JEE Main Physics Question Paper with Solution 2023 January 30th Shift 2 - Evening
A
A-IV, B-I, C-III, D-II
B
A-I, B-IV, C-III, D-II
C
A-IV, B-III, C-I, D-II
D
A-IV, B-I, C-II, D-III
A

B

C

D

Solution
Given circuit represent XOR.
A
Both A and R are true and R is the correct explanation of A
B
A is false but R is true
C
Both A and R are true but R is NOT the correct explanation of A
D
A is true but R is false
Solution
Nuclear density is independent of A.
A
EI=0,EII=0,EIII=0
B
EI≠0,EII=0,EII=0
C
EI=0,EII=0,EIII≠0
D
EI≠0,EII=0,EIII≠0
Solution
Electric field inside material of conductor is zero.
A
√gR
B
2√gR
C
√2gR
D
√gR2
Solution
Loss in PE = Gain in KE
(−GMm2R)−(−GMmR)=12mv2
⇒v2=GMR=gR
⇒v=√gR
A
2.77×10−2mm
B
6.9×10−2mm
C
6.06×10−2mm
D
3.0×10−2mm
Solution
Y=F/AΔℓℓ
⇒F=YAℓΔℓ
(AΔℓℓ)1=(AΔℓℓ)2
⇒Δℓ2Δℓ1=A1A2×ℓ2ℓ1
⇒Δℓ20.2=12.4×2.4×21
⇒Δℓ2=6.9×10−2mm
A
0.6m/s
B
2.5m/s
C
1.5m/s
D
0.02m/s
Solution
20×10−3×18060×100=10V
⇒v=0.6m/s
A
1:5
B
1:3
C
1:2
D
1:4
Solution
F∝I1I2
F1:F2I=1:4
A
2:1
B
1:4
C
1:1
D
4:1
Solution
Kav=52kT
Ratio =1:1
A
4√3×10−4T
B
√3×10−4T
C
4√3×10−5T
D
3√3×10−5T
Solution
dtan60∘=2√3
d=2cm
B=3×μ0i2πdsin60∘
=3×2×10−7×22×10−2×√32
=3√3×10−5
A
25N
B
15N
C
10N
D
20N
Solution
θ=30∘
cosθ=√3gT
⇒√32=√3gT
⇒T=20N
A
23Ω
B
32Ω
C
13Ω
D
12Ω
Solution
1Req=12+112+14+16+12
=6+1+3+2+612=1812=32
⇒Req=23Ω
A
A is true but R is false
B
Both A and R are true and R is the correct explanation of A
C
Both A and R are true but R is NOT the correct explanation of A
D
A is false but R is true
Solution
Both A and R are true and R is the correct explanation of A
A
4
B
94
C
3
D
32
Solution
KE=P22m,
P=hλ
eV1=(hλ)22m
eV2=(h1.5λ)22m
V1V2=(1.5)2=94
A
A-II, B-III, C-IV, D-I
B
A-IV, B-III, C-II, D-I
C
A-I, B-II, C-III, D-IV
D
A-IV, B-III, C-I, D-II
A
12A
B
20A
C
2A
D
2√2A
Solution
z=√1002+(200−100)2
=100√2Ω
irms=Vrmsz=200√2100√2
=2A
A
3.75km/h
B
4.25km/h
C
3.50km/h
D
4.00km/h
Solution
2Vav=13+15=815
⇒Vav=154=3.75km/h
A
7.8∘
B
4.5∘
C
6 circ
D
1.3∘
Solution
δ1=δ2 [for no average deviation]
⇒6∘(1.54−1)=A(1.72−1)
⇒A=6∘×0.540.72
=18∘4=4.5∘
A
140πWm2
B
110πWm2
C
120πWm2
D
12πWm2
Solution
IEF=12×54π×52
=140πW/m2
Answer: 12
Solution
→E=2x2ˆi−4yˆj+6ˆk
ϕnet =−8×3+2×6=−12
−12=qϵ0
|q|=12ϵ0
Answer: 54
Solution
a=−μkg=−3
V=18−3×2
V=12m/s
KE=12mv2+12mr22v2r2
KE=34mv2
KE=3×18=54J
Answer: 313
Solution
41∘−5∘95∘−5∘=C−0∘100∘−0∘
⇒C=3690×100=40∘C=313K
Answer: 4
Solution
12mV2=Pt
V=√2Ptm
dxdt=√2Ptm
x=√2Pm23[t3/2]40
x=16√P3=13×16√P
α=4
Q25. If the potential difference between B and D is zero, the value of x is 1nΩ. The value of n is ______

Answer: 2
Solution
23=xx+1x
⇒23=1x+1
⇒x=0.5=12
n=2
Answer: 125
Solution
a=ω2R=(28×2π60)2×1.8
=(5660×227)2×1.8
=(44)2225×1.8
=1936×1.8225
x=125
Answer: 3
Solution
I=4I0cos2(Δϕ2)
