JEE Advanced 2017 Paper - 2 with Solutions - JEE MCQ

# JEE Advanced 2017 Paper - 2 with Solutions - JEE MCQ

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## 54 Questions MCQ Test National Level Test Series for JEE Advanced 2025 - JEE Advanced 2017 Paper - 2 with Solutions

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*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 1

### A uniform magnetic field B exists in the region between x = 0 and x = 3R/2 (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters region 2 from region 1 at point P1(y = –R). Which of the following options(s) is/are correct ?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 1

(A, B) For the charge +Q to return region 1, the radius of the circular path taken by charge should by 3R/2.
mv² ÷ (3R/2) = QvB
Therefore,
2p / 3R = Q
So,
B = 2p / 3QR
i.e., B should be equal or greater than 2p/2QR
'A' is the correct option.
When B = 8p / 13QR
mv² / r = Qv (8p / 13QR)
Therefore, v = 13R / 8

Also CP2² = CO² + OP2²
= [(5R/8)² + (3R/2)²]²
CP2 = 13R / 8

Thus the particle will enter region 3 through the point P1 on X-axis 'B' is the correct option.
Change in momentum =√2p
Thus, 'C' is incorrect.
Further, mv² / r = qvB
Therefore, r = mv / qB
'D' is incorrect.

JEE Advanced 2017 Paper - 2 with Solutions - Question 2

### A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the sun and the Earth. The Sun is 3 × 105 times heavier than the Earth and is at a distance 2.5 ×104 times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is ve = 11.2 km s–1. The minimum initial velocity (vs) required for the rocket to be able to leave the SunEarth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 2

Given

From energy conservation

where r = distance of rocket from Sun

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JEE Advanced 2017 Paper - 2 with Solutions - Question 3

### Three vectors and  are shown in the figure. Let S be any point on the vector The distance between the points P and S is  The general relation among vectors  and is :

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 3

Let vector from point P to point S be

from triangle rule of vector addition

JEE Advanced 2017 Paper - 2 with Solutions - Question 4

A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is :

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 4

The given points (1, 2, 3, 4, 5, 6) makes 360° angle at 'O'. Hence angle made by vertices 1 & 2 with 'O' is 60°.

Direction of magnetic field at 'O' due to each segment is same. Since it is symmetric star shape, magnitude will also be same.
Magnetic field due to section BC.

JEE Advanced 2017 Paper - 2 with Solutions - Question 5

A photoelectric material having work-function φ0 is illuminated with light of wavelength  The fastest photoelectron has a de-Broglie wavelength λd. A change in wavelength of the incident light by Δλ results in a change Δλd in λd. Then the ratio Δλ/ Δλ is proportional to

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 5

According to photo electric effect equation :

Assuming small changes, differentiating both sides,

JEE Advanced 2017 Paper - 2 with Solutions - Question 6

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT = 0.01 second and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 ms–2and the velocity of sound is 300 ms–1. Then the fractional error in the measurement, δL/L, is closest to

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 6

Total time taken

is also small, so taking binomial approximation

JEE Advanced 2017 Paper - 2 with Solutions - Question 7

Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density r remains uniform throughout the volume. The rate of fractional change in density  is constant. The velocity v of any point on the surface of the expanding sphere is proportional to :

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 7

Density of sphere is

Velocity of any point on the circumfrence V is equal to

(rate of change of radius of outer layer).

JEE Advanced 2017 Paper - 2 with Solutions - Question 8

Consider regular polygons with number of sides n = 3, 4, 5 ..... as shown in the figure. The center of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is Δ. Then Δ depends on n and h as :

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 8

OA = h

Initial height of COM = h
Final height of COM =

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 9

A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is q. Which of the following statements about its motion is/are correct ?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 9

When the bar makes an angle q; the height of its COM (mid point) is

Since force on COM is only along the vertical direction, hence COM is falling vertically downward.
Instantaneous torque about point of contact is

Now;

Path of A is an ellipse.

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 10

Two coherent monochromatic point sources S1 and S2 of wavelength λ = 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is Δθ. Which of the following options is/are correct ?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 10

At point P2; Δ x = d = 1.8 mm = 3000 λ hence a (bright fringe) will be formed at P2

Now,

Δθ increases as θ  decreases At P2, the order of fringe will be maximum.
For total no. of bright fringes d = nλ
⇒  n = 3000
∴ total no. of fringes = 3000

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 11

A source of constant voltage V is connected to a resistance R and two ideal inductors L1 and L2 through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At  t = 0, the switch is closed and current begins to flow. Which of the following options is/ are correct?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 11

Since inductors are connected in parallel

Current through resistor at any time t is given by

I1 + I2 = I ...(i)
L1I1 = L2I2 ...(ii)
From (i) & (ii) we get

(D) value of current is zero at t = 0
value of current is V/R at t = ∞
Hence option (D) is incorrect.

