JEE (Main + Advanced) : LEADER COURSE - ALLEN Career Institute

*1001CJA103517010*

Paper Code ENGLISH

(1001CJA103517010)

CLASSROOM CONTACT PROGRAMME (Academic Session : 2017 - 2018)

JEE (Main + Advanced) : LEADER COURSE PHASE : III, IV, V & (i) Test Type : REVIEW TEST Test Pattern : JEE-Advanced TEST DATE : 05 - 11 - 2017 Time : 3 Hours

PAPER – 2

Maximum Marks : 240

READ THE INSTRUCTIONS CAREFULLY

1.

This sealed booklet is your Question Paper. Do not break the seal till you are told to do so.

2.

Use the Optical Response sheet (ORS) provided separately for answering the questions.

3.

Blank spaces are provided within this booklet for rough work.

4.

Write your name, form number and sign in the space provided on the back cover of this booklet.

5.

After breaking the seal of the booklet, verify that the booklet contains 32 pages and that all the 20 questions in each subject and along with the options are legible. If not, contact the invigilator for replacement of the booklet.

6.

You are allowed to take away the Question Paper at the end of the examination.

OPTICAL RESPONSE SHEET : 7.

The ORS will be collected by the invigilator at the end of the examination.

8.

Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.

9.

Write your name, form number and sign with pen in the space provided for this purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate bubble under each digit of your form number.

DARKENING THE BUBBLES ON THE ORS : 10.

Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.

11.

Darken the bubble

12.

The correct way of darkening a bubble is as :

13.

The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way.

14.

Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase or "un-darken" a darkened bubble.

15.

Take g = 10 m/s2 unless otherwise stated.

COMPLETELY.

Please see the last page of this booklet for rest of the instructions

DO NOT BREAK THE SEALS WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR

GENERAL :

Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2 SOME USEFUL CONSTANTS Atomic No. Atomic masses :

H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17, Br = 35, Xe = 54, Ce = 58, H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127, Xe = 131, Ba=137, Ce = 140,



Boltzmann constant

k = 1.38 × 10–23 JK–1



Coulomb's law constant

1 = 9 ×10 9 4 0

    

Universal gravitational constant Speed of light in vacuum Stefan–Boltzmann constant Wien's displacement law constant Permeability of vacuum

G = 6.67259 × 10–11 N–m2 kg–2 c = 3 × 108 ms–1  = 5.67 × 10–8 Wm–2 –K–4 b = 2.89 × 10–3 m–K µ0 = 4 × 10–7 NA–2



Permittivity of vacuum

0 =



Planck constant

h = 6.63 × 10–34 J–s

1 0 c2

Space for Rough Work

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Leader Course/Phase-III, IV & V/05-11-2017/Paper-2 HAVE CONTROL  HAVE PATIENCE  HAVE CONFIDENCE  100% SUCCESS BEWARE OF NEGATIVE MARKING

PART-1 : PHYSICS SECTION–I(i) : (Maximum Marks : 32) 

This section contains EIGHT questions.



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



For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS



For each question, marks will be awarded in one of the following categories : Full Marks

: +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened.

Zero Marks

: 0 If none of the bubbles is darkened.

Negative Marks : –2 In all other cases. 1.

2.

Heat is added to a substance, but its temperature does not rise. Which of the following statements does not provide the explanation for this observation? (A) The substance must be a gas. (B) The substance must be a non-perfect solid. (C) The substance undergoes a change of phase. (D) The substance has unusual thermal properties. A constant torque C acts for a time t on a lamina rotating in a horizontal plane about a smooth vertical axis. If the moment of inertia of the lamina about that axis is I and the angular velocity changes from 1 to 2 in the time t, (A) the lamina has a constant angular acceleration (B) the work done by the torque is equal to I (22 – 12) (C) the impulse of the torque is Ct, (D) Ct = I (2 – 1) Space for Rough Work

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Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2 3.

As shown below, a boy is using the rope through a fixed pulley to move a box with constant speed v. The kinetic friction coefficient between the box and the ground is  < 1; assume that the fixed pulley is massless and there is no friction between the rope and the fixed pulley. Then, while the box is moving on surface, which of the following statements is/are correct?