I1=4I0cos2(π4)=2I0
I2=4I0cos2(2π3)=I0
⇒I1+I2I0=3
Answer: 1584
Solution
ξmax
=100 \times 14 \times 10^{-2} \times 3 \times \frac{360 \times 2 \pi}{60}
=1584\, V
Answer: 88
Solution
4 v ^2=50- x ^2
\Rightarrow v =\frac{1}{2} \sqrt{50- x ^2}
\omega=\frac{1}{2}
T =\frac{2 \pi}{\omega}=4 \pi=\frac{88}{7}
x =88
Answer: 300
Solution
\frac{ dN _1}{ dt }=-\lambda_1 N
\frac{ dN _2}{ dt }=-\lambda_2 N
\frac{ dN }{ dt }=-\left(\lambda_1+\lambda_2\right) N
\Rightarrow \lambda_{ eq }=\lambda_1+\lambda_2
\Rightarrow \frac{1}{ t _{1 / 2}}=\frac{1}{300}+\frac{1}{30}=\frac{11}{300}
\Rightarrow t _{1 / 2}=\frac{300}{11}
JEE Main Chemistry Question Paper with Solution 2023 January 30th Shift 2 - Evening
A
Strong hydrogen bond in Boric acid
B
Strong van der Waal's interaction in Boric acid
C
Strong covalent bond in BF _3
D
Strong ionic bond in Boric acid
Solution
Boric acid has strong hydrogen bonding while BF_3 does not. Therefore boric acid is solid
A
180^{\circ}
B
90^{\circ}
C
90^{\circ} \& 180^{\circ}
D
90^{\circ} \& 120^{\circ}
Solution
The Cl - Co - Cl bond angle in above octahedral complex is 90^{\circ}
A
IO _3^{-} \& IO _3^{-}
B
I _2 \& IO _3^{-}
C
I _2 \& I _2
D
IO _3^{-} \& I _2
Solution
In acidic medium
2 MnO _4^{-}+10 I ^{-}+16 H ^{+} \rightarrow 2 Mn ^{2+}+5 I _2+8 H _2 O
In neutral/faintly alkaline solution
2 MnO _4^{-}+ I ^{-}+ H _2 O \rightarrow 2 MnO _2+2 OH ^{-}+ IO _3^{-}
A
Water has high amount of fluoride compounds
B
Highly polluted water
C
Very clean water
D
Slightly polluted water
Solution
Clean water as BOD value of < 5 while polluted water has BOD of 15 or more.
A
a>b>c>d
B
b > a > d > c
C
b > d > a > c
D
c > a > d > b
Solution
Due to - M effect of - NO _2 group, it increases acidity + M effect of N \left( CH _3\right)_2 decreases acidity.
Hyperconjugation of isopropyl decrease acidity
\therefore order of acidic strength
(c) > (a) > (d) > (b)
A
A is false but R is true
B
Both A and R are true and R is the correct explanation of A
C
A is true but R is false
D
Both A and R are true but R is not the correct explanation of A
Solution
Antiallergic and antacid drugs work on different receptors
A
A is false but R is true
B
Both A and R are true but R is not the correct explanation of A
C
A is true but R is false
D
Both A and R are true and R is the correct explanation of A
Solution
The acid sensitive alcohol group reacts with HCl,
hence Clemmenson reduction is not suitable for
above conversion.
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
In Electrolytic refining, the pure metal is used as cathode and impure metal is used as anode.
Na _3 AlF _6 is added during electrolysis of Al _2 O _3 to lower the melting point and increase conductivity.
