JEE Main 2022 — 29 Jun Shift 1 (Previous Year Paper)

JEE Main · 2022 · with answers and explanations

questions
90
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 4 marks · −1

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t=0t=0 s. Ball B is thrown vertically down with an initial velocity 'u' at t=2t=2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms−1^{-1}. [use g=10g=10 ms−2^{-2}]:

  1. A.10
  2. B.15
  3. C.20
  4. D.30

Correct answer: D

Explanation: A falls 80 m in 4 s; B covers 80 m in 2 s: 80=2u+20⇒u=3080=2u+20\Rightarrow u=30 m/s.

Question 2

Physics · 4 marks · −1

A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms−1^{-1} and 40 ms−1^{-1} respectively. The velocity of the third piece will be:

  1. A.15 ms−1^{-1}
  2. B.25 ms−1^{-1}
  3. C.35 ms−1^{-1}
  4. D.50 ms−1^{-1}

Correct answer: B

Explanation: 2m v=m302+402=50m⇒v=252m\,v=m\sqrt{30^2+40^2}=50m\Rightarrow v=25 m/s.

Question 3

Physics · 4 marks · −1

The activity of a radioactive material is 2.56×10−32.56\times10^{-3} Ci. If the half life of the material is 5 days, after how many days the activity will become 2×10−52\times10^{-5} Ci?

  1. A.30 days
  2. B.35 days
  3. C.40 days
  4. D.25 days

Correct answer: B

Explanation: Ratio 128=27128=2^7: 7 half-lives =35=35 days.

Question 4

Physics · 4 marks · −1

A spherical shell of 1 kg mass and radius R is rolling with angular speed ω\omega on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is a3R2ω\frac a3R^2\omega. The value of a will be:

Figure for question 4
  1. A.2
  2. B.3
  3. C.5
  4. D.4

Correct answer: C

Explanation: LO=Icmω+mvR=23R2ω+R2ω=53R2ωL_O=I_{cm}\omega+mvR=\frac23R^2\omega+R^2\omega=\frac53R^2\omega, so a=5a=5.

Question 5

Physics · 4 marks · −1

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be: (Given gas constant R = 8.3 JK−1^{-1}-mol−1^{-1})

  1. A.249 J
  2. B.415 J
  3. C.498 J
  4. D.830 J

Correct answer: C

Explanation: n=2n=2, Cv=32RC_v=\frac32R: Q=2×1.5×8.3×20=498Q=2\times1.5\times8.3\times20=498 J.

Question 6

Physics · 4 marks · −1

A wire of length L is hanging from a fixed support. The length changes to L1L_1 and L2L_2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to:

  1. A.L1L2\sqrt{L_1L_2}
  2. B.L1+L22\frac{L_1+L_2}{2}
  3. C.2L1−L22L_1-L_2
  4. D.3L1−2L23L_1-2L_2

Correct answer: C

Explanation: L1=L+xL_1=L+x, L2=L+2x⇒L=2L1−L2L_2=L+2x\Rightarrow L=2L_1-L_2.

Question 7

Physics · 4 marks · −1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal. Reason R: Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal. In the light of the above statements, choose the most appropriate answer from the options given below.

  1. A.Both A and R are correct and R is the correct explanation of A
  2. B.Both A and R are correct but R is not the correct explanation of A
  3. C.A is correct but R is not correct
  4. D.A is not correct but R is correct

Correct answer: B

Question 8

Physics · 4 marks · −1

A particle of mass 500 gm is moving in a straight line with velocity v=bx5/2v=bx^{5/2}. The work done by the net force during its displacement from x=0x=0 to x=4x=4 m is: (Take b=0.25b=0.25 m−3/2^{-3/2} s−1^{-1}).

  1. A.2 J
  2. B.4 J
  3. C.8 J
  4. D.16 J

Correct answer: D

Explanation: v(4)=0.25×32=8v(4)=0.25\times32=8 m/s; W=12(0.5)(64)=16W=\frac12(0.5)(64)=16 J.

Question 9

Physics · 4 marks · −1

A charge particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charge particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charge particle:

  1. A.1 : 1
  2. B.1 : 2
  3. C.2 : 1
  4. D.1 : 4

Correct answer: C

Explanation: r=2mKqB∝Kr=\frac{\sqrt{2mK}}{qB}\propto\sqrt K: ratio 2 : 1.

Question 10

Physics · 4 marks · −1

For a series LCR circuit, I vs ω\omega curve is shown: (a) To the left of ωr\omega_r, the circuit is mainly capacitive. (b) To the left of ωr\omega_r, the circuit is mainly inductive. (c) At ωr\omega_r, impedance of the circuit is equal to the resistance of the circuit. (d) At ωr\omega_r, impedance of the circuit is 0. Choose the most appropriate answer from the options given below:

Figure for question 10
  1. A.(a) and (d) only
  2. B.(b) and (d) only
  3. C.(a) and (c) only
  4. D.(b) and (c) only

Correct answer: C

Explanation: Below resonance XC>XLX_C>X_L (capacitive); at resonance Z=RZ=R.

Question 11

Physics · 4 marks · −1

A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1 kgs−1^{-1} and at a speed of 10 ms−1^{-1}. Then, the initial acceleration of the block, in ms−2^{-2}, will be:

Figure for question 11
  1. A.3
  2. B.6
  3. C.5
  4. D.4

Correct answer: C

Explanation: F=dmdtv=10F=\frac{dm}{dt}v=10 N; a=5a=5 m s−2^{-2}.

Question 12

Physics · 4 marks · −1

In van dar Wall equation [P+aV2][V−b]=RT\left[P+\frac{a}{V^2}\right][V-b]=RT; P is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants ab\frac ab is dimensionally equal to:

  1. A.PV\frac PV
  2. B.VP\frac VP
  3. C.PV
  4. D.PV3^3

Correct answer: C

Explanation: [a]=PV2[a]=PV^2, [b]=V[b]=V, so a/b∼PVa/b\sim PV.

