![JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2)
2- Page 2 -
Answer : (4)
Solution
B B
B
B
B
R
i
[sin90° – sin45°] + R
i
+ R
i
(sin45º + sin90º)
R
i
R
i
R
i
R
i
R
i
R
i
Q2. Position of particle as a function of time given as r costi + sinωt. Choose correct
statement about and where and are velocity and acceleration of particle at time t.
(1) is perpendicular to and is towards origin
(2) and are perpendicular to
(3) is parallel to and parallel to .
(4) is perpendicular to and is away from origin.
Answer : (1)
Solution
r costi sintj
v dt
dr
sinti costj
a dt
dv
costi sintj
a r ∴ a is antiparallel to r
vr (–sinωt cosωt + cosωt sinωt) = 0
so v ⊥r
Q3. A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If
the work done on the system is 10 J, the amount of energy absorbed from the reservoir at
lower temperature is.
(1) 99J
(2) 90J
(3) 1J
(4) 100J](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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- 1. 1- Page 1 - JEE MAIN 2020 Memory Based Answer Key with Solution (08 Jan, 2020 Shift 2) 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. Find magnetic field at O (1) R i (2) R i (3) R i (4) R i
- 2. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 2- Page 2 - Answer : (4) Solution B B B B B R i [sin90° – sin45°] + R i + R i (sin45º + sin90º) R i R i R i R i R i R i Q2. Position of particle as a function of time given as r costi + sinωt. Choose correct statement about and where and are velocity and acceleration of particle at time t. (1) is perpendicular to and is towards origin (2) and are perpendicular to (3) is parallel to and parallel to . (4) is perpendicular to and is away from origin. Answer : (1) Solution r costi sintj v dt dr sinti costj a dt dv costi sintj a r ∴ a is antiparallel to r vr (–sinωt cosωt + cosωt sinωt) = 0 so v ⊥r Q3. A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is. (1) 99J (2) 90J (3) 1J (4) 100J
- 3. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 3- Page 3 - Answer : (2) Solution For Carnot engine using as refrigerator W QT T It is given ⇒ T T ⇒ T T So, Q2 = 90 J (as W = 10 J) Q4. Two uniformly charged solid spheres are such that E1 is electric field at surface of 1st sphere due to itself. E2 is electric field at surface of 2nd sphere due to itself. r1, r2 are radius of 1st and 2nd sphere respectively. If E E r r then ratio of potential at the surface of spheres 1st and 2nd due to their self charges is : (1) r r (2) r r (3) r r (4) r r Answer : (2) Solution E E r r V V Er Er r r × r r r r Q5. Output at terminal Y of given logic circuit. (1) 1 (2) 0 (3) Not determine (4) Oscillating between 0 and 1 Answer : (2) Solution Y = ABA
- 4. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 4- Page 4 - AB A = AB + A = 0 + 0 = 0 Q6. Velocity of a wave in a wire is v when tension in it is 2.06 × 104 N. Find value of tension in wire when velocity of wave become . (1) 5.15 × 103 N (2) 8.24 × 104 N (3) 6 × 104 (4) 5.15 × 104 N Answer : (1) Solution v ∝ T ⇒ v v T T ⇒ v v T × ⇒ T × N = 0.515×104 N Q7. n mole He and 2n mole of O2 is mixed in a container. Then CV CP mix will be. (1) (2) (3) (4) Answer : (1) Solution mix ncv n cv ncp n cp n Rn R n R n R
