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

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

### If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s2 ; radius of earth = 6400 Km]

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

at pole, weight = mg = 196
m = 19.6 kg
at equator, weight = mg – mω2R

= 195.33 N

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

### ​In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is connected to 220 V Mains supply then find minimum fuse current

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

Total power is (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)
= 4325 W
So current is = 4325/220
= 19.66 A
Ans is 20 Amp.

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

### In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial relaxation time. Given

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

relaxation time
and

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

A block of mass 10kg is suspended from string of length 4m. When pulled by a force F along horizontal from midpoint. Upper half of string makes 45° with vertical, value of F is

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

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

The surface mass density of a disc of radius a varies with radial distance as σ = A + Br where A & B are positive constants then moment of inertia of the disc about an axis passing through its centre and perpendicular to the plane

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

σ = A + Br

I =
(A + Br )2πr3 dr

= 2pa4

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

Cascaded Carnot engine is an arrangement in which heat sink of one engine is source for other. If high temperature for one engine is T1, low temperature for other engine is T2 (Assume work done by both engine is same) Calculate lower temperature of first engine.

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

Let, QH : Heat input to Ist engine
QL : Heat rejected from Ist
engine
QL': Heat rejected f rom IInd             engine
Work done by Ist engine = work done by IInd engine
QH – QL = QL – QL'
2QL = QH + QL'

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

Activity of a substance changes from 700 s–1 to 500 s–1 in 30 minute. Find its half-life in minutes

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

=> ln2 = λt1/2...........(i)

(i)/(ii)

(2.06004) 30 = t1/2 = 61.8 min.

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

In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is

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

= 5.9mm

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

An ideal fluid is flowing in a pipe in streamline flow. Pipe has maximum and minimum diameter of 6.4 cm and 4.8 cm respectively. Find out the ratio of minimum to maximum velocity.

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

Using equation of continuity
A1V1 = A2V2

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

There is a electric circuit as shown in the figure. Find potential difference between points a and b.

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

Diode is in forward bias, so it will behave as simple wire so,

So Vab= 10V

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

A particle of mass m and positive charge q is projected with a speed of v0 in y–direction in the presence of electric and magnetic field are in x–direction. Find the instant of time at which the speed of particle becomes double the initial speed.

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

As(magnitude of velocity does not change in y–z plane)

∴

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

Two sources of sound moving with same speed v and emitting frequency of 1400 Hz are moving such that one source s1 is moving towards the observer and s2 is moving away from observer. If observer hears beat frequency of 2 Hz. Then find the speed of source. Given
Vsound >> VSource
Vsound = 350 m/s

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

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

An electron & a photon have same energy E. Find the ratio of de Broglie wavelength of electron to wavelength of photon. Given mass of electron is m & speed of light is C.

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

λd for electron =
λ for photon = hC/E
Ratio =

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

​A ring is rotated about diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively :

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

When ωt = π
∵  φ will have maximum.
∵ e will be minimum.
t =  = 5 sec.

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

Electric field in space is given by A positively charged particle at (0, 0, π/K) is given velocity v0kˆ at t = 0. Direction of f orce acting on particle is

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

Force due to electric field is in direction
because at t = 0, E =
Force due to magnetic field is in direction
∴ it is parallel to
∴ net force is antiparallel to

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

Focal length of convex lens in air is 16 cm (μglass = 1.5). Now the lens is submerged in liquid of refractive index 1.42. Find the ratio of focal length in medium to focal length in air has closest value

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

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

A lift of mass 920 kg has a capacity of 10 persons. If average mass of person is 68 kg. Friction f orce between lift and lift shaft is 6000 N. The minimum power of motor required to move the lift upward with constant velocity 3 m/s is [g = 10 m/s2]

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

Net force on motor will be
Fm = [920 + 68(10)]g + 6000
= 22000 N
So, required power for motor

= 22000 x 3
= 66000 watt

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

The hysteresis curve for a material is shown in the figure. Then for the material retentivity, coercivity and saturation magnetization respectively will be

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

x = retentivity
y = coercivity
z = saturation magnetization

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

​An inductor of inductance 10 mH and a resistance of 5Ω is connected to a battery of 20 V at t = 0. Find the ratio of current in circuit at t = ∞ to current at t = 40 sec.

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

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

Find the dimension of

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

Energy density in magnetic field =

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

A capacitor of 60 pF charged to 20 volt. Now battery is removed and then this capacitor is connected to another identical uncharged capacitor. Find heat loss in nJ.

