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JEE Main 2020 Question Paper with Solution (8th January - Morning)


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JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 1

A block of mass m is connected at one end of spring fixed at other end having natural length l0 and spring constant K. The block is rotated with constant angular speed (ω) about the fixed end in gravity free space. The elongation in spring is- 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 1


JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 2

3 point charges are placed on circumference of a circle of radius 'd' as shown in figure. The electric field along x-axis at centre of circle is:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 2


JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 3

Choose the correct P-V graph of ideal gas for given V-T graph.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 4

Find the coordinates of centre of mass of the lamina, shown in figure.  

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 4



JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 5

Which graph correctly represents v ariation between relaxation time  of gas molecules with absolute temperature (T) of an ideal gas.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 5

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 6

​If two capacitors C1 & C2 are connected in parallel then equivalent capacitance is 10μF. If both capacitor are connected across 1V battery then energy stored by C2 is 4 times that of C1. Then the equivalent capacitance if they are connected in series is -

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 6


Given C1 + C2 = 10μF ........ (i)

⇒ 4C1 = C2 …(ii)
from equation (i) & (ii)
C1 = 2μF
C2 = 8μF
If they are in series

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 7

A rod of mass 4m and length L is hinged at the mid point. A ball of mass 'm' moving with speed V in the plane of rod which is perpendicular to axis of rotation, strikes at one end at an angle of 45º and sticks to it. The angular velocity of system after collision is–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 7


Loi = Lof

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 8

​Two photons of energy 4eV and 4.5eV are incident on two metals A and B respectively. Maximum kinetic energy for ejected electron is TA for A and TB = TA – 1.5eV for metal B. Relation between de-Broglie wavelengths of ejected electron of A and B are λB = 2λA. The work function of metal B is -

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 8

Relation between De-Broglie wav elength and K.E. is


⇒ TA = 2 eV
∴ KEB = 2 – 1.5 = 0.5 eV
φB = 4.5 - 0.5 = 4 eV

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 9

​There is a potentiometer wire of length 1200 cm and 60 mA current is flowing in it. A battery of emf 5V and internal resistance of 20Ω is balanced on potentiometer wire with balancing length 1000 cm. The resistance of potentiometer wire is–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 9


Potential gradient = 5 / 1000 = Vp/1200
 VP = 6V
and RP = = 100Ω

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 10

 A telescope has magnifying power 5 and length of tube is 60cm then the focal length of eye piece is–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 10

m = fo/fe
5 = fo/fe
fo = 5fe
fo + fe = 60
6fe = 60
fe = 10

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 11

Two spherical bodies of mass m1 and m2 have radii 1 m and 2 m respectively. The gravitational field of the two bodies with the radial distance from centre is shown below. The value of is-

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 11


JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 12

​When a proton of KE = 1.0 MeV moving in South to North direction enters the magnetic field (from West to East direction), it accelerates with acceleration a = 1012 m/s2. The magnitude of magnetic field is–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 12


∴ Bqv = ma

= 0.71× 10–3T
So, 0.71 mT

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 13

​If electric field around a surface is given by where 'A' is the normal area of surface and Qin is the charge enclosed by the surface. This relation of gauss's law is valid when

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 13

Magnitude of electric field is constant & the surface is equipotential

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 14

Stopping potential depends on Planks constant (h), current (I), Universal gravitational constant (G) and speed of light (C). Choose the correct option for the dimension of stopping potential (V)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 14

V = K (h)a(I)b(G)c(C)d (V is voltage)
we know [h] = ML2T-1


a – c = 1………………(1)
2a + 3c + d = 2………………(2)
–a –2c –d = –3 ………………(3)
b = –1………………(4)
on solving
c = –1
a = 0
d = 5,
b = –1
V = K (h)° (I)-1(G)-1(C)5

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 15

​A cylinder of height 1m is floating in water at 0°C with 20 cm height in air. Now temperature of water is raised to 4°C, height of cylinder in air becomes 21cm. The ratio of density of water at 4°C to density of water at 0°C is– (Consider expansion of cylinder is negligible)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 15



mg = A(80) ρ0ºg
mg = A(79) ρ4ºg

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 16

Number N of the α-particles deflected in Rutherford's α-scattering experiment varies with the angle of deflection θ. Then the graph between the two is best represented by. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 16


 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 17

If relative permittivity and relative permeability of a medium are 3 and 4/3 respectively. The critical angle for this medium is.

