JEE Main - July 2022
1.
Position of a particle x at time t are related as
=
t
2x + 4 . The velocity of the particle
at t = 4s is equal to (in S.I. units)
(A) 4
(B) 2
(C) 1
(D) 5
Ans. (A)
Sol.=
t
2x + 4
t2 = 2x + 4
Differentiating w.r.t. time
2t = 2
⇒
dx
dt
dx
= t
dt
⇒ v=t
∴
2.
v(t = 4) = 4 ms–1
A projectile with initial kinetic energy E is projected at angle 45°. Its kinetic energy at
top most point is equal to
(A)
E
2
(B)
3E
2
(C)
E
4
(D)
E
3
Ans. (A)
1
mv 02 (v0 is initial speed)
2
Sol. K.E1 = E =
…(i)
At highest point
v = v (horizontal) = v0 cos45°
v=
∴
3.
v0
2
K.E2 =
2
1 v0 E
m
=
2
2
2
For the given circuit find current through battery.
Ans. (2)
JEE Main - July 2022
Sol. All three resistances are in parallel
∴
Rq =
E
6
= = 2 Amp
Req 2
∴ =
I
4.
R 9
=
= 3Ω
3 3
Two rods of identical lengths and cross-sectional area are connected in series. If σ1
and σ2 is the thermal conductivity of material of two rods then equivalent conductivity
of combination is equal to
(A)
(C)
2σ1σ2
σ 1 + σ2
σ 1σ2
σ 1 − σ2
(B)
(D)
σ 1σ2
σ 1 + σ2
2σ1σ2
σ 1 − σ2
Ans. (A)
Sol. Req = R1 + R2
ρeq
∴
2


= ρ 1 + ρ2
A
A
A
2ρeq = ρ1 + ρ2
2
1
1
=
+
σeq σ1 σ2
∴
5.
2σ σ
σeq = 1 2
σ 1 + σ2
A particle thrown at angle 45° with horizontal with speed u has its range equal to R.
At what angle should it be thrown with same speed for its range to be half of its initial
value.
(A) 60°
(B) 30°
(C) 15°
(D) 70°
Ans. (C)
Sol. R =
u2 sin ( 2θ )
g
=
R
R=
u2
sin90° (θ = 45°)
g
u2
g
Now, R′ =
…(i)
R
2
JEE Main - July 2022
u2 sin ( 2θ ) 1 u2
=
g
2 g
1
sin ( 2θ ) =
2
2θ = 30°
θ = 15°
6.
A travelling microscope has Vernier scale with 9 MSD = 10 VSD. If one main scale
division (MSD) is equal to 1 mm, then least count of travelling microscope is
(A) 0.005 m
(B) 0.002 m
(C) 0.0001 m
(D) 0.0005 m
Ans. (C)
Sol. Least count, LC = 1 MSD – 1 VSD
7.
⇒
LC = 1 MSD –
⇒
LC
=
⇒
LC = 0.0001 m
9
MSD
10
1
1
MSD
=
× 0.001m
10
10
A cart is moving down a smooth incline of inclination α. What is the time-period of a
bob hanging from the roof of the cart with a light string?
(A) 2π
l
g cos α
(B) 2π
l
g
(C) 2π
l
g sin α
(D) 2π
l
g cot α
Ans. (A)
Sol. ∴ =
g eff g cos α
T= 2π
T= 2π
l
g eff
l
g cos α
JEE Main - July 2022
8.
If the length of wire is doubled and the radius is halved, then the young's modulus Y
becomes
(A) same
(B) 8 times the original value
(C) 4 times the original value
(D) None of these
Ans. (A)
Sol. Young’s modulus nature of material and it is independent of geometry of the wire.
9.
Find the ratio of energy of electron when it transitions from second to first energy
state in comparison to highest state to first energy state of hydrogen atom
(A)
1
4
(B)
5
36
(C)
8
9
(D)
3
4
Ans. (D)

1
E1 (2 →=
1) 13.6  1 − 
4

Sol. ⇒
= 13.6 ×
3
eV
4
…(i)