Question 13

Physics · 4 marks · −1

Two vectors A⃗\vec A and B⃗\vec B have equal magnitudes. If magnitude of A⃗+B⃗\vec A+\vec B is equal to two times the magnitude of A⃗−B⃗\vec A-\vec B, then the angle between A⃗\vec A and B⃗\vec B will be:

  1. A.sin⁡−1(35)\sin^{-1}\left(\frac35\right)
  2. B.sin⁡−1(13)\sin^{-1}\left(\frac13\right)
  3. C.cos⁡−1(35)\cos^{-1}\left(\frac35\right)
  4. D.cos⁡−1(13)\cos^{-1}\left(\frac13\right)

Correct answer: C

Explanation: 2+2cos⁡θ=4(2−2cos⁡θ)⇒cos⁡θ=352+2\cos\theta=4(2-2\cos\theta)\Rightarrow\cos\theta=\frac35.

Question 14

Physics · 4 marks · −1

The escape velocity of a body on a planet 'A' is 12 kms−1^{-1}. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is:

  1. A.12 kms−1^{-1}
  2. B.24 kms−1^{-1}
  3. C.36 kms−1^{-1}
  4. D.6 kms−1^{-1}

Correct answer: A

Explanation: ve∝Rρv_e\propto R\sqrt\rho: 12×2=1\frac12\times2=1, so 12 km/s.

Question 15

Physics · 4 marks · −1

At a certain place the angle of dip is 30° and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at that certain place, is:

  1. A.13\frac1{\sqrt3}
  2. B.12\frac12
  3. C.3\sqrt3
  4. D.1

Correct answer: A

Explanation: B=BHcos⁡30∘=13B=\frac{B_H}{\cos30^\circ}=\frac{1}{\sqrt3} G.

Question 16

Physics · 4 marks · −1

A longitudinal wave is represented by x=10sin⁡2π(nt−xλ)x=10\sin2\pi\left(nt-\frac x\lambda\right) cm. The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

  1. A.2π2\pi
  2. B.5π5\pi
  3. C.π\pi
  4. D.5π2\frac{5\pi}{2}

Correct answer: B

Explanation: 2πn(10)=4nλ⇒λ=5π2\pi n(10)=4n\lambda\Rightarrow\lambda=5\pi.

Question 17

Physics · 4 marks · −1

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will:

  1. A.increase by 50%
  2. B.decrease by 15%
  3. C.increase by 25%
  4. D.increase by 33%

Correct answer: A

Explanation: At constant V, U=12CV2∝KU=\frac12CV^2\propto K: ×1.5\times1.5.

Question 18

Physics · 4 marks · −1

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1×1051\times10^5 NC−1^{-1}. If the charge on the particle is 40 μ\muC and the initial velocity is 200 ms−1^{-1}, how much distance it will travel before coming to the rest momentarily:

  1. A.1 m
  2. B.5 m
  3. C.10 m
  4. D.0.5 m

Correct answer: D

Explanation: a=qEm=410−4=4×104a=\frac{qE}{m}=\frac{4}{10^{-4}}=4\times10^4 m s−2^{-2}; s=v22a=0.5s=\frac{v^2}{2a}=0.5 m.

Question 19

Physics · 4 marks · −1

Using Young's double slit experiment, a monochromatic light of wavelength 5000 Å produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000 Å is used and the separation between the slits is doubled, then the new fringe width will be:

  1. A.0.5 mm
  2. B.1.0 mm
  3. C.0.6 mm
  4. D.0.3 mm

Correct answer: D

Explanation: β∝λ/d\beta\propto\lambda/d: 0.5×1.2/2=0.30.5\times1.2/2=0.3 mm.

Question 20

Physics · 4 marks · −1

Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission:

  1. A.375×107375\times10^7
  2. B.75×10775\times10^7
  3. C.375×108375\times10^8
  4. D.75×10975\times10^9

Correct answer: B

Explanation: f=3×1014f=3\times10^{14} Hz; 0.02f=6×10120.02f=6\times10^{12}; /8000=7.5×108/8000=7.5\times10^8.

Question 21

Physics · 4 marks · −1

Two coils require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be ________ min.

Correct answer: 15

Explanation: t=t1t2t1+t2=15t=\frac{t_1t_2}{t_1+t_2}=15 min.

Question 22

Physics · 4 marks · −1

The intensity of the light from a bulb incident on a surface is 0.22 W/m2^2. The amplitude of the magnetic field in this light-wave is ________ ×10−9\times10^{-9} T. (Given: Permittivity of vacuum ϵ0=8.85×10−12\epsilon_0=8.85\times10^{-12} C2^2 N−1^{-1}-m−2^{-2}, speed of light in vacuum c=3×108c=3\times10^8 ms−1^{-1})

Correct answer: 43

Explanation: E0=2Icϵ0≈12.9E_0=\sqrt{\frac{2I}{c\epsilon_0}}\approx12.9 V/m; B0=E0/c≈43×10−9B_0=E_0/c\approx43\times10^{-9} T.

Question 23

Physics · 4 marks · −1

As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm2^2 for each plate. The equivalent thermal conductivity of the compound plate is (1+5α)\left(1+\frac5\alpha\right) K, then the value of α\alpha will be ________.

Figure for question 23

Correct answer: 21

Explanation: Keq=6.54/K+2.5/2K=6.55.25K=(1+521)KK_{eq}=\frac{6.5}{4/K+2.5/2K}=\frac{6.5}{5.25}K=\left(1+\frac{5}{21}\right)K.

Question 24

Physics · 4 marks · −1

A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is x\sqrt x cm. The value of xx is ________.

Correct answer: 700

Explanation: A\'^2=x^2+\frac{9v^2}{\omega^2}=25+9\times75=700.

Question 25

Physics · 4 marks · −1

The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x×10−3x\times10^{-3} Ω\Omega cm. The value of xx is ________. [Take π=3.14\pi=3.14]

Figure for question 25

Correct answer: 144

Explanation: R=tan⁡45∘=1 ΩR=\tan45^\circ=1\ \Omega; ρ=RAL=3.14×1.4431.4=0.144 Ω\rho=\frac{RA}{L}=\frac{3.14\times1.44}{31.4}=0.144\ \Omega cm.

Question 26

Physics · 4 marks · −1

300 cal. of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be ________ °C.

Correct answer: 102

Explanation: T2=500×225300=375T_2=500\times\frac{225}{300}=375 K =102∘=102^\circC.

