JEE Main Question Paper with Solution 2023 January 25th Shift 2 - Evening

A

$20 V$

B

$16 V$

C

$8 V$

D

$12 V$

Solution

Induced emf across the ends $= Bv \ell$
$=2 \times 8 \times 1=16 V$

A

$8 \,m s^{-1}$

B

$20\, ms ^{-1}$

C

$25\, ms ^{-1}$

D

$40\, ms ^{-1}$

Solution

$x =4 t ^2$
$v =\frac{ dx }{ dt }=8 t$
At $t =5 \sec$
$v =8 \times 5=40 \,m / s$

A

$\frac{3}{2} R$

B

$\frac{7}{2} R$

C

$\frac{9}{2} R$

D

$\frac{5}{2} R$

Solution

Diatomic gas molecules have three translational degree of freedom, two rotational degree of freedom & it is given that it has one vibrational mode so there are two additional degree of freedom corresponding to one vibrational mode, so total degree of freedom $=7$
$C _{ v }=\frac{ fR }{2}=\frac{7 R }{2}$

A

1.5S

B

3s

C

4s

D

2s

Solution

Let time from 0 to $A / 2$ is $t _1$
$\&$ from $A / 2$ to $A$ is $t _2$
then $\omega t _1=\pi / 6$
$\omega t _2=\pi / 3$
$\frac{ t _1}{ t _2}=\frac{1}{2}$
$t _2=2 t _1=2 \times 2=4 \sec$

A

$125$

B

5

C

25

D

625

Solution

$\because$ Volume of wire is constant in stretching
$V _{ i }= V _{ f }$
$A _{ i } \ell_{ i }= A _{ f } \ell_{ f }$
$A \ell= A ^{\prime}(5 \ell)$
$A ^{\prime}=\frac{ A }{5}$
$R _{ f }=\frac{\rho \ell_{ f }}{ A _{ f }}=\frac{\rho(5 \ell)}{\left(\frac{ A }{5}\right)}$
$=25\left(\frac{\rho \ell}{ A }\right)$
$=25 \times 5=125 \Omega$

A

$2.0$

B

$1.0$

C

$1.5$

D

$0.5$

Solution

$\tau= K \theta$
$NiAB = K \theta$
$A =\frac{ K \theta}{ NiB }=\frac{4 \times 10^{-5} \times 0.05}{200 \times 10 \times 10^{-3} \times 0.01}$
On solving $A =10^{-4} m ^2=1 cm ^2$

A

Statenent I is correct but statement II is incorrect

B

Both Statement I and Statement II are incorrect

C

Statenent I is incorrect but statement II is correct

D

Both Statenent I and statement II are correct

Solution

Statement - I is correct
When P-N junction is formed an electric field is generated form N-side to P-side due to which barrier potential arises & majority charge carrier can not flow through the junction due to barrier potential so current is zero unless we apply forward bias voltage.

A

$1:2$

B

$4: 1$

C

$2: 1$

D

$1:1$

Solution

$\text { Range }=\frac{ u ^2 \sin 2 \theta}{ g }$
Range for projection angle " $\alpha$ "
$R _1=\frac{ u ^2 \sin 2 \alpha}{ g }$
Range for projection angle " $\beta$ "
$R _2=\frac{ u ^2 \sin 2 \beta}{ g }$
$\alpha+\beta=90^{\circ} \text { (Given) }$
$\Rightarrow \beta=90^{\circ}-\alpha$
$R _2=\frac{ u ^2 \sin 2\left(90^{\circ}-\alpha\right)}{ g }$
$R _2=\frac{ u ^2 \sin \left(180^{\circ}-2 \alpha\right)}{ g }$
$R _2=\frac{ u ^2 \sin 2 \alpha}{ g }$
$\Rightarrow \frac{ R _1}{ R _2}=\frac{\left(\frac{ u ^2 \sin 2 \alpha}{ g }\right)}{\left(\frac{ u ^2 \sin 2 \alpha}{ g }\right)}=\frac{1}{1}$

A

$\frac{t_P}{100}=\frac{t_Q-180}{150}$

B

$\frac{t_Q}{150}=\frac{t_P-180}{100}$

C

$\frac{t_p}{180}-\frac{t_Q-40}{100}$

D

$\frac{t_Q}{100}=\frac{t_P-30}{150}$

Solution

$\frac{\text { reading on scale }-\text { Lower fixed point }}{\text { upper fixed po int-lower fixed point }}=\text { constant }$
$\frac{t_p-30}{180-30}=\frac{t_Q-0}{100-0}$
$\frac{t_p-30}{150}=\frac{t_Q}{100}$

A

A and D only

B

$B$ and $C$ only

C

$A$ and $C$ Only

D

C and Donly

Solution

$F =\frac{ Gm _1 m _2}{ r ^2}$
$\Rightarrow F \propto \frac{1}{ r ^2}$
$\Rightarrow F \propto m _1 m _2$
$\Rightarrow$ This force provides centripetal force and acts towards sun
$\Rightarrow T ^2 \propto a ^3 \text { (Kepler's third law) }$

A

Both Statement I and Statement II are incorrect

B

Statement I is correct but statement II is incorrect

C

Both Statement I and statement II are correct

D

Statement I is incorrect but statement II is correct

Solution

Stopping potential $V _{ S }=\frac{ KE _{\max }}{ e }$
$V _{ S }=\frac{\frac{ hC }{\lambda}-\phi}{ e }$
Stopping potential does not depend on intensity or power of light used, it only depends on frequency or wavelength of incident light.
So both statements I and II are correct

A

A-III, B-IV, C-II, D-I

B

A-I, B-II, C-IV, D-III

C

A-III, B-II, C-I, D-IV

D

A-I, B-IV, C-III, D-II

A

$0.50$

B

$0.33$

C

$0.25$

D

$0.60$

Solution

$F _1= mg \sin 45^{\circ}+ f = mg \sin 45^{\circ}+\mu N$
$F _1=\frac{ mg }{\sqrt{2}}+\mu mg \cos 45^{\circ}$
$F _1=\frac{ mg }{\sqrt{2}}(1+\mu)$

$F _2= mg \sin 45^{\circ}- f = mg \sin 45^{\circ}-\mu N$
$=\frac{ mg }{\sqrt{2}}(1-\mu)$
$F _1=2 F _2$
$\frac{ mg }{\sqrt{2}}(1+\mu)=2 \frac{ mg }{\sqrt{2}}(1-\mu)$
$1+\mu=2-2 \mu$
$\mu=1 / 3=0.33$

A

B and C Only

B

A. C, and D Only

C

B and D only

D

A and $D$ Only

Solution

Plane mirror forms erect, same sized, laterally inverted and virtual image of real object.

