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


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JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 1

A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no slipping between string & pulley)

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 1


v = ωR (no slipping)



 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 2

Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is :

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 2

Take 1kg mass at origin


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 3

In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of first minima is  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 3

For 2nd minima
d sinθ = 2λ
sinθ = (given)
......(1)
So for 1st minima is
d sinθ = λ
sinθ = (from equation (i))
θ = 25.65° (from sin table)
θ ≈ 25°

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 4

There are two infinite plane sheets each having uniform surface charge density +σ C/m2. They are inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:
   

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 4


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 5

A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by k = k0(1 + αx). Calculate capacitance of system: (given αd << 1)

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 5

Capacitance of element 

Capacitance of element 

Given αd << 1

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 6

A long solenoid of radius R carries a time dependent current I = I0 t(1 – t). A ring of radius 2R is placed coaxially near its centre. During the time interval 0 ≤ t ≤ 1, the induced current IR and the induced emf VR in the ring vary as:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 6


and 

 

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

If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated such that transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer).

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

So θ > 45° and 90 – θ < 45º so only one option is correct i.e. 18.4º
angle rotated should be = 90° – 71.6° = 18.4°

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

Three moles of ideal gas A with  is mixed with two moles of another ideal gas B with . The of mixture is (Assuming temperature is constant)

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


on rearranging we get




γmix = 1.42

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 9

Given magnetic field equation is B = 3 × 10–8 sin(ωt + kx + φ)then appropriate equation for electric field (E) will be :

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 9

 (speed of light in vacuum)
E0 = B0C = 3 × 10–8 × 3 × 108
= 9 N/C
So E = 9 sin (ωt + kx + φ)

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 10

There is a LCR circuit, If it is compared with a damped oscillation of mass m oscillating with force constant k and damping coefficient 'b'. Compare the terms of damped oscillation with the devices in LCR circuit.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 10

In damped oscillation 
ma + bv + kx = 0


In the circuit

Comparing equation (i) and (ii)
m = L, b = R, k = 1/c

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 11

A lift can hold 2000kg, friction is 4000N and power provided is 60HP. (1 HP = 746W) Find the maximum speed with which lift can move up.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 11

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 12

A H–atom in ground state has time period T = 1.6 × 10–16 sec. find the frequency of electron in first excited state 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 12


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 13

Magnification of compound microscope is 375. Length of tube is 150mm. Given that focal length of objective lens is 5mm, then value of focal length of eyepiece is:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 13

Case-I
If final image is at least distance of clear vision


Case-II
If final image is at infinity

fe = 22 mm 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 14

1 litre of a gas at STP is expanded adiabatically to 3 litre. Find work done by the gas. Given γ = 1.40 and 31.4= 4.65

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 14


now work done 
Closest ans is 90.5 J

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 15

A string of length 60 cm, mass 6gm and area of cross section 1mm2 and velocity of wave 90m/s. Given young's modulus is Y = 1.6 × 1011 N/m2. Find extension in string. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 15


after substituting value of μ,v,l,A and Y we get
Δl = 0.3 mm

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 16

Which of the following gate is reversible 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 16

A logic gate is reversible if we can recover input data from the output eg. NOT gate

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 17

A thin uniform rod is of mass M and length L. Find the radius of gyration for rotation about an axis passing through a point at a distance of L/4 from centre and perpendicular to rod.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 17


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 18

A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When it reaches a height equal to radius of earth, it ejects a rocket of mass M/10 and itself starts orbiting the earth in circular path 
of radius 2R, find the kinetic energy of rocket.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 18




Kinetic energy 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 19

The current 'i' in the given circuit is

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 19


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 20

A current carrying circular loop is placed in an infinite plane if φ1 is the magnetic flux through the inner region and φ0 is magnitude of magnetic flux through the outer region, then  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 20

As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux in the inner region must be passing through the outer region. Since flux in two regions are in opposite direction,
∴ φi = - φ0

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 21

Consider a loop ABCDEFA with coordinates A (0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0) E(0, 5, 5) and F(0, 0, 5). Find magnetic flux through loop due to magnetic field 


