JEE Main Physics Question Paper with Solution 2022 July 25th Shift 1 - Morning
A
[PA−1T0]
B
[PAT−1]
C
[PA−1T]
D
[PA−1T−1]
Solution
Viscosity = pascal.second
PxAyTz=[M1L−1T−1]
[M1L+1T−1]x[L2]y[T1]z=M1L−1T−1
MxL+x+2yT−x+z=M1L−1T−1
x=1x+2y=−1−x+z=−1
y=−1
z=0
Viscosity =P1A−1T0
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=ϵ0→E
[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+R2−2R2cos135
⇒√2R2−2R2(−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=4≅x=0
vF=3(2)2∣4≅x=2
=16
W=ΔK=12m(162−42)
=12×12(256−16)
=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
F=GMMr2+√2GMM(√2r)2
=GMMr2(1+1√2)
=G×104132(1+1√2)
F≃100G
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=3√2
B
T1T2=√32
C
T1T2=√23
D
T1T2=√23
Solution
T1=2π√3m2k
T2=2π√m3k
T1T2=2π√3m2k2π√m3k=3√2
A

B

C

D

Solution
Q=CV
V=1CQ
Straight line with slope =1C
Slope =1C=12×10−6=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=12π√LC
To increase the resonating frequency product of L and C should decrease.
By joining capacitor in series, capacitor will decrease
A
2√2μ0L2πℓ
B
μ0ℓ22√2πL
C
2√2μ0ℓ2πL
D
μ0L22√2πℓ
Solution
Assuming current I in outer loop magnetic field at centre =4×μ0i4π×L2×(2sin45∘)=2√2μ0iπL
M= Flux through inner loop i
M=2√2μ0ℓ2πL
A
5pF
B
50pF
C
100pF
D
200pF
Solution
Current in capacitor I=VXC
I=(V)×(ωC)
C=IVω=6.9×10−6230×600=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
5×10−10vam
B
5×10−10m
C
1.5×10−10m
D
5×10−10vBm
Solution
μAμB=c/VAc/VB=VBVA=12
Let the thickness is d
dvB−dvA=5×10−10
d=5×10−10×vAvBvA−vB
As vA=2vB⇒d=5×10−10×2vB
Or d=5×10−10×vA
A
1.537eV
B
2.46eV
C
0.615eV
D
1.23eV
Solution
k1=1230800−ϕ.....(1)
k2=2k1=1230500−ϕ.....(2)
Eliminating k1 from (1) and (2) we get
0=1230500−1230400+ϕ
ϕ=0.615eV
Q18. The momentum of an electron revolving in nth orbit is given by : (Symbols have their usual meanings)
A
nh2πr
B
nh2r
C
nh2π
D
2πrnh
Solution
Angular momentum is integral multiple of h2πmvr=nh2π
So momentum mv =nh2πr
A
→μL=e→L2m
B
→μL=−e→L2m
C
→μ1=−e→Lm
D
→μ1=2e→Lm
Solution
Ratio of magnetic moment and angular momentum
→μ→L=q2m
For −
→μ=−e2m→L
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

Answer: 3
Solution
Stopping distance =v22a=d
If speed is made 13rd
d′=19d.d′=279=3
Braking acceleration remains same
Answer: 3
Solution
θ=45∘
Taking components along x & y
F1=√2−1√2=2−1√2=1√2
F2=√2+1√2=2+1√2=3√2
F1:F2=1:3
x=3
Answer: 5
Solution
Δℓ1=FℓAY=Fℓπr2Y=5cm
Δℓ2=4F4ℓπ16r2Y=Fℓπr2Y=5cm
Answer: 60
Solution
ℓB(1+αBΔT)−ℓi(1+αiΔT)=ℓB−ℓi
αBℓB=ℓiαi
1.8×10−5×40=ℓi×1.2×10−5
ℓi=1.8×10−5×401.2×10−5=3×402=60
ℓi=60cm
Answer: 20
Solution
VS=0,Vob =5m/s
fdirect =(320−5320)640=630Hz
freflected =(320+5320)640=650Hz
fbeat =650−630=20Hz
Answer: 45
Solution
No. of electric field lines per unit area = electric field.
E=ρr3∈0, for r=R
E=ρR3ϵ0=2×63×8.85×10−12=0.45×1012NC−1
=45×1010N/C
Answer: 4
Solution
By nodal analysis
V0−21kΩ+V0−41kΩ+V0−61kΩ=0
3V0−12=0
V0=4
Answer: 4
Solution
Each wire has resistance =ρ4ℓπd2=r
Eight wire in parallel, then equivalent resistance is r8=ρℓ2πd2
Single copper wire of length 2l has resistance
R=ρ2ℓ×4πd21=ρℓ2πd2
⇒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
3.2
B
4.2
C
5.2
D
6.2
Solution
pH =\frac{ pK _{ w }- pK _{ b }-\log C }{2}
=\frac{14-5+1.48}{2}=5.24
A
(A) - (II), (B) - (IV), (C) - (I), (D) - (III)
B
(A) - (II), (B) - (I), (C) - (IV), (D) - (III)
C
(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
D
(A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Solution
Factual
A
Unnilbium
B
Unnilunium
C
Unnilquadium
D
Unniltrium
Solution
Atomic Number 103
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
Gabriel Pthalimide reaction

A

B

C

D

Solution
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: 2
Solution
Diamagnetic species are: N _2, O _2^{2-}
Answer: 104
Solution
3 C _{( gr )}+4 H _{2( g )} \rightarrow C _3 H _{8( g )}
=-103.7 \,kJ\, mol ^{-1}
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: 4
Solution
E = E ^0-\frac{2.303 RT }{ nF } \log Q
Here, E =+0.801 \,V , E ^0=0.008-(-0.763)
=+0.771\, V
\therefore 0.801=+0.771-\frac{0.06}{ n } \log 10^{-2}
\Rightarrow n =4
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: 0
Solution
\Delta_0 \propto \frac{1}{\lambda}
Here, CN ^{-}being SFL will have maximum CFSE
So, \left[ Co ( CN )_6\right]^{3-} will be d ^2 sp ^3, \mu=0
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
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
\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