A

$\frac{1}{2} m g R_e$

B

$3 m g R_e$

C

$\frac{2}{3} m g R_e$

D

$\frac{1}{3} m g R_e$

Solution

$U =\frac{- GM _{ e } m }{ r }$
$U _{ i }=\frac{- GM _{ e } m }{ R _{ e }}$
$U _{ f }=\frac{- GM _{ e } m }{\left( R _{ e }+ h \right)}=\frac{- GM _{ e } m }{ R _{ e }+2 R _{ e }}$
$\frac{- GM _{ e } m }{3 R _{ e }}$
Increase in internal energy $\Delta U = U _{ f }- U _{ i }$
$=\frac{2}{3} \frac{ GM _{ e } m }{ R _{ e }}$
$\frac{2}{3} \frac{ GM _{ e }}{ R _{ e }^2} mR _{ e }$
$=\frac{2}{3} mgR _{ e }$

A

$x=-4 \,cm$

B

$x=4 \, cm$

C

$x=8 \,cm$

D

$x=6 \,cm$

Solution

$E_P=\frac{K \times 10}{2^2}-\frac{K \times 40}{\left(x_0-2\right)^2}=0$
$\frac{1}{2}=\frac{2}{x_0-2}$
$x_0-2=4$
$x_0=6\, cm$

A

A-I, B-III, C-IV, D-II

B

A-I, B-II, C-III, D-IV

C

A-II, B-III, C-IV, D-I

D

A-III, B-I, C-II, D-IV

Solution

$Y =\frac{\text { Stress }}{\text { Strain }}=\frac{ F / A }{\Delta \ell / \ell}=\frac{\left[ MLT ^{-2}\right]}{\left[ L ^2\right]}=\left[ ML ^{-1} T ^{-2}\right]$
$F =6 \pi \eta rv \Rightarrow \eta=\frac{ F }{6 \pi rv }$
${[\eta]=\frac{\left[ MLT ^{-2}\right]}{[ L ]\left[ LT ^{-1}\right]}=\left[ ML ^{-1} T ^{-1}\right]}$
$E = hv \Rightarrow h =\frac{ E }{v}=\frac{\left[ ML ^2 T ^{-2}\right]}{\left[ T ^{-1}\right]}=\left[ ML ^2 T ^{-1}\right]$
Work function has same dimension as that of energy, so $[\phi]=\left[ ML ^2 T ^{-2}\right]$

A

A-I, B-II, C-IV, D-III

B

A-II, B-I, C-IV, D-III

C

A-I, B-II, C-III, D-IV

D

A-II, B-I, C-III, D-IV

Solution

$\Delta U = nC _{ v } \Delta T$
For isothermal process $T$ is constant
So $\Delta U =0$
$A \longrightarrow II$
Adiabatic process
$\Delta Q = 0$
$\Delta Q =\Delta U +\Delta W$
$\Delta U =-\Delta W$
Work done by gas is positive
So $\Delta U$ is negative
$\text { B } \longrightarrow \text { I }$
For Isochoric process $\Delta W =0$
$C \longrightarrow IV$
For Isobaric process
$\Delta W = P \Delta V \neq 0$
$\Delta U = nC _{ V } \Delta T \neq 0$
Heat absorbed goes partly to increase internal energy and partly do work.

A

A-I, B-II, C-III, D-IV

B

A-IV, B-I, C-II, D-III

C

A-III, B-IV, C-I, D-II

D

A-II, B-III, C-IV, D-I

Solution

Gauss's Law of electrostatic
$\phi=\oint \vec{ E } \cdot d \vec{ s }=\frac{ q }{\epsilon_0}$
Faraday's law $\oint \vec{E} \cdot \vec{d l}=\frac{-d \phi_B}{d t}$
Gauss's law of magnetism $\oint \vec{ B } \cdot d \vec{ A }=0$
Ampere's Maxwell law
$\oint \vec{B} . d \vec{l}=\mu_0 i_C+\mu_0 \in_0 \frac{d \phi_E}{d t}$
Where $i _{ C }$ : Conduction current
$\epsilon_0 \frac{ d \phi_{ E }}{ dt }$ : Displacement current

A

D

B

B

C

A

D

$C$

Solution

$\lambda=\frac{ hc }{\Delta E }$
$\Delta E _{ A }=2.2\, eV$
$\Delta E _{ B }=5.2\, eV$
$\Delta E _{ C }=3\, eV$
$\Delta E _{ D }=10 \, eV$
$\lambda_{ A }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{2.2 \times 1.6 \times 10^{-19}}$
$=\frac{12.41 \times 10^{-7}}{2.2} m$
$=\frac{1241}{2.2} nm =564\, nm$
$\lambda_{ B }=\frac{1241}{5.2} nm =238.65\, nm$
$\lambda_{ C }=\frac{1241}{3} nm =413.66 \, nm$
$\lambda_{ D }=\frac{1241}{10}=124.1 \, nm$

Solution

$E _{ eq }=\frac{\frac{12}{3}-\frac{6}{6}}{\frac{1}{3}+\frac{1}{6}}$
$E _{ eq }=6 V$
$r _{ eq }=2 \Omega$
$R =4 \Omega$

Solution

$f =10\, cm$
$\frac{1}{ v }-\frac{1}{ u }=\frac{1}{ f }$
$\frac{1}{15}-\frac{1}{- u }=\frac{1}{10}$
$\Rightarrow \frac{1}{ u }=\frac{1}{10}-\frac{1}{15}$
On solving we get value of $u$ as $30 \,cm$.