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

V0 = 20 V
Heat loss = Ui – Uf

= 6 × 10–9 J = 6 nJ

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

When m gram of steam at 100°C is mixed with 200 gm of ice at 0°C. it results in water at 40°C. Find the value of m in gram .
(given : Latent heat of fusion (Lf) = 80 cal/gm,  Latent heat of vaporisation (Lv) = 540 cal/gm., specific heat of water (Cw)= 1 cal/gm/°C)

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

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

A solid cube of side 'a' is shown in the figure. Find maximum value of for which the block does
not topple before sliding.

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

For no toppling

For no toppling

0.2a + 0.4b ≤ 0.5a
0.4b ≤ 0.3a

b ≤ 0.75a (in limiting case)
But it is not possible as b can maximum be equal to 0.5a
∴

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

Magnitude of resultant of two vectors is equal to magnitude of .  Find the angle between
and resultant of 2and

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

So angle between (2 + ) and is 90°
Alternate solution

P2 + Q2 + 2PQcosθ = P2
Q + 2Pcosθ =θ
cosθ =

tan α =  = ∞
∵ [2Pcosθ + Q = 0]
α = 90°

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

A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of 'n' is

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

Let the emf of cell is e internal resistance is 'r' and potential gradient is x.
only cell connected :
ε = 560 x ................. (i)
After connecting the resistor
................... (ii)
from (1) and (2)

56 = 50 + 5r

n = 12

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

Which of the following reactions are possible?
(A)

(B)

(C)

(D)

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

Vinyl halides and aryl halides do not give Friedel craft's reaction.

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

A and B are in the given reaction ?

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 27

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

The correct statement about gluconic acid is

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

Gluconic acid
is obtained by partial oxidation of glucose by Tollen's reagent or Fehling solution or Br2,H2O.
Gluconic acid can not form hemiacetal or acetal

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

Stability order of following alkoxide ions is

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

When negative charge is delocalised with electron withdrawing group like (NO2) then stability increases.
(A) Negative charge is delocalised with NO2 group
(B) Negative charge is delocalised with carbon of alkene
(C) Negative charge is localised

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

A and B are –

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 30

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

For the complex [Ma2b2] if M is sp3 or dsp2 hybridised respectiv ely then total number of optical isomers are respectively :

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

Both will not show optical isomerism.

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

Bond order and magnetic nature of CN are respectively

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

CN is a 14 electron system.

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

Which of the following is incorrect?

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

Theory based.

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

NaOH + Cl2 → A + other products
Hot & conc.
Ca(OH) + Cl2 → B + other products
Cold & dil.
A & B are respectively

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

6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
2Ca(OH)2 + Cl2 → Ca(OCl)2 + CaCl2 + H2O

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

There are two beakers (I) having pure volatile solvent and (II) having volatile solvent and non-volatile solute. If both beakers are placed together in a closed container then:

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

There will be lowering in vapour pressure in second beaker.

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

Metal with low melting point containing impurities of high melting point can be purified by

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

Theory based

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

Which of the following statements are correct ?
(I) On decomposition of H2O2, O2 gas is released .
(II) 2-ethylanthraquinol is used in preparation of H2O2
(III) On heating KClO3, Pb(NO3)2, NaNO3, O2 gas is released.
(IV) In the preparation of sodium peroxoborate, H2O2 is treated with sodium metaborate.

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

Theory based

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

Amongs the following which is redox reaction?

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

N2 + O2 → 2NO
3O2 → 2O
2NaOH + H2SO4 → Na2SO4 + 2H2O
AgNO3 + NaCl → NaNO3 + AgCl

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

Select the correct options :

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

Option a is correct. The graph is for no of molecules vs speed. So don’t go for values in Vrms,Vavg and Vmp. It's a derivation according to Maxwell velocity distribution of gas. Most gas particles have the most probable velocity. Some of the particles have avg velocity while few particles have root mean square velocity.

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

Which one of the following amongs each pair will release maximum energy on gaining one electron (A = F, Cl), (B = S, Se), (C = Li, Na)

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

Theory based

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

Which of the following statements are incorrect ?
(A) Co+3 with strong field ligand forms high magnetic moment complex.
(B) For Co+3 if pairing energy(P) > Δo then the complex formed will have t42g, e2g configuration
(C) For [Co(en)3]3+ λabsorbed is less than λabsorbed for [CoF6]3-
(D) If Δo = 18000 cm-1 for Co+3 then with same ligands for it Δt = 16000 cm-1

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 41

Theory based

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

0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?

