2 is PQ, then the empirical formula of the compound 1 is P _5 Q _4

D

If atomic weight of P and Q are 70 and 35 , respectively, then the empirical formula of compound 1 is P _2 Q.

Solution

Compound Weight % of P Weight % of Q
1 50 50
2 44.4 55.6
3 40 60

For option (A)
Let atomic mass of P be M _{ P } and atomic mass of Q be M _{ Q } Molar ratio of atoms P : Q in compound 3 is
\frac{40}{M_p}: \frac{60}{M_Q}=3: 4
\frac{2 M_Q}{3 M_p}=\frac{3}{4} \Rightarrow 9 M_p=8 M_Q
Molar ratio of atoms P : Q in compound 2 is
\frac{44.4}{M_p}: \frac{55.6}{M_Q}
=44.4 M _{ Q }: 55.6 M _{ P }
=44.4 M _{ Q }: 55.6 \times \frac{8 M _{ Q }}{9}
=44.4: 55.6 \times \frac{8}{9}
=9: 10
\Rightarrow Empirical formula of compound 2 is therefore P _9 Q _{10}
Option (A) in incorrect
For option (B)
Molar Ratio of atoms P: Q in compound 3 is \frac{40}{ M _{ p }}: \frac{60}{ M _{ Q }}=3: 2
\frac{2 M _{ Q }}{3 M _{ p }}=\frac{3}{2} \Rightarrow 9 M _{ p }=4 M _{ Q }
If M _{ P }=20 \Rightarrow M _{ Q }=\frac{9 \times 20}{4}=45
Option (B) is correct
For option (C)
Molar ratio of atoms P : Q in compound 2 is
\frac{44.4}{M_p}: \frac{55.6}{M_Q}=44.4 M _Q: 55.6 M _{ p }=1: 1
\Rightarrow \frac{ M _{ p }}{ M _{ Q }}=\frac{44.4}{55.6}
Molar ratio of atoms P : Q in compound 1 is
\frac{50}{ M _{ p }}: \frac{50}{ M _{ Q }}= M _{ Q }: M _{ p }
=55.6: 44.4
\simeq 5: 4
Hence, empirical formula of compound 1 is P _5 Q _4
Hence, option ( C ) is correct
For option (D)
Molar ratio of atoms P : Q in compound 1 is
\frac{50}{ M _{ p }}: \frac{50}{ M _{ Q }} = M _{ Q }: M _{ p }
=35: 70=1: 2
Hence, empirical formula of compound 1 is PQ _2
Hence, option (D) is incorrect

A

For the reaction, M (s)+2 H ^{+}(a q) \rightarrow H _2(g)+ M ^{2+}(a q), if \frac{ dE _{\text {cell }}}{ dT }=\frac{ R }{ F }, then the entropy change of the reaction is R (assume that entropy and internal energy changes are temperature independent).

B

The cell reaction, Pt (s)\left| H _2(g, 1 bar )\right| H ^{+}(a q, 0.01 M ) \| H ^{+}(a q, 0.1 M )\left| H _2(g, 1 bar )\right| Pt (s), is an entropy driven process.

C

For racemization of an optically active compound, \Delta S >0.

D

\Delta S >0, for \left[ Ni \left( H _2 O \right)_6\right]^{2+}+3 en \rightarrow\left[ Ni ( en )_3\right]^{2+}+6 H _2 O (where en = ethylenediamine).