Solution

$f_{\text {app }}=f\left(\frac{v}{v-v_s}\right)$
$=320\left(\frac{330}{330-66}\right)$
$=400 Hz$

Solution

$=\frac{\in_0 A }{\left(\frac{ d }{3}+\frac{ d }{2}\right)}=\frac{6 \in_0 A }{5 d }$
$=\frac{6}{5} \times 5 \mu F =6 \mu F$

Solution

Magnetic fields due to both wires will be perpendicular to each other.
$B_1=\frac{\mu_0 i_1}{2 \pi d } B_2=\frac{\mu_0 i _2}{2 \pi d }$
$B _{ net }=\sqrt{ B _1^2+ B _2^2} \Rightarrow \frac{\mu_0}{2 \pi d } \sqrt{ i _1^2+ i _2^2}$
$\Rightarrow \frac{4 \pi \times 10^{-7}}{2 \pi \times(7 / \sqrt{2}) \times 10^{-2}} \times \sqrt{8^2+15^2}\left( d =\frac{7}{\sqrt{2}} cm \right)$
$\Rightarrow 68 \times 10^{-6} T$

Solution

$I _1=\frac{2}{5} m _1 R ^2+ m _1 R ^2$
$I _1= m _1 R ^2\left(\frac{7}{5}\right)$
$I _1=7 R ^2$

$I _2=\frac{ m _2 R ^2}{4}+ m _2 R ^2$
$I _2=\frac{5}{4} m _2 R ^2$
$I _2=5 R ^2$
$\frac{ I _2}{ I _1}=\frac{5}{7}$
$x =5$

Solution

Surface Tension $= T$
$R$ : Radius of bigger drop
$r$ : Radius of smaller drop
Volume will remain same
$\frac{4}{3} \pi R ^3=1000 \times \frac{4}{3} \pi r ^3$
$R =10 r$
$u _{ i }= T \cdot 4 \pi R ^2$
$u _{ f }= T \cdot 4 \pi r ^2 \times 1000$
$\frac{ u _{ f }}{ u _{ i }}=\frac{1000 r ^2}{ R ^2}$
$\frac{ u _{ f }}{ u _{ i }}=\frac{10}{1}$
$\text { So, } x =1$

Solution

$m _1 v _1= m _2 v _2 \Rightarrow \frac{ m _1}{ m _2}=\frac{2}{3}$
Since, Nuclear mass density is constant
$\frac{ m _1}{\frac{4}{3} \pi r _1^3}=\frac{ m _2}{\frac{4}{3} \pi r _2^3}$
$\left(\frac{ r _1}{ r _2}\right)^3=\frac{ m _1}{ m _2}$
$\frac{ r _1}{ r _2}=\left(\frac{2}{3}\right)^{\frac{1}{3}}$
$\text { So, } x=2$

Solution

$\cos \phi=\frac{R}{Z}=\frac{ R }{\sqrt{ R ^2+\left( X _{ C }- X _{ L }\right)^2}}$
$\cos \phi=\frac{80}{\sqrt{(80)^2+(60)^2}}$
$\cos \phi=\frac{80}{100} \Rightarrow \frac{8}{10}$
So, $x=8$

A

Both Statement I and Statement II are false

B

Both Statement I and Statement II are true

C

Statement I is false but Statement II is true

D

Statement I is true but Statement II is false

Solution

In froth floatation method a rotating paddle draws in air and stirs the pulp

A

A-III, B-IV, C-II, D-I

B

A-III, B-II, C-IV, D-I

C

A-I, B-IV, C-II, D-III

D

A-III, B-II, C-I, D-IV

A

B and C only

B

A and C only

C

A and D only

D

A, B and C only

Solution

$B. \overset{\bullet}{Cl} + O_3 \rightarrow O_2 + Cl\overset{\bullet}{O}$
D. 'Blue baby' syndrome occurs due to the presence of excess of nitrate ions in water.

A

A-III, B-I, C-IV, D-II

B

A-III, B-IV, C-II, D-I

C

A-III, B-II, C-IV, D-I

D

A-III, B-IV, C-I, D-II

A

B

C

D

A

$A$ is not correct but $R$ is correct

B

Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$

C

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

D

$A$ is correct but $R$ is not correct

Solution

The alkali metals and their salts impart characteristic colour to oxidizing flame.

A

B

C

D

A

A-IV, B-III, C-I, D-II

B

A-II, B-III, C-IV, D-I

C

A-IV, B-I, C-II, D-III

D

A-IV, B-III, C-II, D-I

Solution

Cobalt catalyst $\rightarrow$ Methanol production
Syn gas $\rightarrow$ Coal gasification
$\left( C _{(\text {Red hot coke })}+ H _2 O ( g ) \rightarrow CO + H _2\right)$
Nickel catalyst $\rightarrow$ Water gas production
Brine solution $\rightarrow$ Production
$($ aq. $NaCl ) \begin{pmatrix} H _2 \rightarrow \text { Cathode } \\ Cl _2 \rightarrow \text { anode }\end{pmatrix}$

A

A is not correct but $R$ is correct

B

Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$

C

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

D

A is correct but $R$ is not correct

Solution

The oxide which form acid on dissolving in water is acidic oxide.

A

A-III, B-II, C-I, D-IV

B

A-II, B-III, C-IV, D-I

C

A-IV, B-I, C-III, D-II

D

A-III, B-I, C-II, D-IV

Solution

LIST I Coordination entity		LIST II Wavelength of light absorbed in nm
A	$\left[ CoCl \left( NH _3\right)_5\right]^{2+}$	I	535
B	$\left[ Co \left( NH _3\right)_6\right]^{3+}$	II	475
C	$\left[ Co ( CN )_6\right]^{3-}$	III	310
D	$\left[ Cu \left( H _2 O \right)_4\right]^{2+}$	IV	600

$E =\frac{ hc }{\lambda} \Rightarrow E \propto \frac{1}{\lambda}$
$\Rightarrow \Delta( CFSE ) \propto \frac{1}{\lambda_{\text {absorb }}} \propto \text { strength of ligand. }$

A

$1: 2$

B

$1: 1$

C

$3:1$

D

$2:1$

Solution

Assume : Mass of solvent $\approx$ Mass of solution
Case I :-
$0.25=\frac{W_1}{62} \times \frac{1000}{500}$
Case II :-
$0.25=\frac{ W _2}{62} \times \frac{1000}{250}$
$\frac{ W _1}{ W _2}=\frac{2}{1}$

A

$Rb$

B

$Na$

C

$Li$

D

$K$

Solution

Sodium have lowest oxidation potential in alkali metals. Hence it is weakest reducing agent among alkali metals.