1
and E2 (∞ =
→ 1) 13.6  1 − 
∞

E2 = 13.6 eV
⇒
…(ii)
Equation (i) divided by (ii)
E1
So,
10.
E2
=
3
4
If one mole of monoatomic gas and three mols of diatomic gas are mixed, then the
molar heat at constant volume is α2R/4. The value of α is ……….. .
Ans. (3)
Sol. CV
mix
=
(n C
CV
mix
CV
mix
So, α = 3
1
V1
+ n2CV
n1 + n2
2
)

3
5 
 1 × R + 3 × R
2
2 
=
1+ 3
=
9
R
4
JEE Main - July 2022
11.
A wire of length 314 cm is made into a circular coil. Find its magnetic moment (in Am2
if i = 14 A. (π = 3.14)
Ans. (11 Am2)
Sol. µ = iπr 2
 l 
µ = iπ  
 2π 
2
( 2πr =l )
2
 3.14 
14π
µ= 14 × π 
 =
2
×
3.1
4
4


µ = 11 Am2
12.
If the value of electric field at depletion layer in p – n junction if width is 6 × 10–6 m
and potential difference is 0.6 V, is x × 105 V/m. Then find the value of x?
Ans. (1)
Sol. d (width of depletion layer)
= 6 × 10–6 m
…(i)
Potential difference across laye r= 0.6 v …(ii)
13.
v
0.6
=
d 6 × 10−6
∴
=
E
⇒
E = 1 × 105 v/m
Which gas has Cp/Cv (γ) ratio 7/5? Choose most appropriate option.
(A) Monoatomic
(B) Diatomic
(C) Polyatomic Linear
(D) Both (B) and (C)
Ans. (D)
Sol.
γ=
Cp
Cv
= 1+
2
f
Diatomic gases have specific heat ratio of
7
.
5
Polyatomic linear gases have specific heat ratio
14.
7
.
5
Assertion: Grease/oil stains can not be removed by water wash
Reason: The angle of contact between water and oil is obtuse
(A) Both A and R are true and R is correct explanation of A.
(B) Both A and R are true but R is not correct explanation of A.
(C) A is true and B is false
(D) A is false and B is true
Ans. (A)
Sol. Due to greater cohesive force between grease / oil molecules.
JEE Main - July 2022
15.
A solids sphere is charged such that charge density ρ varies with radial distance r as

r
ρ = ρ0  1 −  for r ≤ R. The electric field at a radial distance r (r < R), at point P is
R

(A)
ρ0r  1
r 
 −

0  3 4R 
(B)
(C)
ρ0r 
r 
1 −

0  2R 
(D)
ρ0r  1
r 
 −

0  2 3R 
ρ0r  1
r 
 −

0  2 4R 
Ans. (A)
Sol. Charge enclosed within volume of radius r

r
qin = ∫ 4πr 2dr ρ0  1 − 
R

r
r

1
= 4πρ0  ∫ r 2dr − ∫ r 3dr 
R0
0

 r3 r4 
= 4πρ0  −

 3 4R 
…(i)
Using Gauss law


qi
∫ E ⋅ dA− =∈
…(ii)
0
 r r2 
E × 4πr 2 = 4πρ0r 2  −

 3 4R 
E=
16.
ρ0r  1
r 
 −

0  3 4R 
Statement: Electric potential is constant inside a conductor.
Reason: Electric field just outside of the conductor is perpendicular to its surface.
(A) Only statement is correct.
(B) Only reason is correct.
(C) Both statement & Reason are correct but reason is not the correct explanation of
statement.
(D) Both statement & Reason are correct and reason is the correct explanation of
statement.
Ans. (C)
Sol. Potential inside conductor is constant because net electric field inside conductor is
always zero.
JEE Main - July 2022
17.
ˆ and ( −3ıˆ − 2ˆ
ˆ
Two objects of masses m and 3m are located at (ıˆ + 2ˆ
 + k)
 + k)
respectively. Find position vector of COM of two objects is,
(A) 2ıˆ + 2ˆ
 − kˆ
(B) ı̂ + ˆ − kˆ
(C) −2ıˆ − ˆ + kˆ
(D) ı̂ + 2ˆ
 − 2kˆ
Ans. (C)


m r + m2r2

Sol. rcom = 1 1
m1 + m2
ˆ + 3m( −3ıˆ − 2ˆ
ˆ

m(ıˆ + 2ˆ
 + k)
 + k)
rcom =
m + 3m

−8mˆı − 4mˆ
 + 4mkˆ
rcom =
=−2ıˆ − ˆ + kˆ
4m
18.
Value of acceleration due to gravity above the surface of Earth at height h (h  Re ) is
equal to acceleration due to gravity at depth d below Earth's surface, for d = αh. Then
find the value of α ?
Ans. (2)


