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JEE Advanced Practice Test- 18 - JEE MCQ


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54 Questions MCQ Test JEE Main & Advanced Mock Test Series - JEE Advanced Practice Test- 18

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*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 1

Directions: The following question has FOUR choices, out of which ONE or MORE are correct.

The electrostatic potential (Φr) of a spherical symmetric system, kept at origin, is shown in the adjacent figure, and given as

Which of the following options is/are correct?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 1

We have

The potential distribution shown in the graph is that of a spherical conductor in which, for the value of r < />0, E = 0.

Hence, the electrostatic energy stored is also zero for r < />0

The entire charge q stays on the surface only, for r < />0 ; E = 0

While for r ≥ R0 ,

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 2

Directions: The following question has FOUR choices, out of which ONE or MORE are correct.

A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s immediately after the collision.

Detailed Solution for JEE Advanced Practice Test- 18 - Question 2

Linear momentum along horizontal axis conserved

Final momentum=

As the particle is moving along the venicaldrection after collision, hence

Pf = 0,

Ring will come to rest after collision

Using conservation of angular momentum about the point doodad of the ring

Indal angular momentum,

-ve sign means direction of ‘L’ is into the plane of paper

+ ve sign means direction of ‘L' is out of the plane of paper

Indal angular momentum,

Final angular momentum about the point of contact

Hence, the ring will rotate anti-clockwise and the velocity of the point of contact is towards right. So, friction will act towards left.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 3

Directions: The following question has FOUR choices, out of which ONE or MORE are correct.

A series R-C circuit is connected to AC voltage source. Consider two cases; (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current IR through the resistor and voltage VC across the capacitor are compared in the two cases. Which of the following is/are true?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 3

Effective voltage across the circuit

,

As C increases, Vc decreases

As CB > CA

So, VCA > VCB

V remains same for both the cases, with the increase in C,

decreases current will increase IRA < />RB

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 4

Directions: The question has four choices, out of which ONE or MORE is correct.

In the given circuit, the AC source has ω = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is/are:

Detailed Solution for JEE Advanced Practice Test- 18 - Question 4
The current through the circuit, I is 0.3 A. The voltage across the 100 Ω resistor

= 10√2 V.

In the upper branch, the net impedance, (C = 100 μF, R = 100 Ω)

Therefore, the net impedance will be:

The current flowing the upper branch is:

which leads by (45°) w.r.t. voltage.

In the lower branch, the net impedance is:

L = 0.5 H, R = 50 Ω

Therefore, the net impedance will be:

The current flowing the lower branch is:

which lags by (45°) w.r.t. Voltage.

Thus, total current I is given by summation of phasors (I1) and (I2) which differ by (90°) in phase.

Hence,

Also, voltage across the 100 Ω resistor

=

Similarly, voltage across the 50 Ω resistor

=

So, options (a) and (c) are correct.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 5

The electron in a hydrogen atom makes a transition n1 → n2, where n1 and n2 are the principal quantum numbers of two states. Assume the Bohr’s model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of n1 and n2 are

Detailed Solution for JEE Advanced Practice Test- 18 - Question 5

Time period, (in nth state)

i.e.

But rn ∝ n2 and vn ∝ 1 / n

Therefore, Tn ∝ n3

Given

Hence, n1 = 2n2

Option that satisfy above condition are

( n1 = 4, n2 = 2 and n1 = 6, n2 = 3 )

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 6

A resistance R of thermal coefficient of resistivity = α is connected in parallel with a resistance = 3R, having thermal coefficient of resistivity = 2α. Find the value of αeff.

Detailed Solution for JEE Advanced Practice Test- 18 - Question 6

For parallel combination

From (1) and (2)

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 7

Two cars A and B are approaching each other as shown in figure.

At t = 0, the velocities of A and B are v1 and v2 and they are at x = 0 and x = x0, respectively. The car A is moving with constant velocity while B is moving with constant acceleration a, whose direction is shown. If Xrel, Vrel and arel are displacement, velocity and acceleration of A w.r.t. B respectively, then mark out the correct options.

