Courses

# JEE Advanced 2018 Question Paper with Solutions (20th May - Evening)

## 54 Questions MCQ Test Mock Test Series for JEE Main & Advanced 2021 | JEE Advanced 2018 Question Paper with Solutions (20th May - Evening)

Description
This mock test of JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) for JEE helps you for every JEE entrance exam. This contains 54 Multiple Choice Questions for JEE JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) (mcq) to study with solutions a complete question bank. The solved questions answers in this JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) exercise for a better result in the exam. You can find other JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
*Multiple options can be correct
QUESTION: 1

### A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as dK/dt = γt, where γ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?

Solution:

The kinetic energy of the particle is given by,

Integrate both sides of the above equation.

As γ and m are constant; therefore, v ∝t.
The acceleration of the particle is calculated as,

The force on the particle is given by,
F = ma

F = constant
Find the distance of the particle from the origin.

Thus, distance does not increase linearly with time.

*Multiple options can be correct
QUESTION: 2

### Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity μ0​. Which of the following statements is (are) true?

Solution:

The resistive force on the plate is viscous force, therefore,

From the equation,

The tangential shear stress on the floor is given by,

From the equation,

*Multiple options can be correct
QUESTION: 3

### An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density λ. It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 120° at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is ε0. Which of the following statements is (are) true?

Solution:

Consider the diagram.

Use trigonometry to find PQ.
PQ = 2PM
= 2Rsin60°

The charge enclosed by the portion of wire PQ is given by,

Apply Gauss law,

This flux points radially outward from the wire so that the electric field in the direction of z-component is zero and also it is normal to the wire, not to the spherical shell.

QUESTION: 4

A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale.)

Solution:

The image of line AB is behind the mirror.
Use the mirror formula,

Thus, the line AB is perpendicular to the principle axis behind the mirror. Also the point C is at focus point; therefore, its image is formed at infinity.

*Multiple options can be correct
QUESTION: 5

In a radioactive decay chain, nucleus decays to  nucleus. Let Nα and Nβ be the number of α and β particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

Solution:

Let the number of alpha particles emitted be x and the number of beta particles emitted be y in the decay process.
The reaction for the radioactive decay is shown below:
Equate the mass numbers on both sides of the equation, 232=212+4x+0y
4x=20
x=5
Equate the atomic numbers on both sides of the equation, 90=82+2x-y
2x-y=8
y=2

*Multiple options can be correct
QUESTION: 6

In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7cm and 83.9cm. Which of the following statements is (are) true?

Solution:

The difference of two successive resonances is equal to half of the wavelength, thus,

The level 50.7 cm is corresponding to third harmonic. If end correction is e, then,

The speed of sound is calculated as,
v=fλ
= 500 x 66.4
= 332.00 m/s

*Answer can only contain numeric values
QUESTION: 7

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4kg is at rest on this surface. An impulse of 1.0Ns is applied to the block at time to t = 0 so that it starts moving along the x-axis with a velocity v(t) = v0e–t/τ, where v0 is a constant and τ = 4s. The displacement of the block, in metres, at t = τ is_____________. Take e−1 = 0.37.

Solution:

At time t = 0, due to impulse, the block moves with velocity v0.
Thus, find v0.

= 1/0.4
= 2.5 m/s
The velocity along the x-axis is calculated as,

Integrate both sides of the above equation.

*Answer can only contain numeric values
QUESTION: 8

A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface. The maximum height it reaches after the bounce, in meters is ___________.

Solution:

Consider the diagram showing the given condition for the ball.

For the angle 45°, the maximum height is calculated as,

The velocity of the ball becomes u' after hitting the ground and it loses half of its kinetic energy at that time, that is,

The maximum height for the angle 30° is calculated below,

= 30 m

*Answer can only contain numeric values
QUESTION: 9

A particle, of mass 10−3 kg and charge 1.0 C is initially at rest. At time t = 0 , the particle comes under the influence of an electric field  where E0 = 1.0 NC−1 and ω = 103 rad s−1. Consider the effect of only the electrical force on the particle. Then the maximum speed, in ms−1, attained by the particle at subsequent times is ____________.