Solution

\Delta G =\Delta H - T \Delta S
\Delta G =\Delta H + T \left(\frac{ d \Delta G }{ dT }\right)_{ p }
- nF \left(\frac{ dE _{\text {cell }}}{ dT }\right)=-\Delta S
\left.\frac{ dE _{\text {cell }}}{ dT }=\frac{\Delta S }{ nF }=\frac{ R }{ F } \text { (given }\right)
\Rightarrow \Delta S = nR
For the reaction, M ( g )+2 H ^{\oplus}( aq ) \longrightarrow H _2( g )+ M ^{2 \oplus}( aq )
n =2
\Rightarrow \Delta S =2 R
Hence, option (A) is incorrect
For the reaction, Pt _{( s )} \mid H _{2( g )}, 1 bar \left| H ^{\oplus}{ }_{ aq }(0.01 M )\right|\left| H ^{\oplus}( aq , 0.1 M )\right| H _2( g , 1 bar ) \mid Pt _{( s )}
E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.0591}{1} \log \frac{0.01}{0.1}=0.0591 V
E_{cell} is positive \Rightarrow \Delta G <0 and \Delta S >0(\Delta H =0 for concentration cells )
Hence, option (B) is correct
Racemization of an optically active compound is a spontaneous process.
Here, \Delta H =0 (similar type of bonds are present in enantiomers)
\Rightarrow \Delta S >0
Hence, option (C) is correct.
\left[ Ni \left( H _2 O \right)_6\right]^{2+}+3 en \rightarrow\left[ Ni ( en )_3\right]^{2+}+6 H _2 O is a spontaneous process
more stable complex is formed
\Rightarrow \Delta S >0

A

B

B

B _2 H _6

C

B _2 O _3

D

HBF _4

Solution

(A) 2 B +2 NH _3 \rightarrow 2 BN +3 H _2
Boron produced BN with ammonia but Boron is element not compound. So that this option not involve in answer.
(B) 3 B _2 H _6+6 NH _3 \rightarrow 3\left[ BH _2\left( NH _3\right)_2\right]^{+}\left[ BH _4^{-}\right] \xrightarrow{ T =200^{\circ} C } 2 B _3 N _3 H _6+12 H _2 B _3 N _3 H _6 \xrightarrow{ T >200^{\circ} C }( BN )_{ x }
(C) B _2 O _3(\ell)+2 NH _3 \xrightarrow{1200^{\circ} C } 2 BN _{( s )}+3 H _2 O _{( g )}
(D) HBF _4+ NH _3 \rightarrow NH _4\left[ BF _4\right]

A

Limestone is used to remove silicate impurity.

B

Pig iron obtained from blast furnace contains about 4 \% carbon.

C

Coke (C) converts CO _2 to CO.

D

Exhaust gases consist of NO _2 and CO.

Solution

(A) CaO + SiO _2 \rightarrow CaSiO _3 (in the temperature range 900-1500 K )
(B) In fusion zone molten iron becomes heavy by absorbing elemental impurities and produces Pig iron. (in the temperature range 900-1500 K )
(C) C + CO _2 \rightarrow 2 CO (in the temperature range 900-1500 K )
(D) Exhaust gases does not contain NO _2.

A

Compounds P and Q are carboxylic acids.

B

Compound S decolorizes bromine water.

C

Compounds P and S react with hydroxylamine to give the corresponding oximes.

D

Compound R reacts with dialkylcadmium to give the corresponding tertiary alcohol.

A

The polymerization of chloroprene gives natural rubber.

B

Teflon is prepared from tetrafluoroethene by heating it with persulphate catalyst at high pressures.

C

PVC are thermoplastic polymers.

D

Ethene at 350-570 K temperature and 1000-2000 atm pressure in the presence of a peroxide initiator yields high density polythene.

Solution

(a) The polymerisation of neoprene gives natural rubber.
(b) is correct statement
(c) is correct statement
(d) Ethene at 350-570 \,K temperature and 1000-2000 atm pressure in the pressure of a peroxide initiator yields low density polythene.

A

25

B

35

C

55

D

75

Solution

Atom ' X ' occupies FCC lattice points as well as alternate tetrahedral voids of the same lattice \Rightarrow \frac{1}{4} th distance of body diagonal
=\frac{\sqrt{3} a}{4}=2 r_x
\Rightarrow a =\frac{8 r_x}{\sqrt{3}}
Number of atoms of X per cell
\underset{\text{(FCC lattice points)}}{=4} + \underset{\text{(Alternate tetrahedral voids)}}{4 } = 8
\% packing efficiency =\frac{\text { Volume occupied by } X }{\text { Volume of cubic unit cell }} \times 100
=\frac{8 \times \frac{4}{3} \pi\left( r _{ x }\right)^3}{ a ^3} \times 100
=\frac{8 \times \frac{4}{3} \pi\left( r _{ x }\right)^3}{\left(\frac{\left.8 r _{ x }\right)^3}{\sqrt{3}}\right)} \times 100
=\left(8 \times \frac{4}{3} \times \pi \times \frac{1}{8^3} \times 3 \sqrt{3}\right) \times 100
=\frac{\sqrt{3} \pi}{16} \times 100
=34 \%

