JEE Main Question Paper with Solution 2022 July 25th Shift 1 - Morning

JEE Main Physics Question Paper with Solution 2022 July 25th Shift 1 - Morning

A

[PA1T0]

B

[PAT1]

C

[PA1T]

D

[PA1T1]

Solution

Viscosity = pascal.second
PxAyTz=[M1L1T1]
[M1L+1T1]x[L2]y[T1]z=M1L1T1
MxL+x+2yTx+z=M1L1T1
x=1x+2y=1x+z=1
y=1
z=0
Viscosity =P1A1T0

A

Electric displacement (D) and surface charge density

B

Displacement current and electric field

C

Current density and surface charge density

D

Electric potential and energy

Solution

Electric displacement
D=ϵ0E
[D]=[ϵ0E]=[ϵ0σϵ0]
[D]=[σ]
Surface change density =σ.

A

42 m

B

47 m

C

19 m

D

40 m

Solution

d=Rθ
60=R(3π4)
R=60×43π=80πm
Displacement =R2+R22R2cos135
2R22R2(0.7)
3.4R2=3.4(80π)2
47m

A

64 J

B

60 J

C

120 J

D

128 J

Solution

vi=3(02)+4=4x=0
vF=3(2)24x=2
=16
W=ΔK=12m(16242)
=12×12(25616)
=2404=60J

A

53

B

45

C

35

D

1415

Solution

V=2gH1+k2/R2
Vcylinder Vsphere =(1+k2/R2)sphere (1+k2/R2)cylinder 
=1+2/51+1/2=75×23=1415

A

21 G

B

100 G

C

59 G

D

42 G

Solution

image
F=GMMr2+2GMM(2r)2
=GMMr2(1+12)
=G×104132(1+12)
F100G

A

3.536×105Pa

B

3.536×106Pa

C

1.25×106Pa

D

1.25×105Pa

Solution

P1=2×107Pa
P1V1=P2V2
Since V2=2V1 Hence P2=P1/2 (isothermal expansion)
P2=1×107Pa
P2(V2)γ=P3(2V2)γ
P3=1×10721.5=3.536×106

A

(1) and (4) only

B

(1), (2) and (4) only

C

(2) and (4) only

D

(1), (2) and (5) only

Solution

KEavg =32KT
P=13ρV2rms
Note : Statement (4) is correct only if we consider it at constant volume and not constant pressure. Ideally, this question must be bonus but most appropriate answer is option (A)

A

T1T2=32

B

T1T2=32

C

T1T2=23

D

T1T2=23

Solution

T1=2π3m2k
T2=2πm3k
T1T2=2π3m2k2πm3k=32

A

B

C

D

Solution

Q=CV
V=1CQ
Straight line with slope =1C
Slope =1C=12×106=5×105

A

8:5

B

5:4

C

5:3

D

8:7

Solution

Radius of circular path R=2mkqB
q=2mkRB
q1q2=m1m2×R2R1=94×56=54

A

Source frequency should be increased

B

Another resistance should be added in series with the first resistance.

C

Another capacitor should be added in series with the first capacitor

D

The source frequency should be decreased

Solution

f=12πLC
To increase the resonating frequency product of L and C should decrease.
By joining capacitor in series, capacitor will decrease

A

22μ0L2π

B

μ0222πL

C

22μ02πL

D

μ0L222π

Solution

Assuming current I in outer loop magnetic field at centre =4×μ0i4π×L2×(2sin45)=22μ0iπL
image
M= Flux through inner loop i
M=22μ02πL

A

5pF

B

50pF

C

100pF

D

200pF

Solution

Current in capacitor I=VXC
I=(V)×(ωC)
C=IVω=6.9×106230×600=50pF

A

In primary rainbow, observer sees red colour on the top and violet on the bottom

B

In primary rainbow, observer sees violet colour on the top and red on the bottom

C

In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops

D

Primary rainbow is less bright than secondary rainbow.

