JEE Advanced 2015 Paper -2 with Solutions - JEE MCQ

# JEE Advanced 2015 Paper -2 with Solutions - JEE MCQ

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## 60 Questions MCQ Test National Level Test Series for JEE Advanced 2024 - JEE Advanced 2015 Paper -2 with Solutions

JEE Advanced 2015 Paper -2 with Solutions for JEE 2024 is part of National Level Test Series for JEE Advanced 2024 preparation. The JEE Advanced 2015 Paper -2 with Solutions questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced 2015 Paper -2 with Solutions MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced 2015 Paper -2 with Solutions below.
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JEE Advanced 2015 Paper -2 with Solutions - Question 1

### Section 1 Q. No. 1 - 8 Carry 4 marks each The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. Q. An electron in an excited state of Li2+ ion has angular momentum 3h/2π. The de Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is

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JEE Advanced 2015 Paper -2 with Solutions - Question 2

### A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3 from M, the tension in the rod is zero for m = k(M/288). The value of k is

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JEE Advanced 2015 Paper -2 with Solutions - Question 3

### The energy of a system as a function of time t is given as E(t) = A2exp(-αt), where α = 0.2 s-1. The measurement of A has an error of 1.25 %. If the error in the measurement of time is 1.50 %, the percentage error in the value of E(t) at t = 5 s is

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JEE Advanced 2015 Paper -2 with Solutions - Question 4

The densities of two solid spheres A and B of the same radii R vary with radial distance r as ρA(r) =

, respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centres are IA and IB, respectively. If

the value of n is

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JEE Advanced 2015 Paper -2 with Solutions - Question 5

Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, π/3, 2π/3 and π.
When they are superposed, the intensity of the resulting wave is nI0. The value of n is

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First and fourth wave interfere destructively. So from the interference of 2nd and 3rd wave only,

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JEE Advanced 2015 Paper -2 with Solutions - Question 6

For a radioactive material, its activity A and rate of change of its activity R are defined as   and   , where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life  ) and
Q(mean life  2) have the same activity at t = 0. Their rates of change of activities at t = 2 are RP and RQ,
respectively. If  , then the value of n is

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JEE Advanced 2015 Paper -2 with Solutions - Question 7

A monochromatic beam of light is incident at 600 on one face of an equilateral prism of refractive index n and
emerges from the opposite face making an angle θ(n) with the normal (see the figure). For n = √3 the value of θ is 600 and . The value of m is

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JEE Advanced 2015 Paper -2 with Solutions - Question 8

In the following circuit, the current through the resistor R (=2) is I Amperes. The value of I is

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JEE Advanced 2015 Paper -2 with Solutions - Question 9

Section 2

Q. No. 9 -18 carry 4 marks each and 2 marks is deducted for every wrong answer.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct.

Q.

A fission reaction is given by  where x and y are two particles. Considering    to be at rest, the kinetic energies of the products are denoted by KXe, KSr, Kx(2MeV) and Ky(2MeV),

respectively. Let the binding energies per nucleon of  be 7.5 MeV, 8.5 MeV and 8.5 MeV respectively. Considering different conservation laws, the correct option(s) is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 9

Q value of reaction = (140 + 94) × 8.5 – 236 × 7.5 = 219 Mev
So, total kinetic energy of Xe and Sr = 219 – 2 – 2 = 215 Mev
So, by conservation of momentum, energy, mass and charge, only option (A) is correct

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 10

Two spheres P and Q of equal radii have densities ρ1 and ρ2, respectively. The spheres are connected by a massless string and placed in liquids L1 and L2 of densities σ1 and σ2 and viscosities η1 and
η2, respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal velocity    and Q alone in L1 has terminal velocity ,
ρ then

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JEE Advanced 2015 Paper -2 with Solutions - Question 11

In terms of potential difference V, electric current I, permittivity ε0, permeability μ0 and speed of light c,
the dimensionally correct equation(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 12

Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution, a spherical cavity of radius R2, centred at P with distance OP = a = R1 – R2 (see figure) is made. If the electric field inside the cavity at position   then the correct statement(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 13

In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 14

A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own
gravity. If P(r) is the pressure at r(r < R), then the correct option(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 15

parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When
two dielectrics of different relative permittivities (ε1 = 2 and ε2 = 4) are introduced between the two plates
as shown in the figure, the capacitance becomes C2. The ratio C2/C1 is

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JEE Advanced 2015 Paper -2 with Solutions - Question 16

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the
figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state.
The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the
piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct
statement(s) is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 16

Note: A and C will be true after assuming pressure to the right of piston has constant value P1

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JEE Advanced 2015 Paper -2 with Solutions - Question 17

SECTION 3

Q. No.  17-20 carry  4 marks each and 2 marks is deducted for every wrong answer.

This section contains TWO paragraphs

Based on each paragraph, there will be TWO questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct

PARAGRAPH 1

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

Q.

For two structures namely S1 with  n1 =   and n2= 3 / 2, and S2 with n1 = 8/5 and n2 = 7/5 and
taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 18

PARAGRAPH 1

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

Q.

If two structures of same cross-sectional area, but different numerical apertures NA1 and
NA2 (NA2 < NA1 ) are joined longitudinally, the numerical aperture of the combined structure is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 18

For total internal reflection to take place in both structures, the numerical aperture should be the least one for the combined structure & hence, correct option is D.

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 19

PARAGRAPH 2
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are
, w and d, respectively. A uniform magnetic field  is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

Q.

Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are
w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically located on
the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K
and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic
field strength B, the correct statement(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 20

PARAGRAPH 2
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are
, w and d, respectively. A uniform magnetic field  is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

Q.

Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)
with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in
magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed
between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,
the correct option(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 21

SECTION 1

Q. No. 21 -28 carry 4 marks each.

The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

Q.

In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by . For this reaction, the ratio of the rate of change of [H+] to the rate of change of  is

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JEE Advanced 2015 Paper -2 with Solutions - Question 22

The number of hydroxyl group(s) in Q is

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JEE Advanced 2015 Paper -2 with Solutions - Question 23

Among the following, the number of reaction(s) that produce(s) benzaldehyde is

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JEE Advanced 2015 Paper -2 with Solutions - Question 24

In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe–C bond(s) is

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JEE Advanced 2015 Paper -2 with Solutions - Question 25

Among the complex ions,

number of complex
ion(s) that show(s) cis-trans isomerism is

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JEE Advanced 2015 Paper -2 with Solutions - Question 26

Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing
product formed is

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1 mole of B2H6 reacts with 6 mole of MeOH to give 2 moles of B(OMe)3.
3 mole of B2H6 will react with 18 mole of MeOH to give 6 moles of B(OMe)3

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JEE Advanced 2015 Paper -2 with Solutions - Question 27

The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar
conductivity of a solution of a weak acid HY (0.10 M). If  the difference in their pKa values,
pKa (HX) - pKa (HY), is (consider degree of ionization of both acids to be << 1)

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JEE Advanced 2015 Paper -2 with Solutions - Question 28

closed vessel with rigid walls contains 1 mol of  and 1 mol of air at 298 K. Considering complete
decay of  , the ratio of the final pressure to the initial pressure of the system at 298 K is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 28

In conversion of   particles are ejected.
The number of gaseous moles initially = 1 mol
The number of gaseous moles finally = 1 + 8 mol; (1 mol from air and 8 mol of 2He4)
So the ratio = 9/1 = 9

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JEE Advanced 2015 Paper -2 with Solutions - Question 29

SECTION 2

Q. No. 29-36 carry 4 marks each and 2 marks is deducted for every wrong answer.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct

Q.

One mole of a monoatomic real gas satisfies the equation p(V – b) = RT where b is a constant. The
relationship of interatomic potential V(r) and interatomic distance r for the gas is given by

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 29

At large inter-ionic distances (because a → 0) the P.E. would remain constant.
However, when r → 0; repulsion would suddenly increase.