- 5. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 5- Page 5 - Q8. A uniform solid sphere of radius R has a cavity of radius 1 m cut from it if centre mass of the system lies at the periphery of the cavity then (1) (R2 + R + 1) (2–R) = 1 (2) (R2 – R – 1) (2–R) = 1 (3) (R2 – R + 1) (2–R) = 1 (4) (R2 + R – 1) (2–R) = 1 Answer : (1) Solution M R M Xcom M M MX MX ⇒ R R R R ⇒ R R R (R ≠ 1) RR R R R (R2 + R + 1) (2 – R) = 1 Alternative: Mremaining (2 – R) = Mcavity (1 – R) ⇒ (R3 – 13 ) (2 – R) = 13 [R – 1] ⇒ (R2 + R + 1) (2 – R) = 1 Q9. A solid sphere of mass m= 500gm is rolling without slipping on a horizontal surface. Find kinetic energy of a sphere if velocity of centre of mass is 5 cm/sec. (1) × J (2) × J (3) 35 × 10–4 J (4) 35 × 10–3 J
- 6. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 6- Page 6 - Answer : (2) Solution K.E. of the sphere = Translational K.E + Rotational K.E. mv R K K = Radius of gyration × × × J Q10. Two liquid columns of same height 5m and densities ρ and 2ρ are filled in a container of uniform cross sectional area. Then ratio of force exerted by the liquid on upper half of the wall to lower half of the wall is. (1) (2) (3) (4) Answer : (1) Solution F F
- 7. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 7- Page 7 - Q11. Two square plates of side 'a' are arranged as shown in the figure. The minimum separation between plates is 'd' and one of the plate is inclined at small angle α with plane parallel to another plate. The capacitance of capacitor is (given α is very small) (1) (2) (3) (4) Answer : (1) Solution dc = d x adx ⇒ c = a ln = a ln d a ≈ d a d a Q12. In YDSE path difference at a point on screen is . Find ratio of intensity at this point with maximum intensity. (1) 0.853 (2) 0.533 (3) 0.234 (4) 0.123 Answer : (1) Solution I = I0 cos2 ∆
- 8. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 8- Page 8 - I I cos ×∆x cos I I Q13. In the given circuit switch is closed at t = 0. The charge flown in time t = TC (where TC is time constant). (1) R e L (2) R L (3) R L e (4) R L Answer : (1) Solution q TC idt R t TC e tTC TC R TC TC e TC R × e × R L R e L Q14. A particle is dropped from height h = 100 m, from surface of a planet. If in last sec of its journey it covers 19 m. Then value of acceleration due to gravity that planet is : (1) 8m/s2 (2) m/s2 (3) m/s2 (4) 2 m/s2 Answer : (1) Solution
- 9. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 9- Page 9 - Area of shaded trapezium g t t × ...(i) gt ...(ii) ⇒ t g g t ⇒ g g g = 8 m/s2 Q15. A charge particle of mass m and charge q is released from rest in uniform electric field. Its graph between velocity (v) and distance travelled (x) will be : (1) (2) (3) (4) Answer : (1) Solution V m qE x
- 10. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 10- Page 10 - Q16. An object is moving away from concave mirror of focal length f starting from focus. The distance of an object from pole of mirror is x. The correct graph of magnitude of magnification(m) verses distance x is: (1) (2) (3) (4) Answer : (4) Solution At focus, magnification is ∞ Q17. In full scale deflection current in galvanometer of 100 Ω resistance is 1 mA. Resistance required in series to convert it into voltmeter of range 10 V. (1) 00.99 KΩ (2) 9.9 KΩ (3) 9.8 KΩ (4) 10 KΩ Answer : (2) Solution Vg = ig Rg = 0.1 V V = 10 V R = Rg Vg V = 100 × 99 = 9.9 KΩ
- 11. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 11- Page 11 - Q18. There are two identical particles A and B. One is projected vertically upward with speed ground and other is dropped from height h along the same vertical line. Collision between them is perfectly inelastic. Find time taken by them to reach the ground after collision in terms of . (1) (2) (3) (4) Answer : (1) Solution time for collision = t gh h After t1 VA = 0 – gt1 = gh and VB gh gt gh at the time of collision Pi Pf ⇒ mVA mVB mVf ⇒ gh gh Vf Vf = 0 and height from ground = h gt h h h so time = × g h g h