Solution

In primary rainbow, red colour is at top and violet is at bottom.
Intensity of secondary rainbow is less in comparison to primary rainbow.

A

5×1010vam

B

5×1010m

C

1.5×1010m

D

5×1010vBm

Solution

μAμB=c/VAc/VB=VBVA=12
Let the thickness is d
dvBdvA=5×1010
d=5×1010×vAvBvAvB
As vA=2vBd=5×1010×2vB
Or d=5×1010×vA

A

1.537eV

B

2.46eV

C

0.615eV

D

1.23eV

Solution

k1=1230800ϕ.....(1)
k2=2k1=1230500ϕ.....(2)
Eliminating k1 from (1) and (2) we get
0=12305001230400+ϕ
ϕ=0.615eV

A

nh2πr

B

nh2r

C

nh2π

D

2πrnh

Solution

Angular momentum is integral multiple of h2πmvr=nh2π
So momentum mv =nh2πr

A

μL=eL2m

B

μL=eL2m

C

μ1=eLm

D

μ1=2eLm

Solution

Ratio of magnetic moment and angular momentum
μL=q2m
For
μ=e2mL

A

B

C

D

Solution

When both A and B have logical value ' 1 '
both diode are reverse bias and current will flow in resistor hence output will be 5 volt i.e. logical value ' 1 '.
In all other case conduction will take place, hence output will be zero volt i.e. logical value ' 0 '.
So truth table is
image

Answer: 3

Solution

Stopping distance =v22a=d
If speed is made 13rd
d=19d.d=279=3
Braking acceleration remains same

Answer: 3

Solution

image
θ=45
Taking components along x & y
F1=212=212=12
F2=2+12=2+12=32
F1:F2=1:3
x=3

Answer: 5

Solution

Δ1=FAY=Fπr2Y=5cm
Δ2=4F4π16r2Y=Fπr2Y=5cm

Answer: 60

Solution

B(1+αBΔT)i(1+αiΔT)=Bi
αBB=iαi
1.8×105×40=i×1.2×105
i=1.8×105×401.2×105=3×402=60
i=60cm

Answer: 20

Solution

image
VS=0,Vob =5m/s
fdirect =(3205320)640=630Hz
freflected =(320+5320)640=650Hz
fbeat =650630=20Hz

Answer: 45

Solution

No. of electric field lines per unit area = electric field.
E=ρr30, for r=R
E=ρR3ϵ0=2×63×8.85×1012=0.45×1012NC1
=45×1010N/C

Answer: 4

Solution

image
By nodal analysis
V021kΩ+V041kΩ+V061kΩ=0
3V012=0
V0=4

Answer: 4

Solution

Each wire has resistance =ρ4πd2=r
Eight wire in parallel, then equivalent resistance is r8=ρ2πd2
Single copper wire of length 2l has resistance
R=ρ2×4πd21=ρ2πd2
d1=4d

Answer: 3

Solution

E _{ g }=\frac{ hc }{\lambda}=\frac{1242}{\lambda( nm )}=\frac{1242}{400}=3.105
Answer rounded to 3\, eV

Answer: 150

Solution

d =\sqrt{2 Rh }
d =\sqrt{2 \times 6400 \times h \times 10^{-3}}( h in m )
Area =\pi d ^2
=\left(\pi \times 2 \times 6400 \times h \times 10^{-3}\right) km ^2
6.03 \times 100000=100 \times \pi \times 2 \times 6400 \times 10^{-3} h
h =\frac{6.03 \times 10^5}{10 \times \pi \times 128}
h =150 \,m

JEE Main Chemistry Question Paper with Solution 2022 July 25th Shift 1 - Morning

A

16

B

8

C

4

D

2

Solution

Let n \left( SO _2 Cl _2\right)= x moles
\therefore n \left( H _2 SO _4\right)= x , n ( HCl )=2 x
\Rightarrow n \left( H ^{+}\right)=4 x
For Neutralisation
\Rightarrow n \left( H ^{+}\right)= n \left( OH ^{-}\right)
\Rightarrow 4 x =16
\Rightarrow x =4