Solution:

The velocity of the charged particle is maximum when the electrostatic force becomes zero. Thus,

Due to electric field, charge particles start to be in motion, therefore,

Simplify the above equation,

*Answer can only contain numeric values
QUESTION: 10

A moving coil galvanometer has 50 turns and each turn has an area 2 × 10−4 m2. The magnetic field produced by the magnet inside the galvanometer is 0.02 T. The torsional constant of the suspension wire is 10−4 N m rad−1. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2 rad. The resistance of the coil of the galvanometer is 50Ω . This galvanometer is to be converted into an ammeter capable of measuring current in the range 0−1.0A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is __________.

Solution:

The moment of the deflecting couple is equal to the moment of the restoring couple. That is,

The resistance is connected in shunt with the ammeter for the galvanometer as follows,

From the above circuit the value of S is calculated below,

*Answer can only contain numeric values
QUESTION: 11

A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .

Solution:

According to Hooke’s law,

The least count of the vernier caliper is
Therefore, the third division of vernier scale coincides with the main scale.

*Answer can only contain numeric values
QUESTION: 12

One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.

Solution:

For mono-atomic gas, the adiabatic index is given as,

Thus, the final temperature of the gas is calculated below,

The decrease in internal energy for the mono-atomic gas is given as,

*Answer can only contain numeric values
QUESTION: 13

In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F= n ×10−4N due to the impact of the electrons. The value of n is __________. Mass of the electron me = 9 x 10-31kg and 1.0 eV = 1.6 ×10−19 j.

Solution:

Let N be the number of photons emitted per second, then, P = Nhv
For negligible kinetic energy, the work function is equal to the energy of incident photon. Thus,

The change in kinetic energy is calculated as,
K = qV

As the work efficiency is 100% , thus, the number of electrons is equal to the number of photons per second.
The force experienced by the anode electrons is given as,

n = 24

*Answer can only contain numeric values
QUESTION: 14

Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV The value of Z is __________.

Solution:

Calculate the transition energy for the emitted photon from
n = 2 to n = 1.
...(1)
Calculate the transition energy for the emitted photon from n = 3 to n = 2.
...(2)
Equate the equations (1) and (2), to find the value of z.

z2 = 9
z = 3

QUESTION: 15

The electric field E is measured at a point P (0,0,d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II.

Solution:

(1) For a point charge, the electric field is given as,

(2) For a dipole, the electric field is given as,

(3) For a line charge, the electric field is given as,

(4) Net electric field due to two infinite wires is given as,

(5) The electric field due to sheet is given as,

Hence, E is independent of d.

QUESTION: 16

A planet of mass M, has two natural satellites with masses m1 and m2. The radii of their circular orbits are R1 and R2 respectively. Ignore the gravitational force between the satellites. Define v1, L1,K1 and T1 to be, respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and v2, L2,K2 and T2 to be the corresponding quantities of satellite 2. Given m1/m2 = 2 and R1/R2 = 1/4, match the ratios in List-I to the numbers in List-II.

Solution:

(P) The speed of the natural satellite is given as,

(Q) Angular momentum of the satellite is given as,
L = mvR
Therefore,

(R) The kinetic energy of the satellite is given by,

Therefore,

(S) The Time period of the satellite is given by,

QUESTION: 17

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV -diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II.

