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JEE Advanced Practice Test- 15 - JEE MCQ


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54 Questions MCQ Test JEE Main & Advanced Mock Test Series - JEE Advanced Practice Test- 15

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*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 1

A particle of charge 'q' and mass 'm' moves rectilinearly under the action of an electric field E = α - βx. Here, α and β are positive constants, and 'x' is the distance from the point where the particle was initially at rest.

Chose the correct option(s).

Detailed Solution for JEE Advanced Practice Test- 15 - Question 1
As we have, ma = qE so acceleration of the particle is a = q(α - β)/m

At mean position a = 0

Hence velocity of the particle is zero at x = 0 and x = 2α/β

We can conclude that particle oscillated between these two limits and have amplitude A = α/β

Acceleration of the particle is maximum at x = 0 and x = 2α/β , amax = qα/m

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 2

In the given circuit, the

Detailed Solution for JEE Advanced Practice Test- 15 - Question 2

Current through the 2Ω resistance

Current through the 2Ω resistance

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 3

Let Z1 and Z2 complex numbers such that Z1 ≠ Z2 and |Z1| = |Z2|. If Z1 has positive real part and Z2 has negative imaginary part, then may be

Detailed Solution for JEE Advanced Practice Test- 15 - Question 3
Given Z1 ≠ Z2 and |Z1| and |Z2|

Z1 has positive real part and

Z2 has negative imaginary part

Then there can be following cases

When a, b are positive real numbers

Also,

When a, b are non negative real numbers

Similarly in all the other cases

is always purely imaginary

is either zero or purely imaginary.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 4

Which of the following statements is/are correct?

Detailed Solution for JEE Advanced Practice Test- 15 - Question 4
If field due to a point charge varies are r-25, then E = cr-25; where c is a constant.

Using Gauss's law, we have

Hence, Gauss law is invalid and option 1 is wrong.

Electric field due to a dipole is not symmetric; hence, a Gaussian surface cannot be used to find the electric field due to a dipole using

Gauss law. Hence, option 2 is wrong.

Electric field on the line joining the charges is opposite in direction if they are of the same sign only. Hence, option 3 is correct.

By definition of potential, work done by external force to move a unit positive charge from point A to point B is given by:

W = q(VB -VA) = VB - VA

Hence, option 4 is correct.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 5

A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :

Detailed Solution for JEE Advanced Practice Test- 15 - Question 5
On the curve

y = x2 ⇒ at x = 1/2

y = 1/4

Hence the coordinate

Differentiating : y = x2 ⇒ vy = 2xvx

Which satisfies the line

4x − 4y − 1= 0 (tangent to the curve)

and magnitude of velocity :

As the line 4x - 4y = 1 does not pass through the origin, therefore The magnitude of angular momentum of particle about origin at that position is 0 is not correct.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 6

A particle is moving in a straight line according to the graph given below, assume that at = 0 particle is located at x = 1 m and have initial velocity 1 m s−1.

then :

Detailed Solution for JEE Advanced Practice Test- 15 - Question 6
In v − x graph,

v = (tan 45)x ⇒ v = x ⇒ dv/dx = 1 ,and we know that acceleration, a = dv/dt = vdv/dx

⇒ a = x × 1 ⇒ a = x, so the graph between acceleration and displacement is,

Now we know that rate of change of linear displacement is called velocity,

at t = 0, x0 = 1 (given) ⇒ x = et

∵ v = x ⇒ x = v ⇒ x = (tan 45) v = x = v,

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 7

In the circuit shown, resistance R = 100 Ω, inductance L= 2π and capacitance C = 8/π µF are connected in series with an AC source of 200 volt and frequency f. If the readings of the hot wire voltmeters V1 and V2 are same, then,

Detailed Solution for JEE Advanced Practice Test- 15 - Question 7

= 1000 volt

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 8

A spherical planet has uniform density ρ. The minimum time period T for a satellite in orbit around it,

Detailed Solution for JEE Advanced Practice Test- 15 - Question 8
For minimum time period radius of orbit equals R of planet

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 9

After the switch is closed, the work done by the cell is constant over the time. C (in μF) is


Detailed Solution for JEE Advanced Practice Test- 15 - Question 9
Let the current through the inductor be I1 and the current through the capacitor be I2.

I1 = Io(I - e-t/(L/R)), where Io = E/R

I2 = Ioe -t/RC

Constant current means current I through the battery is constant.