A

Cl _2 O

B

ClO _2

C

Cl _2 O _6

D

Cl _2 O _7

Solution

HClO _3+ HCl \rightarrow \underset{\text { (Paramagnetic) }}{ ClO _2}+\frac{1}{2} Cl _2+ H _2 O
2 ClO _2+2 O _3 \rightarrow Cl _2 O _6+2 O _2

A

B

C

D

Solution

Basic catalyse tautomerism through enediol intermediate

JEE Advanced Mathematics Question Paper with Solution JE Advanced 2022 Paper 2

Answer: 1

Solution

\alpha \in\left(0, \frac{\pi}{4}\right), \beta \in\left(-\frac{\pi}{4}, 0\right) \Rightarrow \alpha+\beta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)
\sin (\alpha+\beta)=\frac{1}{3}, \cos (\alpha-\beta)=\frac{2}{3}
\left(\frac{\sin \alpha}{\cos \beta}+\frac{\cos \alpha}{\sin \beta}+\frac{\cos \beta}{\sin \alpha}+\frac{\sin \beta}{\cos \alpha}\right)^2
\left(\frac{\cos (\alpha-\beta)}{\cos \beta \sin \beta}+\frac{\cos (\beta-\alpha)}{\sin \alpha \cos \alpha}\right)^2
=4 \cos ^2(\alpha-\beta)\left(\frac{1}{\sin 2 \beta}+\frac{1}{\sin 2 \alpha}\right)^2
=4 \cos ^2(\alpha-\beta)\left(\frac{2 \sin (\alpha+\beta) \cos (\alpha-\beta)}{\sin 2 \alpha \sin 2 \beta}\right).....(1)
=\frac{16 \cos ^4(\alpha-\beta) \sin ^2(\alpha+\beta) \times 4}{(\cos 2(\alpha-\beta)-\cos 2(\alpha+\beta))^2}
=\frac{64 \cos ^4(\alpha-\beta) \sin ^2(\alpha+\beta)}{\left(2 \cos ^2(\alpha-\beta)-1-1+2 \sin ^2(\alpha+\beta)\right)^2}
=64 \times \frac{16}{81} \times \frac{1}{9} \frac{1}{\left(2 \times \frac{4}{9}-1-1+\frac{2}{9}\right)^2}
=\frac{64 \times 16}{81 \times 9} \cdot \frac{81}{64}=\frac{16}{9}
{\left[\frac{16}{9}\right]=1 \text { Ans. }}

Answer: 8

Solution

x d y-\left(y^2-4 y\right) d x=0, x>0
\int \frac{d y}{y^2-4 y}=\int \frac{d x}{x}
\int\left(\frac{1}{y-4}-\frac{1}{y}\right) d y=4 \int \frac{d x}{x}
\log _c|y-4|-\log _c|y|=4 \log _e x+\log _e c
\frac{| y -4|}{| y |}= cx ^4 \xrightarrow{(1,2)} c =1
|y-4|=|y| x^4
C -1
y-4=y x^4
y=\frac{4}{1-x^4}
y(1)=\text { ND (rejected) }
C-2
y-4=-y x^4
y=\frac{4}{1+x^4}
y(1)=2
y (\sqrt{2})=\frac{4}{5} \Rightarrow 10 y (\sqrt{2})=8

Answer: 5

Solution

f ( x )=\log _2\left( x ^3+1\right)= y
x ^3+1=2^y \Rightarrow x =\left(2^y-1\right)^{1 / 3}= f ^{-1}( y )
f ^{-1}( x )=\left(2^{ x }-1\right)^{1 / 3}
=\int_1^2 \log _2\left( x ^3+1\right) dx +\int\limits_1^{\log _2 9}\left(2^x-1\right)^{1 / 3} dx
=\int_1^2 f ( x ) dx +\int\limits_1^{\log _2 9} f ^{-1}( x ) dx =2 \log _2 9-1
=8<9<2^{7 / 2} \Rightarrow 3<\log _2 9<\frac{7}{2}
=5<2 \log _2 9-1<6
{\left[2 \log _2 9-1\right]=5}