Question 27

Physics · 4 marks · −1

d1\sqrt{d_1} and d2\sqrt{d_2} are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an α\alpha particle is approaching a gold nucleus. For d1=x d2d_1=x\,d_2, the value of xx will be ________.

Correct answer: 3

Explanation: b∝cot⁡θ2b\propto\cot\frac\theta2: d1d2=cot⁡230∘cot⁡245∘=3\frac{d_1}{d_2}=\frac{\cot^230^\circ}{\cot^245^\circ}=3.

Question 28

Physics · 4 marks · −1

A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by 100 μ\muA, it brings a change of 10 mA in collector current. If the load resistance is 2 kΩ\Omega and input resistance is 1 kΩ\Omega, the value of power gain is x×104x\times10^4. The value of xx is ________.

Correct answer: 2

Explanation: β=100\beta=100; power gain =β2RLRi=2×104=\beta^2\frac{R_L}{R_i}=2\times10^4.

Question 29

Physics · 4 marks · −1

A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is ________ mm.

Correct answer: 225

Explanation: Ball lens: f=μR2(μ−1)=22.5f=\frac{\mu R}{2(\mu-1)}=22.5 cm from the centre =225=225 mm.

Question 30

Physics · 4 marks · −1

For the network shown below, the value of VB−VAV_B-V_A is ________ V.

Figure for question 30

Correct answer: 10

Question 31

Chemistry · 4 marks · −1

Production of iron in blast furnace follows the following equation Fe3O4(s)+4CO(g)→3Fe(l)+4CO2(g)\mathrm{Fe_3O_4(s)+4CO(g)\rightarrow3Fe(l)+4CO_2(g)} when 4.640 kg of Fe3_3O4_4 and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is: [Given: Molar Atomic mass (g mol−1^{-1}): Fe = 56, O = 16, C = 12]

  1. A.1400
  2. B.2200
  3. C.3360
  4. D.4200

Correct answer: C

Explanation: 20 mol Fe3_3O4_4 (limiting) vs 90 mol CO; Fe =60=60 mol =3360=3360 g.

Question 32

Chemistry · 4 marks · −1

Which of the following statements are correct? (A) The electronic configuration of Cr is [Ar] 3d5^5 4s1^1. (B) The magnetic quantum number may have a negative value. (C) In the ground state of an atom, the orbitals are filled in order of their increasing energies. (D) The total number of nodes are given by n−2n-2. Choose the most appropriate answer from the options given below:

  1. A.(A), (C) and (D) only
  2. B.(A) and (B) only
  3. C.(A) and (C) only
  4. D.(A), (B) and (C) only

Correct answer: D

Explanation: Total nodes =n−1=n-1, so (D) is false.

Question 33

Chemistry · 4 marks · −1

Arrange the following in the decreasing order of their covalent character: (A) LiCl (B) NaCl (C) KCl (D) CsCl. Choose the most appropriate answer from the options given below:

  1. A.(A) > (C) > (B) > (D)
  2. B.(B) > (A) > (C) > (D)
  3. C.(A) > (B) > (C) > (D)
  4. D.(A) > (B) > (D) > (C)

Correct answer: C

Question 34

Chemistry · 4 marks · −1

The solubility of AgCl will be maximum in which of the following?

  1. A.0.01 M KCl
  2. B.0.01 M HCl
  3. C.0.01 M AgNO3_3
  4. D.Deionised water

Correct answer: D

Question 35

Chemistry · 4 marks · −1

Which of the following is a correct statement?

  1. A.Brownian motion destabilises sols.
  2. B.Any amount of dispersed phase can be added to emulsion without destabilising it.
  3. C.Mixing two oppositely charged sols in equal amount neutralises charges and stabilises colloids.
  4. D.Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.

Correct answer: D

Question 36

Chemistry · 4 marks · −1

The electronic configuration of Pt (atomic number 78) is:

  1. A.[Xe] 4f14^{14} 5d9^9 6s1^1
  2. B.[Kr] 4f14^{14} 5d10^{10}
  3. C.[Xe] 4f14^{14} 5d10^{10}
  4. D.[Xe] 4f14^{14} 5d8^8 6s2^2

Correct answer: A

Question 37

Chemistry · 4 marks · −1

In isolation of which one of the following metals from their ores, the use of cyanide salt is not commonly involved?

  1. A.Zinc
  2. B.Gold
  3. C.Silver
  4. D.Copper

Correct answer: D

Question 38

Chemistry · 4 marks · −1

Which one of the following reactions indicates the reducing ability of hydrogen peroxide in basic medium?

  1. A.HOCl+H2O2→H3O++Cl−+O2\mathrm{HOCl+H_2O_2\rightarrow H_3O^++Cl^-+O_2}
  2. B.PbS+4H2O2→PbSO4+4H2O\mathrm{PbS+4H_2O_2\rightarrow PbSO_4+4H_2O}
  3. C.2MnO4−+3H2O2→2MnO2+3O2+2H2O+2OH−\mathrm{2MnO_4^-+3H_2O_2\rightarrow2MnO_2+3O_2+2H_2O+2OH^-}
  4. D.Mn2++H2O2→Mn4++2OH−\mathrm{Mn^{2+}+H_2O_2\rightarrow Mn^{4+}+2OH^-}

Correct answer: C

Question 39

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Metal): (A) Li (B) Na (C) Rb (D) Cs. List - II (Emitted light wavelength (nm)): (I) 670.8 (II) 589.2 (III) 780.0 (IV) 455.5. Choose the most appropriate answer from the options given below:

  1. A.(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. B.(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  3. C.(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  4. D.(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

Correct answer: A

Question 40

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Metal): (A) Cs (B) Ga (C) B (D) Si. List - II (Application): (I) High temperature thermometer (II) Water repellent sprays (III) Photoelectric cells (IV) Bullet proof vest. Choose the most appropriate answer from the options given below:

  1. A.(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. B.(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  3. C.(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  4. D.(A)-(I), (B)-(IV), (C)-(II), (D)-(III)

Correct answer: A

Question 41

Chemistry · 4 marks · −1

The oxoacid of phosphorus that is easily obtained from a reaction of alkali and white phosphorus and has two P–H bonds, is:

  1. A.Phosphonic acid
  2. B.Phosphinic acid
  3. C.Pyrophosphorus acid
  4. D.Hypophosphoric acid

Correct answer: B

Question 42

Chemistry · 4 marks · −1

The acid that is believed to be mainly responsible for the damage of Taj Mahal is

  1. A.sulfuric acid.
  2. B.hydrofluoric acid.
  3. C.phosphoric acid.
  4. D.hydrochloric acid.

Correct answer: A

Question 43

Chemistry · 4 marks · −1

Two isomers 'A' and 'B' with molecular formula C4_4H8_8 give different products on oxidation with KMnO4_4 in acidic medium. Isomer 'A' on reaction with KMnO4_4/H+^+ results in effervescence of a gas and gives ketone. The compound 'A' is

  1. A.But-1-ene.
  2. B.cis-But-2-ene.
  3. C.trans-But-2-ene.
  4. D.2-methyl propene.

Correct answer: D

Explanation: (CH3)2C=CH2→(CH3)2C=O+CO2\mathrm{(CH_3)_2C=CH_2\to(CH_3)_2C=O+CO_2}.

Question 44

Chemistry · 4 marks · −1

2-Bromophenol →(CH3)3CLi\xrightarrow{(\mathrm{CH_3})_3\mathrm{CLi}} [A] →(i) CO2, (ii) H3O+\xrightarrow{\text{(i) CO}_2\text{, (ii) H}_3\text{O}^+} salicylic acid (2-hydroxybenzoic acid). In the given conversion the compound A is:

Figure for question 44
  1. A.2-lithiophenol (Li on ring, OH retained)
  2. B.2-lithiophenoxide lithium (Li on ring and OLi)
  3. C.1-bromo-2-tert-butoxybenzene (Br retained, OC(CH3)3)
  4. D.2-tert-butylphenol (C(CH3)3 on ring, OH)

Correct answer: B

Explanation: Excess t-BuLi deprotonates the OH and exchanges Br for Li, giving the dilithio species 2−LiC6H4OLi\mathrm{2-LiC_6H_4OLi}.

Question 45

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The esterification of carboxylic acid with an alcohol is a nucleophilic acyl substitution. Statement II: Electron withdrawing groups in the carboxylic acid will increase the rate of esterification reaction. Choose the most appropriate option:

  1. A.Both Statement I and Statement II are correct.
  2. B.Both Statement I and Statement II are incorrect.
  3. C.Statement I is correct but Statement II is incorrect.
  4. D.Statement I is incorrect but Statement II is correct.

Correct answer: A

Question 46

Chemistry · 4 marks · −1

Aniline →Br2(excess), H2O\xrightarrow{\text{Br}_2\text{(excess), H}_2\text{O}} A (major product); aniline →(i) (CH3CO)2O, (ii) Br2, CH3COOH, (iii) HCl\xrightarrow{\text{(i) (CH}_3\text{CO)}_2\text{O, (ii) Br}_2\text{, CH}_3\text{COOH, (iii) HCl}} B (major product). Consider the above reactions, the product A and product B respectively are

Figure for question 46
  1. A.4-bromoaniline and 4-bromoaniline
  2. B.2,4,6-tribromoaniline and 2,4,6-tribromoaniline
  3. C.2,4,6-tribromoaniline and 4-bromoaniline
  4. D.4-bromoaniline and 2,4,6-tribromoaniline

Correct answer: C

Explanation: Free aniline brominates at all o/p positions; acetylation moderates the ring so only the para product forms, then hydrolysis frees the amine.

Question 47

Chemistry · 4 marks · −1

The polymer, which can be stretched and retains its original status on releasing the force is

  1. A.Bakelite.
  2. B.Nylon 6,6.
  3. C.Buna-N.
  4. D.Terylene.

Correct answer: C

Question 48

Chemistry · 4 marks · −1

Sugar moiety in DNA and RNA molecules respectively are

  1. A.β\beta-D-2-deoxyribose, β\beta-D-deoxyribose.
  2. B.β\beta-D-2-deoxyribose, β\beta-D-ribose
  3. C.β\beta-D-ribose, β\beta-D-2-deoxyribose.
  4. D.β\beta-D-deoxyribose, β\beta-D-2-deoxyribose.

Correct answer: B

Question 49

Chemistry · 4 marks · −1

Which of the following compound does not contain sulfur atom?

  1. A.Cimetidine
  2. B.Ranitidine
  3. C.Histamine
  4. D.Saccharin

Correct answer: C

Question 50

Chemistry · 4 marks · −1

Given below are two statements. Statement I: Phenols are weakly acidic. Statement II: Therefore they are freely soluble in NaOH solution and are weaker acids than alcohols and water. Choose the most appropriate option:

  1. A.Both Statement I and Statement II are correct.
  2. B.Both Statement I and Statement II are incorrect.
  3. C.Statement I is correct but Statement II is incorrect.
  4. D.Statement I is incorrect but Statement II is correct.

Correct answer: C

Explanation: Phenols are stronger acids than alcohols and water, so II is incorrect.

Question 51

Chemistry · 4 marks · −1

Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL−1^{-1} at 257°C and 100 mm Hg. The molar mass of geraniol is ________ g mol−1^{-1}. (Nearest Integer) [Given: R = 0.082 L atm K−1^{-1} mol−1^{-1}]

Correct answer: 152

Explanation: M=dRTP=0.46×0.082×530100/760≈152M=\frac{dRT}{P}=\frac{0.46\times0.082\times530}{100/760}\approx152.

Question 52

Chemistry · 4 marks · −1

17.0 g of NH3_3 completely vapourises at −33.42-33.42°C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol−1^{-1}. The enthalpy change for the vapourisation of 85 g of NH3_3 under the same conditions is ________ kJ.

Correct answer: 117

Explanation: 5 mol ×\times 23.4 = 117 kJ.

Question 53

Chemistry · 4 marks · −1

1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is ________. (Nearest integer) [Given: Density of acetic acid is 1.02 g mL−1^{-1}, Molar mass of acetic acid is 60 g mol−1^{-1}, KfK_f(H2_2O) = 1.85 K kg mol−1^{-1}]

Correct answer: 5

Explanation: m≈0.0102m\approx0.0102; i=0.01981.85×0.0102≈1.05i=\frac{0.0198}{1.85\times0.0102}\approx1.05, so ≈5%\approx5\%.