Solution:

(P) From the thermodynamic process shown, the process (1) represents adiabatic process. The heat transfer is zero in case of adiabatic process that is no heat is exchanged between the gas and its surroundings.
(Q) The second thermodynamic process is isobaric process. Pressure is constant in isobaric process. Thus, the work done by the gas is given as,

(R) The third thermodynamics process is isochoric process. Volume is constant in this process. The work done by the gas is zero.
W = 0
(S) The fourth thermodynamics process is isothermal process. Temperature is constant in this process. ∆T= 0

QUESTION: 18

In the List-I below, four different paths of a particle are given as functions of time. In these functions, α and β are positive constants of appropriate dimensions and α ≠ β . In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned:  is the linear momentum,  is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path.

Solution:

(P) The velocity of the particle is given by,

= constant
The linear momentum is given by,
= constant
The angular momentum of the particle is given by,

The relation between the force and the potential energy is,

U = constant
The kinetic energy of the particle is,

= constant
Total energy is also constant.
(Q) The velocity of the particle is given by,

So, both the linear momentum and the kinetic energy are not conserved.
Linear acceleration of the particle is given by,

Force on the particle is calculated as,

The potential energy of the particle is,

Angular momentum of the particle is,

Total energy is also constant because force is central. (R) The velocity of the particle is given by,

Thus, linear momentum is not conserved.
Angular momentum of the particle is,

But the magnitude of the velocity is constant.

Kinetic energy of the particle is,

Linear acceleration of the particle is given as,

The potential energy of the particle is,

U = constant
Total energy is also constant as force is central.
(S) The velocity of the particle is,

So, linear momentum, angular momentum, and kinetic energy are not conserved.
Linear acceleration of the particle is given as,

Force on the particle is,

The potential energy of the particle is,

Total energy is also constant as force is central.

*Multiple options can be correct
QUESTION: 19

​The correct option(s) regarding the complex

Solution:

(A) The following figure shows the structures of isomers formed.

Therefore, it has two geometrical isomers.
(B) The following figure shows the structures of isomers formed.

Therefore, it has three geometrical isomers.
(C) The oxidation state of Cobalt is +3. It is d6 in configuration and as it is attached to strong field ligand. Therefore, it is diamagnetic.
(D) Ammonia is a strong ligand in comparison to water. Therefore, CFSE is lower in the former case and thus, it will absorb light of longer wavelength.

*Multiple options can be correct
QUESTION: 20

The correct option(s) to distinguish nitrate salts of Mn2+ and Cu2+ taken separately is (are)

Solution:

(A) Both the ions Mn2+ and Cu2+ cannot be distinguished as they both impart green colour to the flame.
(B) Cu2+ belongs to group (II) and thus, on passing H2S through acidic medium it forms precipitate of CuS.
(C) Both the ions Mn2+ and Cu2+ form precipitate on passing H2S through basic medium.
(D) According to the electrochemical series the value of standard reduction potential for Cu2+ is greater than that of Mn2+.

QUESTION: 21

Aniline reacts with mixed acid (conc. HNO3 and conc. H2SO4) at 288 K to give P (51%), Q (47%) and R (2%). The major product(s) of the following reaction sequence is (are)

Solution:

The following figure shows the correct reaction sequence.

QUESTION: 22

The Fischer presentation of D-glucose is given below.

The correct structure(s) of β-L-glucopyranose is (are)

Solution:

The following figure shows the structure of β − L− glucopyranose

QUESTION: 23

For a first order reaction A(g) → 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t=0) and at time t are P0 and Pt, respectively. Initially, only A is present with concentration [A]0, and t1/3 is the time required for the partial pressure of A to reach 1/3rd of its initial value. The correct option(s) is (are).
(Assume that all these gases behave as ideal gases)

Solution:

The final pressures are calculated below.

The first order rate equation is given as,

The rate constant is not dependent on the concentration. Hence, the following figure shows the graph obtained.

*Multiple options can be correct
QUESTION: 24

For a reaction, the plots of [A] and [P] with time at temperatures T1 and T2 are given below.
If T2 >T1, the correct statement(s) is (are)
(Assume ∆Hθ and ∆Sθ are independent of temperature and ratio of lnK at T1 to lnK at T2 is greater than T2/T1. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

Solution:

As the temperature increases the concentration decreases. That is the given reaction is exothermic in nature. Therefore, ∆HΘ < 0 .
It is given that,

With increase in temperature, the value of change in Gibbs free energy increases, this occurs when ∆SΘ< 0 .