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(A) The magnitude of the force on the rope is constant. (B) The magnitude of friction between the ground and the box is decreasing. (C) The magnitude of the normal force of the ground on the box is increasing. (D) The magnitude of the normal force of the ground on the box is decreasing 4.

A block of mass 2m is attached to one end of spring of spring constant k, whose other end is fixed to the ceiling. The block is dropped when spring is at its natural length. (A) The tension in the spring is 4mg when the block comes to rest for the first time

2m k (C) Velocity and acceleration of the block are maximum at the same time (D) Acceleration of block midway between equilibrium and extreme position is g/2 (B) Maximum velocity of the block is g


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Leader Course/Phase-III, IV & V/05-11-2017/Paper-2 5.

A massless bucket is initially at rest next to one end of a chain that lies in a straight line on the floor, as shown in Fig. The chain has uniform mass density  (kg/m). You push on the bucket (so that it gathers up the chain) with the force F(t) that gives the bucket and whatever chain is inside, a constant acceleration a at all times. 't' is time. There is no friction between the bucket and the floor. bucket F(t)

chain a

(A) F(t) at time t is

3 2 2 a t 2

(B) work done by F(t) upto time t is

3 3 4 a t 8

3 2 2 3 3 4 a t (D) work done by F(t) upto time t is a t 4 4 The refractive index of the medium with a certain region, x > 0, y > 0, changes with y. A thin light ray travelling in the x-direction in medium having refractive index 0 = 1 strikes another medium of refractive index at right angles and moves through the medium along a circular arc of radius R as shown in the figure. The material with the greatest known refractive index is diamond, but even the refractive index of this material does not reach the value max = 2.5. It is this limit that sets the maximum angular size of the arc the light ray can cover. Angular size of arc is the angle subtended by the arc at the centre. Which of the following statement(s) is/are CORRECT ? y (C) F(t) at time t is

6.

x

(A) The variation of refractive index with yis given as

R Ry

R 1 3ˆ j is iˆ  2 2 2 (C) The maximum angular size of the arc of light is max. The value of cos max = 2/5 (B) The unit vector in the direction of refracted light at y =

(D) The maximum angular size of the arc of light is max. The value of sin max = 2/5 Space for Rough Work

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A cubical block of mass m and side length b is placed on a smooth floor. A smooth and rigid rod of length L and with negligible mass is leaning against the block. A sphere of mass M is attached to the upper end of the rod. The lower end of the rod is hinged at point O. The rod can rotate freely around the point O in the vertical plane as shown in the figure. Initially the angle between the rod and the floor is  while the system is at rest. Sometime after releasing, the angle between the rod and the floor is Which of the following is/are CORRECT?

 

O

(A) The speed of the block at the instant rod makes angle  with the horizontal is b

2MgL(sin   sin ) mb2  ML2 sin 4 

(B) The speed of the block at the instant rod makes angle  with the horizontal is b sin 


(C) the angular velocity of the rod at the instant it makes angle  with the horizontal is sin 2 


(D) The angular velocity of the rod at the instant it makes angle  with the horizontal is sin 

MgL(sin   sin ) mb2  ML2 sin4 


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Two equal masses are raised at constant velocity by ropes that run over pulleys as shown. Mass B is raised twice as fast as mass A. The magnitudes of the forces are FA and FB, while the power Supplied is respectively PA and PB. Which of the following statements is correct?

B A

(A) FB = FA

(B) PB = PA

(C)PB = 2PA.

(D) FB = 2FA


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Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2     

SECTION–I(ii) : (Maximum Marks : 16) 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. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. Paragraph for Questions 9 and 10 A student wants to find the specific heat of a dark oil. She pours 100 grams of the oil into a glass beaker of negligible mass, and then submerges a small 50 watt light bulb in the oil. Before the student turns on the light bulb, the oil is 20ºC, room temperature. The student then seals the top of the beaker and turns on the light bulb, at time t = 0. She can see some light coming from the bulb, though it looks much dimmer than usual. A small magnetic stirrer ensures that the oil has a uniform temperature throughout the beaker, at all times. A thermometer keeps track of the oil’s temperature. After the light bulb has been on for 40 seconds, the student switches it off. The oil’s temperature at that moment is 30ºC. The student immediately unseals the top of the beaker and removes the light bulb. When the top of the beaker is sealed, assume that negligible heat is absorbed by the beaker or escapes into the environment.

to electrical outlet thermometer light bulb

stirrer

9.

