JEE Main 2020 Question Paper With Solution 08 Jan 2020 Shift 1 Memory Based

1- Page 1 -
JEE MAIN 2020
Memory Based
Answer Key with Solution
(08 Jan, 2020 Shift 1)
Date : 08/01/2020
Time allowed: 3 hours Maximum marks: 300
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 75 questions. The maximum marks are 300.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry
and Mathematics having 25 questions in each part of equal weightage. Each part has
two sections.
(i) Section-I : This section contains 20 multiple choice questions which have only
one correct answer. Each question carries 4 marks for correct answer and
–1 mark for wrong answer.
(ii) Section-II : This section contains 5 questions. The answer to each of the
questions is a numerical value. Each question carries 4 marks for correct
answer and there is no negative marking for wrong answer.
PHYSICS
Q1. A block of mass m is connected at one end of spring fixed at other end having natural length
 and spring constant K. The block is rotated with constant angular speed (ω) in gravity free
space. The elongation in spring is –
(1) Km

m
(2) Km

m
(3) Km

m
(4) Km

m
Answer : (1)
Solution

JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 1)
2- Page 2 -
mω2( + x) = kx
x

  m
k
Q2. 3 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 :
(1) 
d
q
(2) 
d
q
(3) 
d
q
(4) 
d
q
Answer : (3)
Solution
Enet  d
kq
×cos°  
d
q

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Q3. Choose the correct P-V graph of ideal gas for given V-T graph.
(1)
(2)
(3)
(4)
Answer : (1)
Solution
Q4. Find the co-ordinates of centre of mass of the lamina, shown in figure.
(1) 0.75, 1.75
(2) 0.75, 1.5
(3) 0.5, 1.75
(4) 0.5, 1.5

4- Page 4 -
Answer : (1)
Solution
rcm  
×
i
j×i 
j

rcm  
 i  
 j
Q5. Which graph correctly represents variation between relaxation time (τ) of gas molecules with
absolute temperature (T) of gas.
(1)
(2)
(3)
(4)
Answer : (1)
Solution
∝T


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Q6. If two capacitors C1 & C2 are connected in parallel then equivalent capacitance is 10μF. If
both capacitance are connected across 1V battery then energy stored by C2 is 4 times of C1.
Then the equivalent capacitance if they are connected in series is –
(1) 1.6μF
(2) 16μF
(3) 4μF
(4) 

μF
Answer : (1)
Solution
Given C1 + C2 = 10 μF ...(i)


CV
 

CV
⇒ 4C1 = C2 ...(ii)
from equation (i) & (ii)
C1 = 2 μF
C2 = 8 μF
If they are in series
Ceq  C C
C
C
 F
Q7. 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, strikes at the end at an angle of 45º and sticks to it. The
angular velocity of system after collision is –
(1) L
 V
(2) L
 V
(3) L
V
(4) L
V
Answer : (1)
Solution
Loi = Lof

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
mV
× 






mL
 
mL 

×
  L
V
 L
V
Q8. Two photons of energy 4eV and 4.5 eV incident on two metals A and B respectively.
Maximum kinetic energy for ejected electron is TA for A and TB = TA – 1.5 eV for metal B.
Relation between de-Broglie wavelength of ejected electron of A and B are λB = 2λA. The
work function of metal B is –
(1) 3 eV
(2) 1.5 eV
(3) 4.5 eV
(4) 4 eV
Answer : (4)
Solution
Relation between De-Broglie wavelength and K.E. is
  
KEme
h
⇒ ∝KE

B
A
 
KEA
KEB
⇒ 

 

TA
TA 
⇒ TA = 2 eV
∴ KEB = 2 – 1.5 = 0.5 eV
ϕB = 4.5 – 0.5 = 4 eV
Q9. 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 –
(1) 80 Ω
(2) 100 Ω
(3) 120 Ω
(4) 60 Ω
Answer : (2)
Solution
Potential gradient = 

 
VP

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VP = 6 V
and RP = I
VP
 
× 

 Ω
Q10. A telescope has magnification 5 and length of tube 60 cm then the focal length of eye piece
is –
(1) 10 cm
(2) 20 cm
(3) 30 cm
(4) 40 cm
Answer : (1)
Solution
m = fe
fo
5 = fe
fo
fo = 5fe
fo + fe = 60
6fe = 60
fe = 10
Q11. Two spherical bodies of mass m1 & m2 are having radius 1 m & 2 m respectively. The
gravitational field of the two bodies with their radial distance is shown below. The value of
m
m
is –
(1) 