*Answer can only contain numeric values
QUESTION: 25

The total number of compounds having at least one bridging oxo group among the molecules given below is ____.
N2O3, N2O5, P4O6, H4P2O5, H5P3O10, H2S2O3, H2S2O5

Solution:

The structures corresponding to the given compounds are shown below.

The compounds containing A − O− A or A =O are known as oxo compounds. Thus, the compounds containing at least one oxo group among the molecules is six .

*Answer can only contain numeric values
QUESTION: 26

Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is ____. (Atomic weights in g mol−1: O = 16, S = 32, Pb = 207)

Solution:

The given reactions are,
2PbS+3O2 → 2PbO + 2SO2
2PbO + PbS → 3Pb SO2
The number of moles of lead produced by 3 moles of oxygen is 3.
Mass of 1 mole of oxygen is 16 g .
Thus, the mass of 3 moles of oxygen is calculated below.
M= 3 x 16 g
= 96 g
Mass of 1 mole of lead is 207.2 g.
Thus, find the mass of 3 moles of lead.

Mass of lead produced by 96 g oxygen is 621.6 g .
Thus, the mass of lead produced by 1 g oxygen is calculated below,

Mass of lead produced by 1000 g oxygen is calculated below,

Thus, the mass of lead produced by 1 kg oxygen is 6.475 kg .

*Answer can only contain numeric values
QUESTION: 27

To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction,
MnCl2 + K2S2O8 + H2O → KMnO4 + H2SO4 + HCl (equation not balanced).
Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2 (in mg) present in the initial solution is ____.
(Atomic weights in g mol-1: Mn = 55, Cl = 35.5)

Solution:

The given reactions are.

Number of millimoles of oxalic acid are calculated below,

The number of millimoles of KMnO4 that react with oxalic acid is 1.
The number of millimoles of MnCl2 that are required initially is 1.
Thus, calculate the mass of  MnCl2 that is required initially.
M=1x126 mg
= 126 mg

*Answer can only contain numeric values
QUESTION: 28

For the given compound X, the total number of optically active stereoisomers is ____.

Solution:

In the given diagram the solid wedge represents the configuration and geometry is fixed. While the dashed wedge shows that the configuration and geometry at the specified carbon is not fixed.
The number of stereocentres is three.
Thus, the number of stereoisomers is calculated below.
2n = 23
=8
There also occurs the formation of one meso-compound. Therefore, the total number of optically active stereoisomers is 8 −1= 7 .

*Answer can only contain numeric values
QUESTION: 29

In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is ____.
(Atomic weights in g mol-1: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis)

Solution:

The following figure shows the correct sequence of reaction.

Thus, the moles of D formed are calculated below,

The molar mass for tribromoaniline is 330 g / mol .
Mass of D formed is given as,
m = 1.5 mol 330 g / mol
= 495 g

*Answer can only contain numeric values
QUESTION: 30

The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises
copper as per the reaction given below:
2Cu(s) + H2O(g) → Cu2O(s) + H2(g)
is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln () is ____.
(Given: total pressure = 1 bar,
R (universal gas constant) = 8 J K-1mol-1, ln (10) = 2.3. Cu(s) and Cu2O(s) are mutually immiscible.
At 1250 K:

Gibbs energy)

Solution:

The given equations are shown below.

Subtract the equation (2) from (1).

The relation between Gibbs free energy and equilibrium constant is given below.