Which of these graphs best shows the temperature of the oil as a function of time?

temp.

(B) 30

10.

temp.

temp.

(A) time (s)

temp.

(C) 30 time (s)

(D) 30

time (s)

30

time (s)

Starting again from room temperature, the experiment is repeated. But this time, the student uses a 100 watt light bulb, and leaves it on for 60 seconds. The temperature of the oil after those 60 seconds will be about (A) 40ºC

(B) 50ºC

(C) 60ºC

(D) 90ºC


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Leader Course/Phase-III, IV & V/05-11-2017/Paper-2 Paragraph for Question No. 11 and 12 One of the more important and spectacular long range projects currently in development by NASA is the design and building of a permanent space station that would orbit the Earth. This space station would serve as both home and laboratory to scientists and civilians. One problem with living in space is the lack of a strong gravitational force. Although people can get used to living in a weightless condition, there are some biological consequences to prolonged sightlessness, inclined calcium depletion from bone and loss of muscle mass. These conditions are believed to be reversible if the exposure to weightlessness is for a short time, but long exposures to weightlessness may have irreversible effects. Consequently, people who spend a long time in space may never be able to come back to Earth ! To avoid this possibly permanent and undesirable effect, one space station design would provide an artificial gravity. A typical design has the space station built like a wagon wheel. Residents would live and work along the inner rim of the station, as schematically depicted in figure 1. Residents would also have workspace closer to the center of the station, but with the same orientation as along the rim. The artificial gravity is provided by a normal force acting on the residents and pointing towards the central axis of the station. The station rotates at a constant angular velocity about this central axis.

Figure-1


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12.

The space station would mimic the Earth’s gravitational pull : (A) by providing a mass equivalent to that of the Earth. (B) by spinning as fast as the Earth (C) by providing a large coefficient of friction (D) by providing a suitable normal force. Suppose two residents of a rotating space station play catch with a ball. They are positioned as shown below :

r1

direction of rotation

r2

Person 1 is located a distance r1 away from the center of the station and Person 2 is located a distance r2 away from the center of the station. The space lab is spinning in the direction indicated. Person 1 throws a ball directly towards Person 2. As viewed by an observer standing outside the space station and at rest with respect to the central axis, the resultant linear velocity of the ball is BEST represented by which of the following arrows ? (A)

(B)

(C)

(D)


SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type No question will be asked in section II and III

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Leader Course/Phase-III, IV & V/05-11-2017/Paper-2 SECTION–IV : (Maximum Marks : 32) 




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



For each question, darken the bubble corresponding to the correct integer in the ORS.



For each question, marks will be awarded in one of the following categories : Full Marks Zero Marks

1.

: +4 If only the bubble corresponding to the correct answer is darkened. : 0 In all other cases.

A uniform L shaped rigid body AOB of mass 1kg, where AO = OB = 10m and AOB is a right angle, is hinged to a smooth joint at Point A of the body and can swing freely in a vertical plane. Initially, the body is released from rest with AB horizontal. The maximum kinetic energy of the body is K Joule.The value of

K (5 5  25)

is

O

A

B

2.

Smooth joint

The figure shows an optical fibre of diameter DE = 1.0 mm is bent into a circular arc with centre C where CD = 1.5 mm. Light ray only enters into the optical fibre at right angle to DE. the minimum refractive index of the optical fibre so that all light enters into the optical fibre undergo total internal reflection is n.The value of 3n is

C

D

E


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Two identical elastic strings AB and BC of natural length a and elastic constant 2mg/a are fastened together at B. Their other ends A and C are fixed to two points 4a apart in a vertical line (A above C). A particle of mass m is attached to B. The height above C at which the 4h is particle rests in equilibrium is h. Then value of a

4.

A mass M hangs from a light rope which passes over a rough cylinder, the coefficient of friction being and the angle of lap being . The least value of F, the tension in the upper part of the rope required to prevent the mass from falling is F0.The value of

F0 is Mge

F

M


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Two particles are connected by a light string of length 2R. The system is draped over cylinder of radius R as shown in the figure, coefficient of static friction being  =

3 . The angle  0 (in degrees) when the system is in equilibrium and about to slide is  0 .the value of 10

6.