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 21



φ = (3 × 25) + (4 × 25)  = 175 weber

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 22

A Carnot's engine operates between two reservoirs of temperature 900K and 300K. The engine performs 1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is:


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 22

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 23

A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10–5/°C along the x-axis and 5 × 10–6/°C along y-axis and z-axis. If coefficient of volume expansion of the solid is C × 10–6/°C then the value of C is 


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 23

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 24

A particle is released at point A. Find its kinetic energy at point P. (Given m = 1 kg and all surfaces are frictionless)


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 24

As only consevative internal force acts upon the mass and earth system, thus we can say mechanical energy is conserved. Thus we get that net loss in PE = net gain in KE
Loss in PE = mg Δh = 1 x 10 x 1 = 10J
Thus gain in KE = 10J

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 25

On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x  


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 25

Energy of photon. E == 4eV  >  2eV (so photoelectric effect will take place)
= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule
No. of photons falling per second

No. of photoelectron emitted per second

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 26

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 26


2 × 0.34 = + 1 x 0.522
Eº1 = 0.68 – 0.522
Eº1 = 0.158

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 27

Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 28

Arrange the following in order of their pKb value
  (B)  CH3–NH–CH3    (C) CH3–CH=NH  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 28

Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is aliphatic amine, here the 'N' is sp3 whereas in option 'C' the 'N' is sp2, hence B is more basic than C.

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 29

1-Methylethylene oxide Product 'X' will be –

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 29

--excess HBr---> 


 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 30

Correct order of Intermolecular forces

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 31

Hex-3-ynal  (X), formed product X will be: 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 31


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 32

  , formed product 'X' is used as:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 32



Methyl orange is used as an indicator in acid base titration.

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 33

In which of the following Saytzeff product will not be formed as major product ?
(A)                         (B)   
(C)                  (D) 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 33





 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 34

Match the column

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 34

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 35

​Atomic radius of Ag is similar to

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 35

Atomic radius of Ag is closest to Au.

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 36

Correct IUPAC name of [Pt(NH3)2Cl(CH3NH2 )]Cl is:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 36

The IUPAC name of [Pt(NH3)2Cl(CH3NH2 )]Cl is diamminechlorido(methylamine)platinum(II)chloride.
The ligands present are ammine, chloro and methyl amine. The ligands are named according to the alphabetical order.
The prefix di indicates two.
The oxidation state of platinum is +2. The oxidation state is written in roman numerals inside parenthesis.

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 37

Vapour pressure of pure CS2 and CH3COCH3 are 512 mm of Hg and 312 mm of Hg respectively. Total vapour pressure of mixture is 600 mm of Hg then find incorrect statement:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 37

Above mixture of liquids show positive deviation from Raoult's Law

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 38

​Purest form of commercial iron is:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 38

Purest form is wrought iron.

 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 39


Mixture of above three organic compound was subjected to aq NaHCO3 and followed by dil NaOH. compounds which will be soluble in given solvent will be : 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 39


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 40

Which theory can explain bonding of Ni(CO)4:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 40

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 41

​n = 5, ms = + 1/2 How many orbitals are possible:

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 42

In zeolites & synthetic resin method which will be more efficient in removing permanent hardness of water :

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 42

The difference between the zeolite and ion exchange process is that the zeolite process uses the mineral zeolite as the exchanging resin for cations in the hard water whereas the ion exchange process include several different resins for the ion exchange.

∴ Synthetic resins method is more efficient as it can exchange both cations as well as anions.

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 43

Oxidation state of potassium in K2O, K2O2 & KO2 are respectively –

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 43

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 44

Decreasing order of dipole moment in CHCl3, CCl4 & CH4 is –

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 44

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 45

Amongst the following which is not a postulate of Dalton's atomic theory

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 45

1) Elements are composed of extremely small particles called atoms that are indivisible and indestructible
2)All atoms of a given element are identical; they have the same size, mass, and chemical properties
3) Atoms of 1 element are different from the atoms of all other elements
4)Compounds are composed of atoms of more than 1 element. The relative number of atoms of each element in a given compound is always the same.
5)Chemical reactions only involve the rearrangement of atoms. Atoms are not created or destroyed during chemical reactions.