Question 54

Chemistry · 4 marks · −1

A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is ________ cm3^3. (Nearest integer) [Given: Faraday constant F = 96500 C mol−1^{-1} at STP, molar volume of an ideal gas is 22.7 L mol−1^{-1}]

Correct answer: 127

Explanation: 720 C =7.46×10−3=7.46\times10^{-3} F: H2_2 3.73×10−33.73\times10^{-3} + O2_2 1.87×10−31.87\times10^{-3} mol =5.6×10−3=5.6\times10^{-3} mol ≈127\approx127 cm3^3.

Question 55

Chemistry · 4 marks · −1

The activation energy of one of the reactions in a biochemical process is 532611 J mol−1^{-1}. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300=x×10−3k310k_{300}=x\times10^{-3}k_{310}. The value of xx is ________. [Given: ln 10 = 2.3, R = 8.3 J K−1^{-1} mol−1^{-1}]

Correct answer: 1

Explanation: ln⁡k300k310=−5326118.3⋅1093000=−6.9=−3ln⁡10\ln\frac{k_{300}}{k_{310}}=-\frac{532611}{8.3}\cdot\frac{10}{93000}=-6.9=-3\ln10, so x=1x=1.

Question 56

Chemistry · 4 marks · −1

The number of terminal oxygen atoms present in the product B obtained from the following reaction is ________. FeCr2O4+Na2CO3+O2→A+Fe2O3+CO2\mathrm{FeCr_2O_4+Na_2CO_3+O_2\rightarrow A+Fe_2O_3+CO_2}; A+H+→B+H2O+Na+\mathrm{A+H^+\rightarrow B+H_2O+Na^+}

Correct answer: 6

Explanation: A = Na2_2CrO4_4, B = Na2_2Cr2_2O7_7: dichromate has 6 terminal O.

Question 57

Chemistry · 4 marks · −1

An acidified manganate solution undergoes disproportionation reaction. The spin-only magnetic moment value of the product having manganese in higher oxidation state is ________ B.M. (Nearest integer)

Correct answer: 0

Explanation: MnO42−→_4^{2-}\to MnO4−_4^- (Mn7+^{7+}, d0^0) + MnO2_2: 0 B.M.

Question 58

Chemistry · 4 marks · −1

Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55 g of the compound neutralised 12.5 mL of 1 M H2_2SO4_4 solution. The percentage of nitrogen in the compound is ________. (Nearest integer)

Correct answer: 64

Explanation: NH3_3 = 25 mmol; N = 0.35 g; 63.6% ≈64\approx64.

Question 59

Chemistry · 4 marks · −1

Observe structures of the following compounds: 2-chlorobutane, glycine (H2_2N–CH2_2–COOH), alanine (CH3_3–CH(NH2_2)–COOH), styrene (vinylbenzene) and menthol (2-isopropyl-5-methylcyclohexanol). The total number of structures/compounds which possess asymmetric carbon atoms is ________.

Figure for question 59

Correct answer: 3

Explanation: 2-Chlorobutane, alanine and menthol are chiral: 3.

Question 60

Chemistry · 4 marks · −1

C6H12O6→ZymaseA→ΔNaOIB+CHI3\mathrm{C_6H_{12}O_6\xrightarrow{Zymase}A\xrightarrow[\Delta]{NaOI}B+CHI_3}. The number of carbon atoms present in the product B is ________.

Correct answer: 1

Explanation: A = ethanol; iodoform reaction gives HCOONa (1 carbon).

Question 61

Mathematics · 4 marks · −1

The probability that a randomly chosen 2×22\times2 matrix with all the entries from the set of first 10 primes, is singular, is equal to:

  1. A.133104\frac{133}{10^4}
  2. B.18103\frac{18}{10^3}
  3. C.19103\frac{19}{10^3}
  4. D.271104\frac{271}{10^4}

Correct answer: C

Explanation: ad=bcad=bc with primes forces {a,d}={b,c}\{a,d\}=\{b,c\}: 10 cases with all equal plus 90×290\times2 with a≠da\ne d, i.e. 190 of 10410^4.

Question 62

Mathematics · 4 marks · −1

Let the solution curve of the differential equation xdydx−y=y2+16x2x\frac{dy}{dx}-y=\sqrt{y^2+16x^2}, y(1)=3y(1)=3 be y=y(x)y=y(x). Then y(2)y(2) is equal to:

  1. A.15
  2. B.11
  3. C.13
  4. D.17

Correct answer: A

Explanation: Homogeneous: v+v2+16=8xv+\sqrt{v^2+16}=8x; at x=2x=2, v+v2+16=16⇒v=7.5v+\sqrt{v^2+16}=16\Rightarrow v=7.5, y=15y=15.

Question 63

Mathematics · 4 marks · −1

If the mirror image of the point (2,4,7)(2,4,7) in the plane 3x−y+4z=23x-y+4z=2 is (a,b,c)(a,b,c), then 2a+b+2c2a+b+2c is equal to:

  1. A.54
  2. B.50
  3. C.−6-6
  4. D.−42-42

Correct answer: C

Explanation: Image =(−5813,8013,−2113)=\left(-\frac{58}{13},\frac{80}{13},-\frac{21}{13}\right), so 2a+b+2c=−62a+b+2c=-6.