Answer: 1

Solution

x^{16\left(\log _5 x\right)^3-68 \log _5 x}=5^{-16}
Take \log to the base 5 on both sides and put \log _5 x=t
16 t ^4-68 t ^2+16=0
\Rightarrow 4 t ^4-17 t ^2+4=0 \begin{cases} t _1 \\t _2 \\t _3 \\t _4\end{cases}
t _1+ t _2+ t _3+ t _4=0
\log _5 x _1+\log _5 x _2+\log _5 x _3+\log _5 x _4=0
x_1 x_2 x_3 x_4=1

Answer: 5

Solution

\beta=\displaystyle\lim _{x \rightarrow 0} \frac{e^{x^3}-\left(1-x^3\right)^{1 / 3}}{\frac{x \sin ^2 x}{x^2} x^2}+\frac{\left(\left(1-x^2\right)^{1 / 2}-1\right) \sin x}{x \frac{\sin ^2 x}{x^2} x^2}
use expansion
\beta=\displaystyle\lim _{x \rightarrow 0} \frac{\left(1+x^3\right)-\left(1-\frac{x^3}{3}\right)}{x^3}+\displaystyle\lim _{x \rightarrow 0} \frac{\left(\left(1-\frac{x^2}{2}\right)-1\right)}{x^2} \frac{\sin x}{x}
\beta=\displaystyle\lim _{x \rightarrow 0} \frac{4 x^3}{3 x^3}+\displaystyle\lim _{x \rightarrow 0} \frac{-x^2}{2 x^2}
\beta=\frac{4}{3}-\frac{1}{2}=\frac{5}{6}
6 \beta=5

Answer: 3

Solution

A=\begin{pmatrix}\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2\end{pmatrix}|A|=-1
\Rightarrow\left|A^7-(\beta-1) A^6-\beta A^5\right|=0
\Rightarrow|A|^5\left|A^2-(\beta-1) A-\beta I\right|=0
\Rightarrow|A|^5\left|\left(A^2-\beta A \right)+A-\beta I\right|=0
\Rightarrow|A|^5|A(A-\beta I)+I(A-\beta I)|=0
|A|^5|(A+I)(A-\beta I)|=0
A+I=\begin{pmatrix}\beta+1 & 0 & 1 \\ 2 & 2 & -2 \\ 3 & 1 & -1\end{pmatrix} \Rightarrow|A+I|=-4,
Here | A | \neq 0 \&|A+I| \neq 0
A-\beta I=\begin{pmatrix}0 & 0 & 1 \\ 2 & 1-\beta & -2 \\ 3 & 1 & -2-\beta\end{pmatrix}
|A-\beta I|=2-3(1-\beta)=3 \beta-1=0 \Rightarrow \beta=\frac{1}{3}
9 \beta=3

Answer: 7

Solution

image
S P- PP =20
\beta-\frac{\delta}{\sin \frac{\alpha}{2}}=20
\beta^2+\frac{\delta^2}{\sin ^2 \frac{\alpha}{2}}-400=\frac{2 \beta \delta}{\sin \frac{\alpha}{2}}
\frac{1}{ SP }=\frac{\sin \frac{\alpha}{2}}{\delta}
\cos \alpha=\frac{ SP ^2+\beta^2-656}{2 \beta \frac{\delta}{\sin \frac{\alpha}{2}}}
=\frac{\frac{2 \beta \delta}{\sin \frac{\alpha}{2}}-256}{\frac{2 \beta S}{\sin \frac{\alpha}{2}}}=\cos \alpha
\frac{\lambda-128}{\lambda}=\cos \alpha
\lambda(1-\cos \alpha)=128
\frac{\beta \delta}{\sin \frac{\alpha}{2}} \cdot 2 \sin ^2 \frac{\alpha}{2}=128
\frac{\beta \delta}{9} \sin \frac{\alpha}{2}=\frac{64}{9} \Rightarrow\left[\frac{\beta \delta}{9} \sin \frac{\alpha}{2}\right]=7 where [.] denotes greatest integer function