- 12. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 12- Page 12 - Q19. Length of a simple pendulum is 25.0 cm and time of 40 oscillation is 50 sec. If resolution of stop watch is 1 sec then accuracy is g is (in %) (1) 2.4 (2) 3.4 (3) 4.4 (4) 5.4 Answer : (3) Solution T ∆T g ∆g L ∆L g ∆g T ∆T L ∆L = 4.4% Q20. An electron is moving initially with velocity v0 in uniform electric field E = –E0 . If initial wavelength of electron is λ0 and mass of electron is m. Find wavelength of electron as a function time. (1) m v e E t (2) m v e E t (3) eEt mv (4) eEt mv Answer : (1) Solution Initially m v h Velocity as a function of time = v i v j m eE tk so wavelength m v m e E t h m v e E t
- 13. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 13- Page 13 - Q21. An asteroid of mass m (m << mE) is approaching with a velocity 12 km/s when it is at distance of 10 R from the centre of earth (where R is radius of earth). When it reaches at the surface of Earth, its velocity is (Nearest Integer) in km/s. Answer : 16 Solution KEi + PEi = KEf + PEf R GMm mv R GMm v u R GM v u R GM = 16.028 km/s ≃ 16 Q22. In H–spectrum wavelength of 1st line of Balmer series is λ = 6561Å. Find out wavelength of 2nd line of same series in nm. Answer : 486 Solution RZ n n R R R R ×6561Å = 4860Å = 486 nm Q23. There are three containers C1, C2 and C3 filled with same material at different constant temperature. When we mix then for different volume then we get some final temperature as shown in the below table. So find value of final temperature θ as shown in the table. Answer : 50°C Solution
- 14. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 14- Page 14 - 1θ1 + 2θ2 = (1 + 2) 60 θ1 + 2θ2 = 180 ...(i) 0 × θ1 + 1 × θ2 + 2 × θ3 = (1 + 2) 30 θ2 + 2θ3 = 90 ...(ii) 2 × θ1 + 0 × θ2 + 1 × θ3 = (2 + 1) 60 2θ1 + θ3 = 180 ...(iii) and θ1 + θ2 + θ3 = (1 + 1 + 1) θ ...(iv) from (1) + (2) + (3) 3θ1 + 3θ2 + 3θ3 = 450 ⇒ θ1 + θ2 + θ3 = 150 from (iv) equation 150 = 3θ ⇒ θ = 50ºC Q24. Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero. Answer : 50°C Solution V1 = ε1 – i. r1 0 = 10 – i × 20 i = 0.5A V2 = ε2 – ir2 = 10 – 0.5 × 5 V2 = 7.5V 0.5 = x 0.5 = 0.25 + x x = 0.25 x = = 30
- 15. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 15- Page 15 - Q25. An EMW is travelling along z-azis. B × j T C = 3 × 108 m/s & Frequency of wave is 25 Hz, then electric field in volt/m. Answer : 15 Solution B E c E = B × c = 15 N/c
- 16. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 16- Page 16 - CHEMISTRY Q26. Correct bond energy order of following is- (1) C–Cl > C–Br > C–I > C–F (2) C–F > C–Cl > C–Br > C–I (3) C–F < C–Cl < C–Br < C–I (4) C–I < C–Br < C–F < C–Cl Answer : (3) Solution Bond energy ∝ Bond lengh Q27. Determine Bohr's radius of Li2+ ion for n = 2. Given (Bohr's radius of H-atom = a0) (1) (2) (3) (4) Answer : (2) Solution r Z a n For Li2+ r a a Q28. Given the following reaction sequence A & B are respectively (1) Mg, Mg3N2 (2) Na Na3N (3) Mg Mg(NO3)2 (4) Na NaNO3 Answer : (1) Solution Mg N B MgN HO MgOH NH