Detailed Solution for JEE Advanced Practice Test- 18 - Question 7

a1 = 0

a2 = −a

arel = a1 − a2

= 0 − (−a)

= a

arel is constant w.r.t time.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 8

Suppose the potential energy between electron and proton at a distance r is given by . Application of Bohr's theory to hydrogen atom in this case shows that

Detailed Solution for JEE Advanced Practice Test- 18 - Question 8

By Eqs. (ii) and (iii), we get

Total energy = 1/2(potential energy)

=

Total energy ∝ n6

Total energy ∝ m−3

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 9

For the circuit shown in the figure, find the current (in ampere) through the source.


Detailed Solution for JEE Advanced Practice Test- 18 - Question 9

Peak current through

Peak current through R2

Phase difference between I1 and I2 is π/2

∴ Peak current through the source is

∴ I = 5 A

RMS current = 3.53 A

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 10

Suppose the potential energy between an electron and a proton at a distance r is given by Ke2 / 3r3. Application of Bohr's theory to hydrogen atom in this case shows that energy in the nth orbit is proportional to nx then x=


Detailed Solution for JEE Advanced Practice Test- 18 - Question 10

By (ii)ii and (iii)

Total energy = 1/2 (potential energy)

=

Total energy ∝ n6

Total energy ∝ m−3

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 11

A disc of radius R rotating with an angular velocity ω0 is placed on a rough horizontal surface. If initial velocity of centre of disc is zero, then velocity of centre of disc when it ceases to slip, comes out to be ω0R/k, value of k is _____.


Detailed Solution for JEE Advanced Practice Test- 18 - Question 11

Let v be the velocity of centre of mass of disc.

Condition for pure rolling i.e., when slip ceases is v = Rω

From conservation of angular momentum,

Hence, the value of k = 3.

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 12

A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.


Detailed Solution for JEE Advanced Practice Test- 18 - Question 12

Velocity of sound in the string = v = ; where T is the tension and μ is the mass per unit length.

Distance between successive nodes,

λ/2 = 5 cm

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 13

A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is μ. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 μ, then N is


Detailed Solution for JEE Advanced Practice Test- 18 - Question 13

From the figure:

F1 = mg(sinθ − μcosθ)

F2 = mg(sinθ + μcosθ)

Given, F2 ​= 3F1

∴ sinθ + μcosθ = 3sinθ − 3μcosθ

∴ N = 10 μ = 5 N

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 14

Steel wire of length L' at 40°C is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is 10-5/°C, Young's modulus of steel is 1011 N/m2 and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of 'm' in kg is

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Practice Test- 18 - Question 14

Youngs modulus

We know that

Also

From (i) and (ii)

= 3.14

JEE Advanced Practice Test- 18 - Question 15

Answer the following by appropriately matching the lists based on the information given:

One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the P-V diagram. List II gives the characteristics involved in the cycle. Match them with each of the processes given in List I.

Detailed Solution for JEE Advanced Practice Test- 18 - Question 15

(A) It is an isobaric process.

Volume decreases, so temperature decreases. If temperature decreases, internal energy decreases.

From the first law of thermodynamics,

Q = U + W

Since work done is negative, heat is lost. It is the work done on the gas.

(B) It is an isochoric process.

Pressure decreases, so temperature decreases and hence internal energy decreases. Work done is always zero. Heat is lost by the system.

JEE Advanced Practice Test- 18 - Question 16

Answer the following by appropriately matching the lists based on the information given:

List I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as λf . Match each system with statements given in List II describing the nature and wavelength of the standing waves.

Detailed Solution for JEE Advanced Practice Test- 18 - Question 16

(A) λ1/4 = L

λ1 = 4L

Longitudinal wave

(B) λ1/2 = L

λ1 = 2L

Longitudinal wave

JEE Advanced Practice Test- 18 - Question 17

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.

The speed of the first ball is

Detailed Solution for JEE Advanced Practice Test- 18 - Question 17
Let the speeds of the two balls, 1 and 2, be v1 and v2, respectively. Since the speed of the second ball is half of the first, let

v1 = 2v ...(1)

v2 = v ...(2)

If y1 and y2 are the maximum heights reached by the balls 1 and 2, respectively, then

Since it is given that y1 − y2= 15 m

Substituting in equation

v1 = 20 m s−1

JEE Advanced Practice Test- 18 - Question 18

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.