Answer: 6

Solution

x^2+\frac{5}{12}=\frac{2-8 x}{3}
x^2+\frac{8 x}{3}+\frac{5}{12}-2=0
image
12 x^2+32 x-19=0
12 x^2+38 x-6 x-19=0
2 x(6 x+19)-1(6 x+19)=0
(6 x+19)(2 x-1)=0
x=\frac{1}{2}
\alpha=2 A _1+ A _2
\alpha=2\left(\int\limits_0^{1 / 2} x ^2+\frac{5}{12} dx +\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}\right)
\Rightarrow \alpha=2\left[\left(\frac{ x ^3}{3}+\frac{5 x }{12}\right)_0^{1 / 2}+\frac{1}{12}\right]
\Rightarrow \alpha=2\left[\frac{1}{24}+\frac{5}{24}+\frac{1}{12}\right]
\Rightarrow \alpha=2\left[\frac{1+5+2}{24}\right] \Rightarrow \alpha=2 \times \frac{8}{24} \Rightarrow 9 \alpha=9 \times \frac{8}{12}
\Rightarrow 9 \alpha=6

A

(0, \sqrt{2})

B

(1,2)

C

(\sqrt{2}, 3)

D

(2 \sqrt{2}, 3 \sqrt{2})

Solution

\angle PRQ =70^{\circ}-40^{\circ}=30^{\circ}
\angle RQS =70^{\circ}-15^{\circ}=55^{\circ}
\angle QSR =180^{\circ}-55^{\circ}-70^{\circ}=55
\therefore QR = RS =1
\angle QPR =180^{\circ}-70^{\circ}-30^{\circ}=80^{\circ}
image
Apply sine-rule in \triangle PRQ :
\frac{\alpha}{\sin 30^{\circ}}=\frac{1}{\sin 80^{\circ}} \Rightarrow \alpha=\frac{1}{2 \sin 80^{\circ}} .....(1)
Apply sine-rule in \triangle PRS
\frac{\beta}{\sin 40^{\circ}}=\frac{1}{\sin \theta} \Rightarrow \beta \sin \theta=\sin 40^{\circ}.....(2)
4 \alpha \beta \sin \theta=\frac{4 \sin 40^{\circ}}{2 \sin 80^{\circ}}=\frac{4 \sin 40^{\circ}}{2\left(2 \sin 40^{\circ} \cos 40^{\circ}\right)}
=\sec 40^{\circ}
Now \sec 30^{\circ} < \sec 40^{\circ} < \sec 45^{\circ}
\Rightarrow \frac{2}{\sqrt{3}} < \sec 40^{\circ} < \sqrt{2}

A

The minimum value of g(x) is 2^{\frac{7}{6}}

B

The maximum value of g(x) is 1+2^{\frac{1}{3}}

C

The function g( x ) attains its maximum at more than one point

D

The function g( x ) attains its minimum at more than one point

Solution

\alpha=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^6+\ldots
\alpha=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}
\therefore g ( x )=2^{ x / 3}+2^{1 / 3(1- x )}
\therefore g ( x )=2^{ x / 3}+\frac{2^{1 / 3}}{2^{x / 3}}
where g(0)=1+2^{1 / 3} \& g (1)=1+2^{1 / 3}
\therefore g ^{\prime}( x )=\frac{1}{3}\left(2^{ x / 3}-\frac{2^{1 / 3}}{2^{x / 3}}\right)=0
\Rightarrow 2^{2 x / 3}=2^{1 / 3} \Rightarrow x =\frac{1}{2}= critical point
image