JEE Advanced 2017 Paper - 2 with Solutions - Question 12

A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque τ about an axis normal to the plane of the paper passing through the point

Q. Which of the following options is/are correct ?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 12

(A) is incorrect.

If force is applied normal to surface at point X
τ = Fy R sinθ
Thus τ depends on θ & it is not constant

(B) is incorrect

if force applied tangentially at S

but it will climb as mentioned in question.

If force is applied normal to surface at P then line of action of force will pass from Q & thus τ = 0

(D) is incorrect.

if force is applied at P tangentially the

Constant

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 13

The instantaneous voltages at three terminals marked X, Y and Z are given by

An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be:-

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 13

Potential difference between X & Y = VX – VY
Potential difference between Y & Z = VY – VZ
Phasor of the voltages :

similarly
Also difference is independent of choice of two terminals.

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 14

A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct ?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 14

Every point on circumference of flat surface is at equal distance from point charge

Hence circumference is equipotential.
Flux passing through curved surface = – flux passing through flat surface.

∴ Flux through curved surface =

Note : Flux through surface can be calculated using concept of solid angle.

∴ Solid angle subtended
φ for 4π solid angle
∴ φ for  solid angle

JEE Advanced 2017 Paper - 2 with Solutions - Question 15

PARAGRAPH–1
Consider a simple RC circuit as shown in figure 1.
Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e., charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC.
Process 2 : In a different process the voltage is first set to  and maintained for a charging time T >> RC. Then the voltage is raised to   without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1. These two processes are depicted in Figure 2.

Q. In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are related by :-

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 15

When switch is closed for a very long time capacitor will get fully charged & charge on capacitor will be q = CV
Energy stored in capacitor

Work done by battery (w) = Vq = VCV = CV2
dissipated across resistance ∈D = (work done by battery) - (energy store)
.........(ii)

from (i) & (ii)

D = ∈C

JEE Advanced 2017 Paper - 2 with Solutions - Question 16

PARAGRAPH–1
Consider a simple RC circuit as shown in figure 1.
Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e., charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC.
Process 2 : In a different process the voltage is first set to  and maintained for a charging time T >> RC. Then the voltage is raised to   without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1. These two processes are depicted in Figure 2.

Q. In Process 2, total energy dissipated across the resistance ED is :-

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 16

For process (1)
Charge on capacitor =

energy stored in capacitor =

work done by battery =

Heat loss =

For process (2)
Charge on capacitor =

Extra charge flow through battery =

Work done by battery :

Final energy store in capacitor :

energy store in process 2 :

Heat loss in process (2) = work done by battery in process (2) – energy store in capacitor process (2)

For process (3)
Charge on capacitor = CV0
extra charge flow through battery :

work done by battery in this process :

find energy store in capacitor :

energy stored in this process :

Now total heat loss (ED) :

final energy store in capacitor :

so we can say that ED =

JEE Advanced 2017 Paper - 2 with Solutions - Question 17

PARAGRAPH -2
One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity w0. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is µ and the acceleration due to gravity is g.

Q. The total kinetic energy of the ring is :-

JEE Advanced 2017 Paper - 2 with Solutions - Question 18

PARAGRAPH -2
One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity w0. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is µ and the acceleration due to gravity is g.

Q.

The minimum value of w0 below which the ring will drop down is :-

JEE Advanced 2017 Paper - 2 with Solutions - Question 19

Which of the following combination will produce H2 gas ?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 19

(A) Zn + 2NaOH → Na2ZnO2 + H2

(B) 4Au + 8NaCN + O2 + 2H2O  → 4Na[Au(CN)2] + 4NaOH

(C) Cu + 4HNO3 →  Cu(NO3)2 + 2NO2 + 2H2O

(D)  Formation of passive layer of Fe2O3 on the surface of Fe and NO2 gas is evolved.

JEE Advanced 2017 Paper - 2 with Solutions - Question 20

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol–1. The figures shown below represents plots of vapour pressure (V.P.) versus temperature (T).
[Molecular weight of ethanol is 46 g mol–1]

Among the following, the option representing change in the freezing point is -

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 20

Ethanol should be considered non volatile as per given option

JEE Advanced 2017 Paper - 2 with Solutions - Question 21

The order of the oxidation state of the phosphorus atom in H3PO2 , H3PO4 , H3PO3 and H4P2O6 is

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 21

; oxidation state of P = +5

; oxidation state of P = +4

; oxidation state of P = +4

; oxidation state  of P = +1
Hence Ans (A)

JEE Advanced 2017 Paper - 2 with Solutions - Question 22

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
ΔfG° [C(graphite)] = 0 kJ mol -1
ΔfG° [C(diamond)] = 2.9 kJ mol -1

The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)]  reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is

[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 22

Ans. (A)

C(graphite) → C(diamond) ;

As  dGT = V.dP

P2 = 14501  bar

JEE Advanced 2017 Paper - 2 with Solutions - Question 23

The major product of the following reaction is

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 23

JEE Advanced 2017 Paper - 2 with Solutions - Question 24

The order of basicity among the following compounds is

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 24

Basic strength µ stability of conjugated acid. ∝ + M / +H / +I

(I)
Conjugated acid stabilized by 2 equivalent R.S.
(II)

(III)

(IV)

Conjugated acid stabilized by 3 equivalent R.S.