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

2 × mole of Urea = mole of NH3...(1)
mole of NH3 = mole of HCl ........(2)
∴ mole of HCl = 0.02 mole

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 43

Number of sp2 hybrid carbon atoms in aspartame is –

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

All stared carbon atoms of aspartame are sp2 hybrid. Aspartame is methyl ester of dipeptide formed from aspartic acid and phenylalanine.

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 44

3 gram of acetic acid is mixed in 250 mL of 0.1 M HCl. This mixture is now diluted to 500 mL. 20 mL of this solution is now taken is another container mL of 5M NaOH is added to this. Find the pH of this solution.  (log 3 = 0.4771, pKa = 4.74)

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

m mole of acidic acid in 20 mL = 2
m mole of HCl in 20 mL = 1
m mole of NaOH = 2.5

pH = PKa + log
= 4.74 + log 3
=  4.74 + 0.48 = 5.22

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 45

Flocculation value for As2S3 sol by HCl is 30 m mole L–1. Calcualte mass of H2SO4 required in gram for 250 mL sol.

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

For 1L sol 30 m mol of HCl is required
∴ For 1L sol 15 m mol H2SO4 is required
For 250 mL of sol
m mol H2SO4 = 0.3675 g

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

Determine total number of atoms in per unit formula of (A), (B) & (C)

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

(A) = CrO2Cl2
(B) = Na2CrO4
(C) = CrO5

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

Calculate ΔfH° (In kJ/mol) for C2H6(g), if ΔcH° [C(graphite)] = -393.5 kJ/mol,
ΔCHº [H2(g)] = -286 kJ/mol and
ΔCHº [C6H6(g)] = -1560 kJ/mol

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

JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 48

Let A = [aij], B=[bij] are two 3 × 3 matrices such that bij = λi + j - 2aij & |B| = 81. Find |A| if λ = 3.

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

=>81 = 33.3. 32 |A|
=> 3= 36 |A|
=> |A|= 1/9

JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 49

From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find the locus of mid point of PQ.

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

slope of  PQ = = -1
⇒ k – α = - h + 2α
⇒ α = ......... (1)
Also 2h = 2α + β
2k = α + β
⇒  2h = α + 2k
⇒ a = 2h - 2k .....(2)
from (1) & (2)
= 2(h - k)
so locus is 6x –6y = x + y
⇒ 5x = 7y

JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 50

Pair of tangents are drawn from origin to the circle x2 + y2 – 8x – 4y + 16 = 0 then square of length of chord of contact is

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

L = √S1 = √16 = 4
R = = 2
Length of Chord of contact =
square of length of chord of contact = 64/5

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

Contrapositive of if A ⊂ B and B ⊂ C then C ⊂ D

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

Let P = A ⊂ B, Q = B ⊂ C, R = C ⊂ A
Contrapositive of (P ∧ Q) → R is ~ R → ~ (P ∧ Q)
R ν (~ P ν ~ Q)

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

Let y(x) is solution of differential equation (y2 – x) dx/dy = 1 and y(0) = 1, then find the value of x where curve cuts the x-axis

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

c = –e
y = 0, x = 0 – 0 + 2 + (–e) (e–0)
x = 2 –e

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

Let θ1 and θ2 (whereθ1 < θ2)are two solutions of 2cot2θ θ ∈ [0, 2π) then is equal to

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

2 cos2θ - 5sinθ + 4sin2θ = 0, sinθ ≠0

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

Let 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 terms = S. If S = (102)m then m =

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

S = 7 + 17 + 27 + 37 + 47 +......20 terms

= 10[2040] = (102) (20)
⇒ m  = 20

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

If  6 , then number of ordered pairs (r, k) are –(where k ∈ I).

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

Hence number of order pair = 4

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

Let 4α = 5 then a =

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

⇒ 4(2 - ea - e-2a) = 5
Put e = t
⇒ 4t2 + 4t – 3 = 0
⇒(2t + 3) (2t - 1) = 0
⇒
⇒ a = ln2

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

Let f(x) is a five degree polynomial which has critical points x = ±1 and  = 4 then which one is incorrect.

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

f(x) = ax5  +bx4 + cx3
= 4
⇒ 2 + c = 4
⇒ c = 2
f'(x) = 5ax4  + 4bx3 + 6x2

f'(1) = 0  ⇒  5a + 4b + 6 = 0
f'(–1) = 0  ⇒ 5a – 4b + 6 = 0
b = 0,

f'(x) = –6x4 + 6x2
= 6x2 (–x2 +1)
= –6x2 (x+1) (x–1)

Minimal at x = –1
Maxima at x = 1
x = –1
x = 1

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

If are unit vectors such that  and and  then

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

⇒ λ = -3/2

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

Coefficient of xin  is

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

Coefficient of x7 is 11C7 = 11C4 = 330

JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 60

Let α and β are the roots of x2 - x - 1 = 0 such that Pk = αk + βk, k ≥ 1 then which one is incorrect?