- 17. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 17- Page 17 - Q29. Correct order of magnetic moment (spin only) for the following xomplexes (a) [Pd(PPh3)2Cl2] (b) [Ni(CO)4] (c) [Ni(CN)4]2– (d) [Ni(H2O)6]2+ (1) a = b = c < d (2) a < b < c < d (3) a > b > c > d (4) a = b > c > d Answer : (1) Solution [Pd(PPh3)2Cl2] Pd2+ = 4d8 M = 0 [Ni(CO)4] Ni = 3d8 4s2 (SFL) M = 0 [Ni(CN)4]2– Ni2+ = 3d8 (SFL) M = 0 [Ni(H2O)6]2+ Ni2+ = 3d8 (WFL) tg eg So, unpaired electron is 2 M = BM Q30. Determine total number of neutrons in three isotopes of hydrogen (1) 1 (2) 2 (3) 3 (4) 4 Answer : (3) Solution 1 1H 2 1H(D) 3 1H(T) Number of neutrons 0 + 1 + 2 = 3 0 + 1 + 2 = 3 Q31. Compare Ea (activation energy) for a, b, c and d. (1) Eb > Ec > Ed > Ea (2) Ea > Ed > Ec > Eb (3) Ec > Eb > Ea > Ed (4) Ed > Ea > Eb > Ec Answer : (1) Solution log k = log A – RT Ea slope = R Ea ⇒ Eb > Ec > Ed > Ea
- 18. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 18- Page 18 - Q32. Which of the following exhibit both Frenkel & Schottky defect? (1) AgBr (2) KCl (3) CsCl (4) ZnS Answer : (1) Solution Only AgBr can exhibit both Schottky and Frenkel defect. Q33. Given Basicity of B is : (1) 1 (2) 2 (3) 3 (4) 4 Answer : (1) Solution Basicity = 1 Q34. Which reaction does not occurs in the blast furnace in the metallurgy of Fe (a) CaO + SiO2 CaSiO3 (b) Fe2O3 + CO Fe3O4 + CO2 (c) FeO + SiO2 FeSiO3 (d) FeO ∆ Fe + O2 (1) a & b (2) a,b & c (3) c & d (4) a, b, c, d Answer : (3) Solution Theory based
- 19. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 19- Page 19 - Q35. Correct order of radius of elements is: C, O, F, Cl, Br (1) Br > Cl > C > O > F (2) Br < Cl < C < O < F (3) Cl < C < O < F < Br (4) C > F > O > Br > Cl Answer : (1) Solution Q36. Amongs the following which will show geometrical isomerism. (a) [Ni(NH3)5Cl]+ (b) [Ni(NH3)4ClBr] (c) [Ni(NH3)3Cl]+ (d) [Ni(NH3)2(NO2)2] (1) b, d (2) a, b (3) a, b & c (4) a, b, c & d Answer : (1) Solution Ma4bc can show 2 G.I. Ma2b2 can show 2 G.I. (Square planar) Ma4bc, 2 G.I. Ma2b2, 2 G.I. Q37. Assertion: pH of water increases on increasing temperature. Reason: H2O → H+ + OH– is an exothermic process. (1) Both assertion and reason are correct and reason is correct explanation of assertion. (2) Both assertion and reason are correct and reason is not correct explanation of assertion. (3) Assertion is true & reason is false. (4) Both assertion and reason are incorrect. Answer : (4) Solution Theory Based
- 20. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 20- Page 20 - Q38. Assertion: It has been found that for hydrogenation reaction the catalytic activity increases from group5 to group-11 metals with maximum activity being shown by groups 7-9 elements of the periodic table. Reason: For 7-9 group elements adsorption rate is maximum. (1) Both assertion and reason are correct and reason is correct explanation of assertion. (2) Both assertion and reason are correct and reason is not correct explanation of assertion. (3) Assertion is true & reason is false. (4) Both are incorrect Answer : (1) Solution Theory Based Q39. The major product of the following reactions is. (1) (2) (3) (4) Answer : (3) Solution
- 21. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 21- Page 21 - Q40. Find the final major product of the following reactions (1) (2) (3) (4) Answer : (1) Solution Q41. There are two compounds A and B of molecular formula C9H18O3. A has higher boiling point than B. What are the possible structures of A and B? (1) (2) (3) (4) Answer : (2) Solution In (A), extensive inter-molecular H-bonding is possible while in (B) there is no Inter-molecular H-bonding.