The time interval between the throw of balls is

Detailed Solution for JEE Advanced Practice Test- 18 - Question 18

Let the speeds of the two balls, 1 and 2, be v1 and v2, respectively. Since the speed of the second ball is half of the first, let

v1 = 2v ...(1)

v2 = v ...(2)

If y1 and y2 are the maximum heights reached by the balls 1 and 2, respectively, then

Since it is given that y1 − y2= 15 m

Substituting in equation

v1 = 20 m s−1

v2 = 10 m s−1

When maximum heights are reached,

Given that both the balls reach the maximum heights at the same time.

If t2 is the time taken by the ball 2 to cover a distance of 5 m, then from

Since t1 (time taken by ball 1 to cover distance of 20 m) is 2 s, time interval between the two throws,

= t1 − t2 = 2 − 1 = 1 s

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 19

Which of the following statements is/are true?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 19

1) When treated with bromine water, phenol gives 2, 4, 6-tribromophenol.

3) On reaction with concentrated sulphuric acid at 373 K, phenol gives 4-hydroxybenzene sulphonic acid.

4)With cold solution containing sodium nitrite and dilute HCl, phenol gives p-nitrosophenol.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 20

The given graph represents the variations in compressibility factor (z) = pV/nRT versus p, for three real gases A, B and C.

Which of the following statements is/are correct?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 20

Only statement (2) is incorrect.

From the graph, it is clear that the value of z decreases with an increase in pressure.

We can explain on the basis of Van der Waals equation as follows:

At a high pressure, when 'p' is large, V will be small and one cannot ignore 'b' in comparison to V.

However, the term a/V2 may be considered negligible in comparison to 'p' in Van der Waals equation.

p(V - b) = RT

pV - pb = RT

Hence, z is greater than one and it increases linearly with pressure.

Therefore, statement (2) is false.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 21

Which of the following statements is/are true about HCOOH?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 21

Presence of an electron repelling alkyl group reduces the acidity of carboxylic acids due to an increase in electron density at the carbonyl carbon atom. Formic acid contains a hydrogen atom instead of an alkyl group. Hence, it is much stronger than any other carboxylic acid.

Thus, option (a) is correct.

It undergoes dehydration to give carbon monoxide on heating with conc. H2SO4.

HCOOH → CO + H2O

So, option (2) is true.

Due to the presence of -CHO group, it has some reducing properties. It can reduce Tollen`s reagent to give metallic silver.

Thus, option (c) is true.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 22

Which of the following is/are correct statement(s) about [Ni(CO)4]?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 22
In [Ni(CO)4], the valence shell electronic configuration of ground state Ni atom is 3d84s2. All of these 10 electrons are pushed into 3d orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three 4p orbitals undergo sp3 hybridization and form bonds with four CO ligands to give Ni(CO)4 with tetrahedral geometry. Since all the electrons are paired, Ni(CO)4 is diamagnetic.

Outer electronic configuration of Ni atom in ground state

Outer electronic configuration of Ni atom in Ni(CO)4

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 23

Which of the following molecules has as O—O bond ?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 23

H2S2O8 ON of S = 16−2 / 2 = 7 >, 6

H2SO5 ON = 7 > 6

Means these compounds must have an peroxide linkage,

Means these compounds does not have peroxide linkage.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 24

Which of the following statements is/are true for P4S3 molecule -

Detailed Solution for JEE Advanced Practice Test- 18 - Question 24

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 25

If the radius of Na+ ion is 95 pm and that of Cl ion is 181 pm, then

Detailed Solution for JEE Advanced Practice Test- 18 - Question 25

which lies in the range 0.414−0.732. Hence, its coordination number is 6 and structure is octahedral.

Length of the unit cell

= 2r(rNa+ + rCl−) = 2(95 + 181) = 552 pm

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 26

Which of the following is/are correct?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 26
Because of the H- bonding NH4+ is more extensively solvated than Na+ , hence NH4+ salts is more soluble than Na+ salts.

The polarising power of Sn4+ is greater than Sn2+, hence SnCl4 is more covalent

The polarisibility of Cl- is greater than F- because of greater size.

Formal charge

=

In SO2 , formal charge of S = 6 − 2 − 1/2 × 4 = 2

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 27

How many gram equivalents are there per mole of H2S in its oxidation to SO2?


Detailed Solution for JEE Advanced Practice Test- 18 - Question 27
Problem is based on concept of oxidation and change in oxidation number. While solving this problem, students are advised to go through calculation of change in oxidation number, then to calculate number of moles of H2O used.