The axle of a uniform cylinder with mass m and radius R is connected to a spring with spring constant k, as shown in the Fig. A horizontal board with mass m rests on top of the cylinder, and the board also rests on top of a frictionless support near its left end.The system is slightly displaced from equilibrium. There is no slipping between the cylinder and the board, or between the cylinder and the ground. The angular frequency of the oscillatory motion is p. 11p2m The value of is k

m frictionless

m k


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As shown, a large ball of mass M = m is connected on each end by a light thread of length l to small balls of mass m. Initially the three balls are along the straight line on a smooth surface. The large ball is suddenly given an initial velocity v in the direction perpendicular to the line. mv 2 The tension in the thread at the moment the two small balls meet is T. The value of is T v m

8.

m

M

A thin ring of mass 2m and radius R is pivoted at P on a frictionless table, as shown in figure. A bug of mass m runs along the ring with speed v with respect to the ring. The bug starts from the pivot with the ring at rest. When the bug reaches the diametrically opposite point on the ring (point X), it is moving with velocity v with respect to the table. The value of v/v is X

v

P

g bu


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Leader Course/Phase-III, IV & V/05-11-2017/Paper-2

PART-2 : CHEMISTRY SECTION–I(i) : (Maximum Marks : 32) 












Zero Marks



Select the correct statement(s) :   (A) On decreasing pressure to the equilibrium H2O(s)   H2O(l), more ice will be formed

(B) At equilibrium if Kp0 = 1 then always G0 will be zero   (C) For the reaction 2SO2(g) + O2(g)   2SO3(g), Kp is less than KC at 25ºC.   (D) For reaction N2O4(g)   2NO2(g) ; H = 54 kJ if activation energy of forward reaction

is 57.2 kJ then activation energy for reverse reaction is 3.2 kJ 2.

For an adiabatic irreversible process involving an ideal gas. (A)  U = nCv,m  T

(B) W = nCv,m  T

(C) W = –P ext.(V2–V1)

(D) TV–1 = constant Space for Rough Work

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Salect the incorrect statement (s) (A)  Hatomisation of graphite is equal to  Hf [C(g)] (B) Hcomb of ‘H’ atom is equal to  Hf [H2O(l)] (C) Hf [H2O(l)] is zero (D) Hcomb of graphite is equal  Hf [CO(g)]

4.

Consider the following endothermic reaction ) – 2  Fe(3 (aq.)  Cl (aq .)  FeCl (aq.)

Equilibrium may be disturbed by :

5.

(A) Addition of Fe(NO 3)3

(B) Addition of AgNO3

(C) Change in temperature

(D) Addition of catalyst

Which of the following is/are diamagnetic species and it's HOMO is the antibonding molecular orbital : (A) N2

6.

(B) O2

(C) F2

(D) N 22

Which of the complex/es follow sidgwick rule : (A) Mn2(CO)10

(B) [PtCl3(CH2=CH2)] 

(C) Ferrocene

(D) [Fe(CN)6]4– Space for Rough Work

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Choose the correct option(s) for following sequence : CH3 H

OH

H

H

H2SO4

P (Major)

Br2 CCl4

Q (Major)

CH3

(A) Major product (Q) is meso-2,3-dibromobutane (B) Major product (P) is cis-but-2-ene (C) Major product (Q) is racemic-2,3-dibromobutane (D) Major product (P) is trans-but-2-ene 8.

Following order can be incorrect for : F > Cl > Br > I (A) Nucleophilicity in H2O

(B) Leaving group tendency

(C) Stability of anion

(D) Nucleophilicity in DMSO Space for Rough Work

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Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2 SECTION–I(ii) : (Maximum Marks : 16) 

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.



For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.





Zero Marks


Negative Marks : –2 In all other cases. Paragraph for Questions 9 and 10 The size of the energy gap o between t2g and eg levels can be measured easily by recording UV-visible spectrum of the complex. Colour of complex compound can also be compare with energy gap 0 9.

Which of the following complex is/are NOT coloured : 2

(A)  TiCl 6  10.

(B) [Fe(CN)6]3–

(C) [Zn(H2O)4]2+

(D) [Fe(H2O)5NO]SO4

Which of the following ligand strength order is/are CORRECT according to absorption spectra. 