- 22. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 22- Page 22 - Q42. Kjeldahl method cannot be used for : (1) (2) (3) CH3–CH2–CH2–C≡N (4) Answer : (1) Solution Kjeldahl method is not applicable to nitro or diazo groups present in the ring, as nitrogen atom can't be converted to ammonium sulfate under the reaction conditions. Q43. A compound X that adds 2 hydrogen molecules on hydrogenation. The compound X also gives 3-oxohexanedioic acid on oxidative ozonolysis. The compound 'X' is: (1) (2) (3) (4) Answer : (3) Solution
- 23. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 23- Page 23 - Q44. Formation of Bakelite follows : (1) Electrophilic substitution followed by condensation. (2) Nucleophilic addition followed by dehydration. (3) Electrophilic addition followed by dehydration. (4) Hydration followed by condensation. Answer : (1) Solution Formation of Bakelite follows electrophilic substitution reaction of phenol with formaldehyde followed by condensation Q45. Products formed by hydrolysis of maltose are (1) α-D-Glucose, α-D-Glucose (2) α-D-Glucose, β-D-Glucose (3) α-D-Galactose, β-D-Glucose (4) β-D-Galactose, α-D-Glucose Answer : (1) Solution Maltose on hydrolysis gives 2 moles of α-D-glucose.
- 24. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 24- Page 24 - Q46. Temperature of 4 moles of gas increases from 300 K to 500 K find 'Cv' if ΔU = 5000 J. Answer : 06.25 Solution ΔU = nCvΔT 5000 = 4 × Cv (500 – 300) Cv = 6.25 JK–1 mol–1 Q47. Given : Esn sn VEPb Pb V Determine Pb Sn at equilibrium For cell reaction Sn | Sn2+ || Pb2+ | Pb take F RT V Answer : 02.15 Solution At Equilibrium state. Ecell = 0 ; Eºcell = 0.01 V Sn Pb Sn Pb 0 = 0.01 – log Pb Sn 0.01 = log Pb Sn log Pb Sn ⇒ Pb Sb = 101/3 = 2.1544 Q48. Given following reaction, NaClO3 + Fe → O2 + FeO + NaCl In the above reaction 492 L of O2 is obtained at 1 atm & 300 K temperature. Determine mass of NaClO3 required (in kg). (R = 0.082 L atm mol–1 K–1 ) Answer : 02.13 Solution mol of NaClO3 = mol of O2 mol of O2 = RT PV × × = 20 mol mass of NaClO3 = 20 × 106.5 = 2130 g Q49. Complex [ML5] can exhibit trigonal bipyramidal and square pyramidal geometry. Determine total number of 180º, 90º & 120º L-M-L bond angles. Answer : 20.00 Solution
- 25. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 25- Page 25 - Q50. How many atoms lie in the same plane in the major product (C)? Answer : 12.00 Solution Number of atoms in one plane = 12
- 26. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 26- Page 26 - MATHEMATICS Q51. Let = 2 + , = – + and is nonzero vector and × = × = 0 find , (1) (2) (3) (4) Answer : (3) Solution × × × × Q52. Let coefficient of x4 and x2 in the expansion of is α and β then α – β is (1) 48 (2) 60 (3) –132 (4) –60 Answer : (3) Solution 2[6 C0 x6 + 6 C2 x4 (x2 – 1) + 6 C4 x2 (x2 – 1)2 + 6 C6 (x2 – 1)3 ] = 2[x6 + 15(x6 – x4 ) + 15x2 (x4 – 2x2 + 1) + (–1 + 3x2 – 3x4 + x6 )] = 2(32x6 – 48x4 + 18x2 – 1) α = – 96 and β = 36 ∴ α – β = –132 Q53. Differential equation of x2 = 4b(y + b), where b is a parameter, is (1) (2) (3) (4) Answer : (2) Solution 2x = 4by ⇒ b = So, differential equation is x2 =