(2) 

(3) 

(4) 

Answer : (1)
Solution
  

Gm
  

Gm
∴ 

 

m
m

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m
m
 

Q12. When proton of KE = 1.0 MeV moving in South to North direction enters the magnetic field
(from West to East direction), it accelerates with a = 1012
m/s2
. The magnitude of magnetic
field is –
(1) 0.71 mT
(2) 7.1 mT
(3) 71 mT
(4) 710 mT
Answer : (1)
Solution
∵ K.E. = 1.6×10–13
= 

×1.6×10–27
V2
V =  ×107
∴ Bqv = ma
B  
× 
××
× 
×
= 0.71×10–3
T
so 0.71 mT
Q13. If electric field around a surface is given by E A
Qin
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
(1) Surface is equipotential
(2) Magnitude of electric field is constant
(3) Magnitude of electric field is constant & the surface is equipotential
(4) For all Gaussian surfaces.
Answer : (3)
Solution
Magnitude of electric field is constant & the surface is equipotential

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Q14. A 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)
(1) hI–1
G1
C5
(2) h–1
I1
G–1
C6
(3) h0
I1
G1
C6
(4) h0
I–1
G–1
C5
Answer : (4)
Solution
V = K(h)a
(I)b
(G)c
(C)d
(V is voltage)
we know [h] = ML2
T–1
[I] = A
[G] = M–1
L3
T–2
[C] = L T–1
[V] = M L2
T–3
A–1
M L2
T–3
A–1
= (M L2
T–1
)a
(A)b
(M–1
L3
T–2
)c
(LT–1
)d
ML2
T–3
A–1
= Ma–c
L2a+3c+d
T–a–2c–d
Ab
a – c = 1 ...(i)
2a + 3c + d = 2 ...(ii)
–a –2c –d = –3 ...(iii)
b = –1 ...(iv)
on solving
c = –1
a = 0
d = 5, b = –1
V = K (h)° (I)–1
(G)–1
(C)5
Q15. 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)
(1) 1.01
(2) 1.03
(3) 2.01
(4) 1.04
Answer : (1)
Solution

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mg = A(80) ρ0°C g
mg = A(79) ρ4°C g
°C
°C
 

= 1.01
Q16. Number of the α-particle deflected in Rutherford's α-scattering experiment varies with the angle
of deflection. Then the graph between the two is best represented by.
(1)
(2)
(3)
(4)
Answer : (2)
Solution
N∝
sin




Q17. If relative permittivity and relative permeability of a medium are 3 and 

respectively. The
critical angle for this medium is.
(1) 45°
(2) 60°
(3) 30°
(4) 15°
Answer : (3)
Solution
V = 

n = rr = 2
sin c = 

c = 30°

11- Page 11 -
Q18. The given loop is kept in a uniform magnetic field perpendicular to plane of loop. The field
changes from 1000G to 500G in 5 seconds. The average induced emf in loop is –
(1) 15 μV
(2) 28 μV
(3) 30 μV
(4) 48 μV
Answer : (1)
Solution
   dt
d
  dt
AdB

= (16 × 4 – 4 × 2) 

×10–4
×10–4
= 56× 

×10–8
= 56 × 10–6
V
Q19. Choose the correct Boolean expression for the given circuit diagram :
(1) A . B
(2) A  B
(3) A + B
(4) A  B
Answer : (4)
Solution
First part of figure shown OR gate and
Second part of figure shown NOT gate
so Yp = OR + NOT = NOR gate
Y = AB  AB

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Q20. A Solid sphere of density    R
r
, 0 < r ≤ R just floats in a liquid then density of
liquid is – (r is distance from centre of sphere)
(1) 


(2) 


(3) 