A

$+6$ to $+3$

B

$+6$ to $+2$

C

$+3$ to $+1$

D

$+2$ to $+1$

A

A-II, B-III, C-I, D-IV

B

A-III, B-I, C-IV, D-II

C

A-III, B-IV, C-I, D-II

D

A-IV, B-III, C-I, D-II

A

2 - Bromo - 1 - deuterobutane

B

2 - Bromo - 2 - deuterobutane

C

2 - Bromo - 3 - deuterobutane

D

2 - Bromo-1 - deutero - 2 - methylpropane

A

decreases by 3 units

B

decreases by 2 units

C

increases by 2 units

D

increases by 1000 units

Solution

$\Delta\left[ H ^{+}\right]=1000$
$\Delta pH =-\log \Delta\left[ H ^{+}\right]=-\log 10^3$
$=-3$

A

Both Statement I and Statement II are correct

B

Statement I is correct but Statement II is incorrect

C

Statement I is incorrect but Statement II is correct

D

Both Statement I and Statement II are incorrect

Solution

Statement II : The corssed arrow symbolises the direction of the shift of electron density in the molecule.

A

$H \left[ AgCl _3\right] \&\left( NH _4\right)\left[ Ag ( OH )_2\right]$

B

$H \left[ AgCl _3\right] \&\left[ Ag \left( NH _3\right)_2\right] Cl$

C

$AgCl \&\left[ Ag \left( NH _3\right)_2\right] Cl$

D

$AgCl \&\left( NH _4\right)\left[ Ag ( OH )_2\right]$

A

$Be < Si < Mg < K$

B

$K < Mg < Be < Si$

C

$Be < Si < K < Mg$

D

$Si < Be < Mg < K$

Solution

Metallic character increases down the group and decreases along the period.

A

$A$ is correct but $R$ is not correct

B

Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$

C

A is not correct but $R$ is correct

D

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

Solution

Butylated hydroxyl anisole is an antioxidant.

Solution

Element	Percentage	Mole	Mole ratio
C	85.8	$\frac{85.8}{12}=7.15$	1
H	14.2	$\frac{14.2}{1}=14.2$	2

Empirical formula $\left( CH _2\right)$
$14 \times n =84$
$n =6$
$\therefore$ Molecular formula $C _6 H _{12}$

Solution

Overall reaction :-
$H _{2( g )}+ M _{\text {(aq) }}^{3+} > M _{\text {(aq) }}^{+}+2 H _{\text {(aq) }}^{+}$
$E _{\text {Cell }}= E _{\text {Cathode }}^{ o }- E _{\text {anode }}^{\circ}-\frac{0.059}{2} \log \frac{\left[ M ^{+}\right] \times 1^2}{\left[ M ^{+3}\right] 1}$
$0.1115=0.2-\frac{0.059}{2} \log \frac{\left[ M ^{+}\right]}{\left[ M ^{+3}\right]}$
$3=\log \frac{\left[ M ^{+}\right]}{\left[ M ^{+3}\right]}$
$\therefore a =3$

Solution

$\pi= iCRT$
$\pi \propto iC$
$A, B, D$ and $E$ have same value of osmatic pressure.

Solution

$\Rightarrow 28=16.8+ x$
$x =11.2 L$
$n _{ CH _4}=\frac{ PV }{ RT }=\frac{1 \times 5.6}{0.082 \times 298}=0.229 \text { mole }$
$n _{ C _2 H _2}= \frac{11.2}{0.082 \times 298}=0.458 \text { mole }$
$\therefore \text { Heat evolved }=0.229 \times 900+0.458 \times 1400$
$=206.1+641.2$
$= 847.3 \,kJ$

Solution

B and D options are correct.

Solution

$p _{ x }, p _{ y }, p _{ z }, d _{ z ^2} \& d _{ x ^2- y ^2}$ are axial orbitals.

Solution

‘$A$’ water vapours are absorbed by calcium chloride.
C. As the adsorption proceeds, $\Delta H$ becomes less and less negative

Solution

$t _{10 \%}=\frac{1}{ K } \ln \left(\frac{ a }{ a - x }\right)=\frac{1}{ K } \ln \left(\frac{100}{90}\right)$
$t _{10 \%}=\frac{2.303}{ K }(\log 10-\log 9)$
$t _{10 \%}=\frac{2.093}{ K } \times(0.04)$
Similarly
$t _{90 \%}=\frac{1}{ K } \ln \left(\frac{100}{10}\right)$
$t _{90 \%}=\frac{2.303}{ K }$
$\frac{ t _{90 \%}}{ t _{10 \%}}=\frac{1}{0.04}=25$
$e ^{ kt }=\frac{ a }{ a - x }$
$\frac{ a - x }{ a }= e ^{- kt }$
$1-\frac{ x }{ a }= e ^{- kt }$
$x = a \left(1- e ^{- kt }\right)$
$\alpha=\frac{ x }{ a }=\left(1- e ^{- kt }\right)$

Solution

$\left[ Co \left( NH _3\right)_4 Cl _2\right] Cl \Rightarrow$ Gives 1 mole $AgCl$
$\left[ Ni \left( H _2 O \right)_6\right] Cl _2 \Rightarrow$ Gives 2 moles $AgCl$
$\left[ Pt \left( NH _3\right)_2 Cl _2\right] \Rightarrow$ Gives No $AgCl$
$\left[ Pd \left( NH _3\right)_4\right] Cl _2 \Rightarrow$ Gives 2 moles $AgCl$
Total number of moles of $AgCl =5$ mole.