Question 64

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a function defined by f(x)={max⁡t≤x{t3−3t},x≤2x2+2x−6,2<x<3[x−3]+9,3≤x≤52x+1,x>5f(x)=\begin{cases}\max_{t\le x}\{t^3-3t\}, & x\le2\\ x^2+2x-6, & 2<x<3\\ [x-3]+9, & 3\le x\le5\\ 2x+1, & x>5\end{cases} where [t][t] is the greatest integer less than or equal to tt. Let mm be the number of points where ff is not differentiable and I=∫−22f(x) dxI=\int_{-2}^{2}f(x)\,dx. Then the ordered pair (m,I)(m, I) is equal to:

  1. A.(3,274)\left(3,\frac{27}{4}\right)
  2. B.(3,234)\left(3,\frac{23}{4}\right)
  3. C.(4,274)\left(4,\frac{27}{4}\right)
  4. D.(4,234)\left(4,\frac{23}{4}\right)

Correct answer: C

Question 65

Mathematics · 4 marks · −1

Let a⃗=αi^+3j^−k^\vec a=\alpha\hat i+3\hat j-\hat k, b⃗=3i^−βj^+4k^\vec b=3\hat i-\beta\hat j+4\hat k and c⃗=i^+2j^−2k^\vec c=\hat i+2\hat j-2\hat k where α,β∈R\alpha,\beta\in\mathbb{R}, be three vectors. If the projection of a⃗\vec a on c⃗\vec c is 103\frac{10}{3} and b⃗×c⃗=−6i^+10j^+7k^\vec b\times\vec c=-6\hat i+10\hat j+7\hat k, then the value of α+β\alpha+\beta is equal to:

  1. A.3
  2. B.4
  3. C.5
  4. D.6

Correct answer: A

Explanation: α+83=103⇒α=2\frac{\alpha+8}{3}=\frac{10}3\Rightarrow\alpha=2; b⃗×c⃗=(2β−8,10,6+β)⇒β=1\vec b\times\vec c=(2\beta-8,10,6+\beta)\Rightarrow\beta=1.

Question 66

Mathematics · 4 marks · −1

The area enclosed by y2=8xy^2=8x and y=2xy=\sqrt2x that lies outside the triangle formed by y=2xy=\sqrt2x, x=1x=1, y=22y=2\sqrt2, is equal to:

  1. A.1626\frac{16\sqrt2}{6}
  2. B.1126\frac{11\sqrt2}{6}
  3. C.1326\frac{13\sqrt2}{6}
  4. D.526\frac{5\sqrt2}{6}

Correct answer: C

Explanation: Area between curves on [0,4][0,4] is 823\frac{8\sqrt2}{3}; the triangle has area 22\frac{\sqrt2}{2}; difference 1326\frac{13\sqrt2}{6}.

Question 67

Mathematics · 4 marks · −1

If the system of linear equations 2x+y−z=72x+y-z=7, x−3y+2z=1x-3y+2z=1, x+4y+δz=kx+4y+\delta z=k, where δ,k∈R\delta,k\in\mathbb{R} has infinitely many solutions, then δ+k\delta+k is equal to:

  1. A.−3-3
  2. B.3
  3. C.6
  4. D.9

Correct answer: B

Explanation: Δ=−7δ−21=0⇒δ=−3\Delta=-7\delta-21=0\Rightarrow\delta=-3; third equation == first −- second, so k=7−1=6k=7-1=6; δ+k=3\delta+k=3.

Question 68

Mathematics · 4 marks · −1

Let α\alpha and β\beta be the roots of the equation x2+(2i−1)=0x^2+(2i-1)=0. Then, the value of ∣α8+β8∣|\alpha^8+\beta^8| is equal to:

  1. A.50
  2. B.250
  3. C.1250
  4. D.1500

Correct answer: A

Explanation: α2=β2=1−2i\alpha^2=\beta^2=1-2i, so ∣α8+β8∣=2∣1−2i∣4=50|\alpha^8+\beta^8|=2|1-2i|^4=50.

Question 69

Mathematics · 4 marks · −1

Let Δ∈{∧,∨,⇒,⇔}\Delta\in\{\wedge,\vee,\Rightarrow,\Leftrightarrow\} be such that (p∧q) Δ ((p∨q)⇒q)(p\wedge q)\,\Delta\,((p\vee q)\Rightarrow q) is a tautology. Then Δ\Delta is equal to:

  1. A.∧\wedge
  2. B.∨\vee
  3. C.⇒\Rightarrow
  4. D.⇔\Leftrightarrow

Correct answer: C

Explanation: If p∧qp\wedge q is true then qq is true, so the consequent is true: ⇒\Rightarrow gives a tautology.

Question 70

Mathematics · 4 marks · −1

Let A=[aij]A=[a_{ij}] be a square matrix of order 3 such that aij=2j−ia_{ij}=2^{j-i}, for all i,j=1,2,3i,j=1,2,3. Then, the matrix A2+A3+⋯+A10A^2+A^3+\dots+A^{10} is equal to:

  1. A.(310−32)A\left(\frac{3^{10}-3}{2}\right)A
  2. B.(310−12)A\left(\frac{3^{10}-1}{2}\right)A
  3. C.(310+12)A\left(\frac{3^{10}+1}{2}\right)A
  4. D.(310+32)A\left(\frac{3^{10}+3}{2}\right)A

Correct answer: A

Explanation: AA has rank 1 with trace 3, so An=3n−1AA^n=3^{n-1}A; ∑n=2103n−1=310−32\sum_{n=2}^{10}3^{n-1}=\frac{3^{10}-3}{2}.

Question 71

Mathematics · 4 marks · −1

Let a set A=A1∪A2∪⋯∪AkA=A_1\cup A_2\cup\dots\cup A_k, where Ai∩Aj=ϕA_i\cap A_j=\phi for i≠ji\ne j, 1≤i,j≤k1\le i,j\le k. Define the relation R from A to A by R={(x,y):y∈AiR=\{(x,y): y\in A_i if and only if x∈Aix\in A_i, 1≤i≤k}1\le i\le k\}. Then, R is:

  1. A.reflexive, symmetric but not transitive
  2. B.reflexive, transitive but not symmetric
  3. C.reflexive but not symmetric and transitive
  4. D.an equivalence relation

Correct answer: D

Explanation: Being in the same block of a partition is an equivalence relation.

Question 72

Mathematics · 4 marks · −1

Let {an}n=0∞\{a_n\}_{n=0}^{\infty} be a sequence such that a0=a1=0a_0=a_1=0 and an+2=2an+1−an+1a_{n+2}=2a_{n+1}-a_n+1 for all n≥0n\ge0. Then, ∑n=2∞an7n\sum_{n=2}^{\infty}\frac{a_n}{7^n} is equal to:

  1. A.6343\frac{6}{343}
  2. B.7216\frac{7}{216}
  3. C.8343\frac{8}{343}
  4. D.49216\frac{49}{216}

Correct answer: B

Explanation: an=(n2)a_n=\binom n2; ∑(n2)xn=x2(1−x)3=1/49216/343=7216\sum\binom n2x^n=\frac{x^2}{(1-x)^3}=\frac{1/49}{216/343}=\frac{7}{216}.