*Answer can only contain numeric values
QUESTION: 31

Consider the following reversible reaction,

The activation energy of the backward reaction exceeds that of the forward reaction by 2RT (in J mol−1). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of ∆Gθ (in J mol−1) for the reaction at 300 K is ____.
(Given; ln (2) = 0.7, RT = 2500 J mol−1 at 300 K and G is the Gibbs energy)

Solution:

The relation for activation energy is given as,
Eab = Eaf + 2RT
The relation for pre-exponential factor is given below. Af = 4Ab
The equilibrium constant is calculated as,

The relation between equilibrium constant and Gibbs free energy is given below.

∴ The absolute value of ΔG= 8500J/molΔG = 8500J/mol.

*Answer can only contain numeric values
QUESTION: 32

Consider an electrochemical cell: A (s) An+ (aq, 2 M) || B2n+ (aq,1M) | B(s). The value of ∆Hθ for the cell reaction is twice that of ΔGθ at 300 K . If the emf of the cell is zero, the ∆Sθ (in J K-1 mol-1) of the cell reaction per mole of B formed at 300 K is _________.
(Given: ln(2) = 0.7, R (universal gas constant) = 8.3 J K-1 mol-1. H, S and G are enthalpy, entropy and Gibbs energy, respectively.)

Solution:

The following reaction takes place at anode.
A → An++ ne   ....(1)
The following reaction takes place at cathode.
B2n+ + 2ne→ B   ...(2)
Multiply equation (1) with a factor of 2 and then add both the equations.

The final equation obtained is,

The Nernst equation with substituted values is given below.

Now, substitute the value of ∆GΘ.

QUESTION: 33

Match each set of hybrid orbitals from LIST–I with complex(es) given in LIST–II.

The correct option is

Solution:

The hybridisation of [FeF6]4− is given below.

As F is a weak ligand; therefore, no pairing occurs. Therefore, hybridisation is sp3d2.
The hybridisation of  is given  below.

As Cl and water are weak ligands therefore, no pairing occurs. Thus, hybridisation is d2sp3.
The hybridisation of  is given below.

As six ligands are attached, it means six vacant orbitals are required. Thus, pairing will not occur. The, hybridisation is d2sp3.
The hybridisation of  is shown below.

As Cl is a weak ligand; therefore, no pairing occurs. Thus, hybridisation is sp3.
The hybridisation of NiCO4 is shown below.

As CO is a strong ligand; therefore, pairing occurs. Thus, hybridisation is sp3.
The hybridisation of  is shown below.

As CN is a strong ligand; therefore, pairing occurs. Thus, hybridisation is dsp2.

QUESTION: 34

The desired product X can be prepared by reacting the major product of the reactions in LIST-I with one or more appropriate reagents in LIST-II. (given, order of migratory aptitude: aryl > alkyl > hydrogen)

The correct option is

Solution:

(P) The final product is given as shown below.

(Q) The final product is given as shown below.

(R) The final product is given as shown below.

(S) The final product is given as shown below.

QUESTION: 35

LIST-I contains reactions and LIST-II contains major products.

Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option.

Solution:

The product formed in reaction (P) is shown below.

The product formed in reaction (Q) is shown below.

The product formed in reaction (R) is shown below.

The product formed in reaction (S) is shown below.

QUESTION: 36

Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of the solutions on [H+] are given in LIST-II.
(Note: Degree of dissociation (α) of weak acid and weak base is << 1; degree of hydrolysis of salt <<1; [H+] represents the concentration of H+ ions)

Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option is

Solution:

(P) The millimoles of each species are found as shown below.

The given solution is a buffer; therefore, there occurs no change in the concentration of hydrogen ions.
(Q) The millimoles of each species are found as shown below.

The given solution consists of salt of weak acid and strong base. Thus hydrogen ion concentration is given as shown below.

After dilution the concentration decreases by half.

(R) The millimoles of each species are found as shown below.

The given solution consists of salt of strong acid and weak base. Thus hydrogen ion concentration is given as shown below.

After dilution the concentration decreases by half.