Change in oxidation number = +4 − (−2) = +6

∴ 1 mole H2O = 6H2O = 6 equivalents of H2O

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 28

No. of carbons in is X. Then, X/3 is


Detailed Solution for JEE Advanced Practice Test- 18 - Question 28
It is caprolactam, which on heating polymerises to give polyamide Nylon -6. One monomeric unit of Nylon-6 contain 6 carbon so 2-unit will have 12 carbons.

X = 12

12/3 = 4

x/3 = 4

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 29

How many of the following give pink precipitates with NaOH (Precipitate may or may not dissolve in excess of NaOH )?


Detailed Solution for JEE Advanced Practice Test- 18 - Question 29

Co2+, CoOH2 is pink precipitate.

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 30

In a mixture of H-He+ gas (He+ is singly ionised He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He+ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

The quantum number n of the state finally populated in He+ ions is


Detailed Solution for JEE Advanced Practice Test- 18 - Question 30

Energy Level Diagram

In , excitation energy = 10.2 eV

When H atoms transfer their excitation energy (10.2 eV) to the electron in He, it is further excited to say n = n2.

Thus, 10.2 = 13.6 × 22 × (1/22 - 1/n22)

∴ n2 = 4

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 31

EDTA4– is ethylenediaminetetraacetate ion. What is the total number of N–Co–O bond angles in [Co(EDTA)]1– complex ion?


Detailed Solution for JEE Advanced Practice Test- 18 - Question 31

Total No. of N-Co-O bond angle is 8.

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 32

Directions: Read the paragraph and answer the following question.

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Applications of colligative properties are very useful in day to day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.

A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.

Given: Freezing point depression constant of water (Kwater) = 1.86 K kg mol-1

Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1

Boiling point elevation constant of water = 0.52 K kg mol-1

Boiling point elevation constant of ethanol = 1.2 K kg mol-1

Standard freezing point of water = 273 K

Standard freezing point of ethanol = 155.7K

Standard boiling point of water = 373 K

Standard boiling point of ethanol = 351.5 K

Vapour pressure of pure water = 32.8 mm Hg

Vapour pressure of pure ethanol = 40 mm Hg

Molecular weight of water = 18 g mol-1

Molecular weight of ethanol = 46 g mol-1

In answering the following question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.


Detailed Solution for JEE Advanced Practice Test- 18 - Question 32

ΔTb = Kb.molality

= 0.52 x 6.1728

(T0 - TB0) = 3.20

Tb = 373 + 3.20

= 376.2 K

JEE Advanced Practice Test- 18 - Question 33

Arrange the following by appropriately matching the lists based on the information given in the paragraph.

List - 1 represents compounds of Xe of group 18 in the modern periodic table.

List - 2 represents the corresponding shapes of the compounds given in List - 1.

Which of the following options has the correct combination considering List - 1 and List - 2?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 33

JEE Advanced Practice Test- 18 - Question 34

Arrange the following by appropriately matching the lists based on the information given in the paragraph.

List - 1 represents coordination compounds.

List - 2 represents shapes/geometries and magnetic characters of the coordination compounds given in List - 1.

Which of the following options has the correct combination considering List - 1 and List - 2?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 34

[Ni(CI)4]2-

Hybridisation: sp3

Tetrahedral, paramagnetic

[NiCN)4]2-

Hybridisation: dsp2

Square planar, diamagnetic

JEE Advanced Practice Test- 18 - Question 35

A definite amount of reducing agent is oxidised by 20 mL of 1 M KMnO4 in acid medium, then the same amount of reducing agent is oxidised to the same state by how many mL of 1 M KMnO4 in neutral medium itself changing to Mn4+ state?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 35

∴ Milliequivalents of KMnO4 in acid medium = Milliequivalents of KMnO4 in neutral medium.

1 × 5 × 20 = 1 × 3 × V

∴ V = 33.3mL

JEE Advanced Practice Test- 18 - Question 36

Name the end product in the following series of reaction.

Detailed Solution for JEE Advanced Practice Test- 18 - Question 36

P4O10 is a powerful dehydrating agent.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 37

Directions: The following question has four choices, out of which ONE or MORE can be correct.