A

$\frac{11}{6}-\log _e 4$

B

$\frac{11}{12}+\log _{ e } 4$

C

$\frac{11}{6}+\log _{ e } 4$

D

$\frac{11}{12}-\log _e 4$

Solution

$I=16 \int\limits_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}$
$=16 \int\limits_1^2 \frac{d x}{x^3 x^4\left(1+\frac{2}{x^2}\right)^2}$
Let, $1+\frac{2}{x^2}=t \Rightarrow \frac{-4}{x^3} d x=d t$
$I=-4 \int\limits_3^{\frac{3}{2}} \frac{d t}{\left(\frac{2}{t-1}\right)^2 t^2}$
$I=-4 \int\limits_3^{\frac{3}{2}}\left(\frac{t-1}{2}\right)^2 \frac{d t}{t^2}$
$I=-\frac{4}{4} \int\limits_3^{\frac{3}{2}}\left(1-\frac{2}{t}+\frac{1}{t^2}\right) d t$
$I=-1\left[t-2 \ell n |t|-\frac{1}{t}\right]_3^{\frac{3}{2}}$
$I=-1\left[\left(\frac{3}{2}-2 \ell n \frac{3}{2}-\frac{2}{3}\right)-\left(3-2 \ell n 3-\frac{1}{3}\right)\right]$
$I=-1\left[2 \ell n 2-\frac{11}{6}\right]$
$I=\frac{11}{6}-\ell n 4$

A

${ }^{51} C _3-{ }^{45} C _3$

B

${ }^{52} C _4-{ }^{45} C _4$

C

${ }^{52} C _3-{ }^{45} C _3$

D

${ }^{51} C _4-{ }^{45} C _4$

Solution

$\displaystyle\sum_{ k =0}^6{ }^{51- k } C _3$
$={ }^{51} C _3+{ }^{50} C _3+{ }^{49} C _3+\ldots+{ }^{45} C _3$
$={ }^{45} C _3+{ }^{46} C _3+\ldots . .+{ }^{51} C _3$
$={ }^{45} C _4+{ }^{45} C _3+{ }^{46} C _3+\ldots . .+{ }^{51} C _3-{ }^{45} C _4$
$\left({ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right)$
$={ }^{52} C _4-{ }^{45} C _4$

A

Only S 1 is true

B

Both S1 and S2 are false

C

Both S1 and S2 are true

D

Only S2 is true

Solution

Given, $A ^{ T }= A , B ^{ T }=- B , C ^{ T }=- C$
Let $M=A^{13} B^{26}-B^{26} A^{13}$
Then, $M^{ T }=\left( A ^{13} B ^{26}- B ^{26} A ^{13}\right)^{ T }$
$=\left( A ^{13} B ^{26}\right)^{ T }-\left( B ^{26} A ^{13}\right)^{ T }$
$=\left( B ^{ T }\right)^{26}\left( A ^{ T }\right)^{13}-\left( A ^{ T }\right)^{13}\left( B ^{ T }\right)^{26}$
$= B ^{26} A ^{13}- A ^{13} B ^{26}=- M$
Hence, $M$ is skew symmetric
Let, $N = A ^{26} C ^{13}- C ^{13} A ^{26}$
then, $N ^{ T }=\left( A ^{26} C ^{13}\right)^{ T }-\left( C ^{13} A ^{26}\right)^{ T }$
$=-( C )^{13}( A )^{26}+ A ^{26} C ^{13}= N$
Hence, $N$ is symmetric.
$\therefore$ Only S2 is true.

A

2

B

1

C

4

D

3

Solution

$f:\{1,2,3,4\} \rightarrow\{ a \in Z :| a | \leq 8\}$
$f( n )+\frac{1}{ n } f ( n +1)=1, \forall n \in\{1,2,3\}$
$f( n +1)$ must be divisible by $n$
$f(4) \Rightarrow-6,-3,0,3,6$
$f(3) \Rightarrow-8,-6,-4,-2,0,2,4,6,8$
$f(2) \Rightarrow-8, \ldots \ldots \cdots \cdots \cdots, 8$
$f(1) \Rightarrow-8, \ldots \ldots \ldots \ldots \ldots .8$
$\frac{f(4)}{3}$ must be odd since $f(3)$ should be even therefore 2 solution possible.

A

$\frac{107}{17}$

B

$\frac{73}{17}$

C

$-\frac{73}{17}$

D

$-\frac{107}{17}$

Solution

Let $A :(3,-4,2) C :(-2,-1,3)$
$\text { B : }(1,2,-1) \text { D : }(5,-2 \alpha, 4)$
$A, B, C, D$ are coplanar points, then
$\Rightarrow\begin{vmatrix} 1-3 & 2+4 & -1-2 \\-2-3 & -1+4 & 3-2 \\5-3 & -2 \alpha+4 & 4-2\end{vmatrix}=0$
$\Rightarrow \alpha=\frac{73}{17}$

A

59

B

60

C

61

D

58

Solution

$f( x )=2 x ^{ n }+\lambda$
$f(4)=133$
$f(5)=255$
$133=2 \times 4^{ n }+\lambda ...$(1)
$255=2 \times 5^{ n }+\lambda....$(2)
$(2)-(1)$
$122=2\left(5^{ n }-4^{ n }\right)$
$\Rightarrow 5^{ n }-4^{ n }=61$
$\therefore n =3 \& \lambda=5$
Now, $f(3)-f(2)=2\left(3^3-2^3\right)=38$
Number of Divisors is $1, 2, 19, 38$; & their sum is $60$

A

$\Delta=\wedge, \nabla=\wedge$

B

$\Delta=\vee, \nabla=\vee$

C

$\Delta=\vee, \nabla=\wedge$

D

$\Delta=\wedge, \nabla=\vee$

Solution

Given $( p \rightarrow q ) \Delta( p \nabla q )$
Option I $\Delta=\wedge, \nabla=\vee$



Hence, it is tautology.
Option $4 \Delta=\wedge, \nabla=\wedge$

A

$4 y^2-18 y+3 x+18=0$

B

$4 y^2-18 y-3 x+18=0$

C

$4 y^2+18 y+3 x+18=0$

D

$4 y^2-18 y-3 x-18=0$

Solution

$y^2=6 x \& y^2=4 a x$
$\Rightarrow 4 a=6 \Rightarrow a=\frac{3}{2}$

$y = mx +\frac{3}{2 m } ;( m \neq 0)$
$h =\frac{6 m -3}{4 m ^2}, k =\frac{6 m +3}{4 m }$, Now eliminating $m$ and we get
$\Rightarrow 3 h =2\left(-2 k ^2+9 k -9\right)$
$\Rightarrow 4 y ^2-18 y +3 x +18=0$

A

$4 \sqrt{6}+\frac{44}{3}$

B

$4 \sqrt{6}-\frac{28}{3}$

C

$4 \sqrt{6}+\frac{28}{3}$

D

$4 \sqrt{6}-\frac{44}{3}$

Solution

$16\left(x^2+4 x\right)-\left(y^2-4 y\right)+44=0$

$16(x+2)^2-64-(y-2)^2+4+44=0$

$16(x+2)^2-(y-2)^2=16$

$\frac{(x+2)^2}{1}-\frac{(y-2)^2}{16}=1$

$A=\int\limits_{-2}^{\sqrt{6}}\left(2-\left(x^2-4\right)\right) d x$
$A=\int\limits_{-2}^{\sqrt{6}}\left(6-x^2\right) d x=\left(6 x-\frac{x^3}{3}\right)_{-2}^{\sqrt{6}}$
$A=\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right)$
$A=\frac{12 \sqrt{6}}{3}+\frac{28}{3}$
$A=4 \sqrt{6}+\frac{28}{3}$