Question 73

Mathematics · 4 marks · −1

The distance between the two points A and A' which lie on y=2y=2 such that both the line segments AB and A'B (where B is the point (2,3)(2,3)) subtend angle π4\frac{\pi}{4} at the origin, is equal to:

  1. A.10
  2. B.485\frac{48}{5}
  3. C.525\frac{52}{5}
  4. D.3

Correct answer: C

Question 74

Mathematics · 4 marks · −1

A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is:

  1. A.229+43\frac{22}{9+4\sqrt3}
  2. B.669+43\frac{66}{9+4\sqrt3}
  3. C.224+93\frac{22}{4+9\sqrt3}
  4. D.664+93\frac{66}{4+9\sqrt3}

Correct answer: B

Explanation: With 4a+3t=224a+3t=22, minimise a2+34t2a^2+\frac{\sqrt3}{4}t^2: −3(22−3t)8+32t=0⇒t=669+43-\frac{3(22-3t)}{8}+\frac{\sqrt3}{2}t=0\Rightarrow t=\frac{66}{9+4\sqrt3}.

Question 75

Mathematics · 4 marks · −1

The domain of the function cos⁡−1(2sin⁡−1(14x2−1)π)\cos^{-1}\left(\frac{2\sin^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi}\right) is:

  1. A.R−{−12,12}\mathbb{R}-\left\{-\frac12,\frac12\right\}
  2. B.(−∞,−1]∪[1,∞)∪{0}(-\infty,-1]\cup[1,\infty)\cup\{0\}
  3. C.(−∞,−12)∪(12,∞)∪{0}\left(-\infty,\frac{-1}{2}\right)\cup\left(\frac12,\infty\right)\cup\{0\}
  4. D.(−∞,−12]∪[12,∞)∪{0}\left(-\infty,\frac{-1}{\sqrt2}\right]\cup\left[\frac{1}{\sqrt2},\infty\right)\cup\{0\}

Correct answer: D

Explanation: Need ∣14x2−1∣≤1\left|\frac1{4x^2-1}\right|\le1: 4x2−1≥14x^2-1\ge1 or 4x2−1≤−14x^2-1\le-1, i.e. ∣x∣≥12|x|\ge\frac1{\sqrt2} or x=0x=0.

Question 76

Mathematics · 4 marks · −1

If the constant term in the expansion of (3x3−2x2+5x5)10\left(3x^3-2x^2+\frac{5}{x^5}\right)^{10} is 2k⋅l2^k\cdot l, where ll is an odd integer, then the value of kk is equal to:

  1. A.6
  2. B.7
  3. C.8
  4. D.9

Correct answer: D

Question 77

Mathematics · 4 marks · −1

∫05cos⁡(π(x−[x2]))dx\int_0^5\cos\left(\pi\left(x-\left[\frac x2\right]\right)\right)dx, where [t][t] denotes greatest integer less than or equal to tt, is equal to:

  1. A.−3-3
  2. B.−2-2
  3. C.2
  4. D.0

Correct answer: D

Explanation: On [0,2)[0,2), [2,4)[2,4), [4,5][4,5] the integrand is ±cos⁡πx\pm\cos\pi x, whose integral over each piece is 0.

Question 78

Mathematics · 4 marks · −1

Let PQ be a focal chord of the parabola y2=4xy^2=4x such that it subtends an angle of π2\frac{\pi}{2} at the point (3,0)(3,0). Let the line segment PQ be also a focal chord of the ellipse E:x2a2+y2b2=1E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a2>b2a^2>b^2. If ee is the eccentricity of the ellipse E, then the value of 1e2\frac{1}{e^2} is equal to:

  1. A.1+21+\sqrt2
  2. B.3+223+2\sqrt2
  3. C.1+231+2\sqrt3
  4. D.4+534+5\sqrt3

Correct answer: B

Question 79

Mathematics · 4 marks · −1

Let the tangent to the circle C1:x2+y2=2C_1: x^2+y^2=2 at the point M(−1,1)M(-1,1) intersect the circle C2:(x−3)2+(y−2)2=5C_2:(x-3)^2+(y-2)^2=5, at two distinct points A and B. If the tangents to C2C_2 at the points A and B intersect at N, then the area of the triangle ANB is equal to:

  1. A.12\frac12
  2. B.23\frac23
  3. C.16\frac16
  4. D.53\frac53

Correct answer: C

Question 80

Mathematics · 4 marks · −1

Let the mean and the variance of 5 observations x1,x2,x3,x4,x5x_1,x_2,x_3,x_4,x_5 be 245\frac{24}{5} and 19425\frac{194}{25} respectively. If the mean and variance of the first 4 observation are 72\frac72 and aa respectively, then (4a+x5)(4a+x_5) is equal to:

  1. A.13
  2. B.15
  3. C.17
  4. D.18

Correct answer: B

Explanation: ∑x=24\sum x=24, ∑x2=154\sum x^2=154; x5=10x_5=10; first four: ∑x2=54\sum x^2=54, a=544−494=54a=\frac{54}4-\frac{49}{4}=\frac54; 4a+x5=154a+x_5=15.

Question 81

Mathematics · 4 marks · −1

Let S={z∈C:∣z−2∣≤1, z(1+i)+zˉ(1−i)≤2}S=\{z\in\mathbb{C}: |z-2|\le1,\ z(1+i)+\bar z(1-i)\le2\}. Let ∣z−4i∣|z-4i| attains minimum and maximum values, respectively, at z1∈Sz_1\in S and z2∈Sz_2\in S. If 5(∣z1∣2+∣z2∣2)=α+β55(|z_1|^2+|z_2|^2)=\alpha+\beta\sqrt5, where α\alpha and β\beta are integers, then the value of α+β\alpha+\beta is equal to ________.