I = I1 + I2 is constant with time.

This is possible, if

L/R = RC

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 10

A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.

The focal length of the lens is (in cm)


Detailed Solution for JEE Advanced Practice Test- 15 - Question 10
Using

u = -15 m

v = mu

v = (-2) × (-15) = 30 cm

Hence,

f = 10 cm

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 11

An 820-turn wire coil of resistance 24 Ω is placed on a top of a-12,500 turn, 7 cm long solenoid as shown in the figure. Both coil and solenoid have a cross-section area of 10-4 m2. The magnetic field produced by the solenoid at the location of the coil is one half as strong as the field at the centre of the solenoid. The solenoid is connected to a battery of emf 60 V and resistance 14 Ω with the help of a switch. The switch is closed at t = 0.

Determine the average emf (in V) induced in the circuit which opposes the current in the circuit to grow for a time interval 0 to t0.


Detailed Solution for JEE Advanced Practice Test- 15 - Question 11
Average emf induced in interval 0-t0 sec,

Therefore, the average emf induced in the circuit is 37.9 V.

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 12

In YDSE, if incident light consists of two wavelengths λ1 = 4000 Å and λ2 = 5600 Å and is parallel to line SO. The minimum distance yy upon screen, measured from point O, will be where the bright fringe due to two wavelengths coincide is nλ1D/d. Find n.


Detailed Solution for JEE Advanced Practice Test- 15 - Question 12

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 13

An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. If the order of accelerating voltage (for electrons) is required in such a tube is (10x) kV. Then the value of x(integer) is .......


Detailed Solution for JEE Advanced Practice Test- 15 - Question 13
In X - ray tube, accelerating voltage provides the energy to the electrons which produce X - rays. For getting X - rays, photons of 27.61 keV is required that the incident electrons must process kinetic energy at length 27.61 keV.

Energy = eV = E

eV = 27.61 keV

V = 27.61 kV

So, the order of accelerating voltage is 30 kV.

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 14

Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:


Detailed Solution for JEE Advanced Practice Test- 15 - Question 14

or g ∝ ρR

Now escape velocity,

= 3 km s−1

JEE Advanced Practice Test- 15 - Question 15

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

List I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S1 and S2. In each of these cases S1P0 = S2P0, S1P1 - S2P1 = λ /4 and S1P2 - S2P2 = λ/3, where λ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index μ and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by δ (P) and the intensity by I(P). Match each situation given in List I with the statement(s) in List II valid for that situation.

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

When path difference increases from 0 to λ/2 , intensity will decrease from maximum zero. Hence, in this case, I(P2) > I(P1) > I(P0)

In this case I(P1) = 0.

JEE Advanced Practice Test- 15 - Question 16

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

Two transparent media of refractive indices μ1 and μ3 have a solid lens shaped transparent material of refractive index μ2 between them as shown in figures in List II. A ray traversing these media is also shown in the figures. In List I different relationship between μ1 , μ2 and μ3 are given. Match them to the ray diagrams shown in List II.

Detailed Solution for JEE Advanced Practice Test- 15 - Question 16
In the above equation, as sine is an increasing function in 0 to π/2 , if μ2 > μ1, i > r and if μ2 < />1, i < />

Thus light rays going from rarer medium to denser medium will bend towards the normal and vice versa.

It can be seen from the image how the light rays bend when they refract at a curved surface. (Line along the radius is the normal here as shown in the figure)

If the ray goes undeviated, then the refractive indexes are same.

So in 1, μ1 < />2 and μ2 = μ3

So in 2, μ1 > μ2​ and μ2 > μ3

So in 3, μ1 < />2 and μ2 = μ3

So in 4, μ1 > μ2​ and μ2 > μ3

So in 5, μ1 > μ2 and μ2 = μ3

JEE Advanced Practice Test- 15 - Question 17

Four identical uniform rods of mass m = 6 kg each are welded their ends to form a square and then welded to a uniform ring having mass m = 4 kg and radius R = 1m. The system is allowed to roll down he incline of inclination θ = 30

Detailed Solution for JEE Advanced Practice Test- 15 - Question 17

JEE Advanced Practice Test- 15 - Question 18

Four capacitors C1(= 1 μF), C2(= 2 μF), C3(= 3 μF), and C4(=4 μF), are connected in a network as shown in the diagram. The e.m.f of the battery is E = 12 V and its internal resistance is negligible. The keys S1 and S2 can be independently put on or off. Indicate the charge on the capacitors C1, C2, C3 and C4 by q1, q2, q3 and q4 respectively and the potential drops across them by V1, V2, V3 and V4 respectively.