(4)  S
Answer : (1)
Solution
   R
r
 0 < r ≤ R
mg = B
 r
dr  L

R
  R
r
r
dr  L

R

R
r
 R
r
dr  
r
 
R
r

R
 L

R


  L
Q21. A Two masses each with mass 0.10 kg are moving with velocities 3 m/s along x axis and 5
m/s along y-axis respectively. After an elastic collision one of the mass moves with a velocity
i  j . The energy of other mass after collision is 
x
then x is.
Answer : 1
Solution
For elastic collision KEi = KEf


m × 25 + 

× m × 9 = 

m × 32 + 

mv2
34 = 32 + v2
KE = 

× 0.1 × 2 = 0.1 J = 

x = 1
Q22. A plano convex lens of radius of curvature 30 cm and refractive index 1.5 is kept in air.
Find its focal length (in cm).
Answer : 60 cm
Solution
f

 
R

 R


R1 = ∝
R2 = –30 cm

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f

 ∞

 


f

 

f = 60 cm
Q23. 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.
Answer : 580 cm
Solution
XA = –3t2
+ 8t + c
VA = (–6t + 8)i
YB = 10 – 8t3
V  tj
V  VB  VA  ji
V  
 
V = 580
Q24. An open organ pipe of length 1m contains a 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.
Answer : 105.75 Hz
Solution
V = 


B
Vair
Vpipe
 



B



B
 

Vpipe  
Vair
fn  
nVpipe
f f  
Vpipe
 


  Hz If  
=   Hz If  

14- Page 14 -
Q25. Four resistors of 15Ω, 12Ω, 4Ω and 10Ω given in cyclic order to form a wheat stone bridge.
What resistance (in Ω) should be connected in parallel across the 10Ω resistor to balance the
wheat stone bridge.
Answer : 10
Solution
  R
R
× 12 = 15 × 4
on solving

15- Page 15 -
CHEMISTRY
Q26. Number of S–O bond in S2O8
2–
and number of S–S bond in Rhombic sulphur are respectively:
(1) 8,8
(2) 6,8
(3) 2,4
(4) 4,2
Answer : (1)
Solution
Q27. Following vanderwaal forces are present in ethyl acetate liquid
(1) H-bond, london forces.
(2) dipole-dipole interation, H-bond
(3) dipole –dipole interation, London forces
(4) H-bond, dipole-dipole interation, London forces
Answer : (3)
Solution
Ethyl acetate is polar molecule so dipole-dipole interaction will be present there.
Q28. Given, for H-atom v RH


n


 n

 

 Select the correct options regarding this formula for
Balmer series.
(i) n1 = 2
(ii) Ionization energy of H atom can be calculated from above formula.
(iii) maximum is for n2 = 3.
(iv) If λ decreases then spectrum lines will converse.
(1) A,B
(2) C,D
(3) A & C
(4) A,B,C & D
Answer : (3)
Solution
Theory based.

16- Page 16 -
Q29. Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol–1
respectively are:
(1) 497, 737, 577, 786
(2) 497, 577, 737, 786
(3) 786, 739, 577, 497
(4) 739, 577, 786, 487
Answer : (3)
Solution
Correct order of ionisation energy will be : Na < Al < Mg < Si
Q30. Select the correct stoichiometry and its ksp value according to given graphs.
(1) XY, Ksp = 2×10–6
(2) XY2, Ksp = 4×10–9
(3) X2Y, Ksp = 9×10–9
(4) XY2, Ksp = 1×10–9
Answer : (3)
Solution
XYsX
aq  Y
aq
 ×  
 
Ksp = [X+
] [Y–
]
or Ksp = 2 × 10–3
× 10–3
or Ksp = 2 × 10–6
Q31. According to Hardy Schultz rule, correct order of flocculation value for Fe(OH)3 sol is
(1) K2CrO4 > K3[Fe(CN)6] > KNO3 > KBr = AlCl3
(2) K3[Fe(CN)6] > K2CrO4 > KNO3 = KBr = AlCl3
(3) K3[Fe(CN)6] < K2CrO4 < KNO3 = KBr = AlCl3
(4) KNO3 > KBr = K2CrO4 > AlCl3 = K3[Fe(CN)6]
Answer : (3)
Solution
According to hardy – schultz rule,
Coagulation value of flocculation value ∝Coagulationpower


