JEE Main Question Paper with Solution 2023 January 30th Shift 2 - Evening

JEE Main Physics Question Paper with Solution 2023 January 30th Shift 2 - Evening

A

EI=0,EII=0,EIII=0

B

EI0,EII=0,EII=0

C

EI=0,EII=0,EIII0

D

EI0,EII=0,EIII0

Solution

Electric field inside material of conductor is zero.

A

gR

B

2gR

C

2gR

D

gR2

Solution

Loss in PE = Gain in KE
(GMm2R)(GMmR)=12mv2
v2=GMR=gR
v=gR

A

2

B

1/2

C

2

D

1/2

Solution

ω=km
ω2ω1=m1m2=12

A

2.77×102mm

B

6.9×102mm

C

6.06×102mm

D

3.0×102mm

Solution

Y=F/AΔ
F=YAΔ
(AΔ)1=(AΔ)2
Δ2Δ1=A1A2×21
Δ20.2=12.4×2.4×21
Δ2=6.9×102mm

A

0.6m/s

B

2.5m/s

C

1.5m/s

D

0.02m/s

Solution

20×103×18060×100=10V
v=0.6m/s

A

43×104T

B

3×104T

C

43×105T

D

33×105T

Solution

dtan60=23
d=2cm
B=3×μ0i2πdsin60
=3×2×107×22×102×32
=33×105

A

25N

B

15N

C

10N

D

20N

Solution

image
θ=30
cosθ=3gT
32=3gT
T=20N

A

23Ω

B

32Ω

C

13Ω

D

12Ω

Solution

image
1Req=12+112+14+16+12
=6+1+3+2+612=1812=32
Req=23Ω

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

12A

B

20A

C

2A

D

22A

Solution

z=1002+(200100)2
=1002Ω
irms=Vrmsz=20021002
=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(1.541)=A(1.721)
A=6×0.540.72
=184=4.5

A

140πWm2

B

110πWm2

C

120πWm2

D

12πWm2

Solution

IEF=12×54π×52
=140πW/m2

Answer: 12

Solution

E=2x2ˆi4yˆj+6ˆk
image
ϕnet =8×3+2×6=12
12=qϵ0
|q|=12ϵ0

Answer: 313

Solution

415955=C01000
C=3690×100=40C=313K

Answer: 4

Solution

12mV2=Pt
V=2Ptm
dxdt=2Ptm
x=2Pm23[t3/2]40
x=16P3=13×16P
α=4

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

image
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

image
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

image
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

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

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

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

image
\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

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
image
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}
image
a _1^2+ b _1^2+ a _2^2+ b _2^2= OP ^2+ OQ ^2
=2(5+7)=24

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

image
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

Solution

Two equations have common root
\therefore(4 a)(26 a)=(-6)^2=36
\Rightarrow a^2=\frac{9}{26}
\therefore a=\frac{3}{\sqrt{26}}
\Rightarrow \beta=13