Initially both the keys are open. The the key S1 is closed. Then the charges on the capacitors are

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

With both keys S1 ans S2 initially open, if now S1 is closed, the capacitors C1 and C3 are in series as also the capacitors C2 and C4.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 19

The compound formed upon combustion of sodium metal in the excess of air is

Detailed Solution for JEE Advanced Practice Test- 15 - Question 19

When sodium metal is burnt in excess of air, mainly sodium peroxide (Na2O2), with little sodium oxide (Na2O), is formed.

If the oxidation is carried at high pressure, then sodium superoxide (NaO2) may be formed. The question does not mention the condition of increased pressure.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 20

For the first order reaction , 2N2O6 (g) → 4NO2(g) + 5O2(g) , the

Detailed Solution for JEE Advanced Practice Test- 15 - Question 20
2N2O6 (g) → 4NO2(g) + 5O2(g)

For a first order reaction, [A] = [A]0e-kt , concentration of A decreases with time exponentially.

Half-life for the first order reaction implies that half-life is independent of initial concentration.

From Arhennius Law, k =

As temperature increases value of k increases, as can be seen from the equation.

As the value of k increases we conclude that the half-life decreases.

For 99.6% completion, [A] = 0.4 and [A]0 = 100

Therefore, number of half-lives used

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 21

Select from following molecules or ions, which is/are aromatic.

Detailed Solution for JEE Advanced Practice Test- 15 - Question 21
For aromatic compounds

For a compound to be aromatic, it should follow Huckle's (4n + 2)πerule, i.e., 2, 6, 10, 14....πe

They should show delocalised, conjugate π system with alternate single and double bonds. In addition to this, the compound should be cyclic and planar or almost planar.

The number of πe is even, but not a multiple of 4.

For anti-aromatic compounds:

4nπe, i.e., 4, 8, 12....πe should be delocalized in a cyclic, planar, or almost planar system with alternate single and double bonds.

Planar conjugated carbocyclic polyene, carbocyclic polyene structure is less stable than its open structure.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 22

According to the kinetic theory of gases,

Detailed Solution for JEE Advanced Practice Test- 15 - Question 22
All the above statements are true according to the kinetic theory of gases.

The collisions of the gas molecules are elastic and the energy lost by one molecule is equal to the energy gained by the other.

Hence, at a given temperature, the average energy of the gas is constant.

Heavier molecules transfer more momentum to the walls of the container. But this is not the postulate of the kinetic theory of gases.

Only a small fraction of the molecules possess high velocity (Vr.m.s), whereas the maximum number of molecules possess a lower velocity called the most probable velocity (Vmp).

The molecules move in random, zig-zag straight lines. The motion is referred to as Brownian motion.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 23

Choose the correct reason(s) for the stability of the lyophobic colloidal particles.

Detailed Solution for JEE Advanced Practice Test- 15 - Question 23
Lyophobic colloids are stable due to preferential adsorption of ions common in their lattice from the solution.

Thus, all the colloidal particles acquire a charge and repel the other colloidal particles as they have like charges.

And potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles stablises the charges acquired by adsorption and makes the lyophobic sol stable.

Hence, options (1) and (4) are correct.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 24

Disaccharides are carbohydrates those contain two monosaccharides molecules, each in the hemiacetal form, joined together by the elimination of a water between two hydroxyl groups. Dehydration involves the anomeric carbon of one monosaccharide and may or may not involve the anomeric carbon of the other monosaccharide when the hemiacetal hydroxyl group on an anomeric carbon is involved in a dehydration, the resulting product is an acetal (in common) and glycoside (in carbohydrate).

Sucrose is a non-reducing sugar (while its hydrolysis products glucose and fructose are reducing sugars) because :

Detailed Solution for JEE Advanced Practice Test- 15 - Question 24
Sucrose is having structure as

In above structure there is no free carbonyl centre of an aldehydic group present ; It is a non-reducing sugar.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 25

Pick out the correct statement(s)

Detailed Solution for JEE Advanced Practice Test- 15 - Question 25

2. Decomposition of acidified KMnO4 is catalysed by sunlight.

3.