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 30

In the following reactions, the product S is

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JEE Advanced 2015 Paper -2 with Solutions - Question 31

The major product U in the following reactions is

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JEE Advanced 2015 Paper -2 with Solutions - Question 32

In the following reactions, the major product W is

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JEE Advanced 2015 Paper -2 with Solutions - Question 33

The correct statement(s) regarding, (i) HClO, (ii) HClO2, (iii) HClO3 and (iv) HClO4, is (are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 34

The pair(s) of ions where BOTH the ions are precipitated upon passing H2S gas in presence of dilute HCl,
is(are)

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Cu2+ , Pb2+ , Hg2+ , Bi3+ give ppt with H2S in presence of dilute HCl.

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JEE Advanced 2015 Paper -2 with Solutions - Question 35

Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain
termination, respectively, are

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JEE Advanced 2015 Paper -2 with Solutions - Question 36

When O2  is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The TRUE

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 36

* Adsorption of O2 on metal surface is exothermic.
* During electron transfer from metal to O2 electron occupies π*2p orbital of O2.
* Due to electron transfer to O2 the bond order of O2 decreases hence bond length increases

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 37

SECTION 3

Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answer

This section contains TWO paragraphs
Based on each paragraph, there will be TWO questions
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct

PARAGRAPH 1

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC  as measured.

(Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)

Q.

Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 37

HCl + NaOH→NaCl + H2O
n = 100 x1 = 100 m mole = 0.1 mole
Energy evolved due to neutralization of HCl and NaOH = 0.1 x 57 = 5.7 kJ = 5700 Joule
Energy used to increase temperature of solution = 200 x 4.2 x 5.7 = 4788 Joule
Energy used to increase temperature of calorimeter = 5700 – 4788 = 912 Joule
ms.Δt = 912
m.sx5.7 = 912
ms = 160 Joule/oC [Calorimeter constant]
Energy evolved by neutralization of CH3COOH and NaOH
= 200x 4.2x5.6 +160x5.6 = 5600 Joule
So energy used in dissociation of 0.1 mole CH3COOH = 5700 - 5600 = 100 Joule
Enthalpy of dissociation = 1 kJ/mole

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JEE Advanced 2015 Paper -2 with Solutions - Question 38

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC  as measured.

(Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)

Q.

The pH of the solution after Expt. 2 is

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JEE Advanced 2015 Paper -2 with Solutions - Question 39

PARAGRAPH 2
In the following reactions

Q.

Compound X is

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JEE Advanced 2015 Paper -2 with Solutions - Question 40

In the following reactions

Q.

The major compound Y is

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JEE Advanced 2015 Paper -2 with Solutions - Question 41

Section 1

Q. No. 41- 48 carry 4 marks each.

The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

Q.

Suppose that   are three non-coplanar vectors in  R3. Let the components of a vector along  be 4, 3 and 5, respectively. If the components of this vector  and  are x, y and z, respectively, then the value of 2x + y + z is

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JEE Advanced 2015 Paper -2 with Solutions - Question 42

For any integer k, let  . The value of the expression

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JEE Advanced 2015 Paper -2 with Solutions - Question 43

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of
the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130
and 140, then the common difference of this A.P. is

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JEE Advanced 2015 Paper -2 with Solutions - Question 44

The coefficient of x9 in the expansion of  (1 + x) (1 + x2) (1 + x3) ….. (1 + x100) is

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JEE Advanced 2015 Paper -2 with Solutions - Question 45

Suppose that the foci of the ellipse  are (f1, 0) and (f2, 0) where f1 > 0 and f2 < 0. Let P1 and P2
be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and (2f2, 0), respectively. Let T1 be
a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through (f1, 0). The m1
is the slope of T1 and m2 is the slope of T2, then the value of

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JEE Advanced 2015 Paper -2 with Solutions - Question 46

Let m and n be two positive integers greater than 1. If

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JEE Advanced 2015 Paper -2 with Solutions - Question 47

If

where tan-1x takes only principal values, then the value of

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JEE Advanced 2015 Paper -2 with Solutions - Question 48

Let   be a continuous odd function, which vanishes exactly at one point and f(1) =1/2.  Suppose
that  and  then the value of f(1/2) is

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JEE Advanced 2015 Paper -2 with Solutions - Question 49

Section 2

Q. No. 49 - 56 carry 4 marks eah and 2 mark is deducted forr every wrong answer.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct.