(S) The given salt is sparingly soluble. It means there occurs no change in hydroxide ion concentration and thus

QUESTION: 37

For any positive integer n, define  as (Here, the inverse trigonometric function tan−1 x assumes values in
Then which of the following statement(s) is (are) TRUE?

Solution:

(A) According to the definition of function.

The first order derivative of the function is given as.

The value of the function is calculated below,

From the above equation, it is clear that 0 is not in the domain of the function, Hence, option (A) and (B) are false.
According to the definition of function,

Apply the limits in the above expression.

The trigonometric identity in terms of secant and tangent of an angle is as follows.

Substitute the values of tan (fn(x)) in the above equation.

*Multiple options can be correct
QUESTION: 38

Let T be the line passing through the points P(−2,7) and Q (2,−5). Let F1 be the set of all pairs of circles (S1,S2) such that T is tangent to S1 at P and tangent to S2 at Q, and also such that S1 and S2 touch each other at a point, say, M. Let E1 be the set representing the locus of M as the pair (S1,S2) varies in F1. Let the set of all straight line segments joining a pair of distinct points of E1 and passing through the point R (1,1) be F2. Let F2 be the set of the mid-points of the line segments in the set F2. Then, which of the following statement(s) is (are) TRUE?

Solution:

The following figure shows the required condition in the question.

The values of different angles are given as.

The expression for ∠QC2M and ∠PC1M is as follows.

The equation of locus for the variable point M is given as.
(x + 2)(x − 2)+ (y − 7)(y + 5) =0
From the equation of locus for the variable point M, it is clear that the equation does not contain point P and Q. The point P will be included when the radius of circle S1 is zero and consequently Q will be included when the radius of circle S2 is zero. If radius of both the circles is zero then, the circle becomes a straight line, which is not possible.
Therefore, the set E1 does not contain P and Q.
The locus of mid point chords passing through (1,1) is given as,

It is clear from the above equation, that line passing through P (−2, 7 ) and R (1,1) is as follows.
y + 2x − 3=0
The centre of locus of M is given as,
C3 ( 0,1)
The foot of the locus of M on the equation (A) is  which
is also the mid point of chord PR of circle x2 +y2−2y−39 = 0 .
Since, point P is not a part of the locus M then PQ is not the chord of locus of M.

Hence, option (B) and (D) are correct.

*Multiple options can be correct
QUESTION: 39

Let S be the set of all column matrices  such that b1, b2, b3 ∈ R and the system of equations (in real variables)
-x + 2y + 5z = b1
2x - 4y + 3z = b2
x - 2y + 2z = b3
Has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each

Solution:

The condition for at least one solution is as follows.

...(1)
(A) The determinant of the equation is given as,
∆≠ 0
Therefore, the equations have a unique solution.
Hence, option (A) is correct. (B) The determinant of the equation is calculated as,
The value of ∆1 is also zero.
The condition for infinite solution is that ∆2 and ∆3 must be equal to zero.

From the above equation it can be concluded, that the values will not satisfy the equation (1).
Hence, option (B) is wrong. (C) The determinant of the equation is given as, ∆= 0
Therefore, the equations of the plane represents that the planes are parallel to each other.
The relationship between b1,b2 and b3 is given as,

It is clear from the above relationship, that the values will not satisfy the equation (1).
Hence, option (C) is wrong. (D) The determinant of the equation is as follows. ∆≠ 0
Thus, the equations have a unique solution.
Hence, option (D) is correct.
Hence, option (A) and option (D) are correct.

*Multiple options can be correct
QUESTION: 40

Consider two straight lines, each of which is tangent to both the circle x2 +y2 = 1/2 and the parabola y2 = 4x. Let these lines intersect at the point Consider the ellipse whose center is at the origin O(0,0) and whose semi-major axis is OQ. If the length of the minor axis of this ellipse is √2, then which of the following statement(s) is (are) TRUE?

Solution:

(A) The required figure for the problem is given below.