(A) CO > CN



(B) NO2 > ON O



(C) CN > NO2



(D) NH3 > CN


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Leader Course/Phase-III, IV & V/05-11-2017/Paper-2 Paragraph for Questions 11 and 12 CH3 Cl Cl

11.

12.

CH3

CH3 Br Br

Br

Cl Br

Cl

Br

CH3

CH3

Cl

(P)

(Q)

CH 3

Br

CH3 Cl

Br

Br

Cl CH3

Cl (R)

(S)

Which of the following relation is incorrect ? (A) P & Q  Diastereomers

(B) Q & R  Enantiomers

(C) Q & S  Diastereomers

(D) P & R  Enantiomers

Choose the incorrect option for above compounds ? (A) Q is optically inactive due to internal compensation (B) All P, Q, R & S can show optical isomerism (C) Both R & S contains two fold (C2) axis of symmetry (D) P is asymmetric compound Space for Rough Work


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Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2 SECTION–IV : (Maximum Marks : 32) 










For each question, marks will be awarded in one of the following categories :

1.

Full Marks

: +4 If only the bubble corresponding to the correct answer is darkened.

Zero Marks

: 0 In all other cases.

What is the normal boiling point of mercury (Hg) ? Given : fH0 (Hg, g) = 67 kJ/mol S0 (H,g,l) = 77.4 ; S0 (Hg,g) = 177.4 J/K/mol Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

2.

What is the pH of a solution in which 25 ml of 0.1M NaOH is added to 25 mL of 0.08 M HCl and final solution is diluted to 500 ml Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer. Space for Rough Work

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Total number of boron atom in anionic part of borax which form -coordinate bond is x and total number of boron atom which can form p-p coordinate bond is y then calculate value of x –y

4.

Total number of complex compound which have low spin as well as Diamagnetic. [Fe(CN)6]3–, [Co(H2O)6]3, [NiF6]2–, [Cr(H2O)6]3, [Fe(CN)6]4–, [Co(C2O4)3]3–, [Ni(en)3]2+, [PtCl4]2–

5.

Total number of complex compound which have only two geometrical isomer and does not have any plane of symmetry in any geometrical isomer. [Pt(en)3]4+, [Cr(gly)3]3, [PtBrCl I(Py)], [ZnBrClI(Py)], [PtCl2Br2]2–, [Co(H2O)4Br2]

6.

Total number of meso stereisomers of following compound are : CH3–CH–CH–CH–CH–CH3 OH OH OH OH Space for Rough Work

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How many of the following are more reactive than CH3–CH2–Br towards SN1 reaction ? CH2–Br

(i) CH 3–Br

(ii) CH3–O–CH2–Br (iii)

(iv) CH2=CH–CH2–Br O

H3C

(v) H3C ——C–Br H3C

(vi) CH3–CH2–I

(vii) H–C–CH2–Br

(viii) CH3–NH–CH2–Br

Alc. KOH

8. H3C

Br

How many alkene(s) are produced in above reaction ? Space for Rough Work

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PART-3 : MATHEMATICS SECTION–I(i) : (Maximum Marks : 32) 











: +4 If only the bubble(s) corresponding to all the correct option(s) is (are)

Zero Marks

darkened. : 0 If none of the bubbles is darkened.


Let , ,  are roots of 36x3 – 189x2 + 320x – 175 = 0 such that || < || < || then which of the following option(s) is/are correct ? (A) , ,  are real (B) ,  imaginary and  is real (C) ( – )( – ) is less than 363 – 812 – 58 + 145 (D) ( – )( – ) is greater than 363 – 812 – 58 + 145

2.

Let 0 < a < b such that 2, 2x 

3.

mean of a, b respectively then which of the following option(s) is/are correct ? (A) number of possible values of x is 2 (B) number of value of x is 1 2 2 (C) a + b can be 63 (D) a2 + b2 can be 3 If cos – 4sin = 1,   (0, 2) then sin + 4cos can be (A)

15

(B) 4

7 x , 2 – 5 are harmonic mean, arithmetic mean and geometric 2

(C) –4

(D)  15


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Let ƒ(x) = cosx and g(x) = n|x|. If the ranges of the composition functions fog(x) and gof(x) are R1 and R2 respectively then which of the following option(s) is/are correct ?