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JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 18

​The given loop is kept in a uniform magnetic field perpendicular to plane of loop. The field changes from 1000G to 500G in 5seconds. The average induced emf in loop is–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 18

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 19

​Choose the correct Boolean expression for the given circuit diagram:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 19

First part of figure shown is OR gate and Second part of figure shown is NOT gate
So Yp = OR + NOT = NOR gate

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 20

A solid sphere of density just floats in a liquid. The density of liquid is – 
(r is distance from centre of sphere)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 20

 0 < r ≤ R
mg = B

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 21

Two particles each of mass 0.10kg are moving with velocities 3m/s along x-axis and 5m/s along y-axis respectively. After an elastic collision one of the mass moves with a velocity . The energy of other particle after collision is x/10 , then x is.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 21

For elastic collision KEi = KEf

34 = 32 + v2

x = 1

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 22

A plano-convex lens of radius of curvature 30 cm made of refractive index 1.5 is kept in air. Find its focal length (in cm).  


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 22


R1 = ∝
R2 = –30 cm


f = 60 cm

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 23

Position of two particles A and B as a function of time are given by XA = – 3t2 + 8t + c and YB = 10 – 8t3. The velocity of B with respect to A at t = 1 is √v . Find v.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 23



= 580

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 24

An open organ pipe of length 1m contains an ideal gas whose density is twice the density of atmosphere at STP. Find the difference between fundamental and second harmonic frequencies if speed of sound in atmosphere is 300 m/s.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 24


*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 25

Four resistors of 15Ω, 12Ω, 4Ω and 10Ω are arranged in cyclic order to form a wheatstone bridge. What resistance (in Ω) should be connected in parallel across the 10Ω resistor to balance the wheatstone bridge.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 25



= 15 × 4
on solving
R = 10 Ω

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 26

Number of S–O bond in S2O82– and number of S–S bond in Rhombic sulphur are respectively:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 26

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 27

Following vanderwaal forces are present in ethyl acetate liquid

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 27

Ethyl acetate is polar molecule so dipole-dipole interaction will be present there.

*Multiple options can be correct
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 28

Given, for H-atom
 
Select the correct options regarding this formula for Balmer series. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 28

Correct Answer : a,c

Explanation :  (a) For balmer series always started ground state n 1 = 2

(c) At the longest wavelength, the higher state energy will be minimum therefore the excited state will be n2 = 3 next to the ground always.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 29

Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol-1 respectively are:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 29

The electronic configurations are as follows:
 11Na [Ne]3s1, 12Mg [Ne]3s2, 13Al [Ne] 3s23p1, 14Si [Ne]3s23p2
The ionization energy of Mg will be larger than that of Na due to fully filled configuration [Ne] 3s2
The ionization of Al will be smaller than that of Mg due to one electron extra than the stable configuration but smaller than Si due to the increase in effective nuclear charge of Si.
⇒ Na < Mg > Al < S

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 30

Select the correct stoichiometry and its ksp value according to given graphs. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 30


Ksp = [X+] [Y]
or Ksp = 2 × 10–3 × 10–3
or Ksp = 2 × 10–6

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 31

According to Hardy Schultz rule, correct order of flocculation value for Fe(OH)3 sol is : 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 31

According to hardy-schultz rule,
Coagulation value or flocculation value 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 32

Which of the following complex exhibit facial meridional geometrical isomerism.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 32

[Ma3b3] type complex shows facial and meridional isomerism 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 33


(A) Intermolecular force of attraction of X > Y.
(B) Intermolecular force of attraction of X < Y.
(C) Intermolecular force of attraction of Z < X.
Select the correct option(s).

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 33

At a particular temperature as intermolecular force of attraction increases vapour pressure decreases.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 34

​Rate of a reaction increases by 106 times when a reaction is carried out in presence of enzyme catalyst at same temperature. Determine change in activation energy. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 34


or 6 ln 10 = (E - Ec ) / RT
or 
or E–EC = 2.303 × 6RT
or 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 35

Gypsum on heating at 393K produces

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 35

Theory based.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 36

​Among the following least 3rd ionization energy is for

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 36

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 37

Accurate measurement of concentration of NaOH can be performed by following titration:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 37

Oxalic acid is a primary standard solution while H2SO4 is a secondary standard solution. 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 38

Arrange the following compounds in order of dehydrohalogenation (E1) reaction.
(A)        (B)       (C)    (D) 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 38

E1 reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster the E1 reaction.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 39



Product A and B are respectively : 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 39



[A] is more stable radical and undergoes Markovnikov addition to form [B].