2h 
d
Sol. g o  1 −
 = g o  1 −


Re 
Re 


2h d
=
R
R
d=2h
∴
19.
α=2
If two sound waves of intensities 4 I and 9 I interfere at two points. At one point, phase
difference between them is 0 and π/2 at the other. The difference between resultant
intensity at the points is x I. The value of x is
Ans. (12)
Sol. I1 (ϕ= 0)= 4I + 9I + 2 4I × 9I cos 0°= 25I

π
π
I2  ϕ=
= 4I + 9I + 2 4I × 9I cos = 13I
2
2

∴
I1 – I2 = 12 I
∴
x = 12
JEE Main - July 2022
1.
Ionic radius for A+ and B– are 281 pm and 108 pm respectively forming a ccp structure.
If B– forms a ccp lattice and A+ fills the octahedral voids, then what is the value of
edge length in pm?
(A) 461 pm
(B) 778 pm
(C) 922 pm
(D) None
Ans. (B)
Sol. When cation is present in octahedral void of CCP structure then
a
r+ + r– =
2
a
281 + 108 =
2
A = 389 × 2 = 778 Pm
Hence (B) is the correct Answer.
2.
A compound which react with NaOH and is insoluble in NaHCO3 but turns neutral FeCl3
solution purple?
(A) Cyclohexanol
(B) Phenol
(C) Cyclohexane
(D) None
Ans. (B)
Sol. Phenol react with NaOH and form purple colour with neutral FeCl3.
Hence (B) is correct.
3.
Product for the given reaction is
Zn + NaOH →
(A) ZnO
(B) ZnO2
(C) [ZnO3]4–
(D) [Zn(OH)4]2–
Ans. (D)
Sol. Zn react with NaOH to form [(Zn(OH)4)]2–
4.
Which of the following is the strongest Bronsted base?
(A)
(B)
(C)
(D)
Ans. (A)
JEE Main - June 2022
Sol. 3° amine is most basic in polar aprotic solvent out of compound in option A and B,
compound B can undergo flipping as follows:
Hence is becomes Less Base.
Therefore option (A) is correct.
5.
Which of the following is an example of expanded octet
(A) BCl3
(B) SF6
(C) H2S
(D) PCl3
Ans. (B)
Sol. SF6 contain 12 electrons in outermost shell of sulphur hence (B) is correct.
6.
Which pair among the following is colourless?
(A) Sc3+, Zn2+
(B) Ti2+, Cu2+
(C) Fe3+, Mn2+
(D) Fe3+, Cu2+
Ans. (A)
Sol. Sc [Ar] 4s2 3d
Sc3+ [Ar] 4s° 3d°
Zn2+ [Ar] 4s° 3d10
All electron are paired no d-d Transition, hence Sc3+ and Zn2+ are colourless. option
(A) is correct.
7.
Consider a complex [Fe(OH)6]3– which act as an inner orbital complex. If the CFSE
value after ignoring pairing energy is represented as –xΔ0, then x is:
(Δ0 is splitting energy in octahedral complex)
(A) 2
(B) 1.2
(C) 0.6
(D) 0.8
Ans. (A)
Sol. In[Fe(OH)6]3–, Fe is present in +3 form.
Fe3+ = [Ar] 4s° 3d5 : t2g5 eg°
CFSE = –0.4 Δ0 × 5
= –2Δ0
Hence correction option is (A).
JEE Main - June 2022
8.
Find limiting reagent and moles of NH3 produced.
N2 + 3H2 → 2NH3
10g
5g
(A) N2, 1.42
(B) N2, 0.71
(C) H2, 1.42
(D) H2, 0.71
Ans. (B)
Sol. Mole of N2 =
Mole of H2=
10
= 0.357
28
5
= 2.5
2
N2
+
Stoichiometric value 0.357
3H2 ⎯→ 2NH3
2.5
3
Since stoichiometric value of N2 is less.
Hence N2 is limiting
Mole of NH3 formed = 2 × mole of N2
= 2 × 0.357 = 0.714
Hence option (B) is correct.
9.
The magnitude of change in oxidation state of manganese in KMnO4 in faintly alkaline