JEE Advanced 2017 Paper - 2 with Solutions - Question 25

For the following cell :

Zn(s) | ZnSO4 (aq.) || CuSO4 (aq.) | Cu(s)
when the concentration of Z n2+ is 10 times the concentration of Cu2+ , the expression for ΔG ( in J mol–1) is [F is Faraday constant , R is gas constant, T is temperature , Eº(cell) = 1.1V]

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 25

ΔG = ΔG0 + 2.303RT logQ
ΔG = - nFE0 + 2.303RT logQ
Given : Eº = 1.1 V and n = 2

ΔG = -2.2 F + 2.303RT

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 26

In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is(are):

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 26

K = P . A . e-Ea / RT
(A) If P < 1 Aarr. > Aexpt
(D) If P > 1 Aarr. < Aexpt
(C) If P is very small, then catalyst is required to carry out the reaction at measurable rate.

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 27

Compound P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C8H8O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction.

(i)

(ii)

The option(s) with suitable combination of P and R, respectively, is(are)

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 27

Ans. (A,C)

(B) Product of ozonolysis of R is having 9 carbon.

(D) Product of ozonolysis of R is having 9 carbon.

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 28

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 28

Ans. (BC)

Sol.

(Surrounding is unfavourable)
(Surrounding is favourable)

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 29

The option(s) with only amphoteric oxides is (are):

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 29

(C) Cr2O3 , BeO , SnO , SnO2
all are amphoteric oxides
(D) ZnO , Al2O3 , PbO , PbO2
all are amphoteric oxides

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 30

Among the following, the correct statement(s) is are

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 30

Ans. (ACD)
(A)

(D) Lewis acidic strength decreases down the group. The decrease in acid strength occurs because as size increases, the attraction between the incoming electron pair and the nucleus weakens. Hence Lewis acidic strength of BCl3 is more than AlCl3.

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 31

The correct statement(s) about surface properties is (are)

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 31

(A) Emulsion is liquid in liquid type colloid.
(B) For adsorption, ΔH < 0 & ΔS < 0
(C) Smaller the size and less viscous the dispersion medium, more will be the brownian motion.
(D) Higher the TC , greater will be the extent of adsorption.

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 32

For the following compounds, the correct statement(s) with respect of nucleophilic substitution reactions is(are);

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 32

(A) I and II follow SN2 mechanism as they are primary

(B) Reactivity order IV > I > III

(C) I and III follows SN1 mechanism as they form stable carbocation

(D) Compound IV undergoes inversion of configuration.

JEE Advanced 2017 Paper - 2 with Solutions - Question 33

Upon heating KClO3 in the presence of catalytic amount of MnO2, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO3 gives Y and Z.

Q. W and X are, respectively

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 33

W and X are respectively

W = O2 and X = P4O10

JEE Advanced 2017 Paper - 2 with Solutions - Question 34

Upon heating KClO3 in the presence of catalytic amount of MnO2, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO3 gives Y and Z.

Q. Y and Z are , respectively

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 34

Y and Z are respectively

Y = N2O5 and Z = HPO3

JEE Advanced 2017 Paper - 2 with Solutions - Question 35

The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group ]

Q.

The reactions, Q to R and S to S, are -

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 35

JEE Advanced 2017 Paper - 2 with Solutions - Question 36

The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]

Q. The product S is -

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 36

JEE Advanced 2017 Paper - 2 with Solutions - Question 37

Three randomly chosen non negative int egers x, y and z are found to satisfy the equat ion x + y + z = 10. Then the probability that z is even, is

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 37

Let z = 2k, where k = 0, 1, 2, 3, 4, 5

∴ x + y = 10 – 2k

Number of non negative integral solutions

JEE Advanced 2017 Paper - 2 with Solutions - Question 38

Let S = {1, 2, 3,.....,9}. For k = 1,2, ....., 5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1 + N2 + N3 + N5  =

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 38

N1 + N2 + N3 + N5

= Total ways – {when no odd}
Total ways = 9C5 Number of ways when no odd, is zero        (∵ only available even are 2, 4, 6, 8)