Correct answer: 26

Question 82

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation dydx+2y2cos⁡4x−cos⁡2x=xetan⁡−1(2cot⁡2x)\frac{dy}{dx}+\frac{\sqrt2y}{2\cos^4x-\cos2x}=xe^{\tan^{-1}(\sqrt2\cot2x)}, 0<x<π20<x<\frac{\pi}{2} with y(π4)=π232y\left(\frac\pi4\right)=\frac{\pi^2}{32}. If y(π3)=π218e−tan⁡−1(α)y\left(\frac\pi3\right)=\frac{\pi^2}{18}e^{-\tan^{-1}(\alpha)}, then the value of 3α23\alpha^2 is equal to ________.

Correct answer: 2

Question 83

Mathematics · 4 marks · −1

Let dd be the distance between the foot of perpendiculars of the points P(1,2,−1)P(1,2,-1) and Q(2,−1,3)Q(2,-1,3) on the plane −x+y+z=1-x+y+z=1. Then d2d^2 is equal to ________.

Correct answer: 26

Explanation: PQ⃗=(1,−3,4)\vec{PQ}=(1,-3,4) is perpendicular to the normal (−1,1,1)(-1,1,1) (dot product 0), so d2=∣PQ∣2=26d^2=|PQ|^2=26.

Question 84

Mathematics · 4 marks · −1

The number of elements in the set S={θ∈[−4π,4π]:3cos⁡22θ+6cos⁡2θ−10cos⁡2θ+5=0}S=\{\theta\in[-4\pi,4\pi]: 3\cos^22\theta+6\cos2\theta-10\cos^2\theta+5=0\} is ________.

Correct answer: 32

Explanation: Using 10cos⁡2θ=5+5cos⁡2θ10\cos^2\theta=5+5\cos2\theta: 3c2+c=03c^2+c=0 with c=cos⁡2θc=\cos2\theta, so c=0c=0 or −13-\frac13; each has 16 solutions for 2θ∈[−8π,8π]2\theta\in[-8\pi,8\pi]: 32.

Question 85

Mathematics · 4 marks · −1

The number of solutions of the equation 2θ−cos⁡2θ+2=02\theta-\cos^2\theta+\sqrt2=0 in R\mathbb{R} is equal to ________.

Correct answer: 1

Explanation: f(θ)=2θ+2−cos⁡2θf(\theta)=2\theta+\sqrt2-\cos^2\theta has f=ˊ2+sin⁡2θ>0f\'=2+\sin2\theta>0: exactly one root.

Question 86

Mathematics · 4 marks · −1

50tan⁡(3tan⁡−1(12)+2cos⁡−1(15))+42tan⁡(12tan⁡−1(22))50\tan\left(3\tan^{-1}\left(\frac12\right)+2\cos^{-1}\left(\frac1{\sqrt5}\right)\right)+4\sqrt2\tan\left(\frac12\tan^{-1}(2\sqrt2)\right) is equal to ________.

Correct answer: 29

Explanation: 2cos⁡−115=2tan⁡−12=π−2tan⁡−1122\cos^{-1}\frac1{\sqrt5}=2\tan^{-1}2=\pi-2\tan^{-1}\frac12, so the first angle is π+tan⁡−112\pi+\tan^{-1}\frac12: 50⋅12=2550\cdot\frac12=25. tan⁡θ2=sec⁡θ−1tan⁡θ=12\tan\frac\theta2=\frac{\sec\theta-1}{\tan\theta}=\frac{1}{\sqrt2}: 44. Total 29.

Question 87

Mathematics · 4 marks · −1

Let c,k∈Rc,k\in\mathbb{R}. If f(x)=(c+1)x2+(1−c2)x+2kf(x)=(c+1)x^2+(1-c^2)x+2k and f(x+y)=f(x)+f(y)−xyf(x+y)=f(x)+f(y)-xy, for all x,y∈Rx,y\in\mathbb{R}, then the value of ∣2(f(1)+f(2)+f(3)+⋯+f(20))∣|2(f(1)+f(2)+f(3)+\dots+f(20))| is equal to ________.

Correct answer: 3395

Explanation: 2(c+1)=−12(c+1)=-1, k=0k=0: f(x)=−x22−54xf(x)=-\frac{x^2}{2}-\frac54x; 2∑f=−2870−5252\sum f=-2870-525, absolute value 3395.

Question 88

Mathematics · 4 marks · −1

Let H:x2a2−y2b2=1H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, a>0,b>0a>0,b>0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(22+14)4(2\sqrt2+\sqrt{14}). If the eccentricity H is 112\frac{\sqrt{11}}{2}, then the value of a2+b2a^2+b^2 is equal to ________.

Correct answer: 88

Explanation: b=72ab=\frac{\sqrt7}{2}a and a+b=2(22+14)⇒a=42a+b=2(2\sqrt2+\sqrt{14})\Rightarrow a=4\sqrt2, a2+b2=32+56=88a^2+b^2=32+56=88.

Question 89

Mathematics · 4 marks · −1

Let P1:r⃗⋅(2i^+j^−3k^)=4P_1:\vec r\cdot(2\hat i+\hat j-3\hat k)=4 be a plane. Let P2P_2 be another plane which passes through the points (2,−3,2)(2,-3,2), (2,−2,−3)(2,-2,-3) and (1,−4,2)(1,-4,2). If the direction ratios of the line of intersection of P1P_1 and P2P_2 be 16,α,β16,\alpha,\beta, then the value of α+β\alpha+\beta is equal to ________.

Correct answer: 28

Explanation: Normal of P2P_2: (−5,5,1)(-5,5,1); (2,1,−3)×(−5,5,1)=(16,13,15)(2,1,-3)\times(-5,5,1)=(16,13,15); α+β=28\alpha+\beta=28.

Question 90

Mathematics · 4 marks · −1

Let b1b2b3b4b_1b_2b_3b_4 be a 4-element permutation with bi∈{1,2,3,…,100}b_i\in\{1,2,3,\dots,100\} for 1≤i≤41\le i\le4 and bi≠bjb_i\ne b_j for i≠ji\ne j, such that either b1,b2,b3b_1,b_2,b_3 are consecutive integers or b2,b3,b4b_2,b_3,b_4 are consecutive integers. Then the number of such permutations b1b2b3b4b_1b_2b_3b_4 is equal to ________.

Correct answer: 18915

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