A

72

B

120

C

6

D

12

Solution

Numbers between $5000 \& 10000$
Using digits $1,3,5,7,9$

A

$\frac{3}{2}$

B

3

C

2

D

$\frac{5}{2}$

Solution

$\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}} \text { and } \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$
$\Rightarrow \text { Shortest distance }=\frac{(\vec{ b }-\vec{ a }) \cdot(\vec{ p } \times \vec{ q })}{|\vec{ p } \times \vec{ q }|}$
$\text { S.D. }=(-\hat{ i }+2 \hat{ j }-\hat{ k }) \cdot \frac{(\vec{ p } \times \vec{ q })}{|\vec{ p } \times \vec{ q }|}$
$\left\{\vec{ p } \times \vec{ q } \equiv \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & \frac{1}{2} & \frac{-1}{12} \\ 1 & 1 & \frac{1}{6}\end{vmatrix}=\frac{1}{6} \hat{ i }-\frac{1}{4} \hat{ j }+\frac{1}{2} \hat{ k } \text { or } 2 \hat{ i }-3 \hat{ j }+6 \hat{ k }\right\}$
$\text { S.D. }=\frac{(-\hat{ i }+2 \hat{ j }-\hat{ k }) \cdot(2 \hat{ i }-3 \hat{ j }+6 \hat{ k })}{\sqrt{2^2+3^2+6^2}}=\left|\frac{-14}{7}\right|=2$

A

$(2,0)$

B

$(0,2)$

C

$(0,0)$

D

$(0,-2)$

Solution

$( z -2 i )(\overline{ z }+2 i )=4( z + i )(\overline{ z }- i )$
$z \overline{ z }+4+2 i ( z -\overline{ z })=4( z \overline{ z }+1+ i (\overline{ z }- z ))$
$3 z \overline{ z }-6 i ( z -\overline{ z })=0$
$x ^2+ y ^2-2 i (2 iy )=0$
$x ^2+ y ^2+4 y =0$

A

$2 e^4+8$

B

11

C

8

D

10

Solution

$\Rightarrow \displaystyle\lim _{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}=\displaystyle\lim _{x \rightarrow \frac{\pi^{+}}{2}} e ^{\frac{\sin 4 x \times \cos 6 x}{\sin 6 x \cdot \cos 4 x}}=e^{2 / 3}$
$\Rightarrow \displaystyle\lim _{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^{\frac{\lambda}{\cos x} \mid}=e^\lambda$
$\Rightarrow f(\pi / 2)=\mu$
For continuous function $\Rightarrow e ^{2 / 3}= e ^\lambda=\mu$
$\lambda=\frac{2}{3}, \mu= e ^{2 / 3}$
Now, $9 \lambda+6 \log _{ e } \mu+\mu^6- e ^{6 \lambda}=10$

A

$\begin{bmatrix}1 & -2023 i \\ 0 & 1\end{bmatrix}$

B

$\begin{bmatrix}1 & 0 \\ 2023 i & 1\end{bmatrix}$

C

$\begin{bmatrix}1 & 2023 i \\ 0 & 1\end{bmatrix}$

D

$\begin{bmatrix}1 & 0 \\ -2023 i & 1\end{bmatrix}$

Solution

$AA ^{ T }=\begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$
$B ^2=\begin{bmatrix}1 & - i \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & - i \\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & -2 i \\ 0 & 1\end{bmatrix}$
$B ^3= \begin{bmatrix}1 & -3 i \\ 0 & 1\end{bmatrix}$
:
:
$B ^{2023}=\begin{bmatrix}1 & -2023 i \\ 0 & 1\end{bmatrix}$
$M = A ^{ T } B A$
$M ^2= M \cdot M = A ^{ T } BA A ^{ T } BA = A ^{ T } B ^2 A$
$M ^3= M ^2 \cdot M = A ^{ T } B ^2 AA ^{ T } BA = A ^{ T } B ^3 A$
:
:
$M ^{2023}=\ldots \ldots \ldots \ldots . . A ^{ T } B ^{2023} A$
$AM ^{2023} A ^{ T }= AA ^{ T } B ^{2023} AA ^{ T }= B ^{2023}$
$=\begin{bmatrix} 1 & -2023 i \\0 & 1\end{bmatrix}$
Inverse of $\left( AM ^{2023} A ^{ T }\right)$ is $\begin{bmatrix}1 & 2023 i \\ 0 & 1\end{bmatrix}$

A

$\frac{\gamma}{\alpha}=\frac{5}{8}$

B

$\frac{\beta}{\gamma}=-5$

C

$\frac{\alpha \beta}{\gamma}=\frac{4}{15}$

D

$\frac{\alpha}{\beta}=-8$

Solution

$L : \frac{ x +1}{2}=\frac{ y -1}{5}=\frac{ z +1}{-1}=\lambda$

Let foot of perpendicular is
$P (2 \lambda-1,5 \lambda+1,-\lambda-1)$
$\overrightarrow{ PA }=(3-2 \lambda) \hat{ i }-(5 \lambda+1) \hat{ j }+(6+\lambda) \hat{ k }$
Direction ratio of line $\Rightarrow \vec{b}=2 \hat{i}+5 \hat{j}-\hat{k}$
$\text { Now, } \Rightarrow \overrightarrow{ PA } \cdot \overrightarrow{ b }=0$
$\Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0$
$\Rightarrow \lambda=\frac{-1}{6}$
$P (2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv P (\alpha, \beta, \gamma)$
$\Rightarrow \alpha=2\left(-\frac{1}{6}\right)-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3}$
$\Rightarrow \beta=5\left(-\frac{1}{6}\right)+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6}$
$\Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6}$
$\therefore$ Check options