A
c
B
d
C
b
D
a
Solution
The +M effect of NH_2 is stabilizing the
carbocation
A
d>b>c>a
B
b > d > c > a
C
a > c > d > b
D
a > b > c > d
Solution
The rate of S_N1 reaction depends upon stability of carbocation which follows the order

A
D > C > B > A
B
B > A > C > D
C
A>B>D>C
D
A>B>C>D
Solution
Bond dissociation energy of E-H bond in hydrides of group 16 follows the order
H _2 O > H _2 S > H _2 Se > H _2 Te
A
72
B
16
C
32
D
50
Solution
The number of electrons in the orbitals of sub-shell of n = 4 are

A
(i) Fe , H ^{+} (ii) Br _2( aq ) (iii) HNO _2 (iv) CuBr
B
(i) Fe , H ^{+} (ii) Br _2( aq ) (iii) HNO _2 (iv) H _3 PO _2
C
(i) Br _2, Fe (ii) Fe , H ^{+} (iii) LiAIH _4
D
(i) Br _2( aq ) (ii) LiAIH _4 (iii) H _3 O ^{+}

A
A-I, B-II, C-III, D-IV
B
A-II, B-I, C-III, D-IV
C
A-I, B-II, C-IV, D-III
D
A-II, B-I, C-IV, D-III
Solution
For \left[ Fe \left( NH _3\right)_6\right]^{+2}, \Delta_0< P, hence the pairing of electrons does not occur in t _{2 g }. Therefore complex is outer orbital and its hybridisation is sp ^3 d ^2.
Match List I with List II
List I (Complexes)
List II (Hybridisation)
[Ni(CO)_4]
sp^{3}
[Cu(NH_{3})_{4}]^{2+}
dsp ^{2}
[Fe(NH_{3})_{6}]^{2+}
sp^{3}d^2
[Fe(H_{2}O)_{6}]^{2+}
sp^{3} d^2
List I (Complexes) | List II (Hybridisation) |
---|---|
[Ni(CO)_4] | sp^{3} |
[Cu(NH_{3})_{4}]^{2+} | dsp ^{2} |
[Fe(NH_{3})_{6}]^{2+} | sp^{3}d^2 |
[Fe(H_{2}O)_{6}]^{2+} | sp^{3} d^2 |
A
A-IV, B-I, C-III, D-II
B
A-III, B-IV, C-I, D-II
C
A-II, B-I, C-III, D-IV
D
A-III, B-I, C-IV, D-II
Solution
List I (Mixture)
List II (Separation Technique)
CHCl_3 + C_6H_5NH_2
Distillation
C_6H_{14} + C_5H_{12}
Fractional
distillation
C_6H_5NH_2 + H_2O
Steam distillation
Organic compound in H_2O
Differential
extraction
List I (Mixture) | List II (Separation Technique) |
---|---|
CHCl_3 + C_6H_5NH_2 | Distillation |
C_6H_{14} + C_5H_{12} | Fractional distillation |
C_6H_5NH_2 + H_2O | Steam distillation |
Organic compound in H_2O | Differential extraction |
A
KHgI _3
B
HgI _2
C
K _2 HgI _4
D
KHg _2 I _2
Solution
Nessler's reagent is K _2 HgI _4.
A
r_0=\frac{a_0}{2}
B
r_0=2 a_0
C
r_0=4 a_0
D
r_0=a_0
Solution
At node \Psi_{2 s}=0
\therefore 2-\frac{r_0}{a_0}=0
\therefore r_0=2 a_0
A
2 LiNO _3 \xrightarrow{\Delta} 2 NaNO _2+ O _2
B
4 LiNO _3 \xrightarrow{\Delta} 2 Li _2 O +2 N _2 O _4+ O _2
C
2 LiNO _3 \longrightarrow 2 Li +2 NO _2+ O _2
D
4 LiNO _3 \xrightarrow{\Delta} 2 Li _2 O +4 NO _2+ O _2
Solution
4 LiNO _3 \stackrel{ A }{\longrightarrow} 2 Li _2 O +4 NO _2+ O _2
A
Mg
B
Ca
C
K
D
Be
Solution
BeCl_2 having covalent nature is soluble in organic solvent
Answer: 150
Solution
\text { Molarity }=\frac{50}{11.35}
\therefore Strength in gm / L =\frac{50}{11.35} \times 34
Answer: 243
Solution
\Delta T _{ f }= i . K _{ f } \cdot m
\Rightarrow \Delta T _{ f }=2.67 \times 1.8 \times \frac{38}{98} \times \frac{1000}{62}