- 27. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 27- Page 27 - x2 = ⇒ Q54. Image of (1, 2, 3) w.r.t a plane is then which of the following points lie on the plane (1) (–1,1,–1) (2) (–1, –1, –1) (3) (–1, –1,1) (4) (1, 1, –1) Answer : (4) Solution d.r of normal to the plane midpoint of P and Q ia equation of plane x + y + z = 1 Q55. lim → sin is equal to (1) 1 (2) 10 (3) 5 (4) 0 Answer : (4) Solution Using L’Hospital lim → sin Q56. Let P be the set of points (x, y) such that x2 ≤ y ≤ – 2x + 3. Then area of region bounded by points in set P is (1) (2) (3) (4) Answer : (2) Solution Point of intersection of y = x2 & y = – 2x + 3 is obtained by x2 + 2x – 3 = 0 ⇒ x = – 3, 1 So, Area
- 28. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 28- Page 28 - Q57. Let f(x) = → R then range of (x) is (where [.] denotes greatest integer function) (1) ∪ (2) ∪ (3) ∪ (4) ∪ Answer : (2) Solution ∈ ∈ ∴ f(x) is a decreasing function ∴ y∈ ∪ ⇒ y∈ ∪ Q58. Let A and I = then value of 10 A–1 is– (1) 4I – A (2) 6I – A (3) A – 4I (4) A – 6I Answer : (4) Solution ⇒ x2 – 6x – 10 = 0 ∴ A2 – 6A – 10 I = 0 ⇒ 10A–1 = A – 6I
- 29. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 29- Page 29 - Q59. Solution set of 3x (3x –1) + 2 = |3x –1| + |3x – 2| contains (1) A singleton set (2) two elements (3) at least four elements (4) infinite elements Answer : (1) Solution Let 3x = t t(t –1) + 2 = |t –1| + |t –2| t2 – t + 2 = |t –1| + |t –2| are positive solution t = a 3x = a x = log3 so singleton set Q60. Mean and variance of 20 observation are 10 and 4. It was found, that in place of 11, 9 was taken by mistake find correct variance. (1) 3.99 (2) 3.98 (3) 4.01 (4) 4.02 Answer : (1) Solution ...(i) ...(ii) ∑xi 2 = 104 × 20 = 2080 Actual mean = Variance = =
- 30. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 30- Page 30 - Q61. λx + 2y + 2z = 5 2λx + 3y + 5z = 8 4λx + λy + 6z = 10 for the system of equation check the correct option. (1) Infinite solutions when λ = 8 (2) Infinite solutions when λ = 2 (3) no solutions when λ = 8 (4) no solutions when λ = 2 Answer : (4) Solution D = D = (λ + 8) (2 – λ) for λ = 2 D1 = = 5[18 – 10] – 2 [48 – 50] + 2 (16 – 30) = 40 + 4 – 28 ≠ 0 No solutions for λ = 2 Q62. For an A. P. T10 = T20 = Find sum of first 200 term. (1) 201 (2) 101 (3) 301 (4) 100 Answer : (4) Solution T10 = = a + 9d ...(i) T20 = = a + 19d ...(ii) ⇒ a = d = ⇒ S200 =
- 31. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 31- Page 31 - Q63. Let α = and a = (1 + α) . If a and b are roots of quadratic equation the quadratic equation is (1) x2 – 102x + 101 = 0 (2) x2 – 101x + 100 = 0 (3) x2 + 101x + 100 = 0 (4) x2 + 102x + 100 = 0 Answer : (1) Solution α = ω, b = 1 + ω3 + ω6 + ... = 101 a = (1 + ω) (1 + ω2 + ω4 + …..ω198 + ω200 ) = (1 + ω) Equation x2 – (101 +1)x + (101) × 1 = 0 ⇒ x2 – 102x + 101 = 0 Q64. Let f(x) is a three degree polynomial for which f ' (–1) = 0, f '' (1) = 0, f(–1) = 10, f(1) = 6 then local minima of f(x) exist at (1) x = 3 (2) x = 2 (3) x = 1 (4) x = –1 Answer : (1) Solution Let f(x) = ax3 + bx2 + cx + d a = d = b = c = ⇒ f(x) = a(x3 – 3x2 – 9x) + d f’(x) = ⇒ f’(x) = 0 ⇒ x = 3 – 1 local minima exist at x = 3