or neutral medium is:
(A) 2
(B) 3
(C) 4
(D) 5
Ans. (B)
Sol. In faintly alkaline medium or neutral medium
Mn+7 of KMnO4 gets converted into MnO2.
Hence change in oxidation number = 7 – 4 = 3
10.
Which of the following pairs will give different products on ozonolysis?
(A)
(B)
(C)
(D)
Ans. (C)
JEE Main - June 2022
Sol. Ozonolysis products are as follows:
In A, B and D Arrangement of bond are symmetric hence same product of ozonolysis
is obtained.
11.
Find C
(A)
(B)
(C)
(D)
Ans. (B)
Sol.
JEE Main - June 2022
12.
Find A and B respectively.
(A)
(B)
(C)
(D)
Ans. (A)
Sol. SN2 reaction with carbon atom as a Nucleophile.
SN2 reaction with N atom as a Nucleophile because AgCN is Covalent does not exist
as CN–.
Hence option (A) is correct.
13.
Which of the following is a hypnotic drug?
(A) Seldane
(B) Terpineol
(C) Amytal
(D) Histamine
Ans. (C)
Sol. Amytal is a hypnotic drug. Hence option (C) is correct.
14.
Ksp of PbS is given as 9 × 10–30 at a given temperature. Its solubility is x × 10–15. Find the
value of x.
(A) 15
(B) 30
(C) 3
(D) 1.5
Ans. (A)
Sol. PbS ⇒ Pb2+
s
+
S2–
s
Ksp = s
2
s=
Ksp =×
9 10–30 =
3 × 10–15
Hence (A) is correct Answer.
15.
Find the number of lone pairs present in following species:
SCl2, CIF3, O3, SF6
(A) 4, 2, 2, 1
(B) 6, 4, 4, 9
(C) 6, 2, 1, 0
(D) 8, 11, 6, 18
Ans. (D)
JEE Main - June 2022
Sol. Structure of the given compounds is as follows:
Hence (D) is correct Answer.
16.
Identify the products formed in the following reaction:
∆
Lithium nitrate + NaNO3 
→ Products
(A) Li2O and NaNO2
(B) LiNO2 NaNO2
(C) Li2O and Na2O
(D) LiNO2 and Na2O
Ans. (A)
Sol. The reaction is as follows:
2LiNO3 ⎯→ Li2O + 2NO2 +
NaNO3 ⎯→ NaNO2 +
1
O
2 2
1
O
2 2
Hence (A) is correct option.
17.
Which of the followings are herbicides?
(A) Sodium chlorate and sodium arsenate
(B) Aldrin and dieldrin
(C) Aldrin and sodium chlorate
(D) Dieldrin and sodium arsenate
Ans. (A)
Sol. Sodium chlorite and sodium Arsenite is herbicides.
18.
In a 5% w/V NaCl solution, we add albumin of egg and stir well. The resultant colloidal
solution is:
(A) Lyophobic
(B) Lyophilic
(C) Emulsion
(D) Precipitate
Ans. (B)
Sol. When NaCl is Added to Albumin of Egg, a Lyophilic. Sol is obtained.
Hence (B) option is correct.
JEE Main - June 2022
19.
The first ionization energy of Na, Mg and Si are 496, 737 and 786 Kj/mol. Determine
the value of first ionization enthalpy of Al.
(A) 788 kj/mol
(B) 747 kj/mol
(C) 577 kj/mol
(D) 840 kj/mol
Ans. (C)
Sol. Order of first ionization is
Hence 1E1 of Al is greater than that of Na and less than that of Mg.
Less than that of Mg.
Hence option (C) is correct.
20.
Calculate the ratio of energy in an H atom when electron jump from infinity to ground
state and 1st excited state to ground state.
(A)
3
2
(B)
4
3
(C)
4
5
(D)
7
9
Ans. (B)
1
1  –13.6
Sol. ΔE(1- ∞) = –13.6  2 – 2  =
1
∞ 
1
1