A

\left(\frac{43+3 \sqrt{205}}{2}\right)^{\frac{1}{4}}

B

\left(\frac{7+\sqrt{33}}{4}\right)^{\frac{1}{4}}

C

\left(\frac{9+\sqrt{65}}{4}\right)^{\frac{1}{4}}

D

\left(\frac{7+\sqrt{13}}{6}\right)^{\frac{1}{4}}

Solution

Let (\overline{ z })^2+\frac{1}{ z ^2}= m + in , m , n \in Z
(\overline{ z })^2+\frac{\overline{ z }^2}{| z |^4}= m + in
\Rightarrow\left( x ^2- y ^2\right)\left(1+\frac{1}{| z |^4}\right)= m ....(1)
\&-2 xy \left(1+\frac{1}{| z |^4}\right)= n......(2)
Equation (1)^2+(2)^2
\left(1+\frac{1}{|z|^4}\right)^2\left[\left( x ^2+ y ^2\right)^2\right]= m ^2+ n ^2
\left(1+\frac{1}{| z |^4}\right)^2(| z |)^4= m ^2+ n ^2
\Rightarrow| z |^4+\frac{1}{| z |^4}+2= m ^2+ n ^2
Now for option (A)
|z|^4=\frac{43+3 \sqrt{205}}{2}
\Rightarrow m ^2+ n ^2=45
\Rightarrow m =\pm 6, n =\pm 3
Option (B)
|z|^4+\frac{1}{|z|^4}+2=\frac{7+\sqrt{33}}{4}+\frac{7-\sqrt{33}}{4}+2=\frac{7}{2}+2=\frac{11}{2}
Option (C)
|z|^4+\frac{1}{|z|^4}+2=\frac{9+\sqrt{65}}{4}+\frac{9-\sqrt{65}}{4}+2=\frac{18}{4}+2=\frac{9}{2}+2=\frac{13}{2}
Option (D)
|z|^4+\frac{1}{|z|^4}+2=\frac{7+\sqrt{13}}{6}+\frac{7-\sqrt{13}}{6}+2=\frac{14}{6}+2=\frac{7}{3}+2=\frac{13}{2}

A

If n =4, then (\sqrt{2}-1) r < R

B

If n =5, then r < R

C

If n =8, then (\sqrt{2}-1) r < R

D

If n =12, then \sqrt{2}(\sqrt{3}+1) r > R

Solution

2( R + r ) \sin \frac{\pi}{ n }=2 r
\frac{ R + r }{ r }=\operatorname{cosec} \frac{\pi}{ n }
image
(A) n =4, R + r =\sqrt{2} r
(B) n =5, \frac{ R + r }{ r }=\operatorname{cosec} \frac{\pi}{5}<\operatorname{cosec} \frac{\pi}{6}
R + r <2 r \Rightarrow r > R
(C) n =8, \frac{ R + r }{ r }=\operatorname{cosec} \frac{\pi}{8}>\operatorname{cosec} \frac{\pi}{4}
R + r >\sqrt{2} r
(D) n =12, \frac{ R + r }{ r }=\operatorname{cosec} \frac{\pi}{12}=\sqrt{2}(\sqrt{3}+1)
R + r =\sqrt{2}(\sqrt{3}+1) r
\sqrt{2}(\sqrt{3}+1) r > R