A

n =3,1=2, m _{ l }=0, s =+\frac{1}{2}

B

n =3,1=2, m _1=-2, s =+\frac{1}{2}

C

n =3,1=3, m _{ l }=-3, s =-\frac{1}{2}

D

n =3,1=0, m _1=0, s =-\frac{1}{2}

Solution

1=0,1,2 \ldots \ldots(n-1)
\therefore \text { for } n =3
1=0,1,2
\Rightarrow 1=3
not possible for n =3

A

0.8

B

1.1

C

1.9

D

2.4

Solution

{[ HCOOH ]=0.5 \,ml ^{-1}}
\Rightarrow\left(0.5\, ml \times 1.05 \,g \,ml ^{-1}\right) HCOOH \text { in } 1 L
\Rightarrow 0.525 \,g \,HCOOH in 1 L
m =\frac{(0.525 / 46)}{1 \,kg } mol
[Assuming dilute solution]
\therefore \Delta T _{ f }= iK _{ f } m \Rightarrow i =\frac{\Delta T _{ f }}{ k _{ f } m }=\frac{0.0405 \times 46}{1.86 \times 0.525}=1.9

A

(3) (4) Only

B

(1), (2), (5) Only

C

(1), (2) Only

D

(2) Only

Solution

FeO + SiO _2 \rightarrow FeSiO _3

A

Mn ^{2+}

B

Mn ^{4+}

C

Mn ^{3+}

D

Mn ^{6+}

Solution

H _2 O _2+ MnO _4^{-} \rightarrow Mn ^{2+}+ O _2 (unbalanced )

A

Li < K < Na < Rb < Cs

B

Li < Na < K < Rb < Cs

C

Cs < Rb < K < Na < Li

D

Li < Na < K < Cs < Rb

Solution

Factual

A

trigonal planar

B

tetrahedral

C

pyramidal

D

square planar

Solution

BF_3 + NaH \xrightarrow{450\,K} \underset{\text{(diborane)}}{B_2H_6} + NaF
B_2H_6 + NMe_3 \rightarrow \underset{\text{symmetrical cleavage}}{2[BH_3 \leftarrow NMe_3]}

A

hypohalite

B

halate

C

perhalate

D

halite

Solution

Br_2 + \underset{\text{(excess)}}{ 5F_2} \rightarrow \overset{+5}{2BrF_5} \xrightarrow{H_2O} HBrO_3 (Forms bromate)

A

NO

B

NO _2

C

SO _2

D

HCHO

Solution

Factual

A

KMnO _4 / H ^{+}and dil. KMnO _4, 273 \,K

B

KMnO _4, (dilute), 273 K and KMnO _4 / H ^{+}

C

KMnO _4 / H ^{+}and O _3, H _2 O / Zn

D

O _3, H _2 O / Zn and KMnO _4 / H ^{+}

A

B

C

D

Solution

image
Formaldehyde HCHO
Melamine formaldehyde Resin is melamine polymer

A

Primary

B

Secondary

C

Tertiary

D

Quaternary

Solution

Primary structure remains intact during denaturation of proteins

A

Agonists

B

Antagonists

C

Allosterists

D

Anti histaminists

Solution

Factual

A

Both Statement I and Statement II are correct.

B

Both Statement I and Statement II are incorrect

C

Statement I is correct but Statement II is incorrect.

D

Statement I is incorrect but Statement II is correct.

Solution

Acrolein has a pungent, suffocating odour. Acrolein is used to detect presence of glycerol

Answer: 22

Solution

{\left[ P _{ gas }\right]_0+\text { V.P. }=4}
{\left[ P _{ gas }\right]_0=4-0.4=3.6}
As volume is doubled, \left[ P _{ gas }\right]_{\text {new }}=1.8\, atm
New Total Pressure =1.8+0.4=2.2 a\,tm

Answer: 1

Solution

\left( t _{1 / 2}\right)_{500 \text { torr }}=240 \,\sec =4\,\min.
\left( t _{1 / 2}\right)_{250 \text { torr }}=4 \,\min.
t _{1 / 2} \propto a ^{1- n }
As t_{1 / 2} is independent of initial pressure. Hence, order is 1 st order.