The equation of the common tangent is given as,

Hence, option (A) is correct and option (B) is wrong. (C) The following figure shows the ellipse.

The area of ellipse is calculated below.

Hence, option (C) is correct and option (D) is wrong.

*Multiple options can be correct
QUESTION: 41

Let s,t,r be non-zero complex numbers and L be the set of solutions  of the equation  Then, which of the following statement(s) is (are) TRUE?

Solution:

(A) The given condition is,
...(1)
Take conjugate on both sides as shown below,
...(2)

Subtract equation (1) from (2),

The condition for unique solution is given as,
...(A)
Simplify further.

Hence, option (A) is correct.
(B) The conditions for the lines to coincide is,
...(B)
On comparing equation (A) and equation (B),

It is clear that the lines are either parallel or coincidental. Thus, there is no concrete outcome.
Hence, option (B) is not correct.
(C) It is clear that L is either a single line or it represents a circle.
The intersection of L is atmost 2 .
Hence, option (C) is correct.
(D)
Assume,

The L will have ∞ elements if L has more than 1 element.
Thus, option (D) is correct.

*Multiple options can be correct
QUESTION: 42

be a twice differentiable function such that
then which of the following statement(s) is (are)
TRUE?

Solution:

According to the definition of the function,

Integrate both sides of the above equation,

Hence, option (A) is not correct.
(B) According to the definition of the function.
f (x) = − x sin x
Also,

Hence, option (B) is correct. (C) The function is,

For tan x = − x , there exist α ∈ (0,π) for which f(α) = 0.
Hence, option (C) is correct.
(D) The first order derivative of the function is given as,
f′′(x) = −2 cos x+x sin x

Hence, option (D) is correct.

*Answer can only contain numeric values
QUESTION: 43

The value of the integral

Solution:

The simplified integral is shown below,
...(1)
Substitute,

Substitute the values in the equation (1),

*Answer can only contain numeric values
QUESTION: 44

Let P be a matrix of order 3 × 3 such that all the entries in P are from the set (−1, 0,1). Then, the maximum possible value of the determinant of P is __________.

Solution:

The determinant of P can be written as,

For the value to be equal to the determinant of 6, it is only possible when,

The further conclusion from the values is shown below.

This is not possible. Thus, the maximum value of determinant cannot be 6.
For the value of determinant equal to 5, Similar contradiction will occur as in case when the values of determinant were 6.
It is clear that for value of determinant to be 5, assume one term as zero. But the equation makes two terms zero.

*Answer can only contain numeric values
QUESTION: 45

Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If α is the number of one-one functions from X to Y and β is the number of onto functions from Y to X , then the value of

Solution:

The numbers of one-one function is calculated as,

The number of onto function Y to X is,
The required values is,

*Answer can only contain numeric values
QUESTION: 46

Let f : R → R be a differentiable function with f(0) =0. If y = f(x) satisfies the differential equation  , then the value of

Solution:

The required function is given as,

...(1)
At x = 0 , y = 0 .
Substitute the value in equation (1).
c = 0
Substitute the values.

*Answer can only contain numeric values
QUESTION: 47

If f : R → R be a differentiable function with f (0) = 1 and satisfying the equation f (x + y) = f (x) f' (y) + f' (x) f (y) for all x,y ∈ R. Then the value of loge (f (4)) is ____.

Solution:

According to the definition of function.

Substitute x = 0 in the function.

Substitute y = 0 in the function.

Take log on both sides,

*Answer can only contain numeric values
QUESTION: 48

Let P be a point in the first octant, whose image Q in the plane x + y= 3 (that is, the line segment PQ is perpendicular to the plane x + y= 3 and the mid-point of PQ lies in the plane x + y= 3)  lies on the z-axis. Let the distance of P from the x- axis be 5. If R is the image of P in the xy-plane, then the length of PR is ____.