A

3

B

5

C

4

D

2

Solution

Since,
$-\sqrt{2} \leq \sin x-\cos x \leq \sqrt{2}$
$\therefore -2 \leq \sqrt{2}(\sin x-\cos x) \leq 2$
$(\text { Assume } \sqrt{2}(\sin x-\cos x)=k)$
$-2 \leq k \leq 2 \ldots \text { (i) }$
$f(x)=\log _{\sqrt{ m }}( k + m -2)$
$\text { Given, }$
$0 \leq f( x ) \leq 2$
$0 \leq \log _{\sqrt{ m }}( k + m -2) \leq 2$
$1 \leq k + m -2 \leq m$
$- m +3 \leq k \leq 2 \ldots \text { (ii) }$
From eq. (i) & (ii), we get $-m+3=-2$
$\Rightarrow m=5$

A

$\left(\frac{9}{2}, \infty\right)$

B

$\left(0, \frac{9}{2}\right)$

C

$\left(-\frac{9}{2}, \frac{9}{2}\right)$

D

$\left(-\infty, \frac{9}{2}\right)$

Solution

$f(x)=2 x^3+(2 p-7) x^2+3(2 p-9) x-6$
$f^{\prime}(x)=6 x^2+2(2 p-7) x+3(2 p-9)$
$f^{\prime}(0)<0$
$\therefore 3(2 p -9)<0$
$p <\frac{9}{2}$
$p \in\left(-\infty, \frac{9}{2}\right)$

A

$3(\hat{i}-\hat{j}-\hat{k})$

B

$3(\hat{i}-\hat{j}+\hat{k})$

C

$3(\hat{i}+\hat{j}-\hat{k})$

D

$3(\hat{i}+\hat{j}+\hat{k})$

Solution

$\vec{ a } \times \vec{ b }=(\hat{ i }-\hat{ j })$
Taking cross product with $\vec{a}$
$\Rightarrow \vec{a} \times(\vec{a} \times \vec{b})=\vec{a} \times(\hat{i}-\hat{j})$
$\Rightarrow (\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$
$\Rightarrow \vec{a}-3 \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$
$\Rightarrow 2 \vec{a}-6 \vec{b}=2 \hat{i}+2 \hat{j}+4 \hat{k}$
$\Rightarrow \vec{a}-6 \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k}$

A

is$-1$

B

is 0

C

is 1

D

does not exist

Solution

$\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}$
$\text { I.F. }=e^{\int \alpha d t}=e^{\alpha t}$
Solution $\Rightarrow y \cdot e^{\alpha t}=\int \gamma c^{-\beta T } \cdot c^{\alpha t} d t$
$\Rightarrow y e^{\alpha t}=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+c$
$\Rightarrow y=\frac{\gamma}{e^{\beta t}(\alpha-\beta)}+\frac{c}{e^{\alpha t}}$
So, $\displaystyle\lim _{t \rightarrow \infty} y(t)=\frac{\gamma}{\infty}+\frac{c}{\infty}=0$

A

16

B

2

C

8

D

4

Solution

$n(s)=36$
Given : $N -2, \sqrt{3 N }, N +2$ are in G.P.
$3 N =( N -2)( N +2)$
$3 N = N ^2-4$
$\Rightarrow N ^2-3 N -4=0$
$( N -4)( N +1)=0 \Rightarrow N =4 \text { or } N =-1 \text { rejected }$
$( Sum =4) \equiv\{(1,3),(3,1),(2,2)\}$
$n ( A )=3$
$P ( A )=\frac{3}{36}=\frac{1}{12}=\frac{4}{48} \Rightarrow k =4$

Solution

$(1,1)(1,2)-(1,14) \Rightarrow 14$ pts.
If $x=2, y=\frac{27}{2}=13.5$
$(2,2)(2,4) \ldots(2,12) \Rightarrow 6$ pts.
If $x =3, y =\frac{51}{4}=12.75$
$(3,3)(3,6)-(3,12) \Rightarrow 4$ pts.
If $x=4, y=12$
$(4,4)(4,8) \Rightarrow 2$ pts.
If $x=5 \cdot y=\frac{45}{4}=11.25$
$(5,5),(5,10) \Rightarrow 2$ pts.
If $x=6, y=\frac{21}{2}=10.5$
$(6,6) \Rightarrow 1 pt$
If $x=7, y=\frac{39}{4}=9.75$
$(7,7) \Rightarrow 1 pt$
If $x=8, y=9$
$(8,8) \Rightarrow 1 pt$
If $x =9 y =\frac{33}{4}=8.25 \Rightarrow$ no pt.
Total $=31$ pts.

Solution

$\int\limits_{\frac{1}{3}}^3|\ell n x| d x=\int\limits_{\frac{1}{3}}^1(-\ell n x) d x+\int\limits_1^3(\ell n x) d x$
$=-[x \ell n x-x]_{1 / 3}^1+[x \ell n x-x]_1^3$
$=-\left[-1-\left(\frac{1}{3} \ell n \frac{1}{3}-\frac{1}{3}\right)\right]+[3 \ell n 3-3-(-1)]$
$=\left[-\frac{2}{3}-\frac{1}{3} \ell n \frac{1}{3}\right]+[3 \ell n 3-2]$
$=-\frac{4}{3}+\frac{8}{3} \ell n 3$
$=\frac{4}{3}(2 \ell n 3-1)$
$=\frac{4}{3}\left(\ell n \frac{9}{ e }\right)$
$\therefore m =4, n =3$
Now, $m ^2+ n ^2-5=16+9-5=20$

Solution

$7$ Red apple(RA),5 white apple(WA), 8 oranges $( O )$
5 fruits to be selected (Note:- fruits taken different)
Possible selections :- $(2 O , 1 RA , 2 WA )$ or $(2 O$,
$2 RA , 1 WA )$ or $(3 O , 1 RA , 1 WA )$
$\Rightarrow{ }^8 C _2{ }^7 C _1{ }^5 C _2+{ }^8 C _2{ }^7 C _2{ }^5 C _1+{ }^8 C _3{ }^7 C _1{ }^5 C _1$
$\Rightarrow 1960+2940+1960$
$\Rightarrow 6860$

Solution

$m _{ PQ } \cdot m _{ QR }=-1$
$\Rightarrow \frac{10-2}{9+3} \times \frac{10-4}{9-\alpha}=-1 \Rightarrow \alpha=13$
$m _{ OP } \cdot m _{ QS }=-1 \Rightarrow m _{ QS }=-\frac{4}{7}$