Answer: 5

Solution

Co ^{3+} can't liberate H _2.
It has d ^6 configuration,
Number of unpaired electrons =4
\mu=\sqrt{4 \times 6}=4.92 \text { B.M. }

Answer: 56

Solution

\% N =\frac{1.4\left( N _1 V _1\right)}{\text { massof organic compound }}
\% N =\frac{1.4(2.5 \times 2 \times 2)}{0.25}=56

JEE Main Mathematics Question Paper with Solution 2022 July 25th Shift 1 - Morning

A

60

B

90

C

108

D

126

Solution

A =\{1,2,3,4\}
B =\{1,2,3,4,5,6\}
Here f(3) can be 2,3,4,5,6
f (3)=2,( f (1), f (2)) \rightarrow(1,1) \rightarrow 6 cases
f (3)=3,( f (1), f (2)) \rightarrow(1,2),(2,1)
\rightarrow 2 \times 6=12 cases
f (3)=4,( f (1), f (2)) \rightarrow(1,3),(3,1),(2,2)
\rightarrow 3 \times 6=18 cases
f (3)=5,( f (1), f (2)) \rightarrow(1,4),(4,1),(2,3)
(3,2)
\rightarrow 4 \times 6=24 cases
f (3)=6, ( f (1), f (2))
(1,5),(5,1),(2,4),(4,2),(3,3)
\rightarrow 5 \times 6=30 cases
Total number of cases =6+12+18+24+ 30=90

A

-4

B

-1

C

1

D

4

Solution

\alpha, \beta, \gamma, \delta root of the equation
x^4+x^3+x^2+x+1=0
Which are 5^{\text {th }} roots of unity except 1 . then \alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}= \alpha+\beta+\gamma+\delta=-1

A

0

B

2

C

3

D

4

Solution

S_n:|z-(3-2 i)|=\frac{n}{4} is a circle center C _1(3,-2) and radius n / 4
T _{ n }:| z -(2-3 i )|=\frac{1}{ n } is a circle center C _2 (2,-3)
and radius 1 / n
Here S_n \cap T_n=\phi
Both circles do not intersect each other
Case-1 : C _1 C _2> n / 4+1 / n
\sqrt{2}>\frac{ n }{4}+\frac{1}{ n }
then n =1,2,3,4
Case-2 : C _1 C _2<\left|\frac{ n }{4}-\frac{1}{ n }\right|
\Rightarrow \sqrt{2} < \left|\frac{n^2-4}{4 n}\right|
\Rightarrow n has infinite solutions for n \in N

A

6

B

7

C

8

D

9

Solution

The system of equation has no solution.
D = \begin{vmatrix}3 \sin 3 \theta & -1 & 1 \\3 \cos 2 \theta & 4 & 3 \\6 & 7 & 7\end{vmatrix}=0
21 \sin 3 \theta+42 \cos 2 \theta-42=0
\sin 3 \theta+2 \cos 2 \theta-2=0
Number of solution is 7 in (0,4 \pi)

A

4

B

-8

C

-4

D

8

Solution

\displaystyle\lim _{n \rightarrow \infty} n\left(1-\frac{n+1}{n^2}\right)^{\frac{1}{2}}+\alpha n+\beta=0
\displaystyle\lim _{n \rightarrow \infty}\left\{11-\frac{1}{2}\left(\frac{n+1}{n^2}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2 !}\left(\frac{n+1}{n^2}\right)^2+\ldots .\right\}+\alpha n+\beta=0
\displaystyle\lim _{n \rightarrow \infty} n-\frac{1}{2}+\frac{1}{n}+\ldots .+n \alpha+\beta=0
\alpha=-1, \beta=\frac{1}{2}
8(\alpha+\beta)=-4