A

\vec{a} \cdot \vec{c}=0

B

\overrightarrow{ b } \cdot \overrightarrow{ c }=0

C

|\vec{b}|>\sqrt{10}

D

|\vec{c}| \leq \sqrt{11}

Solution

\vec{ a }=3 \hat{ i }+\hat{ j }-\hat{ k }
\vec{ b }=\hat{ i }+ b _2 \hat{ j }+ b _3 \hat{ k }
\vec{ c }= c _1 \hat{ i }+ c _2 \hat{ j }+ c _3 \hat{ k }
\begin{pmatrix} 0 & -c_3 & c_2 \\c_3 & 0 & -c_1 \\-c_2 & c_1 & 0\end{pmatrix}\begin{pmatrix} 1 \\b_2 \\b_3\end{pmatrix}=\begin{pmatrix} 3-c_1 \\1-c_2 \\-1-c_3\end{pmatrix}
multiply & compare
b_2 c_3-b_3 c_2=c_1-3 .....(1)
c _3- b _3 c _1=1- c _2.......(2)
c _2- b _2 c _1=1+ c _3 ......(3)
(1) \hat{i}-(2) \hat{j}+(3) \hat{k}
\hat{ i }\left(b_2 c_3-b_3 c_2\right)-\hat{j}\left(c_3-b_3 c_1\right)+\hat{k}\left(c_2-b_2 c_1\right)
=c_1 \hat{i}+c_2 \hat{j}+c_2 \hat{k}-3 \hat{i}-\hat{j}+\hat{k}
\vec{ b } \times \vec{ c }=\vec{ c }-\vec{ a }
Take dot product with \vec{ b }
0=\vec{ c } \cdot \vec{ b }-\vec{ a } \cdot \vec{ b }
\vec{ b } \cdot \vec{ c }=0
\vec{ b } \perp \vec{ c }
\vec{ b }^{\wedge} \vec{ c }=90^{\circ}
Take dot product with \vec{ c }
0=|\vec{ c }|^2-\vec{ a } \cdot \vec{ c }
\vec{ a } \cdot \vec{ c }=|\vec{ c }|^2
\vec{ a } \cdot \vec{ c } \neq 0
\vec{ b } \times \vec{ c }=\vec{ c }-\vec{ a }
Squaring
|\vec{b}|^2|\vec{c}|^2=|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}
|\vec{b}|^2|\vec{c}|^2=|\vec{c}|^2+11-2|\vec{c}|^2
|\vec{b}|^2|\vec{c}|^2=11-|\vec{c}|^2
|\vec{c}|^2\left(|\vec{b}|^2+1\right)=11
|\vec{c}|^2=\frac{11}{|\vec{b}|^2+1}
|\vec{c}| \leq \sqrt{11}
given \vec{a} \cdot \vec{b}=0
b _2- b _3=-3 also
b _2^2+ b _3^2-2 b _2 b _3=9 \,\, b _2 b _3>0
b _2^2+ b _3^2=9+2 b _2 b _3
b _2^2+ b _3^2=9+2 b _2 b _3>9
b _2^2+ b _3^2>9
|\vec{ b }|=\sqrt{1+ b _2^2+ b _3^2}
|\vec{ b }|>\sqrt{10}

A

y(x) is an increasing function

B

y(x) is a decreasing function

C

There exists a real number \beta such that the line y =\beta intersects the curve y = y ( x ) at infinitely many points

D

y ( x ) is a periodic function

Solution

\frac{ dy }{ dx }+12 y =\cos \left(\frac{\pi}{12} x\right)
Linear D.E.
\text { I.F. }= e ^{\int 12 . d x}= e ^{12 x}
Solution of DE
y \cdot e^{12 x}=\int e^{12 x} \cdot \cos \left(\frac{\pi}{12} x\right) d x
y \cdot e^{12 x}=\frac{e^{12 x}}{(12)^2+\left(\frac{\pi}{12}\right)^2}\left(12 \cos \frac{\pi}{12} x+\frac{\pi}{12} \sin \frac{\pi}{12} x\right)+C
\Rightarrow y=\frac{(12)}{(12)^4+\pi^2}\left((12)^2 \cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)\right)+\frac{C}{e^{12 x}}
Given y(0)=0
\Rightarrow 0=\frac{12}{12^4+\pi^2}\left(12^2+0\right)+ C \Rightarrow C =\frac{-12^3}{12^4+\pi^2}
\therefore y =\frac{12}{12^4+\pi^2}\left[(12)^2 \cos \left(\frac{\pi x }{12}\right)+\pi \sin \left(\frac{\pi x }{12}\right)-12^2 \cdot e ^{-12 x }\right]
Now \frac{ dy }{ dx }=\frac{12}{12^4+\pi^2}[\underbrace{-12 \pi \sin \left(\frac{\pi x }{12}\right)+\frac{\pi^2}{12} \cos \left(\frac{\pi x }{12}\right)}_{\text {min. value }}+12^3 e ^{-12 x }]
\left(-\sqrt{144 \pi^2+\frac{\pi^4}{144}}=-12 \pi \sqrt{\left.1+\frac{\pi^2}{12^4}\right)}\right.
\Rightarrow \frac{ dy }{ dx }>0 \forall x \leq 0 \text { \& may be negative/positive for } x >0
So, f ( x ) is neither increasing nor decreasing
For some \beta \in R, y=\beta intersects y=f(x) at infinitely many points
So option C is correct

A

21816

B

85536

C

12096

D

156816

Solution

image
Case-I : when exactly one box provides four balls ( 3 R 1 B or 2 R 2 B )
Number of ways in this case { }^5 C _4\left({ }^3 C _1 \times{ }^2 C _1\right)^3 \times 4
Case-II : when exactly two boxes provide three balls ( 2 R 1 B or 1 R 2 B ) each
Number of ways in this case \left({ }^5 C _3-1\right)^2\left({ }^3 C _1 \times{ }^2 C _1\right)^2 \times 6
Required number of ways =21816
Language ambiguity : If we consider at least one red ball and exactly one blue ball, then required number of ways is 9504 . None of the option is correct.