\Rightarrow \Delta T _{ f }=30.05
\therefore F . P .=243\, K
Answer: 275
Solution
X + Y ^{2+} \rightarrow Y + X ^{2+}
E _{\text{ Cell} }^0=0.36-(-2.36)=2.72 V
E _{\text {Cell }}=2.72-\frac{0.06}{2} \log \frac{0.001}{0.01}
=2.72+0.03=2.75 V
=275 \times 10^{-2} V
Answer: 16
Solution
\log \frac{x}{m}=\log k +\frac{1}{n} \log P
\frac{1}{ n }=\tan 45^{\circ}=1
\log k =0.6020=\log 4
\Rightarrow K =4
\therefore \frac{ x }{ m }= K \cdot P ^{1 / n }
\frac{ x }{ m }=4(0.4)=1.6
\frac{ x }{ m }=1.6=16 \times 10^{-1}
Answer: 1350
Solution
\frac{ t _1}{ t _2}=\frac{\frac{1}{ K } \ln \frac{ a _0}{0.4 a _0}}{\frac{1}{ K } \ln \frac{ a _0}{0.1 a _0}}
\frac{540}{ t _2}=\frac{\ln \frac{10}{4}}{\ln 10}
\frac{540}{ t _2}=\frac{\log 10-\log 4}{\log 10}
\frac{540}{ t _2}=\frac{1-0.6}{1}
\Rightarrow \frac{540}{ t _2}=0.4
\Rightarrow t _2=\frac{540}{0.4}=1350 \sec
Answer: 150
Solution
q =0
\Delta U = w
1 \times 20 \times\left[ T _2-300\right]=-3000
T _2-300=-150
T _2=150 \,K
Answer: 3
Solution
The yield of SO _3 at equilibrium will be due to :
B. Increasing pressure
C. Adding more SO _2
D. Adding more O _2
Answer: 6
Solution
Number of peptide linkage = (amino acid - 1)
= 7 - 1 = 6
Answer: 4
Solution
d =\frac{ Z \times M }{ N _0 \times a ^3}
4=\frac{ Z \times 72}{6 \times 10^{23} \times 125 \times 10^{-24}}
Z=4.166 \simeq 4
JEE Main Mathematics Question Paper with Solution 2023 January 30th Shift 2 - Evening
A
[x] is even but [y] is odd
B
[x]+[y] is even
C
[x] and [y] are both odd
D
[x] is odd but [y] is even
Solution
x=(8 \sqrt{3}+13)={ }^{13} C_0 \cdot(8 \sqrt{3})^{13}+{ }^{13} C_1(8 \sqrt{3})^{12}(13)^1+\ldots
x^{\prime}=(8 \sqrt{3}-13)^{13}={ }^{13} C_0(8 \sqrt{3})^{13} - {}^{13} C_1(8 \sqrt{3})^{12}(13)^1+\ldots
x-x^{\prime}=2\left[{ }^{13} C_1 \cdot(8 \sqrt{3})^{12}(13)^1+{ }^{13} C_3(8 \sqrt{3})^{10} \cdot(13)^3 \ldots\right]
therefore, x-x^{\prime} is even integer, hence [x] is even
\text { Now, } y =(7 \sqrt{2}+9)^9={ }^9 C _0(7 \sqrt{2})^9+{ }^9 C _1(7 \sqrt{2})^8(9)^1 +{ }^9 C _2(7 \sqrt{2})^7(9)^2 \ldots \ldots
y ^{\prime}=(7 \sqrt{2}-9)^9={ }^9 C _0(7 \sqrt{2})^9-{ }^9 C _1(7 \sqrt{2})^8(9)^1 +{ }^9 C _2(7 \sqrt{2})^7(9)^2 \ldots \ldots
y - y ^{\prime}=2\left[{ }^9 C _1(7 \sqrt{2})^8(9)^1+{ }^9 C _3(7 \sqrt{2})^6(9)^3+\ldots\right]
y - y ^{\prime}= Even integer, hence [ y ] is even
A
\sqrt{2} a-b+c=1
B
a+\sqrt{2} b+c=1
C
\sqrt{2} a+b+c=1
D
a+b+\sqrt{2} c=1
Solution
\hat{ v }=\cos 60^{\circ} \hat{ i }+\cos 45^{\circ} \hat{ j }+\cos \gamma \hat{ k }
\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^2 \gamma=1 \quad(\gamma \rightarrow \text { Acute })
\Rightarrow \cos \gamma=\frac{1}{2}
\Rightarrow \gamma=60^{\circ}
Equation of plane is
\frac{1}{2}(x-\sqrt{2})+\frac{1}{\sqrt{2}}(y+1)+\frac{1}{2}(z-1)=0
\Rightarrow x+\sqrt{2} y+z=1
(a, b, c) lies on it.