17- Page 17 -
Q32. Which of the following complex exhibit facial meridional geometrical isomerism.
(1) K[Pt(NH3)Cl3]–
(2) [PtCl2(NH3)2]
(3) [Ni(CO)4]
(4) [Co(NO2)3 (NH3)3]
Answer : (4)
Solution
[Ma3b3] type complex shows facial and meridional isomerism
Q33.
(i) Intermolecular force of attraction of X > Y.
(ii) Intermolecular force of attraction of X < Y.
(iii) Intermolecular force of attraction of Z < X.
Select the correct option(s).
(1) A and C
(2) A and B
(3) B only
(4) B and C
Answer : (3)
Solution
At a particular temperature as intermolecular force of attraction increases vapour pressure decreases
Q34. 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.
(1) –6 × 2.303 RT
(2) +6×2.303RT
(3) + 6RT
(4) –6RT
Answer : (1)
Solution
K = Ae–E/RT
...(i)
equation 
equation 
⇒ 
 eE  EcRT
or
In  E  ECRT
RT
E  EC
  × 

18- Page 18 -
or E – Ec = 2.303 × 6 RT
or ΔEa = Ec – E = –2.303 × 6RT
Q35. Gypsum on heating at 393K produces
(1) dead burnt plaster
(2) Anhydrous CaSO4
(3) CaSO4. 

HO
(4) CaSO4.5H2O
Answer : (3)
Solution
Theory based.
Q36. Among the following least 3rd ionization energy is for
(1) Mn
(2) Co
(3) Fe
(4) Ni
Answer : (3)
Solution
Fe  Ard
s
Q37. Accurate measurement of concentration of NaOH can be performed by following titration:
(1) NaOH in burette and oxalic acid in conical flask
(2) NaOH in burette and concentrated H2SO4 in conical flask
(3) NaOH in volumetric flask and concentrated H2SO4 in conical flask
(4) Oxalic acid in burette and NaOH in conical flask
Answer : (4)
Solution
Oxalic acid is a primary standard solution while H2SO4 is a secondary standard solution.
Q38. Arrange the following compounds in order of dehydrohalogenation (E1) reaction.
(a)
(b)
(c)
(d)
(1) C > B > D > A
(2) C > D > B > A
(3) B > C > D > A
(4) A > B > C > D
Answer : (4)
Solution

19- Page 19 -
E1 reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster
the E1 reaction.
Q39.
∆
peroxide
 A
[A]+  B Product A and B are respectively:
(1)
(2)
(3)
(4)
Answer : (3)
Solution
[A] is more stable radical and undergoes Markovnikov addition to form [B].

20- Page 20 -
Q40. Major product in the following reaction is
dil H
SO

(1)
(2)
(3)
(4)
Answer : (3)
Solution

21- Page 21 -
Q41. Arrange the order of C—OH bond length of the following compounds.
Methanol Phenol p-Ethoxyphenol
(A) (B) (C)
(1) A > B > C
(2) A > C > B
(3) C > B > A
(4) B > C > A
Answer : (2)
Solution
There is not any resonance CH3–OH. Resonance is poor in p-Ethoxyphenol than phenol.
Q42. Which of the following are "green house gases" ?
(a) CO2
(b) O2
(c) O3
(d) CFC
(e) H2O
(1) a, b and d
(2) a, b, c and d
(3) a, c and d
(4) a, c, d and e
Answer : (4)
Solution
CO2, O3, H2O vapours and CFC's are green house gases.
Q43. Two liquids isohexane and 3-methylpentane has boiling point 60°C and 63°C. They can be
separated by
(1) Simple distillation and isohexane comes out first.
(2) Fractional distillation and isohexane comes out first.
(3) Simple distillation and 3-Methylpantane comes out first.
(4) Fractional distillation and 3-Methylpantane comes out first.
Answer : (2)
Solution
Liquid having lower boiling point comes out first in fractional distillation. Simple distillation can't
be used as boiling point difference is very small.
Q44. Which of the given statement is incorrect about glucose?
(1) Glucose exists in two crystalline form α and β.
(2) Glucose gives schiff’s test.
(3) Penta acetate of glucose doest not from oxime.
(4) Glucose forms oxime with hydroxyl amine
Answer : (2)
Solution
Open chain form of glucose is very very small, hence does not gives Schiff's test.