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

= 5(1) + 6
P5 = 11 and P5 = α2 + β2 = α + 1 + β + 1
P2 = 3 and
P3 = α3 + β3 = 2α + 1 + 2β + 1
= 2(1) +2 = 4
P2 × P3 = 12 and P5 = 11
⇒ P5 ≠ P2 × P3

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

Let f(x) = x3 –4x2 + 8x + 11, if LMVT is applicable on f(x) in [0, 1], value of c is :

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

f(x) is a polynomial function
∴ it is continuous and differentiable in [0, 1]
Here f(0) = 11, f(1) = 1 – 4 + 8 + 11 = 16  f '(x) = 3x2 – 8x + 8
∴
= 3c2 – 8c + 8
⇒ 3c2 – 8c + 3 = 0

∴

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

The area bounded by  4x≤ y ≤ 8x + 12 is -

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

4x2 = y
y = 8x + 12
4x2 = 8x + 12
x2 – x – 3 = 0
x2 – 2x – 3 = 0]
x2 – 3x + x – 3 = 0
(x + 1)(x – 3) = 0
x = –1

= 128/3

JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 63

There are 5 machines. Probability of a machine being faulted is 1/4 Probability of atmost two machines s faulted, is   k then value of k is

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

Required probability = when no. machine has fault + when only one machine has fault + when only two machines have fault.

∴ k = 17/8

JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 64

3x + 4y = 12√2 is the tangent to the ellipse then the distance between focii of ellipse is-

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

condition of tangency c2 = a2m2 + b2

a2 = 16
1a = 4

∴ focus are (± 7 , 0)
∴ distance between foci = 2√7

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

If z = is purely real and θ = then arg(sinθ + I cosθ) is -

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

as z is purely real
⇒ 3 cosθ + 4sinθ = 0
⇒ tanθ =
arg(sinθ + icosθ) =

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

a1, a2, a3 …..a9 are in GP where a1 < 0,
a1 + a2 = 4, a3 + a4 = 16, if = 4λ, then λ is equal to

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

a1 + a2 = 4
⇒ a1 + a1r = 4 ................... (i)
⇒ r= 4
r  = ±2
r = 2, a1(1+2) = 4
⇒ a1= 4/3
r = –2,  a1(1–2) = 4
⇒ a1 = - 4

λ = –171

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

If  and  .Then dy/dx at x = 1/2 is

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

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

Let  X = {x : 1 ≤ x ≤ 50, x ∈ N}
A = {x: x is multiple of 2}
B = {x: x is multiple of 7}
Then find number of elements in the smallest subset of X which contain elements of both A and B

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

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 69

If Qis foot of perpendicular drawn from P(1, 0, 3) on a line L and if line L is passing through (α, 7, 1), then value of α is

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

Since PQ is perpendicular to L, therefore

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 70

If f(x) is defined in x ∈
f(x) =
Find k such that f(x) is continuous

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

= 3 + 2  = 5
∴ f(x) will be continuous if f(0) =

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 71

If system of equation
x + y + z = 6
x  + 2y + 3z = 10
3x + 2y + λz = μ has more than two solutions. Find (μ – λ2

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

x + y + z = 6  …….. (1)
x  + 2y + 3z = 10 …….. (2)
3x + 2y + λz = μ ..... (3)
from (1) and (2)
if z = 0
⇒ x + y = 6 and x + 2y = 10
⇒ y = 4, x = 2
(2, 4, 0)
if y = 0 ⇒ x + z = 6 and x + 3z = 10
⇒ z = 2 and x = 4
(4, 0, 2)
so, 3x + 2y + λz = μ must pass through (2, 4, 0) and (4, 0, 2)
so, 6 + 8 = μ
⇒ μ= 14 and 12 + 2λ = μ
12 + 2λ = 14
⇒ λ = 1
so μ – λ2 = 14 - 1 = 13

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Evening) - Question 72

If mean and variance of 2, 3, 16, 20, 13, 7, x, y are 10 and 25 respectively then find xy

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

mean =
⇒  x + y = 19.............(i)
variance =  25

⇒  x2 + y2 = 113………(ii)
(x+y)2 = (19)2
⇒ x2 + y2 + 2xy = 361 ⇒ xy = 124   (exact data is not retrieved so ans. can vary)

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