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 46

Half life of 90Sr is 6.93 years. In a child body 1 μg of 90Sr dopped in place of calcium, how many years will it take to reduce its concentration by 90% (Assume no involvement of Sr in metabolism).


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 46

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 47

Each of solution A and B of 100 L containing 4 g NaOH and 9.8 g H2SO4. Find pH of solution which is obtain by mixing 40 L solution of A and 10 L solution of B.


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 47


*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 48

A(l) → 2B(g)
ΔU = 2.1 kcal, ΔS = 20 cal/k, T = 300 K.
Find ΔG (in kcal)

ΔU = 2.1 kcal, ΔS = 20 cal/k, T = 300 K.


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 48

ΔH = ΔU + ΔngRT
= 2.1 × 103 + 2(2) (300)
= 2100 + 1200
= 3300 cal
ΔG = ΔH - TΔS
= 3300 – (300) (20)
= 3300 – 6000
= –2700 cals = –2.7 kcal

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 49

Cl2 on reaction with hot & conc. NaOH gives two chlorine having products X and Y. On treatment with AgNO3, X gives precipitate. Determine average bond order of Cl and O bond in 'Y' ?


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 49

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 50

Number of chiral centers in chloramphenicol is : 


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 50

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 51

If f(x) is continuous and differentiable in x ∈ [ -7, 0] and f'(x) ≤ 2∈ [-7, 0], also f(-7) = -3 then range of f(-1) + f(0) 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 51

Lets use LMVT for x ∈ [-7, -1]

Also use LMVT for x ∈ [-7, 0]

∴ f(0) + f(-1) ≤ 20

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 52

If y = mx + 4 is common tangent to parabolas y2 = 4x and x2 = 2by. Then value of b is 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 52

y = mx + 4  ……(i)
y2 = 4x tangent
 y = mx + a/m
⇒ y = mx + 1/m ………(ii)
from (i) and (ii)
4 = 1/m ⇒ m =1/4
So line y = + 4 is also tangent to parabola x2 = 2by, so solve

⇒ 2x2 – bx – 16b = 0
⇒ D = 0
⇒ b2 - 4 × 2 × (-16b) = 0
⇒ b2 + 32 × 4b = 0 
b = –128, b = 0 (not possible)

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 53

If α and β are the roots of equation (k+1) tan2x - √2λ tanx = 1 - k and tan2 (α+β) = 50. Find value of λ

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 53



λ = 10

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 54

Find image of point (2, 1, 6) in the plane containing points (2, 1, 0), (6, 3, 3) and (5, 2, 2) 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 54

Plane is x + y – 2z = 3

⇒ (x, y, z) = (6, 5, -2)

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 55

Let (x)k + (y)k = (a)where a, k > 0 and then find k

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 55


k – 1 =
 k == 2/3

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 56

If  g(x) = x2 + x – 1 and g(f(x)) = 4x2 – 10x + 5, then find f.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 56


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 57

If z = x + iy and  real part  = 1 then locus of z is  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 57

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 58

Let y = f(x) is a solution of differential equation and f(0) = 0 then f(1) is equal to :  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 58

ey = t


Put x = 0, y = 0 then c = 1
ey–x = x + 1
y = x + ln(x + 1)
at x = 1 , y = 1 + ln(2) 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 59

​If α is a roots of equation x2 + x + 1 = 0 and A = then A31 equal to : 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 59


=> A4 = I 
=> A30 = A28 × A3 = A3 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 60

The six digit numbers that can be formed using digits 1, 3, 5, 7, 9 such that each digit is used at least once.