A

\alpha=0

B

\alpha=-3

C

\alpha \in(-1,0)

D

\alpha \in(-3,-1)

Solution

f ^{\prime}( x )= e ^{\left(4 x ^3-12 x ^2-180 x +31\right)}\left(12\left( x ^2-2 x +7\right)( x +3)( x -5)+2( x -1)\right)
\text { for } x \in[-3,0]
\Rightarrow f ^{\prime}( x ) < 0
f(x) is decreasing function on [-3, 0]
The absolute maximum value of the function f(x) is at x=-3
\Rightarrow \alpha=-3

A

\frac{27}{4}

B

\frac{29}{4}

C

\frac{37}{4}

D

\frac{9}{2}

Solution

y(x)=a x^3+b x^2+c x+5 is passing through (-2,0) then 8 a-4 b+2 c=5 \ldots \ldots (1)
y^{\prime}(x)=3 a x^2+2 b x+c touches x-axis at (- 2,0)
12 a-4 b+c=0 .....(2)
again, for x=0, y^{\prime}(x)=3
c =3 .....(3)
Solving eq. (1), (2) \& (3) a =-\frac{1}{2}, b =-\frac{3}{4}
y^{\prime}(x)=-\frac{3}{2} x^2-\frac{3}{2} x+3
y(x) has local maxima at x=1
y(1)=\frac{27}{4}

A

\frac{31}{8}

B

\frac{17}{6}

C

\frac{19}{6}

D

\frac{27}{8}

Solution

image
A=\int\limits_{-1}^{\frac{1}{2}}\left(x+2-x^2\right) d x+\int\limits_{\frac{1}{2}}^1\left(4-3 x-x^2\right) d x=\frac{17}{6}

A

4

B

2

C

1

D

0

Solution

f(x) is periodic function whose period is 2
\frac{\pi^2}{10} \int\limits_{-10}^{10} f(x) \cos \pi x d x=\frac{\bar{\pi}^2}{10} \times 10 \int\limits_0^2 f(x) \cos \pi x d x
=\pi^2\left(\int\limits_0^1(1-x) \cos \pi x d x+\int\limits_1^2(x-1) \cos \pi x d x\right)
Using by parts
=\pi^2 \times \frac{4}{\pi^2}=4

A

\frac{3+\sqrt{2}}{3-\sqrt{2}}

B

\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)

C

\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)

D

\frac{\sqrt{2}+1}{\sqrt{2}-1}

Solution

\frac{d y}{d x}=\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}
\frac{d y}{d x}=e^{2 x}-\frac{6 e^x}{2 e^{2 x}+9}
y=\frac{e^{2 x}}{2}-\tan ^{-1}\left(\frac{\sqrt{2} e^x}{3}\right)+c
If C passes through the point \left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)
c=-\frac{\pi}{4}-\tan ^{-1} \frac{\sqrt{2}}{3}
Again C passes through the point \left(\alpha, \frac{1}{2} e ^{2 \alpha}\right)
then e ^\alpha=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)