Equation of QS
$y-10=-\frac{4}{7}(x-9)$
$\Rightarrow 4 x+7 y=106 \ldots .(1)$
$m_{O R} \cdot m_{R S}=-1 \Rightarrow m_{R S}=-8$
Equation of RS
$y-4=-8(x-13)$
$\Rightarrow 8 x+y=108...$(2)
Solving eq. (1) & (2)
$x _1=\frac{25}{2} y _1=8$
$S \left( x _1, y _1\right) \text { lies on } 2 x - ky =1$
$25-8 k =1$
$\Rightarrow 8 k =24$
$\Rightarrow k =3$

Solution

$\cos 2 \theta \cdot \cos \frac{\theta}{2}=\cos 3 \theta \cdot \cos \frac{9 \theta}{2}$
$\Rightarrow 2 \cos 2 \theta \cdot \cos \frac{\theta}{2}=2 \cos \frac{9 \theta}{2} \cdot \cos 3 \theta$
$\Rightarrow \cos \frac{5 \theta}{2}+\cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2}+\cos \frac{3 \theta}{2}$
$\Rightarrow \cos \frac{15 \theta}{2}=\cos \frac{5 \theta}{2}$
$\Rightarrow \frac{15 \theta}{2}=2 k \pi \pm \frac{5 \theta}{2}$
$5 \theta=2 k \pi \text { or } 10 \theta=2 k \pi$
$\theta=\frac{2 k \pi}{5} \theta=\frac{ k \pi}{5}$
$\therefore \theta=\left\{-\pi, \frac{-4 \pi}{5}, \frac{-3 \pi}{5}, \frac{-2 \pi}{5}, \frac{-\pi}{5}, 0, \frac{\pi}{5}, \frac{2 \pi}{5}, \frac{3 \pi}{5}, \frac{4 \pi}{5}, \pi\right\}$
$m =5, n =5$
$\therefore m \cdot n =25$

Solution

$(2023)^{2023}$
$=(2030-7)^{2023}$
$=(35 K -7)^{2023}$
$={ }^{2023} C _0(35 K )^{2023}(-7)^0+{ }^{2023} C _1(35 K )^{2022}(-7)+\ldots .+$
$\ldots \cdots+{ }^{2023} C _{2022}(-7)^{2023}$
$=35 N -7^{2023} .$
$\text { Now },-7^{2023}=-7 \times 7^{2022}=-7\left(7^2\right)^{1011}$
$=-7(50-1)^{1011}$
$=-7\left({ }^{1011} C _0 50^{1011}-{ }^{1011} C _1(50)^{1010}+\ldots \ldots \cdot{ }^{1011} C _{1011}\right)$
$=-7(5 \lambda-1)$
$=-35 \lambda+7$
$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7

Solution

$E _1: \text { Smokers }$
$P \left( E _1\right)=\frac{1}{4}$
$E _2: \text { non-smokers }$
$P \left( E _2\right)=\frac{3}{4}$
E : diagnosed with lung cancer
$P\left(E / E_1\right)=\frac{27}{28}$
$P\left(E / E_2\right)=\frac{1}{28}$
$P\left(E_1 / E\right)=\frac{P\left(E_1\right) P\left(E / E_1\right)}{P(E)}$
$=\frac{\frac{1}{4} \times \frac{27}{28}}{\frac{1}{4} \times \frac{27}{28}+\frac{3}{4} \times \frac{1}{28}}=\frac{\not{27}^9}{\not{30}_{10}}=\frac{9}{10}$
$K=9$

Solution

$\vec{ r }=(\hat{ i }+2 \hat{ j }+3 \hat{ k })+\lambda(\hat{ i }+\hat{ j }+\hat{ k }) \vec{ r }=\vec{ a }+\lambda \vec{ p }$
$\vec{ r }=(+\hat{ i }-\hat{ j }+2 \hat{ k })+\mu(2 \hat{ i }-\hat{ j }) \vec{ r }=\vec{ b }+\mu \vec{ q }$
$\vec{ p } \times \vec{ q }=\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 1 & 1 \\ 2 & -1 & 0\end{vmatrix}=\hat{ i }+2 \hat{ j }-3 \hat{ k }$
$d =\left|\frac{(\vec{ b }-\vec{ a }) \cdot(\vec{ p } \times \vec{ q })}{|\vec{ p } \times \vec{ q }|}\right|$
$d =\left|\frac{(-3 \hat{ j }-\hat{ k }) \cdot(\hat{ i }+2 \hat{ j }-3 \hat{ k })}{\sqrt{14}}\right|$
$=\left|\frac{-6+3}{\sqrt{14}}\right|=\frac{3}{\sqrt{14}}$
$\alpha=\frac{3}{\sqrt{14}}$
Now, $28 \alpha^2=\not{28} ^2 \times \frac{9}{\not{14}}=18$

Solution

$\alpha+\beta=-60^{\frac{1}{4}} \& \alpha \beta= a$
Given $\alpha^4+\beta^4=-30$
$\Rightarrow\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2=-30$
$\Rightarrow\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}^2-2 a^2=-30$
$\Rightarrow\left\{60^{\frac{1}{2}}-2 a \right\}^2-2 a ^2=-30$
$\Rightarrow 60+4 a ^2-4 a \times 60^{\frac{1}{2}}-2 a ^2=-30$
$\Rightarrow 2 a ^2-4.60^{\frac{1}{2}} a +90=0$
Product $=\frac{90}{2}=45$

Solution

$a , b , \frac{1}{18} \rightarrow GP$
$\frac{ a }{18}= b ^2 ....$(i)
$\frac{1}{ a }, 10, \frac{1}{ b } \rightarrow AP$
$\frac{1}{ a }+\frac{1}{ b }=20$
$\Rightarrow a + b =20 ab \text {, from eq. (i); we get }$
$\Rightarrow 18 b ^2+ b =360 b ^3$
$\Rightarrow 360 b ^2-18 b -1=0 \{\because b \neq 0\}$
$\Rightarrow b =\frac{18 \pm \sqrt{324+1440}}{720}$
$\Rightarrow b =\frac{18+\sqrt{1764}}{720} \{\because b >0\}$
$\Rightarrow b =\frac{1}{12}$
$\Rightarrow a =18 \times \frac{1}{144}=\frac{1}{8}$
Now, $16 a+12 b=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3$