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 40

Major product in the following reaction is

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JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 41

​Arrange the order of C—OH bond length of the following compounds.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 41

CH3–OH
                 
        A           B                   C
There is not any resonance in CH3–OH. Resonance is poor in p-Ethoxyphenol than phenol

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 42

​Which of the following are "green house gases"?
(a) CO2           
(b) O2             
(c) O3             
(d) CFC               
(e) H2O

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 42

CO2, O3, H2O vapours and CFC's are green house gases.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 43

Two liquids isohexane and 3-methylpentane has boiling point 60°C and 63°C. They can be separated by

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 43

Liquid having lower boiling point comes out first in fractional distillation. Simple distillation can't be used as boiling point difference is very small.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 44

Which of the given statement is incorrect about glucose?

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 44

Open chain form of glucose is very very small, hence does not gives Schiff's test.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 45

​Reagent used for the given conversion is: 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 45

B2H6 is very selective and usually used to reduce acid to alcohol.

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 46

0.3 g [ML6]Cl3 of molar mass 267.46 g/mol is reacted with 0.125 M AgNO3(aq) solution, calculate volume of AgNO3 required in ml. 


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 46


 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 47

Given : 2H2O → O2 + 4H+ + 4e
Eº = –1.23 V Calculate electrode potential at pH = 5.


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*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 48

Calculated the mass of FeSO4.7H2O, which must be added in 100 kg of wheat to get 10 PPM of Fe.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 48


*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 49

A gas undergoes expansion according to the following graph. Calculate work done by the gas.


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*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 50

Number of chiral centres in Pencillin is


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 50


Star marked atoms are chiral centers. 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 51

If f(x) =  g(x) = then find the area bounded by f(x) and g(x) from x = 1/2 to x =  

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 51



Required area = Area of trapezium ABCD – 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 52

z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 52

z = x + iy
|x| + |y| = 4
Minimum value of |z| = 2√2
Maximum value of |z| = 4

So |z| can't be √7

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 53

If f(x) =  and a – 2b + c = 1 then

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 53

Apply R1 = R1 + R3 – 2R2

⇒ f(x) = 1
⇒ f(50) = 1  

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 54

​Let an is a positive term of a GP and 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 54

Let GP is a, ar, ar2 …….. 
= 200
 ................. (i)

................... (ii)

Form (1) and (2) r = 2
add both
⇒ a2 + a3 + ........... a200 + a201 = 300 ⇒ r(a1 +...........a200) = 300
= 300/r
= 150 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 55

If y(1) = 1 and y(x) = e then x = ? 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 55

Put y = vx





When x = 1, y = 1 then

x2 = y2(1 + 2lny)
x2 = e2(3)
x = ± √3 e
So x = √3e

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 56

Let probability distribution is then value of p(x > 2) is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 56

 ⇒ 6k2 + 5k = 1
6k2 + 5k – 1 = 0
6k2 + 6k – k – 1 = 0
(6k – 1) (k + 1) = 0 ⇒ k = - 1
(rejected) ; k = 1/6
P(x > 2) = k + 2k + 5k2

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 57

 = λtanθ + 2log f(x) + c then ordered pair (λ,f(x)) is 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 57



= – t + 2 log (1+t) + C
= –tanθ + 2 log (1 + tanθ) + C
⇒ λ = -1 and f(x) = 1 + tanθ

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 58

If p → (p ∧ ~ q) is false. Truth value of p & q will be 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 58

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 59

If = A then the value of x at which f(x) = [x2] sinπx is discontinuous 
(where [.] denotes greatest integer function)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 59


4 – 0 = A check when
(A) x = ⇒ x = √5 ⇒ discontinuos
(B) x = ⇒ x = 5 ⇒ continuous
(C) x = √A ⇒ x = 2 ⇒ continuous
(D) x = ⇒ x = 3 ⇒ continuous 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 60

Let one end of focal chord of parabola y2 = 8x is , then equation of tangent at other end of this focal chord is  

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 60

Let is (2t2, 4t) ⇒ t = -1/2
Parameter of other end of focal chord is 2
⇒ point is (8, 8)
⇒ equation of tangent is 8y - 4(x+8) = 0
⇒ 2y - x = 8