A

\left(y^2+x\right)^4=C\left|\left(y^2+2 x\right)^3\right|

B

\left( y ^2+2 x \right)^4= C \left|\left( y ^2+ x \right)^3\right|

C

\left|\left(y^2+x\right)^3\right|=C\left(2 y^2+x\right)^4

D

\left|\left(y^2+2 x\right)^3\right|=C\left(2 y^2+x\right)^4

Solution

\left(x-y^2\right) d x+y\left(5 x+y^2\right) d y=0
\frac{d y}{d x}=\frac{y^2-x}{y\left(5 x+y^2\right)} . \text { Let } y^2=v
\frac{2 y d y}{d x}=2\left(\frac{y^2-x}{5 x+y^2}\right)
\frac{d v}{d x}=2\left(\frac{v-x}{5 x+v}\right) v=k x
k + x \frac{ dk }{ dx }=2\left(\frac{ kx - x }{5 x + kx }\right)
x \frac{ dk }{ dx }=-\frac{\left( k ^2+3 k +2\right)}{ k +5}
\int \frac{(5+k)}{(k+1)(k+2)} d k=\int-\frac{d x}{x}
\int\left(\frac{4}{k+1}-\frac{3}{k+2}\right) d k=-\int \frac{d x}{x}
4 \ln (k+1)-3 \ln (k+2)=-\ln x+\ln c
\frac{(k+1)^4}{(k+2)^3}=-\ln x+\ln c
c\left(y^2+2 x\right)^3=\left(y^2+x\right)^4

A

2 x+y=9

B

3 x-2 y=7

C

x+2 y=6

D

2 x -3 y =3

Solution

Let B\left(x_1, x_1-2\right)
\sqrt{\left(x_1-4\right)^2+\left(x_1-2-3\right)^2}=\frac{\sqrt{29}}{3}
Squaring on both side 18 x _1^2-162 x _1+340=0
x _1=\frac{51}{9} or x _1=\frac{10}{3}
y _1=\frac{33}{9} or y _1=\frac{4}{3}
Option (C) will satisfy \left(\frac{10}{3}, \frac{4}{3}\right)

A

\frac{32 \sqrt{2}}{3}

B

\frac{40 \sqrt{2}}{3}

C

\frac{64}{3}

D

\frac{32}{3}

Solution

(\alpha-0)^2+(\beta-1)^2=(\beta+1)^2
\alpha^2=4 \beta
x^2=4 y
A=2 \int\limits_0^4\left(4-\frac{x^2}{4}\right) d x=\frac{64}{3}

A

\frac{147}{2}

B

96

C

\frac{32}{3}

D

54

Solution

a ( x -3)+ b ( y +4)+ c ( z -7)=0
P : 9 a - b -5 c =0
-11 a - b +5 c =0
After solving DR's \propto(1,-1,2)
Equation of plane
x - y +2 z =21
d =\frac{8}{\sqrt{6}}
d ^2=\frac{32}{3}

A

both (S1) and (S2) are true

B

only (S1) is true

C

only (S2) is true

D

both (S1) and (S2) are false

Solution

\overrightarrow{ a }+\overrightarrow{ b }+\overrightarrow{ c }=0
\overrightarrow{ b }+\overrightarrow{ c }=-\overrightarrow{ a }
|\overrightarrow{ b }|^2+|\overrightarrow{ c }|^2+2 \overrightarrow{ b } \cdot \overrightarrow{ c }=|\overrightarrow{ a }|^2
|\overrightarrow{ c }|^2=36
|\overrightarrow{ c }|^2=6
S1: |\vec{a} \times \vec{b}+\overrightarrow{ c } \times \overrightarrow{ b }|-|\overrightarrow{ c }|
|(\overrightarrow{ a }+\overrightarrow{ c }) \times \overrightarrow{ b }|-|\overrightarrow{ c }|
|-\overrightarrow{ b } \times \overrightarrow{ b }|-|\overrightarrow{ c }|
0-6=-6
S 2: \overrightarrow{ a }+\overrightarrow{ b }+\overrightarrow{ c }=0
\overrightarrow{ b }+\overrightarrow{ c }=-\overrightarrow{ a }
|\overrightarrow{ a }|^2+|\overrightarrow{ b }|^2-2|\overrightarrow{ a }||\overrightarrow{ b }| \cos (\angle ACB )=|\overrightarrow{ c }|^2
\cos (\angle ACB )=\sqrt{\frac{2}{3}}