22- Page 22 -
Q45. Reagent used for the given conversion is:
(1) H2, Pd
(2) B2H6
(3) NaBH4
(4) LiAlH4
Answer : (2)
Solution
B2H6 is very selective and usually used to reduce acid to alcohol.
Q46. 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.
Answer : 26.92
Solution
ML Cl
g

AgNO
vml M
AgCl
or, V =  × 
 ×  × 
  ml.
Q47. Given : 2H2O → O2 + 4H+
+ 4e–
E
= –1.23V Calculate electrode potential at pH = 5.
Answer : –00.93
Solution
E     

log H 

     × pH      × 
        V   V
Q48. Calculated the mass of FeSO4.7H2O, which must be added in 100 kg of wheat to get 10 PPM
of Fe.
Answer : 04.96
Solution
  × 
Mass of Fe LEFT ing
× 
or mass Fe = 1g
FeSO4. 7H2O (M = 278)
56 g in 1 mole
g  

mole 

× g  g

23- Page 23 -
Q49. A gas undergoes expansion according to the following graph. Calculate work done by the gas.
Answer : 48.00
Solution
W  

   ×   J
Q50. Number of chiral centres in Pencillin is
Answer : 03.00
Solution
Star marked atoms are chiral centers.

24- Page 24 -
MATHEMATICS
Q51. Let  
sin
×   sin
×


cos  
 fx  sin



  then find value of λ f


(1) 4
(2) –2
(3) 8
(4) –4
Answer : (2)
Solution
sin x = t
cos x dx = dx
I =  
t
t



dt
  
t
 t




dt
Put 1 + t

= r3
⇒ t
dt
 

r
dr
 

 r
r
dr
  

r  c   

sin
x
sin
x  



 c   
sin
x

sin



 
f(x) =  

cosec2
x and λ = 3
λf

 = –2
Q52. If y(x) is a solution of differential equation   


   
  , such that


 

, then
(1) 

  

(2) 

 

(3) 

 

(4) 

 

Answer : (3)
Solution

y
dy
 
x
dx
  ⇒ sin 
y  sin 
x  c
At x  

 y  

⇒ c  

⇒ sin 
y  cos 
x

25- Page 25 -
Hence y


 sin
cos 


 

Q53. lim
 →


 

 




is equal to
(1) e–2
(2) e2
(3) e2/7
(4) e3/7
Answer : (1)
Solution
Let L  lim
x→ 
x

x


x

 e
lim
x→
x


x
 
x
 
   e
lim
x→
x


x
 
 x
  e

 
 e 
Q54. In a bag there are 5 red balls, 3 white balls and 4 black balls. Four balls are drawn from the
bag. Find the number of ways of in which at most 3 red balls are selected
(1) 450
(2) 360
(3) 490
(4) 510
Answer : (3)
Solution
0 Red, 1Red, 2Red 3 Red
Number of ways = 7
C4 + 5
C1.7
C3 + 5
C2.7
C2 + 5
C3.7
C1 = 35 + 175 + 210 + 70 = 490
Q55. Let f(x) = {(sin (tan–1
x) + sin (cot–1
x)}2–1 where |x| > 1 and dx
dy
 

dx
d
sin 

(1) 

(2) 

(3) 

(4) 

Answer : (2)
Solution
2y = sin–1
f(x) + C = sin–1
(sin(2tan–1
x)) + C ⇒ 

 sin 
sin

C


 

C ∴ C  
for x =  , 2y = sin–1sin

  ⇒ y  

y  



26- Page 26 -
Q56. If 21–x
+ 21+x
, f(x), 3x
+ 3–x
are in A.P. then minimum value of f(x) is
(1) 1
(2) 2
(3) 3
(4) 4
Answer : (3)
Solution
f(x) = 
 x
  x
 x
  x

Using AM ≥ GM
f(x) ≥ 3
Q57. Which of the following is tautology
(1) (p ∧ (p → q)) → q
(2) q → p ∧ (p → q)
(3) p v (p ∧ q)
(4) (p ∧ (p ∨ q))
Answer : (1)
Solution
p q p→q p∧(p→q) (p∧(p→q))→q q→p∧(p→q) p∧q p∨(p∧q) p∨q p∧(p∨q)
T T T T T T T T T T
F F F F T T F T T T
F T T F T F F F T F
F F T F T T F F F F
Q58. A is a 3 × 3 matrix whose elements are from the set {–1, 0, 1}. Find the number of
matrices A such that tr (AAT
) = 3. Where tr (A) is sum of diagonal elements of matrix A.
(1) 572
(2) 612
(3) 672
(4) 682
Answer : (3)
Solution
Let A = [aii]3×3
tr(AAT
) = 3
a

 a

 a

 a

  a

 
possible cases
 →
 →
 →
 →






C ×  ×  
Q59. Mean and standard deviations of 10 observations are 20 and 2 respectively. If p(p ≠ 0) is
multiplied to each observation and then q (q ≠ 0) is subtracted then new mean and standard
deviation becomes half of original value. Then find q