 3   (B) R1  sin  :     2 2  1 (C) R2 = {–e : – <  < } (D) R1  R 2  n :    1 e   n 1 n 1 n n Let A n   2 and Bn   2 for n = 1, 2, 3, ....., then which of 2 2 k 0 n  kn  k k 0 k  n  (n  2)k  n  1 (A) R1 = {cos: 0    }

5.

the following is/are true ? (A) An < Bn 6.

(B) An > Bn

(C) A n 

 3

(D) Bn 

 9

 dy  Let ƒ: [0, )  R, y = ƒ(x) is continuous function and (x2  x)    y  1 , ƒ(1) = 0, then select the  dx  correct option(s) (A) ƒ(2) 

1 3

(B) ƒ(3) 

5 2 2

(C) Area bounded by curve y = ƒ(x), x = 0, x = 2 and x axis is

 ƒ(x) dx 0

1

(D) Area bounded by curve y = ƒ(x), x = 0, x = 2 and x-axis is

2

 t2  3t  2 dt 0


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1001CJA103517010


Consider a differentiable function ƒ(x) for x  [1, 5] if |ƒ'(x)| < 2  x  [1, 5] & ƒ(2) = 4 then which of the following options can be true ?

8.

(A) ƒ(5) – ƒ(1) = –6

(B) ƒ(5) + ƒ(1) = 3

(C) ƒ2(5) – ƒ2(1) = –2

(D) ƒ2(5) – ƒ2(1) = –8

Let 1 <  < 2 <  < 3 and ƒ(x) = (x + )(x – ) + 2 then select the correct option(s) : (A) equation ƒ(x) = 0 has real and distinct roots (B) equation 5|ƒ(x)| = 1 has 4 solutions (C) equation 3|ƒ(x)| = 25 can have 4 solutions (D) equation ƒ(|x|) = 1 can have 4 solutions Space for Rough Work

1001CJA103517010

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Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2



SECTION–I(ii) : (Maximum Marks : 16) 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. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.




  

Full Marks

: +4 If only the bubble(s) corresponding to all the correct option(s) is (are)

Zero Marks

darkened. : 0 If none of the bubbles is darkened.

Negative Marks : –2 In all other cases. Paragraph for Questions 9 and 10 36

Let ƒ(x) is continuous function such that

 ƒ(x)dx  12

and ƒ(x + 4) = ƒ(x + 2) – ƒ(x)  x  R.

0

1 2 30

I   (x  x ) [x

40

40

 (1  x) ]dx

0

120

 ƒ(x)dx

9.

is greater than or equal to

0

(A) 24 10.

(B) 40

(C) 36

(D) 48

I is equal to (A)

2 101. 100C30

(B)

2 71. 101C30

(C)

2 101

101.

C30

(D)

2 71.

100

C30


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1001CJA103517010

Leader Course/Phase-III, IV & V/05-11-2017/Paper-2 Paragraph for Questions 11 and 12 (2  x)3 ,  2/3 x , Let ƒ(x)   x 1 2  e , x  e, 

11.

12.

3  x  1 1  x  1 1x2 2x3

x

and g(x)   ƒ(t)dt , x  (1, 3) 0

Select the correct option(s) (A) ƒ(x) has local maxima at x = 1

(B) ƒ(x) has 2 points of local minima

(C) ƒ(x) is decreasing function in (1, 2)

(D) ƒ(x) is concave downward in (–1, 0)

g(x) has (A) local maxima at x = 1 + n2

(B) local minima at x = e

(C) no local maxima

(D) g(x) is concave upward in (2, 3) Space for Rough Work


1001CJA103517010

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Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2 SECTION–IV : (Maximum Marks : 32) 











1.

Full Marks

: +4 If only the bubble corresponding to the correct answer is darkened.

Zero Marks

: 0 In all other cases.

For a point P(x, y) 0  x  3 in x – y plane moves on path y = |kx – 1| + |kx – 2| + kx  k  [0, 1]. If area of region R consisting of all points P lying in the first quadrant of the plane is A when k takes it's all values k  [0, 1], then value of [A] is [where [x] denotes greatest integer less than or equal to x]

2.

Let

1 e5x  2e3x  ex  e6x  1 dx , x > 0 = g(x). If g(2) – g(1) = –An(e2 – e + 1) + B n(e4 – e2 + 1),

where A and B are coprime numbers then minimum value of A + B is Space for Rough Work

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1001CJA103517010


Let ƒ : R  R, ƒ(x) = xn|x| then number of non-negative integral values of n for which ƒ(x) is non twice differentiable function in it's domain, is

4.