4. KMnO4 also acts as an oxidizing agent in alkaline medium:

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 26

Select equations having endothermic step :

Detailed Solution for JEE Advanced Practice Test- 15 - Question 26

Second electron gain enthalpy always be +ve

Na+ (g) + Cl(g) ⟶ Na+Cl(s); ΔHL.E. = (−)ve

Lattice energy is -ve when two ion combine & from 1 mole solid

N(g) ⟶ N(g); ΔHgain = (+)Ve

On electron addition in half filled (more stable configuration)energy will be added +ve

Al2+ (g) ⟶ Al3+ (g); ΔHI.E. =(+)Ve

to electron remove energy have to gives [+ve]

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 27

Consider a reaction: aG + bH → products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is


Detailed Solution for JEE Advanced Practice Test- 15 - Question 27
Since the rate doubles when [G] is doubled, the order of the reaction with respect to the reactant G is one.

Since the rate becomes 8 times when both [G] and [H] are doubled, the order of the reaction with respect to the reactant H is two.

Rate of reaction = k[G][H]2

Overall order of the reaction = 1 + 2 = 3

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 28

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is


Detailed Solution for JEE Advanced Practice Test- 15 - Question 28
Mohr's salt is FeSO4(NH4)2SO4.

The equation can be written as:

K2SO4 + 6FeSO4 + 7H2SO4 → Cr2(SO4)3 + 3Fe2(SO4)3 + K2SO4 + 7H2O

One mole of potassium dichromate reacts with 6 moles of ferrous sulphate. Thus, the number of moles of Mohr's salt required per mole of dichromate is 6.

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 29

The number of structural isomers for C6H14 is


Detailed Solution for JEE Advanced Practice Test- 15 - Question 29
Structural isomers of C6H14

CH3 - CH2 - CH2 - CH2 - CH2 - CH3

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 30

The Thorium Series:

Find the number of correct statements about the given series.

  1. It is one of the natural series found on earth.
  2. The number of alpha particles obtained on disintegration of one 90Th232 atom to get the stable product is 6.
  3. The number of beta particles obtained on disintegration of one 90Th228 atom to get the stable product is 2.
  4. 90Th232 and 88Ra228 are isodiaphers.
  5. The nuclear isomers of 83Bi212 gives 84P2124 on beta decay and 81Tl208 on alpha decay.


Detailed Solution for JEE Advanced Practice Test- 15 - Question 30
The correct statements will be-

(A) It is one of the natural series found on earth.

(B)

(C)

(D) Isodiaphers are atoms, having the different atomic number and mass number but have the same difference between a number of neutron and number of a proton.

90Th232 and 88Ra228 are isodiaphers,

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 31

When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol-1), a freezing point depression of 2 K is observed. The van't Hoff factor (i) is

(Round off up to 1 decimal place)


Detailed Solution for JEE Advanced Practice Test- 15 - Question 31
Molar mass of naphthoic acid (C11H8O2) = 172 g mol-1
*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 32

20% of the surface sites adsorbed N2. On heating, N2 gas evolves from sites and is collected at 0.001 atm and 298 K, in a container of the volume 2.46 cm3. The density of surface sites is 6.023 × 1014 cm−2. Area of the surface site is 1000 cm2. Find out the number of surface sites occupied per molecule of N2?


Detailed Solution for JEE Advanced Practice Test- 15 - Question 32
For adsorbed N2​ on surface sites,

Now, to find the moles of N2 gas adsorbed using the ideal gas equation:

Molecules of adsorbed,

Surface site density = 6.023 × 1014 cm−2Area of surface site = 1000 cm3Total surface sites available = Number of sites per cm2 × Area

Total surface sites available = 6.023 × 1014 × 1000 = 6.023 × 1017

Surface sites on which N2 is adsorbed = 20% × Available sites

JEE Advanced Practice Test- 15 - Question 33

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

List - 1 represents ions of 3d series of d-block elements in the modern periodic table.

List - 2 represents values of magnetic moment corresponding to the ions present 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- 15 - Question 33
Magnetic moment where n = no. of unpaired electrons

Ni2+ (Z = 28), electronic configuration: 3d8; no. of unpaired electrons: 2

Mn2+ (Z = 25), electronic configuration: 3d5; no. of unpaired electrons: 5

JEE Advanced Practice Test- 15 - Question 34

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

List - 1 represents ions of 3d series of d-block elements in the modern periodic table.