Let z1 and z2 be two distinct complex numbers and let z = (1 - t) z1 + tz2 for some real number t with 0 < t < 1. If Arg(w) denotes the principal argument of a non-zero complex number w, then:

Detailed Solution for JEE Advanced Practice Test- 18 - Question 37

Given z = (1 -t)z1 + tz2

AP + PB = AB

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 38

If eccentricity 'e' of conic satisfies the equation , the curve may be

Detailed Solution for JEE Advanced Practice Test- 18 - Question 38
The given equation is

∴ The given equation becomes

And

Equation (1)

From equations (3) and (4), we get

∴ The conic is an ellipse or a parabola.

So, this option is correct.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 39

Let l, m and n be three consecutive natural numbers (l > m > n). If the angle A of △ABC be given by and the perimetre of the triangle is 12 units, then the area of the triangle ABC is

Detailed Solution for JEE Advanced Practice Test- 18 - Question 39

(a2 + b2 + c2)(l2 + m2 + n2) - (al + bm + cn)2

= (am - bl)2 + (bn - cm)2 + (cl - an)2 > 0

As, l = m + 1 and n = m - 1

⇒ a + b + c = λ(3 m)

⇒ λm = 4

⇒ a = λ(m + 1) and b = 4

c = λ(m - 1) = 4 - λ

a = 4 + λ

As sin A = 1

⇒ λ = 1

Hence, a = 5, b = 4 and c = 3

Area of △ABC

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 40

Directions: The question has four choices, out of which ONE or MORE may be correct.

Let E and F be two independent events. The probability that exactly one of them occurs is 11/25 and the probability of none of them occurring is 2/25. If P(T) denotes the probability of occurrence of the event T, then

Detailed Solution for JEE Advanced Practice Test- 18 - Question 40

Given, P(E, F') or P(F, E') = 11/25

⇒ P(E) P(F') + P(F) P(E') = 11/25

Let P(E) = x, P(F) = y

⇒ x(1 - y) + y(1 - x) = 11/25 ...(1)

P(E', F') = 11/25

⇒ P(E')P(F') = 2/25

Or (1 - x)(1 - y) = 2/25 ...(2)

Solving (1) and (2), we get

x = 3/5 , y = 4/5 or x = 4/5 , y = 3/5

P(E) = 3/5, P(F) = 4/5 or P(E) = 4/5 , P(F) = 3/5

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 41

The function f(x) = xex(1−x) is strictly increasing for every point in

Detailed Solution for JEE Advanced Practice Test- 18 - Question 41
Given function is

f(x) = xex(1−x)

On differentiating both sides w.r.t. x, we get

Thus, f is strictly increasing for every point in the interval (−1/2, 1)

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 42

A fair coin is tossed repeatedly between person A and B, they are called alternately for winning a prize of Rs. 30. The person getting Head will be the winner, and game will continue till a person wins. A starts the call. If expectation of A means Rs. 30×Probability of A to win, then the expectation of

Detailed Solution for JEE Advanced Practice Test- 18 - Question 42

The probability of A winning

⇒ The probability of BB winning the prize is ⅓

∴ Expectation of A = 30 × 2/3 = Rs. 20

and the expectation of B = 30 × 1/3 = Rs. 10

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 43

If one of the vertices of the triangle of maximum possible area that can be inscribed in the curve |z − 2i| = 2, is 2 + 2i, then the remaining two vertices are given by

Detailed Solution for JEE Advanced Practice Test- 18 - Question 43

Let us assume that z1 = 2 + 2i & z2 & z3 are the remaining two vertices of the inscribed triangle. The curve |z − 2i| = 2 obviously represents a circle.

For the fixed circle, the triangle with maximum possible area should be an equilateral triangle. Hence, the inscribed triangle is an equilateral triangle.

Let us assume that z0 represents the Centroid/circumcentre/orthocentre/incentre of inscribed triangle.

Applying rotation about the point z0. we get

And

∴ Options (−1 + i(2 + √3) and (−1 + i(2 - √3) are correct.

*Multiple options can be correct
JEE Advanced Practice Test- 18 - Question 44

Directions: The question has four choices, out of which ONE or MORE may be correct.

If then

Detailed Solution for JEE Advanced Practice Test- 18 - Question 44

We have,

Clearly f(x) is continuous at x = -π/2 and f(x) is not differentiable at x = 0.

Also, f(x) is differentiable at x = -3/2

Now

Thus, f(x) is differentiable at x = 1.