Q.

Let   for all  , then the possible values of m and M are

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JEE Advanced 2015 Paper -2 with Solutions - Question 50

Let S be the set of all non-zero real numbers α such that the quadratic equation αx2 - x + α = 0 has two
distinct real roots x1 and x2 satisfying the inequality x1 - x2 < 1. Which of the following intervals is(are) a
subset(s) of S ?

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JEE Advanced 2015 Paper -2 with Solutions - Question 51

If  , where the inverse trigonometric functions take only the principal
values, then the correct option(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 52

Let E1 and E2 be two ellipses whose centers are at the origin. The major axes of E1 and E2 lie along the
x-axis and the y-axis, respectively. Let S be the circle x2 + (y - 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R, respectively. Suppose that PQ = PR =  . If e1 and e2 are the
eccentricities of E1 and E2, respectively, then the correct expression(s) is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 52

For the given line, point of contact for

and for

Point of contact of x + y = 3 and circle is (1, 2)
Also, general point on x + y = 3 can be taken as

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 53

Consider the hyperbola H : x2 - y2 = 1 and a circle S with center N(x2, 0). Suppose that H and S touch each
other at a point P(x1, y1) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the x-axis at
point M. If (l, m) is the centroid of the triangle ΔPMN, then the correct expression(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 54

The option(s) with the values of a and L that satisfy the following equation is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 55

Let   be continuous functions which are twice differentiable on the interval (-1, 2). Let the
values of f and g at the points -1, 0 and 2 be as given in the following table:

In each of the intervals (-1, 0) and (0, 2) the function (f - 3g)" never vanishes. Then the correct
statement(s) is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 55

Let H (x) = f (x) – 3g (x)
H (- 1) = H (0) = H (2) = 3.
Applying Rolle’s Theorem in the interval [- 1, 0]
H'(x) = f'(x) – 3g'(x) = 0 for atleast one c (- 1, 0).
As H"(x) never vanishes in the interval
Exactly one c (- 1, 0) for which H'(x) = 0
Similarly, apply Rolle’s Theorem in the interval [0, 2].
H'(x) = 0 has exactly one solution in (0, 2)

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JEE Advanced 2015 Paper -2 with Solutions - Question 56

Let f (x) = 7tan8x + 7tan6x - 3tan4x - 3tan2x for all  Then the correct expression(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 57

SECTION 3

Q. No. 57 - 60 carry  4 marks each and 2 mark is deducted for every wrong answer.

This section contains TWO paragraphs.
Based on each paragraph, there will be TWO questions
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct

PARAGRAPH 1

Let   be a thrice differentiable function. Suppose that F(1) = 0, F(3) = -4 and F'(x) < 0 for all x  (1/2, 3). Let f (x) = xF(x) for all

Q.

The correct statement(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 58

PARAGRAPH

Let   be a thrice differentiable function. Suppose that F(1) = 0, F(3) = -4 and F'(x) < 0 for all x  (1/2, 3). Let f (x) = xF(x) for all

Q.

If  then the correct expression(s) is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 58

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 59

PARAGRAPH 2

Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II.

Q.

One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 1/3 then the correct option(s) with the possible values of n1, n2, n3 and n4 is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 59

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 60

PARAGRAPH 2

Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II.

Q.

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from
box I, after this transfer, is 1/3,  then the correct option(s) with the possible values of n1 and n2 is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 60

P (Red after Transfer) = P(Red Transfer) . P(Red Transfer in II Case)
+ P (Black Transfer) . P(Red Transfer in II Case)

Of the given options, option C and D satisfy above condition.

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