∴ Ans : 9C5 – zero = 126

JEE Advanced 2017 Paper - 2 with Solutions - Question 39

If ƒ :R → R is a twice differentiable function such that f''(x) > 0 for all x ∈ R, and  then

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 39

∵f'(x)  is an increasing function
∴ f'(1) > 1

JEE Advanced 2017 Paper - 2 with Solutions - Question 40

If y = y(x) satisfies the differential equation and y(0) = then y(256) =

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 40

JEE Advanced 2017 Paper - 2 with Solutions - Question 41

How many 3 × 3 matrices M with entries from {0,1,2} are there, for which the sum of the diagonal entries of MTM is 5 ?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 41

Let =
∴ tr(MTM) = a2 + b2 + c2 + d2 + c2 + f+ g2 + h2 + i2 = 5,

where entries are {0,1,2} Only two cases are possible.
(I) five entries 1 and other four zero
9C5 × 1
(II) One entry is 2, one entry is 1 and others are 0.
9C2 × 2!
Total = 126 + 72 = 198.

JEE Advanced 2017 Paper - 2 with Solutions - Question 42

Let O be the or igin and let PQR be an arbitrary triangl e. The point S issu ch that Then the triangle PQR has S as its

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 42

Let position vector of

with respect to
Now

Δ Triangle PQR has S as its orthocentre
∴ option (B) is correct.

JEE Advanced 2017 Paper - 2 with Solutions - Question 43

The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 43

The normal vector of required plane is parallel to vector

∴ The equation of required plane passing through (1, 1, 1) will be
–14(x – 1) – 2(y – 1) – 15(z – 1) = 0

⇒ 14x + 2y + 15z = 31
∴ Option (A) is correct

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 44

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 44

(Difference series)
.... (1)

For option (C) :

(on integration)

Hence option (C) is correct.

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 45

If ƒ : R → R is a differentiable function such that f"(x) > 2f(x) for all x ∈ R, and f(0) = 1, then

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 45

Given that,

∴ option (A) is correct

⇒ f(x) is strictly increasing on x ∈ (0, ∞)
⇒ option (C) is correct
As, we have proved above that

⇒ option (D) is incorrect

∴  options (A) and (C) are correct

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 46

If f(x) =  then

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 46

Ans. (B,D)

Expansion of determinant

⇒ more than two solutions

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 47

Let a and b be nonzero real numbers such that 2(cosβ - cosα) + cosα cosβ = 1. Then which of the following is/are true ?

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 47

Ans. (A,C)

We get

Hence (A, C)

JEE Advanced 2017 Paper - 2 with Solutions - Question 48

If g(x) =

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 48

No option matches the result

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 49

If the line x = a divides the area of region R = [(x, y) ∈  x3 < y < x, 0 < x < 1} into two equal parts, then

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 49

Area between y = x3 and y = x in x ∈ (0,1) is

Area of curve linear triangle

*Multiple options can be correct
JEE Advanced 2017 Paper - 2 with Solutions - Question 50

Let f(x) =  Then

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 50

JEE Advanced 2017 Paper - 2 with Solutions - Question 51

Let O be the origin, and    be three unit vectors in the directions of the sides   respectively, of a triangle PQR.

Q.

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 51

= sin R sin (P Q)

JEE Advanced 2017 Paper - 2 with Solutions - Question 52

Let O be the origin, and    be three unit vectors in the directions of the sides   respectively, of a triangle PQR.

Q. If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R + P) is

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 52

–(cos P + cosQ + cosR) ≥ -3/2 as we know cosP + cosQ + cosR will take its maximum value when
P = Q = R = p/3

JEE Advanced 2017 Paper - 2 with Solutions - Question 53

Let p,q be integers and let α, β be the roots of the equation, x2 - x - 1 = 0, where α ≠ β. For n = 0,1,2, ........, let an = pαn + qβn.
FACT : If a and b are rational numbers and   then a = 0 = b.
Q. If a4 = 28, then p + 2q =

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 53

α2 = α + 1 ⇒ α4 = 3 α + 2
∴ α4 = 28 ⇒ pα4 + qβ4 = p(3α + 2) + q(3β + 2) = 28
⇒ p(3α + 2) + q(3 - 3α + 2) = 28
⇒ α(3p - 3q) + 2p + 5q = 28  (αs α ∈ Qc) ⇒ p = q, 2p + 5q = 28 ⇒ p = q = 4
∴ p + 2q = 12

JEE Advanced 2017 Paper - 2 with Solutions - Question 54

Let p,q be integers and let α, β be the roots of the equation, x2 - x - 1 = 0, where α ≠ β. For n = 0,1,2, ........, let an = pαn + qβn.
FACT : If a and b are rational numbers and   then a = 0 = b.

Q. a12 =

Detailed Solution for JEE Advanced 2017 Paper - 2 with Solutions - Question 54

## National Level Test Series for JEE Advanced 2025

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