\Rightarrow a+\sqrt{2} b+c=1
A
|Adj P| >1
B
|A d j P|=\frac{1}{2}
C
P is a singular matrix
D
|A d j P|=1
Solution
P ^{ T }= aP +( a -1) I
\Rightarrow P =a P ^{ T }+( a -1) I
\Rightarrow P ^{ T }- P = a \left( P - P ^{ T }\right)
\Rightarrow P = P ^{ T }, \text { as } a \neq-1
\text { Now, } P = aP +( a -1) I
\Rightarrow P =- I \Rightarrow| P |=1
\Rightarrow|\text{Adj} P |=1
A
164
B
243
C
\frac{125}{3}
D
25
Solution
x^2-p x+\frac{5 p}{4}=0
D=p^2-5 p=p(p-5)
\therefore q=9
0 \leq y \leq(x-9)^2
Area =\int\limits_0^9(x-9)^2 dx =243
A
1
B
0
C
\sin (1)
D
-1
Solution
LHL =\displaystyle\lim _{ k \rightarrow 0} g ( h (- k )) , k >0
=\displaystyle\lim _{ k \rightarrow 0} g (-2+1)
\because f ( x )=-1 \forall x <0
= g (-1)=1
RHL =\displaystyle\lim _{ k \rightarrow 0} g ( h ( k )) , k >0
=\displaystyle\lim _{ k \rightarrow 0} g (-1) ,
\because f ( x )=1, \forall x >0
=1
A
\cot ^{-1}(2022)-\frac{\pi}{4}
B
\frac{\pi}{4}-\cot ^{-1}(2022)
C
\tan ^{-1}(2022)-\frac{\pi}{4}
D
\frac{\pi}{4}-\tan ^{-1}(2022)
Solution
a _2- a _1= a _3- a _2=\ldots . .= a _{2022}- a _{2021}=1.
\therefore \tan ^{-1}\left(\frac{ a _2- a _1}{1+ a _1 a _2}\right)+\tan ^{-1}\left(\frac{ a _3- a _2}{1+ a _2 a _3}\right)+\ldots . .+\tan ^{-1}\left(\frac{ a _{2022}- a _{2021}}{1+ a _{2021} a _{2022}}\right)
=\left[\left(\tan ^{-1} a_2\right)-\tan ^{-1} a_1\right]+\left[\tan ^{-1} a_3-\tan ^{-1} a_2\right]+\ldots . . +\left[\tan ^{-1} a _{2022}-\tan ^{-1} a _{2021}\right]
=\tan ^{-1} a _{2022}-\tan ^{-1} a _1
=\tan ^{-1}(2022)-\tan ^{-1} 1=\tan ^{-1} 2022-\frac{\pi}{4} \text { (option 3) }
=\left(\frac{\pi}{2}-\cot ^{-1}(2022)\right)-\frac{\pi}{4}
=\frac{\pi}{4}-\cot ^{-1}(2022)(\text { option 1) }
A
\frac{13}{6}
B
\frac{5}{17}
C
\frac{17}{5}
D
\frac{6}{13}
Solution
\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}
\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}
\text { Points : } A(-1, k , 0), B (2, k ,-1), C (1,1,2)
\overrightarrow{ CA }=-2 \hat{ i }+( k -1) \hat{ j }-2 \hat{ k }
\overrightarrow{ CB }=\hat{ i }+( k -1) \hat{ j }-3 \hat{ k }
\overrightarrow{ CA } \times \overrightarrow{ CB }=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & k -1 & -2 \\ 1 & k -1 & -3\end{vmatrix}
=\hat{ i }(-3 k +3+2 k -2)-\hat{ j }(6+2)+\hat{ k }(-2 k +2- k +1)
=(1- k ) \hat{ i }-8 \hat{ j }+(3-3 k ) \hat{ k }
The line \frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1} is perpendicular to normal vector.
\therefore 1 \cdot(1- k )+1(-8)+(-1)(3-3 k )=0
\Rightarrow 1- k -8-3+3 k =0
\Rightarrow 2 k =10 \Rightarrow k =5
\therefore \frac{ k ^2+1}{( k -1)( k -2)}=\frac{26}{4 \cdot 3}=\frac{13}{6}
A
\frac{3}{2}
B
3
C
6
D
4
Solution
f^{\prime}(x)=x^2+2 b+a x
g^{\prime}(x)=x^2+a+2 b x
(2 b-a)-x(2 b-a)=0
\therefore x=1 is the common root
Put x=1 \text { in } f^{\prime}(x)=0 \text { or } g^{\prime}(x)=0
1+2 b+a=0
7+2 b+a=6
A
\frac{d}{a}, \frac{e}{b}, \frac{f}{c} are in G.P.