A

\frac{33}{2^{32}}

B

\frac{33}{2^{29}}

C

\frac{33}{2^{28}}

D

\frac{33}{2^{27}}

Solution

n p+n p q=24 ....(1)
n p \cdot n p q=128....(2)
Solving (1) and (2):
We get p =\frac{1}{2}, q =\frac{1}{2}, n =32.
Now,
P ( X =1)+ P ( X =2)
={ }^{32} C _1 pq ^{31}+{ }^{32} C _2 p ^2 q ^{30}
=\frac{33}{2^{28}}

A

\frac{17}{36}

B

\frac{4}{9}

C

\frac{1}{2}

D

\frac{19}{36}

Solution

x ^2+\alpha x +\beta>0, \forall x \in R
D =\alpha^2-4 \beta<0
\alpha^2<4 \beta
Total cases = 6 \times 6=36
Fav. cases =\beta=1, \alpha=1
\beta=2, \alpha=1,2
\beta=3, \alpha=1,2,3
\beta=4, \alpha=1,2,3
\beta=5, \alpha=1,2,3,4
\beta=6, \alpha=1,2,3,4
Total favourable cases =17
P ( x )=\frac{17}{36}

A

5(2 \sqrt{3}+3) m

B

5(\sqrt{3}+3) m

C

10(\sqrt{3}+1) m

D

10(2 \sqrt{3}+1) m

Solution

image
\frac{15}{ AQ }=\tan 60^{\circ}
\frac{15+x}{ AQ }=\tan 75^{\circ}
\frac{(1)}{(2)} \Rightarrow x=10 \sqrt{3}So, P Q=5(2 \sqrt{3}+3) m

A

((\sim p) \vee q) \Rightarrow p

B

p \Rightarrow((\sim p ) \vee q )

C

((\sim p ) \vee q ) \Rightarrow q

D

q \Rightarrow((\sim p) \vee q)

Answer: 7

Solution

A= \begin{bmatrix} 2 & -1 & -1 \\1 & 0 & -1 \\ 1 & -1 & 0\end{bmatrix} \Rightarrow A^2=A \Rightarrow A^n=A
\forall n \in\{1,2, \ldots, 100\}
Now, B = A - I =\begin{bmatrix} 1 & -1 & -1 \\1 & -1 & -1 \\1 & -1 & -1\end{bmatrix}
B ^2 =- B
\Rightarrow B ^3 =- B ^2= B
\Rightarrow B ^5= B
\Rightarrow B ^{99}= B
Also, \omega^{3 k }=1
So, n = common of \{1,3,5, \ldots, 99\} and \{3,6,9, \ldots, 99\}=17

Answer: 6006

Solution

\left( t ^2 x ^{\frac{1}{5}}+\frac{(1- x )^{\frac{1}{10}}}{ t }\right)^{15}
T _{ r +1}={ }^{15} C _{ r }\left( t ^2 x ^{\frac{1}{5}}\right)^{15- r } \cdot \frac{(1- x )^{\frac{ r }{10}}}{ t ^{ r }}
For independent of t,
30-2 r - r =0
\Rightarrow r =10
So, Maximum value of { }^{15} C _{10} x (1- x ) will be at
x =\frac{1}{2}
i.e. 6006

Answer: 38

Solution

x^2-8 a x+2 a=0
p+r=8 a
p r=2 a
\frac{1}{p}+\frac{1}{r}=4
\frac{2}{q}=4
q=\frac{1}{2}
p=\frac{1}{5}
x^2+12 b x+6 b=0
q+s=-12 b
q s=6 b
\frac{1}{q}+\frac{1}{s}=-2
\frac{2}{r}=-2
r =-1
s=\frac{-1}{4}
Now, \frac{1}{ a }-\frac{1}{ b }=\frac{2}{ pr }-\frac{6}{ qs }=38