1 
1
ΔE(1-2) = –13.6  2 – 2  = –13.6  1 – 
4
2 
1

= –13.6 ×
∆E (1–∞ )
=
∆E (1–2)
3
4
–13.6
4
=
3 3
–13.6 ×
4
Hence (B) is the correct Answer.
JEE Main - June 2022
21.
The product B is:
(A)
(B)
(C)
(D)
Ans. (B)
Sol. The product of given reaction ore as follows—
Hence option (B) is correct.
22.
What is gangue?
(A) Impurity in ore
(B) High quality ore
(C) Mineral
(D) Flux
Ans. (A)
Sol. Gangue is an impurity present in the ore.
Hence option (A) is correct.
JEE Main - July 2022
1.
π
2
0
∫
dx
is equal to:
3 + 2 sin x + cos x
(A) tan−1 (2)
(C)
Ans. (B)
Sol.
π
2
0
∫
1
π
tan−1 (2) −
2
8
(D)
π
4
π
− tan−1 (2)
3
dx
3 + 2 sin x + cos x
(1 + t 2 )
x
dx = dt
=t
2
2
Put tan
⇒
(B) tan−1 (2) −
I=
1
∫
0
2dt

4t
1 − t2 
+
(1 + t 2 )  3 +

1 + t2 1 + t2 

1
dt
2
0 3 + 3t + 4t + 1 − t
= 2∫
2
1
1
dt
dt
=∫
2
2
0 2t + 4t + 4
0 (t + 1) + 1
= 2∫
1
= tan−1 (t + 1)
= tan−1 (2) −
0
2.
π
4
Let z = 2 + 3i, then value of (z)5 + (z)5 is:
(A) 246
(C) 248
Ans. (B)
(B) 244
(D) 258
Sol. (z)5 + (z)5 = (2 + 3i)5 + (2 – 3i)5
= 2{25 + 5C2 23(3i)2 + 5C4 2(3i)4}
= 2{32 – 720 + 810}
= 244
3.
A matrix of 3 × 3 order, should be filled either by 0 or 1 and sum of all elements should
be prime number. Then the number of such matrix is equal to
Ans. 282
Sol. sum of elements of matrix lies between 0 and 9 of which 2, 3, 5 and 7 are
Prime
Cases
Sum 2 = (11, 0, 0, 0, 0, 0, 0, 0)
9!
7 !2!
Sum 3 = (111, 0, 0, 0, 0, 0, 0)
9!
6!3!
Sum 5 = (1, 1, 1, 1, 1, 0, 0, 0, 0)
9!
5!4!
Sum 7 = (1, 1, 1, 1, 1, 1, 1, 0, 0)
9!
7 !2!
Total cases = 36 + 84 + 126 + 36 = 282
JEE Main - July 2022
 dx 
x + 4 then   = ?
 dt t =4
(A) 4
(C) 8
Ans. (B)
4.
=
t
(B) zero
(D) 16
Sol. =
t
x +4
dt
1
1
⇒ = =
dx 2 x 2(t + 4)
dx
= 2(t − 4)
dt
dx
At t = 4
=0
dt
⇒
5.
If a1, a2, a3 …….a∞ are in A.P. then
Ans. 16
S
=
Sol.
∞
ar
∞
∞
ar
∑2
r=1
r
= 4 . Then 4a2 =?
 a 1 + (r − 1)d 

2r

= ∑ 
∑
2

r
=r 1 =r 1
∞
a 1 + (r − 1)d
S
=
2
∑
S−
S a1
=
+
2
2
=
=
2r + 1
r =1
a1
2
+
∞
∞
 a 1 + (r)d
∑ 
r=1
2r + 1
−
a 1 + (r − 1)d 

2r + 1

d
∑2
r=1
r+1
 1 


+ d 4 
2
1− 1 


2

a1
S a1 d
=
+
2
2 2
⇒
S = a1 + d = 4
4a2 = 4(a1 + d) = 16
6.