27- Page 27 -
(1) –10
(2) –20
(3) –5
(4) 10
Answer : (2)
Solution
If each observation is multiplied with p & then q is subtracted
New mean x  pxq
⇒ 10 = p(20) – q ...(i)
and new standard deviations
  p ⇒   p ⇒ p  

⇒ p  ± 

If p = 

then q = 0 (from equation (1))
If p =  

q = –20
Q60. If maximum value of 19
Cp is a, 202020Cq is b, 21
Cr is c, then relation between a, b, c is
(1) 

 

 

(2) 

 

 

(3) 

 

 

(4) 

 

 

Answer : (1)
Solution
We know n
Cr is max at middle term
a = 19
Cp = 19
C10 = 19
C9
b = 20
Cq = 20
C10
c = 21
C6 = 21
C10 = 21
C11

C
a
 



C
b
 



 
C
c

a
 
b
 
c

a
 
b
 
c

28- Page 28 -
Q61. Let P(A) 

 P(B) = 

where A and B are independent events then
(1) PB′
A
 

(2) PB′
A
 

(3) PB
A
 

(4) PB
A
 

Answer : (2)
Solution
A & B are independent events so
PB′
A
 

Q62. Let f(x) = 
x
  x
x
  x
then inverse of (x) is
(1) 

log   
  

(2) 

log  
  

(3) 

log  
  

(4) 

log  
  

Answer : (1)
Solution
y  
x
 x
x
 x
y
y
 
 x
x
x
 y
y
x  logy
y

x  

logy
y

f 
x  

logy
y


29- Page 29 -
Q63. Roots of the equation x2
+ bx + 45 = 0, ∈ R lie on the curve |z + 1| =  , where z is
a complex number then
(1) b2
+ b = 12
(2) b2
– b = 30
(3) b2
– b = 36
(4) b2
+ b = 30
Answer : (2)
Solution
Let z = α ± iβ be roots of the equation
So 2α = –b and α2
+ β2
= 45, (α + 1)2
+ β2
= 40
So (α + 1)2
– α2
= –5
⇒ 2α + 1 = –5 ⇒ 2α = –6
So b = 6
hence b2
– b = 30
Q64. For f(x) = n

 
. Rolle’s theorem is applicable on [3,4] the value of f`` (c) is equal to
(1) 

(2) 

(3) 

(4) 

Answer : (1)
Solution
f(3) = f(4) ⇒ α = 12
f ′(x) = xx

x

∴ f ′(c) = 0
∴ c = 
∴ f ′′(c) = 

Q65. Let f(x) = x cos–1
(sin(–|x|)), x ∈ 

 

 then
(1) f`(0) =  

(2) f`(x) is not defined at x = 0
(3) f’(x) is increasing in 
 
 and f’(x) is decreasing in  


(4) f’(x) is decreasing in 

′ and f’(x) is increasing in  



30- Page 30 -
Answer : (4)
Solution
f ′(x) = x(π – cos–1
(sin|x|))
 x

sin 
sinx
 x

x
fx 





x

x x ≥ 
x

x x  
f′x 







x x ≥ 


x x  
f ′(x) is increasing in  

 and decreasing in 


Q66. Let P be a point on x2
= 4y. The segment joining A (0,–1) and P is divided by point Q in
the ration 1:2, then locus of point Q is
(1) 9x2
= 3y + 2
(2) 9x2
= 12y + 8
(3) 9y2
= 12x + 8
(4) 9y2
= 3x + 2
Answer : (2)
Solution
Let point P be (2t, t2
) and Q be (h, k).
h  
t
 k  
t
Hence locus is 3k + 2 = 
h


⇒ 9x2
= 12y + 8
Q67. Ellipse 2x2
+ y2
= 1 and y = mx meet a point A in first quadrant. Normal to the ellipse at P
meets x-axis at  


 and y–axis at (0,β) is
(1) 