B
d, e, f are in A.P.
C
d, e, f are in G.P.
D
\frac{d}{a}, \frac{e}{b}, \frac{f}{c} are in A.P.
Solution
ax ^2+2 bx + c =0
\Rightarrow ax ^2+2 \sqrt{ ac x}+ c =0\left(\because b ^2= ac \right)
\Rightarrow( x \sqrt{ a }+\sqrt{ c })^2=0
x ^2-\frac{\sqrt{ c }}{\sqrt{ a }} \ldots \ldots(1)
\text { Now, } d x ^2+2 ex + f =0
\Rightarrow d \left(\frac{ c }{ a }\right)+2 e \left[-\frac{\sqrt{ c }}{\sqrt{ a }}\right]+ f =0
\Rightarrow \frac{d c}{ a }+ f =2 e \sqrt{\frac{ c }{ a }}
\Rightarrow \frac{ d }{ a }+\frac{ f }{ c }=2 e \sqrt{\frac{1}{ ac }}
\Rightarrow \frac{ d }{ a }+\frac{ f }{ c }=\frac{2 e }{ b }[\text { as } b =\sqrt{ ae }]
\therefore \frac{ d }{ a }, \frac{ e }{ b }, \frac{ f }{ c } \text { are in A.P. }
A
-60
B
-48
C
-84
D
-24
Solution
\vec{ c }=(2 \vec{ a } \times \vec{ b })-3 \vec{ b }
\vec{ b } \cdot \vec{ c }=\vec{ b } \cdot(2 \vec{ a } \times \vec{ b })-3 \vec{ b } \cdot \vec{ b }
=-3| b |^2
=-48
A
\frac{19}{3}
B
12
C
0
D
19
Solution
\displaystyle\lim _{n \rightarrow \infty} \frac{3}{n} \displaystyle\sum_{r=0}^{n-1}\left(2+\frac{r}{n}\right)^2
=3 \int\limits_0^1(2+x)^2 d x=27-8=19
A
343
B
216
C
\frac{343}{8}
D
\frac{125}{8}
Solution
As a ^3, b ^3, c ^3 be in A.P. \rightarrow a ^3+ c ^3=2 b ^3.... (1)
\log _a^b, \log _c^a, \log _b^c are in G.P.
\therefore \frac{\log b }{\log a } \cdot \frac{\log c }{\log b }=\left(\frac{\log a }{\log c }\right)^2
\therefore(\log a )^3=(\log c )^3 \Rightarrow a = c......(2)
From (1) and (2)
a = b = c
T _1=\frac{ a +4 b + c }{3}=2 a ; d =\frac{ a -8 b + c }{10}=\frac{-6 a }{10}=\frac{-3}{5} a
\therefore S _{20}=\frac{20}{2}\left[4 a +19\left(-\frac{3}{5} a \right)\right]
=10\left[\frac{20 a -57 a }{5}\right]
=-74 a
\therefore-74 a =-444 \Rightarrow a =6
\therefore a b c=6^3=216
A
[2 \sqrt{2}, \sqrt{11}]
B
[\sqrt{5}, \sqrt{10}]
C
[\sqrt{5}, \sqrt{13}]
D
[\sqrt{2}, \sqrt{7}]
Solution
y^2=3-x+2+x+2 \sqrt{(3-x)(2+x)}
=5+2 \sqrt{6+x-x^2}
y^2=5+2 \sqrt{\frac{25}{4}-\left(x-\frac{1}{2}\right)^2}
y_{\max }=\sqrt{5+5}=\sqrt{10}
y_{\min }=\sqrt{5}
A
\{99\}
B
\phi
C
N
D
\{9\}
Solution
let a_1 be any natural number
a_1, a_1+1, a_1+2, \ldots ., a_1+99 \text { are values of } a_i{ }^{\prime} S
\bar{x}=\frac{a_1+\left(a_1+1\right)+\left(a_1+2\right)+\ldots . .+a_1+99}{100}
=\frac{100 a_1+(1+2+\ldots . .+99)}{100}=a_1+\frac{99 \times 100}{2 \times 100}
=a_1+\frac{99}{2}
Mean deviation about mean =\frac{\displaystyle\sum_{ i =1}^{100}\left| x _{ i }-\overline{ x }\right|}{100}
=\frac{2\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots .+\frac{1}{2}\right)}{100}
=\frac{1+3+\ldots .+99}{100}
=\frac{\frac{50}{2}[1+99]}{100}
=25
So, it is true for every natural no. ' a _1 '
A
76
B
81
C
72
D
64
Solution
y = mx +\frac{4}{ m }
\frac{\left|\frac{4}{ m }\right|}{\sqrt{1+ m ^2}}=2 \sqrt{2}
\therefore m =\pm 1
y =\pm x \pm 4 Point of contact on parabola
Let m =1,\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right)
R (4,8)
Point of contact on circle Q(-2,2)