Solution:

The coordinates for point P and point Q are given as,

The equation of image of Q on the plane is given as,

The coordinates of point Q are (3 − β ,3−α ,γ) .
Since, point Q lies on the z - axis. Therefore,

The coordinates of point P are (3, 3,γ) and the distance of point P from x- axis is 5. Therefore, find the value of γ.

The coordinates of point P are (3, 3, 4) and R are (3, 3, −4) .
Calculate the distance between points P and R.

*Answer can only contain numeric values
QUESTION: 49

Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively, where O (0, 0, 0) is the origin. Let   be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal and then the value of

Solution:

The following figure shows the cube.

The cross product of vector  is calculated as follows,

The cross product of vector is calculated as follows,
The cross product of vector  calculated below,

*Answer can only contain numeric values
QUESTION: 50

Let
Where 10Cr, r ∈ {1, 2, ....,10} denote coefficients. Then, the value of

Solution:

The required value is calculated as,

Divide both sides of the equation by 1430,

QUESTION: 51

(Here, the inverse trigonometric function sin−1 x assumes values in
Let f :E1 → R be the function defined by  and g :E2 → R be the function defined by

The correct option is:

Solution:

According to the definition of the function E1 and E2.

From the condition of function E2.

Hence, the range of function
From the definition of function E1,

Hence, the range of function
Hence, option (A) is correct.

QUESTION: 52

In a high school, a committee has to be formed from a group of 6 boys M1, M2,M3, M4,M5, M6 and 5 girls G1,G2, G3, G4, G5.
(i) Let α1 be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls.
(ii) Let α2 be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls.
(iii) Let α3 be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls.
(iv) Let α4 be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls and such that both M1 and G1 are NOT in the committee together.

The correct option is:

Solution:

The number of ways of selecting 3 boys and 2 girls is calculated as below,

The number of ways of selecting boys and girls equal in number is,

The number of ways of selecting 5 having atleast 2 girls is,

The number of ways of selecting students in such a way that G1 and M2 are always included.
Hence, option (C is correct).

QUESTION: 53

where a >b> 0, be a hyperbola in the xy-plane whose conjugate axis LM subtends an angle of 60° at one of its vertices N. Let the area of the triangle LMN be

The correct option is:

Solution:

The figure below shows the required hyperbola.

The area of ∆LMN is calculated as follows.

Hence, the length of conjugate axis of hyperbola is4.
The value of a is calculated as follows.
The eccentricity of hyperbola is,

Hence, the eccentricity of hyperbola is
The distance between foci of hyperbola is,

Hence, the distance between foci of hyperbola is 8.
The length of latus rectum of hyperbola is,

Hence, the length of latus rectum of hyperbola is
Hence, option (B) is correct.

QUESTION: 54

Let f1 : R → R,  and f4 : R → R, be functions defined by

where the inverse trigonometric function tan−1 x assumes values in
denotes the greatest integer less than or equal to t,

The correct option is:

Solution:

(P)
The definition of function f1 is,

The continuity of function f1 is checked as follows.

It is clear that the function f1 is continuous at x = 0 and NOT differentiable at x = 0 .
Hence, for option (P), (2) is correct.
(Q)
The definition of function f2 is,

The continuity of function f2 is checked as follows.

It is clear from the above expression that the function f2 is NOT continuous at x = 0 .
Hence, for option (Q), (1) is correct.
(R)
The definition of function f3 is,

The continuity of function f2 is checked as follows.

From above expression, it is clear that, as x → 0 then, (x + 2) → 2 and thus, loge (x + 2) → loge2 and this is less than 1.
The range of function f3 can be expressed as,

The definition of function f3 can be written as,

The conclusion from the definition is,

Hence, for option (R), (4) is correct.
(S)
The definition of function f4 is,
The continuity of function f1 is checked as follows.

The definition of function f4 becomes.

It is clear that the function f4 is differentiable at x = 0 and its derivative is NOT continuous at x = 0 .
Hence, option (D) is correct.