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 45

Let ABCD be a quadrilateral with area 18, with side AB parallel to the side CD and AB = 2CD.

Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the sides, then its radius is


Detailed Solution for JEE Advanced Practice Test- 18 - Question 45

9a2 + b2 - 6ab = a2 + b2

8a2 = 6ab

4a = 3b

Also, ab + 1/2ab = 3/2ab = 18

⇒ ab = 12

a = 3, b = 4

Radius = 2

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 46

The total number of local maxima and local minima of the function


Detailed Solution for JEE Advanced Practice Test- 18 - Question 46

Local maximum at x = -1

Local minimum at x = 0

Hence, the total number of local maxima and local minima of the function is 2.

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 47

Let A, B, C be three complex numbers as defined below:

A = {z : Im z ≥ 1}

B = {z : |z - 2 - i| = 3}

C = {z : Re((1 - i)z) = 3√2}

The number of elements in the set A ∩ B ∩ C is


Detailed Solution for JEE Advanced Practice Test- 18 - Question 47

A = |z : |mz ≥ 1|

B = {z : |z - 2 - i| = 3 }

C = {z : Re ((1 - i)2 = √2}

Let z = x + iy

Number of element in set A ∩ B ∩ C from the graph = 1

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 48

If


Detailed Solution for JEE Advanced Practice Test- 18 - Question 48

Given:

Applying L-hospital rule, we get

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 49

Let then the value of is


Detailed Solution for JEE Advanced Practice Test- 18 - Question 49

Given:

Therefore

= 7

*Answer can only contain numeric values
JEE Advanced Practice Test- 18 - Question 50

The number of rational terms in is


Detailed Solution for JEE Advanced Practice Test- 18 - Question 50

Given the term as

Using multinomial theorem for the term

where p + q + r = 10

General term in

Where α + β + γ = 10

For rational terms α = 0, 2, 4, 6, 8,10; β = 0, 3, 6, 9 and γ = 0, 6

Since α and γ are even numbers and α + β + γ = 10

∴β should also be an even number,

So the possible values of β = 0, 6

Possible values of α, β,γ are

α = 4, β = 6, γ = 0; α = 4, β = 0, γ = 6; α = 10, β = 0, γ = 0

So the number of rational terms for the given term is 33.

JEE Advanced Practice Test- 18 - Question 51

Let (x, y) be such that sin-1(ax) + cos-1(y) + cos-1(bxy) = π/2

Which of the following options correctly matches the statements in Column I with the statements in Column II?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 51

B → q

a = 1, b = 1

a = 1, b = 2

C → p

a = 2, b = 2

JEE Advanced Practice Test- 18 - Question 52

Consider the lines given by:

L1 : x + 3y - 5 = 0, L2 : 3x - ky - 1 = 0, L3 : 5x + 2y - 12 = 0

Which of the following options correctly matches the statements in Column I with the statements in Column II?

Detailed Solution for JEE Advanced Practice Test- 18 - Question 52

A → s

B → p, q

C → r

D → p, q, s

(A) Solving equation L1 and L3,

∴ x = 2, y = 1

L1, L2, L3 are concurrent, if point (2, 1) lies on L2

∴ 6 - k - 1 = 0

⇒ k = 5

(B) Either L1 is parallel to L2, or L3 is parallel to L2, then

⇒ k = -9 or k = -6/5

(C) L1, L2, L3 form a triangle, if they are not concurrent, or not parallel.

(D) L1, L2, L3 do not form a triangle, if

k = 5, -9, -6/5

JEE Advanced Practice Test- 18 - Question 53

Consider the equation

The number of real values of x for which the equation has solution is

Detailed Solution for JEE Advanced Practice Test- 18 - Question 53

Given that

Right hand side of the equation shows that x ≠ 0.

Integrating L.H.S., we get

But , which is possible only for x=6.

Thus, number of real values of xx for which equation has solution is 1.

JEE Advanced Practice Test- 18 - Question 54

Consider the equation

If x takes the values for which the equation has a solution, then the number of values of a∈[0,100] is

Detailed Solution for JEE Advanced Practice Test- 18 - Question 54

We have got x = 6 ⇒ sin(a/6) = 1

Hence, x = 6, a = 12nπ + 3π, n ∈ z

For a ∈ [0, 100], there are exactly 3 values of aa i.e. a = 3π, 15π and 27π.

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