Answer: 27560

Solution

a _1= b _1=1
a _2= a _1+2=3
a _3= a _2+2=5
a _4= a _2+2=7
\Rightarrow a _{ n }=2 n -1
b _2= a _1+ b _1=4
b _3= a _3+ b _2=9
b _4= a _4+ b _3=16
b _{ n }= n ^2
\displaystyle\sum_{n=1}^{15} a_n b_n
\displaystyle\sum_{n=1}^{15}(2 n-1) n^2
\displaystyle\sum_{n=1}^{15}\left(2 n^3-n^2\right)
=2 \frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2 n+1)}{6}
Put n=15
=\frac{2 \times 225 \times 16 \times 16}{4}-\frac{15 \times 16 \times 31}{6}=27560

Answer: 5

Solution

LHS
\displaystyle\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[n k \cdot n+1+2+\ldots+n]
= \displaystyle\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} \cdot\left[n^2 k+\frac{n(n+1)}{2}\right]
= \displaystyle\lim _{n \rightarrow \infty} \frac{( n +1)^{ k -1} \cdot n ^2\left( k +\frac{\left(1+\frac{1}{ n }\right)}{2}\right)}{ n ^{ k +1}}
\Rightarrow \displaystyle\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)
\Rightarrow\left(k+\frac{1}{2}\right)
\text { RHS }
\Rightarrow \displaystyle\lim _{ n \rightarrow \infty} \frac{1}{n^{ k +1}}\left(1^{ k }+2^{ k }+\ldots+ n ^{ k }\right)=\frac{1}{ k +1}
\text { LHS }=\text { RHS }
\Rightarrow k +\frac{1}{2}=33 \cdot \frac{1}{ k +1}
\Rightarrow(2 k +1)( k +1)=66
\Rightarrow( k -5)(2 k +13)=0
\Rightarrow k =5 \text { or }-\frac{13}{2}

Answer: 2

Solution

2 p + f -1=0 .....(1)
2-p f -4 p =0 .....(2)
2=p( f +4)
p =\frac{2}{ f +4}
2 p =1- f
\frac{4}{ f +4}=1- f
f ^2+3 f =0
f =0 \text { or }-3
Hyperbola 3 x^2-y^2=3, x^2-\frac{y^2}{3}=1
y = mx \pm \sqrt{ m ^2-3}
It passes (1,0)
o = m \pm \sqrt{ m ^2-3}
m tends \infty
It passes (1,3)
3=m \pm \sqrt{m^2-3}
(3-m)^2=m^2-3
m=2

Answer: 10

Solution

x^2=\frac{64.5}{75}\left(y-\frac{3}{5}\right)
equation of tangent at \left(\frac{8}{5}, \frac{6}{5}\right)
x \cdot \frac{8}{5}=\frac{64}{15}\left(\frac{y+\frac{6}{5}}{2}-\frac{3}{5}\right)
3 x-4 y=0
equation of family of circle is
\left(x-\frac{8}{5}\right)^2+\left(y-\frac{6}{5}\right)^2+\lambda(3 x-4 y)=0
It touches y axis so f ^2= c
x^2+y^2+x\left(3 \lambda-\frac{16}{5}\right)+y\left(-4 \lambda-\frac{12}{5}\right)+4=0
\frac{\left(4 \lambda+\frac{12}{5}\right)^2}{4}=4
\lambda=\frac{2}{5} \text { or } \lambda=-\frac{8}{5}
\lambda=\frac{2}{5}, r=1
\lambda=-\frac{8}{5}, r=4
d_1+d_2=10

Answer: 3

Solution

DR's of line of shortest distance
\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\0 & 1 & 1 \\2 & 2 & 1\end{vmatrix}=-\hat{ i }+2 \hat{ j }-2 \hat{ k }
angle between line and plane is \cos ^{-1} \sqrt{\frac{2}{27}}=\alpha
\cos \alpha=\sqrt{\frac{2}{27}}, \sin \alpha=\frac{5}{3 \sqrt{3}}
DR's normal to plane (1,-1,-1)
\sin \alpha=\left|\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^2+1+1}}\right|=\frac{5}{3 \sqrt{3}}
\sqrt{3}| a |=5 \sqrt{a^2+2}
3 a^2=25 a^2+50
No value of (a)