  


Let a =3ıˆ + ˆ,b =ˆı + 2ˆ
 + kˆ and a × (b × c)= b + λc is non parallel to c then value of λ is.
(A)
(C)
Ans. (B)

Sol. a ×
⇒
5
1

(B) –5
(D) –1




(b × c)= b + λc= (a.c) b − (a.b) c

  
b ( 1 − a.c ) = − c ( a .b + λ )


Since b and c are non-parallel
⇒
a.c = 1 & a .b = −λ
⇒
λ = −5
JEE Main - July 2022
7.
αex + βe− x + γ sin x 2
then which of the following option is incorrect?
=
x →0
3
x sin2 x
If lim
(A) α2 + β2 + γ2 = 1
(B) αβ + βγ + γα + 1 = 0
(C) αβ2 + βγ2 + γα2 + 3 = 0
(D) α2 – β2 + γ2 + 4 = 0
Ans. (B)
 αex + βe− x + γ sin x   x2  2

Sol. 
=

  sin2 x  3
x3



⇒
lim
αe+ x + βe− x + γ sin x
x3
x →0
should exist
 
 
1  
x2
x2 x3   x3
1− x + − .... +γ x − +.....  
α 1+ x + +... +β
 

3  
21
21 31   31
 
 
⇒ x
Should exist in neighbourhood of zero.
⇒
α + β= 0, α − β + γ= 0
α β γ 2
− − =
6 6 6 3
Solving
we
get
α = 1, β = −1, γ = − 2
substituting
in
options
we
get
that
αβ + βγ + γα + 1 = 0 is incorrect
8.
( x −2)
3
3
2
=
f(x)
If
+ 4 and
P: f(x) attains maximum value at x = 0
Q: f(x) have point of inflection at x = 2
R: f(x) is increasing for x > 2 , then which of the following statement are correct?
(A) P & R
(B) Q & R
(C) P & Q
(D) P, Q, R all
Ans. (B)
( x −2)
3
2
Sol.=
f(x) 3
+4
( x −2)
f '(x) = 3
2
3
(
)
2
ln 3. 3 x2 − 2 .(2x) assumes zero
values 1 at x = ± 2, 0
At x = 0 derivative changes its sign from negative to positive
Also
at x
=
=
2 f '(x) 0 but its not changing the sign hence point of inflesion
f '(x) > 0 for x > 2 ⇒ f is increasing therefore Q & R are true.
JEE Main - July 2022
9.
If
1
1
1
k
+
++
= then 34k is equal to ……
2⋅3⋅4 3⋅4⋅5
100 ⋅ 101 ⋅ 102 101
Ans. 286
Sol.
1
1
1
k
+
++
=
2⋅3⋅4 3⋅4⋅5
100 ⋅ 101 ⋅ 102 101
∑ (r + 1)(r + 2)(r + 3)
1
2
99
=
=
=
10.
99
1
2
2
r=1
∑
r=1
(r + 3) − (r + 1)
(r + 1)(r + 2)(r + 3)
99
1
2
∑
1
2
 1