List - 2 represents values of magnetic moment corresponding to the ions present 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- 15 - Question 34
Magnetic moment where n = no. of unpaired electrons

Ti3+ (Z = 22), electronic configuration: 3d1; no. of unpaired electrons: 1

Cr3+ (Z = 24), electronic configuration: 3d3; no. of unpaired electrons: 3

JEE Advanced Practice Test- 15 - Question 35

On the bases of following information answer the given two questions.

Which amongst the following is correct statement.

Detailed Solution for JEE Advanced Practice Test- 15 - Question 35

Product S is CH3COOH and product T is HCOOH.

Acetic acid react with calcium hydroxide to form calcium acetate which on dry distillation gives acetone and calcium carbonate.

JEE Advanced Practice Test- 15 - Question 36

Valence shell electron-pair repulsion theory predicts the shape of a molecule by considering the most stable configuration of the bond angles in the molecule. The main points of the theory are

(i) Electron pairs in the valence shell of central atom of a molecule, whether bonding or lone pairs are regarded as occupying localised orbitals. These orbitals arrange themselves so as to minimize the mutual electronic repulsions.

(ii) The magnitude of the different types of electronic repulsions follow the order given below:

Lone pair-lone pair> lone pair-bonding pair > bonding pair-bonding pair. These repulsive forces alter the bond angles of the molecule or ion.

(iii) The electron repulsion between two pairs of electrons will be minimum if they stay as far as possible. On this basis, several studied. have been done

Which of the following statements is correct ?

Detailed Solution for JEE Advanced Practice Test- 15 - Question 36
This problem includes the concept of bond angle the basis of Pπ − dπ back bonding and bond stifness, read all statements carefully and answer according to question using above concept.

Due to presence of Pπ − dπ back bonding, NCl3 is sp2 hybridised but due to the absence of vacant d orbital in NH3 it can't form Pπ − dπ back bonding. So bond anlge of NCl3 is greater than NH3. Due to Pπ − dπ back bonding (d-orbitals available in Cl). The lone pair on N not available for exerting repulsion and hence bond angle in NCl3 > NH3.

In BF3 there is Pπ − dπ back bonding from filled orbitals of F creating a partial double bond character making it a poor Lewis acid than BCl3.

Bond multiplicity leads to more repulsion - double bond - lone pair repulsion is too high and hence bond angle increases.

SiCl4 and CCl4 bond angles are both tetrahedral (same).

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 37

Consider the quadratic equation (a + c − b)x2 + 2cx + (b + c − a) = 0, where a, b and c are distinct rational numbers and a + c − b ≠ 0. Then,

Detailed Solution for JEE Advanced Practice Test- 15 - Question 37
For the given quadratic equation (a + c − b)x2 + 2cx + (b + c − a) = 0, discriminant = 4c2 − 4(a + c − b)(b + c − a) = 4(a − b)2, which is greater than equal to zero and is a perfect square of a rational number.

Hence, the roots are real.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 38

Let f(0, 1) → R be defined by f(x) = b-x/1-bx , where b is a constant such that 0 < b < 1.

Detailed Solution for JEE Advanced Practice Test- 15 - Question 38

Therefore, f is always a decreasing function, and thus it is invertible.

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 39

If the adjoint of a 3 × 3 matrix P is then the possible value(s) of the determinant of P is (are)

Detailed Solution for JEE Advanced Practice Test- 15 - Question 39

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 40

The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose

Detailed Solution for JEE Advanced Practice Test- 15 - Question 40
Equation of tangent and normal at point P(at2, 2at) is

ty = x + at2 and y = −tx + 2at + at2

Let the centroid of △PTN be R(h, k).

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 41

Given that the 4th term in the expansion of has the maximum numerical value, then x can lie in the interval(s)

Detailed Solution for JEE Advanced Practice Test- 15 - Question 41
Let tr be the rth term in binomial expansion of

Since t4 is numerically the greatest term, so |t3| < />4| & |t5| < />4|

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 42

If both the roots of the equation ax2 + x + c − a = 0 are imaginary and c > – 1, then

Detailed Solution for JEE Advanced Practice Test- 15 - Question 42

⇒ 2 + 4c > 3a

f(0) = c − a > 0 ⇒ c > a

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 43

If ABC is a triangle and tan A/2, tan B/2, tan C/2 are in H.P., then the minimum value of cot B/2 is equal to

Detailed Solution for JEE Advanced Practice Test- 15 - Question 43
∵ In a triangle ABC, A + B + C = π

Hence equation (1)1 becomes

Equality holds when A = C

*Multiple options can be correct
JEE Advanced Practice Test- 15 - Question 44

If the vectors are coplanar, then -

Detailed Solution for JEE Advanced Practice Test- 15 - Question 44

Also, scalars exist such that

Taking the dot product with

Eliminating l, m, n , we have

Taking dot product with and eliminating l, m, nl, m, n we get the other determinant equal to zero.