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 61

Let x + 6y = 8 is tangent to standard ellipse where minor axis is , then eccentricity of ellipse is 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 61


Equation of tangent ≡ y = mx ± 
comparing with ≡ 
m = -1/6 and a2m2 + b2 = 16/9


a2 = 16


 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 62

if f(x) and g(x) are continuous functions, fog is identity function, g'(b) = 5 and g(b) = a then f'(a) is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 62

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 63

If 7x + 6y – 2z = 0 
3x + 4y + 2z = 0
x – 2y – 6z = 0
then which option is correct

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 63


= 7(–20) – 6(–20) – 2(–10)  
= – 140 + 120 + 20 = 0 
so infinite non-trivial solution exist
now equation (1) + 3 equation (3)
10x – 20z = 0
x = 2z

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 64

Let  x = 2sinθ - sin2θ  and y = 2 cosθ - cos2θ
find θ = π    

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 64




JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 65


g(x) = dt then correct choice is 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 65

F'(x) = x2g(x) 
⇒ F'(1) = 1.g(1) = 0  ......(1)
(∵ g(1) = 0) 
Now F''(x) = 2xg(x) + x2g'(x)

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 66

​Let both root of equation ax2 – 2bx + 5 = 0 are α and root of equation x2 - 2bx - 10 = 0 are α and β. Find the value of α2 + β2

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 66



⇒ b2 = 5a ..... (i) (a ≠ 0)
α + β  = 2b ............ (ii)
and αβ = – 10 ............. (iii)
is also root of x2 – 2bx – 10 = 0
⇒ b2 - 2ab2 - 10a2 = 0 by (i)
⇒ 5a – 10a2 – 10a2 = 0
⇒ 20a2 = 5a
⇒ a = 1/4 and b2 = 5/4
α2 = 20 and β2 = 5
Now α2 + β2
= 5 + 20 
   = 25

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 67

Let A = and B =  then

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 67

A = { x : x ∈ (-2, 2) }
B = { x : x ∈ (-∞, -1] ∪ [5, ∞) }
A ∩ B = { x : x ∈ (-2, -1] }
A ∪ B = { x : x ∈ (-∞, 2) ∪ [5, ∞) }
A - B = { x : x ∈ (-1, 2) }
B - A = { x : x ∈ (-∞, -2] ∪ [5, ∞) }

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 68

Let x = and y = where θ ∈ (0,π/4), then

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 68

y = 1 + cos2θ + cos4θ + ....... 


cos2θ = x

y (1 – x) = 1
  

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 69

Let the distance between plane passing through lines and and plane 23x – 10y – 2z + 48 = 0 is then k is equal to

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 69

Lines must be intersecting

distance of plane contains given lines from given plane is same as distance between point (–3, –2,1) from given plane.
Required distance equal to  = = k = 3 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 70

If 25C0 + 525C1 + 925C2 ……… 101 25C25 = 225k find k = ? 


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 70



= 100 .224 + 225 = 225(50 + 1) = 51.225
So k = 51 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 71

Let circles (x – 0)2 + (y – 4)2 = k and (x – 3)2 + (y – 0)2 = 12 touches each other than find the maximum value of 'k' 


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 71

Two circles touches each other if C1 C2 = |r1 ± r2|
Distance between C2(3, 0) and C1(0, 4) is either  (C1C2 = 5)
=> √k + 1 = 5 or |k – 1| = 5
⇒ k = 16 or k = 36
⇒ maximum value of k is 36 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 72

Let angle between equal to π/3
If vector a is perpendicular to vector b x c then find the value of |vector a x (b x c)|


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 72


 ⇒ 
Also,

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 73

Number of common terms in both sequence 3, 7, 11, ………….407 and 2, 9, 16, ……..905 is   


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 73

First common term = 23
common difference = 7 × 4 = 28
Last term ≤ 407

So n = 14 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 74

If minimum value of term free from x for (x/sinθ + 1/xcosθ) is  L1 in θ ∈ [π/8, π/4] and L2 in θ ∈ [π/16, π/8]

find L2/L1.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 74


for r = 8 term is free from 'x' 

 L1 = 16C82
 
∵ {Min value of L1 at θ = π/4}

{∵ min value of L2 at θ = π/8]
 

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