(2) 

(3) 

(4) 

Answer : (3)
Solution

31- Page 31 -
Let P be (x1, y1)
Equation of normal at P is  


⇒ x

  

⇒  

⇒ x  

So y  

(as P lies in 1st
quadrant)
So β = 
y
 

Q68. If y2
= ax and x2
= ay intersect at A & B. Area bounded by both curves is bisected by line
x = b(given a > b > 0). Area of triangle formed by line AB, x = b and x-axis is 

. then
(1) a6
– 12a3
– 4 = 0
(2) a6
+ 12a3
– 4 = 0
(3) a6
– 12a3
+ 4 = 0
(4) a6
+ 12a3
+ 4 = 0
Answer : (3)
Solution

b
ax


 a
x
dx  
a
⇒ 

a b


 a
b
 
a
...(i)
also area of ΔOQR = 



b
 

⇒ b = 1
Put in (i)
⇒ aa  a
⇒ a
 a
   a
⇒ a
 a
   
Q69. Let ABC is triangle whose vertices are A(1, – 1), B(0, 2), C(x`, y`) and area of ΔABC is 5
and C (x`, y) lie on 3x + y – 4λ = 0, then
(1) λ = 3
(2) λ = –3
(3) λ = 4
(4) λ = 2
Answer : (1)
Solution
D = 


  
  
x′ y′ 
–2(1 – x′) + (y′ + x′) = ± 10
–2 + 2x′ + y′ + x′ = ± 10
3x′ + y′ = 12 or 3x′ + y′ = – 8
λ = 3, –2

32- Page 32 -
Q70. The system of equation 3x + 4y + 5z = μ
x + 2y + 3z = 1
4x + 4y + 4z = δ
is inconsistent, then (δ,μ) can be
(1) (4,6)
(2) (3,4)
(3) (4,3)
(4) (1,0)
Answer : (3)
Solution
Note D =

  
  
  
R →R  R  R


  
  
  
 
Now let P3 ≡ 4x + 4y + 4z – δ = 0. If the system has solutions it will have infinite solution,
so P3 ≡ αP1 + βP2
Hence 3α + β = 4 & 4α + 2β = 4 ⇒ α = 2 & β = –2
So for infinite solution 2μ –2 = δ ⇒ for 2μ ≠ δ + 2
system inconsistent
Q71. Shortest distance between the lines 
x  
 
y  
 
z  
 
y  
 
z  
is
(1) 3
(2) 2
(3) 
(4) 
Answer : (1)
Solution
AB  i  j  k
p  i  j  k
q  i  j  k
p × q 

i j k
  
  
 i  j  k
S.D. = p×q
ABp×q
     
    
 

33- Page 33 -
Q72. If volume of parallelopiped whose there coterminous edges are            
&      is 1 cubic unit then cosine of angle between  and  is
(1) 

(2) 

(3) 

(4) 

Answer : (2)
Solution
± 

  
  
  
⇒      ± ⇒   or   
For λ = 4
cosθ =  

 


Q73. Find the sum   

     
Answer : 1540
Solution
k  


kk
 

k  

k
 k
 
 




 
 


 
 



×
 
× 


 

  
= 1540
Q74. If normal at P on the curve y2
– 3x2
+ y + 10 = 0 passes through the point (0, 3/2) then
slope of tangent at P is n. The value of |n| is equal to
Answer : 4
Solution
P ≡ (x1, y1)
2yy′ – 6x + y′ = 0 ⇒ y′ = y
x


34- Page 34 -



x


y



 
x
y

9 – 6y1 = 1 + 2y1 ⇒ y1 = 1
∴ x1 = ± 2
∴ Slope of tangent = 
±

= ± 4
∴ |n| = 4
Q75. If 2x2
+ (a – 10) x + 

  a∈Z+
has real roots, then minium value of ‘a’ is equal to
Answer : 8
Solution
D ≥ 0
(a – 10)2
– 4(2) 

a ≥ 0
(a – 10)2
– 4(33 – 4a) ≥ 0
a2
– 4a – 32 ≥ 0 ⇒ a∈ (–∞, –4] ∪ [8, ∞)

JEE Main 2020 Question Paper With Solution 08 Jan 2020 Shift 1 Memory Based

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