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 45

The smallest value of k, for which both the roots of the equation x2 - 8kx + 16(k2 - k + 1) = 0 are real, distinct and have values at least 4 , is


Detailed Solution for JEE Advanced Practice Test- 15 - Question 45
The equation is x2 - 8kx + 16(k2 - k + 1) = 0

We have,

D = 64(k2 - (k2- k + 1)) = 64(k - 1) > 0,

⇒ k > 1

f(4) ≥ 0,

⇒ 16 - 32k + 16(k2 - k + 1)) ≥ 0

Which yield: k2 - 3k + 2 ≥ 0

⇒ k ≤ 1 or k ≥ 2.

Hence, k = 2

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 46

Let f be a positive function. Let where 2k - 1 > 0. Then, l1/l2 is


Detailed Solution for JEE Advanced Practice Test- 15 - Question 46

We have a + b = 1

⇒ I1 = I2 - I1

⇒ 2I1 = I2

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 47

A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then what is the value of k – 20?


Detailed Solution for JEE Advanced Practice Test- 15 - Question 47

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 48

If y = sin x − cos x, f(x) and the value of then the value of K is equal to


Detailed Solution for JEE Advanced Practice Test- 15 - Question 48
∵Differentiation of sinx & cosx after fourth order will give sinx & cosx again

Hence, K = 9.

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 49

If the area bounded by the curve y = sin−1(sin x), y = {[|x|]} from x = 0 to x = π is A sq. units, where [⋅] & {⋅} are the greatest integer function and fractional part function respectively, then the value of 8A/π2 is


Detailed Solution for JEE Advanced Practice Test- 15 - Question 49
y = {[|x|]} = 0

The graph of y = sin−1(sin x)can be shown as following

A = the required area

*Answer can only contain numeric values
JEE Advanced Practice Test- 15 - Question 50

z1 and z2 are two complex numbers such that is unimodular, while z2 is not unimodular, then |z1| must be equal to


Detailed Solution for JEE Advanced Practice Test- 15 - Question 50

JEE Advanced Practice Test- 15 - Question 51

Consider the following linear equations:

ax + by + cz = 0

bx + cy + az = 0

cx + ay + bz = 0

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

Detailed Solution for JEE Advanced Practice Test- 15 - Question 51

(A) For a + b + c ≠ 0

⇒ [(a − b)2 + (b − c)2 + (c − a)2] = 0

⇒ a = b = c ≠ 0

And the given equation represents three identical planes x + y + z = 0.

(B) For |A| = 0

⇒ The solution of equation is not unique.

ax + by + cz = 0

⇒ by + cz = (b + c)x and cy + az = (c + a)x

⇒ (b + c)y + (c + a)z = (b + c)x + (c + a)x

⇒ x = y = z

(C) |A| ≠ 0; the equations have a unique solution x = y = z = 0 and three planes meet at a point.

(D) ⇒ a = b = c = 0; x, y and z can take any values, so the given equation represent the whole of three-dimensional plane.

JEE Advanced Practice Test- 15 - Question 52

In the following question, [x] denotes the greatest integer less than or equal to x.

Which of the following options correctly matches the functions in Column I with their properties listed in Column II?

Detailed Solution for JEE Advanced Practice Test- 15 - Question 52

JEE Advanced Practice Test- 15 - Question 53

If α is a repeated root of ax2 + bx + c = 0, then the value of is equal to

Detailed Solution for JEE Advanced Practice Test- 15 - Question 53
Since, α is a repeated root.

JEE Advanced Practice Test- 15 - Question 54

If f(x) is an even function and satisfies the relation , where g(x) is an odd function, then the value of f(5) is equal to

Detailed Solution for JEE Advanced Practice Test- 15 - Question 54
Given that,

Replacing, in the equation (i)i, we get

Adding the equations (i) & (ii)i & ii, we get

⇒ f(x) is odd function. But it is given that f(x) is even function.

⇒ f(x) = 0 ∀x ∈ R.

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