1
−


 2 ⋅ 3 101 ⋅ 102 
r=1


1
1
−


 (r + 1)(r + 2) (r + 2)(r + 3) 
1  101 × 102 − 6 
k
=


2  6 × 101 × 102  101
286
34
⇒
k=
⇒
34k = 286
Let f(x) = | (x − 1) | cos | x − 2 | sin | x − 1 | + | x − 3 | x2 − 5x + 4 . The number of points
where the function is not differentiable is:
(A) 3
(B) 4
(C) 5
(D) 6
Ans. (A)
Sol. |x-1| cos (|x-2|) sin |x-1|
Since cos(=
−θ) cos(θ); cos (| x − =
2 |) cos(x − 2) & |x-1| sin |x-1| is not changing its
definition due to |.| function Þ f1(x) is throughout differentiable
f2 (x) =
| x − 1|| x −4 || x −3|
= |(x–1) (x –4) (x – 3)|
has 3 sharp edges at x = 1, 3, 4
Hence (3) points of non-differentially
11.
If | x − 1 |≤ y ≤ 5 − x2 , then the area of region bounded by the curves is:
(A)
5π 1
−
4 2
(B)
(C)
3π 1
−
4 2
(D) cos−1
Ans. (A)
5π 3
−
4 2
1 1
−
3 2
JEE Main - July 2022
Sol. | x − 1 | ≤ y ≤ 5 − x2
Solving the equations we get point of intersection as (–1, 2) & (2, 1)
Area =
∫(
)
2
2 × 2 1 
5 − x2 dx − 
+ 
2
 2
−1
=
1
2
(
)
5 − x2 x +
5
x
sin−1
2
5
2
−
−1
5
2

5
2  
5
5
−1 1 
=  1 + sin−1
 −  −1 − sin
−
2
2
5 
5 2

=
=
=
12.
5  −1 2
1  1
+ sin−1
 sin
−
2
5
5 2
 2
5
2
1
1  1
×
+
×
sin−1 
−
2
5
5
5 2
 5
5 π 1 5π 1
× − =
−
2 2 2
4 2
The straight line y = mx + c is a focal chord of parabola y 2 = 4x , which also touches
the hyperbola x2 − y 2 =
4, then the value of 'm' is:
2
(A) m = ±
3
(B) m = ±
3
2
(C) m = ±
2
3
(D) m = ±
3
2
Ans. (A)
Sol. y = mx + c passes through the focus (1, 0), hence c + m = 0
⇒
y = mx - m
x2 – m2 (x2 –2x + 1) = 4
(1 –m)2x2 – 2m2x – (m2 + 4) = 0
has equal roots
⇒
4m4 + 4 (1 – m2) (m2 + 4) = 0
m4 – m4 – 3m2 + 4 = 0
⇒
m= ±
2
3
JEE Main - July 2022
13.
The straight line ‘L’ passing through point of intersection of line bx + 10y − 8 =
0 and
2x = 3y . The line also passes through the point (1,1) and is tangent to circle
(
)
17 x2 + y 2 =
16, then the value of 9b2 + 16 is:
Ans. 34
Sol. 4 + λL2 =0 passes through (1, 1)
(bx + 10y –8) + λ(2x –3y) =0
⇒
(b + 2) − λ = 0
⇒
(b + 2) =
λ
Line becomes
(bx + 10y –8) + (b + 2) (2x –3y) = 0
4
Distance from origin is
⇒
17
8
(3b + 4)2 + ( 4 − 3b )
2
=
4
17
9b2 + 24b + 16 + 9b2 – 24b + 16 = 68
⇒
14.
9b2 + 16 = 34
Let S = {1, 2, 3,  , 2022} and
=
A {(a,b) : a,b ∈ S and HCF of b and 2022 is 1}. If an ordered pair (α, β) such that
α, β ∈ S is selected, then the probability that (α, β) ∈ A is:
(A)
112
337
(B)
674
1011
(C)
526
1011
(D)
112
1011
Ans. (A)
Sol. HCF (b, 2022) = 1
Sample space = 2022 × 2022
Favourable cases are those where b & 2022 are coprime
⇒
cases = 2022 × (possible value of b)
2022 = 2 × 3 × 337
Possible values of b = 2022 – n
p ≡ numbers less than equal to 2022 and not coprime to 2022
n(p) = {1011 + 674 + 6 – 337 – 2 – 3 +1} = 1350
Probability
=
2022 × (2022 − 1350)
672
112
= =
2022 × 2022
2022 337
JEE Main - July 2022
15.
Tower of height h, angle of elevation from point A is α. Let B be the point 9m to the
 3 
south of A from where angle of elevation is cos−1 
 . Distance of tower from B is
 13 
15 m. The value of cot α is:
(A) 4/3
(B) 6/5
(C) 5/7
(D) 7/6
Ans. (B)
Sol. cos β =
⇒
3
13
h2 cot 2 α + 81 = 225
Also h cot β = 15 ⇒
=
h
⇒
cot α
=
15.2
= 10
3
225 − 81 12 6
= =
100
10 5