\therefore( QR )^2=36+36=72
A
(P \vee Q) \wedge((\sim P) \vee R)
B
(P \vee \sim Q) \wedge(P \vee \sim R)
C
((\sim P) \vee \sim Q) \wedge((\sim P) \vee \sim R)
D
((\sim P) \vee \sim Q) \wedge((\sim P) \vee R)
Solution
P \rightarrow(\sim Q \wedge R)
\sim P \vee(-Q \wedge R)
(-P \vee \sim Q) \wedge(-P \vee R)
A
54
B
108
C
268
D
186
Solution
a \in\{2,4,6,8,10, \ldots ., 100\}
b \in\{1,3,5,7,9, \ldots \ldots, 99\}
Now, a+b \in\{25,71,117,163\}
(i) a+b=25, no. of ordered pairs (a, b) is 12
(ii) a + b =71, no. of ordered pairs (a, b) is 35
(iii) a + b =117, no. of ordered pairs (a, b) is 42
(iv) a+b=163, no. of ordered pairs (a, b) is 19
\therefore total =108 pairs
A
\log _e|x+y|-\frac{x y}{(x+y)^2}=0
B
\log _e|x+y|-\frac{2 x y}{(x+y)^2}=0
C
\log _e|x+y|+\frac{x y}{(x+y)^2}=0
D
\log _e|x+y|+\frac{2 x y}{(x+y)^2}=0
Solution
Put y=v x
v+x \frac{d v}{d x}=-\left(\frac{1+3 v^2}{3+v^2}\right)
x \frac{d v}{d x}=-\frac{(v+1)^3}{3+v^2}
\frac{\left(3+v^2\right) d v}{(v+1)^3}+\frac{d x}{x}=0
\int \frac{4 d v}{(v+1)^3}+\int \frac{d v}{v+1}-\int \frac{2 d v}{(v+1)^2}+\int \frac{d x}{x}=0
\frac{-2}{(v+1)^2}+\ln (v+1)+\frac{2}{v+1}+\ln x=c
\frac{-2 x^2}{(x+y)^2}+\ln \left(\frac{x+y}{x}\right)+\frac{2 x}{x+y}+\ln x=c
\frac{2 x y}{(x+y)^2}+\ln (x+y)=c
\therefore c=0, \text { as } x=1, y=0
\therefore \frac{2 x y}{(x+y)^2}+\ln (x+y)=0
A
136
B
132
C
140
D
144
Solution
\vec{ a }=\lambda \hat{ i }+2 \hat{ j }-3 \hat{ k }
\vec{ b }=\hat{ i }-\lambda \hat{ j }+2 \hat{ k }
\Rightarrow(\vec{ b }-\vec{ a }) \times((\vec{ a }+\vec{ b }) \times(\vec{ a } \times \vec{ b }))=8 \hat{ i }-40 \hat{ j }-24 \hat{ k }
\Rightarrow((\vec{ a }-\vec{ b }) \cdot(\vec{ a }+\vec{ b }))(\vec{ a } \times \vec{ b })=8 \hat{ i }-40 j -24 \hat{ k }
\Rightarrow 8(\vec{ a } \times \vec{ b })=8 \hat{ i }-40 \hat{ j }-24 \hat{ k }
Now, \vec{a} \times \vec{b}=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2\end{vmatrix}
=(4-3 \lambda) \hat{ i }-(2 \lambda+3) \hat{ j }+\left(-\lambda^2-2\right) \hat{ k }
\Rightarrow \lambda=1
\therefore \vec{ a }=\hat{ i }+2 \hat{ j }-3 \hat{ k }
\vec{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }
\Rightarrow \vec{ a }+\vec{ b }=2 \hat{ i }+\hat{ j }-\hat{ k }, \vec{ a }-\vec{ b }=3 \hat{ j }-5 \hat{ k }
\Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -1 \\ 0 & 3 & -5\end{vmatrix}=2 \hat{i}+10 \hat{ j }+6 \hat{ k }
\therefore required answer =4+100+36=140
A
x^2-18 x+56=0
B
x^2+14 x+24=0
C
x^2-10 x+16=0
D
x^2+18 x+56=0
Solution
\begin{vmatrix}1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4\end{vmatrix}=0 ; 8+\alpha-2(-4+1)+3(-\alpha-2)=0
8+\alpha+6-3 \alpha-6=0
\alpha=4
Answer: 158
Solution
\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 3 & -2 \\ 1 & -1 & 2\end{vmatrix}=4 \hat{ i }-4 \hat{ j }-4 \hat{ k }
\therefore Equation of line is \frac{ x -2}{1}=\frac{ y -3}{-1}=\frac{ z -1}{-1}
Let Q be (5,3,8) and foot of \perp from Q on this line be R.