- 32. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 32- Page 32 - Q65. Let A and B are two events such that P(exactly one) = P (A ∪ B) = then P (A ∩ B) = (1) (2) (3) (4) Answer : (1) Solution P(exactly one) = ⇒ P(A) + P(B) – 2P(A ∩ B) = P(A∪B) = ⇒ P(A) + P(B) – P(A ∩ B) = ∴ P(A∩B) = Q66. Let I = then (1) < I2 < (2) < I2 < (3) < I < (4) < I < Answer : (1) Solution f(x) = f’(x) = = f(1) = f(2) =
- 33. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 33- Page 33 - Q67. Normal at (2, 2) to curve x2 + 2xy – 3y2 = 0 is L. Then perpendicular distance from origin to line L is (1) 4 (2) 2 (3) 2 (4) 4 Answer : (3) Solution x2 + 2xy – 3y2 = 0 x2 + 3xy – xy – 3y2 = 0 (x – y) (x + 3y) = 0 x – y = 0 x + 3y = 0 (2, 2) satisfy x – y = 0 Normal x + y = λ λ = 4 Hence x + y = 4 perpendicular distance from origin = Q68. Which of the following is tautology (1) ~(p ∨ ~q) → (p ∨ q) (2) (~p ∨ q) → (p ∨ q) (3) ~(p ∧ ~q) → (p ∨ q) (4) ~(p ∨ ~q) → (p ∧ q) Answer : (1) Solution (~ p ∧ q) → (p ∨ q) ~{(~p ∧ q) ∧ (~p ∧ ~q)} ~{~p ∧ f} Q69. If a hyperbola has vertices (±6, 0) and P(10, 16) lies on it, then the equation of normal at P is (1) 2x + 5y = 100 (2) 2x + 5y = 10 (3) 2x – 5y = 100 (4) 5x + 2y = 100 Answer : (1) Solution Vertex is at (±6, 0) ∴ a = 6 Let the hyperbola is Putting point P(10, 16) on the hyperbola ⇒ b2 = 144 ∴ hyperbola is
- 34. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 34- Page 34 - ∴ equation of normal is ∴ Putting we get 2x + 5y = 100 Q70. If y = mx + c is a tangent to the circle (x – 3)2 + y2 = 1 and also the perpendicular to the tangent to the circle x2 + y2 = 1 at , then (1) c2 + 6c + 7 = 0 (2) c2 – 6c + 7 = 0 (3) c2 + 6c – 7 = 0 (4) c2 – 6c – 7 = 0 Answer : (1) Solution Slope of tangent to x2 + y2 = 1 at x2 + y2 = 1 2x + 2yy` = 0 y` = – = –1 y = mx + c is tangent of x2 + y2 = 1 So m = 1 y = x + c now distance of (3, 0) from y = x + c is c2 + 6c + 9 = 2 c2 + 6c + 7 = 0 Q71. Let cos sin and cos where α, β ∈ . Then tan (α + 2β) is equal to Answer : (1) Solution cos sin and sin tan α = sin β = sin β = tan2β =
- 35. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 35- Page 35 - tan (α + 2β) = tantan tan tan Q72. The number of four letter words that can be made from the letters of word "EXAMINATION" is Answer : 2454 Solution EXAMINATION 2N, 2A, 2I, E, X, M, T, O Case I All are different so 8 P4 = Case II 2 same and 2 different so 3 C1. 7 C2. Case III 2 same and 2 same so 3 C2. ∴ Total = 1680 + 756 + 18 = 2454 Q73. Let the line y = mx intersects the curve y2 = x at P and tangent to y2 = x at P intersects x-axis at Q. If area (ΔOPQ) = 4, find m (m > 0) Answer : 0.5 Solution 2ty = x + t2 Q(–t2 , 0) = 4 |t|3 = 8 t = ± 2 (t > 0) m =
- 36. JEE MAIN 2020 Question Paper (08 Jan, 2020 Shift 2) 36- Page 36 - Q74. is equal to Answer : 504 Solution × × ×