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 60

1, 3, 5, 7, 9
For digit to repeat we have 5C1 choice And six digits can be arrange in ways.
Hence total such numbers = 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 61

The area that is enclosed in the circle x2 + y2 = 2 which is not common area enclosed by y = x & y2 = x is

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 61

Total area – enclosed area 



JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 62

If sum of all the coefficient of even powers in (1– x + x2 – x3 ……..x2n) (1 + x + x2 +x3 ……….+ x2n) is 61 then n is equal to

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 62

Let (1– x + x2 …..) (1 + x + x2 ……) = a0 + a1 x + a2x2 +……
put  x = 1
1(2n+1) = a0 + a1 + a2 +...,.a2n ...(i)
put  x = –1
(2n+1) x 1 = a0 - a1 + a2 +....a2n.....(ii)
Form (i) + (ii)
4n + 2= 2(a0 + a2 +….)
= 2 x 16
⇒  2n+1 = 61
⇒ n = 30

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 63

If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16 then the value of m + n is


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 63


⇒ n2 - 1 = 120
⇒ n = 11
Var (2, 4, 6,.....,2m) = 16
⇒ var (1, 2,....,m) = 4
⇒ m2 - 1 = 48
⇒ m = 7
⇒ m + n = 18 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 64

Evaluate 


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 64

Put 3x/2 = t



 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 65

​If A(1, 1), B(6, 5), C are vertices of ΔABC. A point P is such that area of ΔPAB, ΔPAC, ΔPBC are equal, also , then length of PQ is 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 65

P will be centroid of ΔABC
 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 66

​(p → q) ∧ (q → ~p) is equivalent to

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 66


Clearly (p → q) ∧ (q → ~p) is equivalent to ~p

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 67

​Find greatest value of k for which 49k + 1 is factor of 1 + 49 + 492 …..(49)125

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 67


 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 68

If f(x) = |2 – |x – 3|| is non differentiable in x ∈ S.Then value of  is   


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 68

∵ f(x) is non differentiable at x = 1,3,5
∑ f(f(x)) = f(f(1) + f(f(3)) + f (f(5))
= 1 + 1 + 1
= 3 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 69

If  system of equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0
have non-trivial solution
then

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 69

For non-trivial solution

(3bc – 4bc) – (2ac – 4ac) + (2ab – 3ab) = 0
–bc + 2ac – ab = 0
ab + bc = 2ac 
a, b, c in H.P.
 in A.P

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 70

If sum of 5 consecutive terms of 'an A.P is 25 & product of these terms is 2520. If one of the terms is – 1/2 then the value of greatest term is

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 70

Let terms be a – 2d, a –d, a, a + d , a + 2d .
sum = 25 

⇒  5a = 25
⇒ a = 5
Product = 2520
⇒ (5–2d) (5 – d) 5(5+d)  (5+2d) =2520
⇒ (25 - 4d2) (25 - d2) = 504
⇒ 625 -100d- 25d2 + 4d4 = 504
⇒ 4d4 - 125d2 + 625 -504 = 0
⇒ 4d4 - 125d2 + 121 = 0
⇒ 4d4 - 121d2 - 4d2 +121 = 0
⇒ (d2-1) (4d2 - 121) = 0
d = ±1, d = ± 11/2
d = ±1,  does not give -1/2 as a term
∴ d = 11/2
∴ Largest term = 5 + 2d = 5 + 11 = 16    

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 71

Let
lies in plane of

of a bisectors angle between& then  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 71

angle bisector can be or 



Compare with 


Not in option so now consider



Compare with 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 72

Given f(a + b + 1 – x) = f(x) R then the value of is equal to 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 72


x → a + b - x
[∵ put x → x + 1 in given equation]
(1) + (2)
2I = 


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 73

If distance between the foci of an ellipse is 6 and distance between its directrices is 12, then length of its latus rectum is 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 73

2ae = 6 and 
⇒ ae = 3 and a/e = 6
⇒ a2 = 18
⇒ b2 = a2 - a2e2 = 18 - 9 = 9

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 74

An unbiased coin is thrown 5 times. Let X be a random variable and k be the value assigned to X for  k = 3, 4, 5 times Head occurs consecutively and otherwise the value of X is assigned –1. What is value of expectation.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 74


k = no. of times head occur consecutively
Now expectation

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 75

If  when then find dy/dα at α = 5π/6

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 75



= –1 – cota 
 
⇒  will be = 4

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