Now, R \equiv( k +2,- k +3,- k +1)
DR of QR are ( k -3,- k ,- k -7)
\therefore(1)( k -3)+(-1)(- k )+(-1)(- k -7)=0
\Rightarrow k =-\frac{4}{3}
\therefore \alpha^2=\left(\frac{13}{3}\right)^2+\left(\frac{4}{3}\right)^2+\left(\frac{17}{3}\right)^2=\frac{474}{9}
\therefore 3 \alpha^2=158
Answer: 24
Solution
\frac{1}{2} \times PC \times \sqrt{5}=\frac{\sqrt{35}}{2} ; PC =\sqrt{7}
a _1^2+ b _1^2+ a _2^2+ b _2^2= OP ^2+ OQ ^2
=2(5+7)=24
Answer: 151
Solution
T_8=11+(8-1) \times 20
=11+140=151
Answer: 23
Solution
x + y =12^{50}+18^{50}=(150-6)^{25}+(325-1)^{25}
=25 K -\left(6^{25}+1\right)=25 K -\left((5+1)^{25}+1\right)
=25 K _1-2
Remainder =23
Answer: 14
Solution
p =\frac{{ }^6 C _1}{6 \times 6}=\frac{1}{6}
q =\frac{{ }^6 C _1 \times{ }^5 C _1 \times 4}{6 \times 6 \times 6 \times 6}=\frac{5}{54}
\therefore p : q =9: 5
\Rightarrow m + n =14
Answer: 1
Solution
\int \sqrt{\sec 2 x-1} d x=\int \sqrt{\frac{1-\cos 2 x}{\cos 2 x}} d x
=\sqrt{2} \int \frac{\sin x}{\sqrt{2 \cos ^2 x-1}} d x
\text { put } \cos x=t \Rightarrow-\sin x d x=d t
=-\sqrt{2} \int \frac{ dt }{\sqrt{2 t^2-1}}
=-\ln |\sqrt{2} \cos x+\sqrt{\cos 2 x}|+c
=-\frac{1}{2} \ln \left|2 \cos { }^2 x+\cos 2 x+2 \sqrt{\cos 2 x} \cdot \sqrt{2} \cos x\right|+c
=-\frac{1}{2} \ln \left|\cos 2 x+\frac{1}{2}+\sqrt{\cos 2 x} \cdot \sqrt{1+\cos 2 x}\right|+c
\because \beta=\frac{1}{2}, \alpha=-\frac{1}{2}
\Rightarrow \beta-\alpha=1
Answer: 240
Solution
Digits are 1,2,2,2,3,3,5
If unit digit 5 , then total numbers =\frac{6 !}{3 ! 2 !}
If unit digit 3 , then total numbers =\frac{6 !}{3 !}
If unit digit 1 , then total numbers =\frac{6 !}{3 ! 2 !}
\therefore total numbers =60+60+120=240
Answer: 432
Solution
f(1)=1 ; f(9)=f(3) \times f(3)
i.e., f(3)=1 or 3
Total function =1 \times 6 \times 2 \times 6 \times 6 \times 1=432
Answer: 25
Solution
A =2 \int\limits_{\frac{1}{3}}^{\frac{1}{2}}\left(2 x-2 x ^2-(1- x )^2\right) dx
=2\left[2 x ^2- x ^3- x \right]_{1 / 3}^{1 / 2}
\therefore A =\frac{5}{108} \Rightarrow 540 A =\frac{5}{108} \times 540=25
Answer: 13