A

\begin{pmatrix}{cc}3034 & 3033 \\ -3033 & -3032\end{pmatrix}

B

\begin{pmatrix}3034 & -3033 \\ 3033 & -3032\end{pmatrix}

C

\begin{pmatrix}3033 & 3032 \\ -3032 & -3031\end{pmatrix}

D

\begin{pmatrix}3032 & 3031 \\ -3031 & -3030\end{pmatrix}

Solution

M =\begin{bmatrix}\frac{5}{2} & \frac{3}{2} \\ \frac{-3}{2} & \frac{-1}{2}\end{bmatrix}
M =\begin{bmatrix}\frac{3}{2}+1 & \frac{3}{2} \\ \frac{-3}{2} & \frac{-3}{2}+1\end{bmatrix}
M = I +\frac{3}{2}\begin{bmatrix}1 & 1 \\ -1 & -1\end{bmatrix}
Let A =\begin{bmatrix}1 & 1 \\ -1 & -1\end{bmatrix}
A ^2= \begin{bmatrix}1 & 1 \\ -1 & -1\end{bmatrix}\begin{bmatrix}1 & 1 \\ -1 & -1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
M ^{2022} =\left( I +\frac{3}{2} A \right)^{2022}
= I +3033 A
=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+3033 \begin{bmatrix}1 & 1 \\ -1 & -1\end{bmatrix}
=\begin{bmatrix}3034 & 3033 \\ -3033 & -3032\end{bmatrix}

A

\frac{15}{256}

B

\frac{3}{16}

C

\frac{5}{52}

D

\frac{1}{8}

Solution

Box I 8( R ) 3( B ) 5( G )
Box II 24(R) 9(B) 15( G )
Box III 1(B) 12( G ) 3( y )
Box IV 10(G) 16(o) 6(w)
A (one of the chosen balls is white)
B (at least one of the chosen ball is green)
P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}
A \cap B \rightarrow(w G)
=\frac{5}{16} \times 1+\frac{8}{16} \times \frac{15}{48}+\frac{3}{16} \times \frac{12}{16}
=\frac{15}{156}=\frac{5}{52}

A

3+\frac{4}{3} \log _e 7

B

4-\frac{3}{4} \log _e\left(\frac{7}{3}\right)

C

4-\frac{4}{3} \log _e\left(\frac{7}{3}\right)

D

3+\frac{3}{4} \log _{ e } 7

Solution

f(n)=n+\displaystyle\sum_{r=1}^n \frac{16 r+(9-4 r) n-3 n^2}{4 r n+3 n^2}
f(n)=n+\displaystyle\sum_{r=1}^n \frac{(16 r+9 n)-\left(4 r n+3 n^2\right)}{4 r n+3 n^2}
f(n)=n+\left(\displaystyle\sum_{r=1}^n \frac{16 r+9 n}{4 r n+3 n^2}\right)-n
\displaystyle\lim _{n \rightarrow \infty} f(n)=\displaystyle\lim _{n \rightarrow \infty} \sum \frac{16 r+9 n}{4 r n+3 n^2}
=\displaystyle\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{\left(16\left(\frac{r}{n}\right)+9\right) \frac{1}{n}}{4\left(\frac{r}{n}\right)+3}
=\int\limits_0^1 \frac{16 x+9}{4 x+3} d x=\int\limits_0^1 4 d x-\int\limits_0^1 \frac{3 d x}{4 x+3}
=4-\frac{3}{4}(\ell n |4 x+3|)_0^1
=4-\frac{3}{4} \ell \operatorname{n} \frac{7}{3}