Number of ordered pair (x, y) which satisfying system of equations

log x (xy)  log y x 2 y

5.

2log y x

 4y  log 2 8 , is

Let ƒ be a n th order derivable function everywhere, n  4 such that ƒ() = ƒ() = 0, ƒ() = ƒ() = 1, ƒ() = ƒ(µ) = –2 where  <  <  <  < µ < , then minimum number of roots of g(x) = ƒ'(x)[ƒ"'(x) + ƒ""(x)] + ƒ"(x)[ƒ"(x) + ƒ"'(x)] is Space for Rough Work

1001CJA103517010

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Number of real solution(s) of the equation cos–1x + cos–1x2 = /4 is

7.

Consider curve y = sin(x – y), –5  x  6 then number of points on the curve where tangent is perpendicular to line 2x + y = , is

8.

Let Ry denotes set of values of y where y = ƒ(x), x > 0 and ƒ(x)  lim

x n  x n then sum of all n  x n  x  n

elements of set Ry is Space for Rough Work

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1001CJA103517010

Leader Course/Phase-III, IV & V/05-11-2017/Paper-2 Space for Rough Work

1001CJA103517010

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Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2 QUESTION PAPER FORMAT AND MARKING SCHEME : 16. The question paper has three parts : Physics, Chemistry and Mathematics. 17. Each part has two sections as detailed in the following table. Que. Type

Section

No. of Que.

I(i)

One or more correct option(s)

I(ii)

Paragraph Based (One or more correct option(s))

4

IV

Single digit Integer (0-9)

8

8

Category-wise Marks for Each Question Maximum Full Partial Zero Negative Marks of the Marks Marks Marks Marks section +4 0 –2 If only the bubble(s) If none In all corresponding of the other — 32 to all the correct bubbles is cases option(s) is(are) darkened darkened +4 0 –2 If only the bubble(s) If none In all corresponding to all of the other — 16 the correct option(s) bubbles is cases is(are) darkened darkened +4 0 If only the bubble In all corresponding — other — 32 to correct answer cases is darkened

NAME OF THE CANDIDATE ................................................................................................ FORM NO. .............................................

I have read all the instructions and shall abide by them.

I have verified the identity, name and Form number of the candidate, and that question paper and ORS codes are the same.

____________________________

____________________________

Signature of the Candidate

Signature of the Invigilator

Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100

E-32/32

[email protected]

www.allen.ac.in

Your Target is to secure Good Rank in JEE 2018

1001CJA103517010

*1001CJA103517010*

Paper Code (1001CJA103517010)

HINDI

CLASSROOM CONTACT PROGRAMME (Academic Session : 2017 - 2018)

JEE (Main + Advanced) : LEADER COURSE PHASE : III, IV & V Test Type : REVIEW TEST Test Pattern : JEE-Advanced TEST DATE : 05 - 11 - 2017 Time : 3 Hours

PAPER – 2

Maximum Marks : 240

  

2.

(ORS)  

3.



4.



5.

 32  20 

  

1.

 6.



 7.



8.

 

9.

  

 : 10.

  

11.

   

12.

:  

13.



14.

 

15.

g = 10 m/s2  

           

Target : JEE (Main + Advanced) 2018/05-11-2017/Paper-2 SOME USEFUL CONSTANTS Atomic No. Atomic masses :

H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17, Br = 35, Xe = 54, Ce = 58, H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127, Xe = 131, Ba=137, Ce = 140,



Boltzmann constant

k = 1.38 × 10–23 JK–1



Coulomb's law constant

1 = 9 ×10 9 4 0

    

Universal gravitational constant Speed of light in vacuum Stefan–Boltzmann constant Wien's displacement law constant Permeability of vacuum

G = 6.67259 × 10–11 N–m2 kg–2 c = 3 × 108 ms–1  = 5.67 × 10–8 Wm–2 –K–4 b = 2.89 × 10–3 m–K µ0 = 4 × 10–7 NA–2



Permittivity of vacuum

0 =



Planck constant

h = 6.63 × 10–34 J–s

1 0 c2



H-2/32

1001CJA103517010

Leader Course/Phase-III, IV & V/05-11-2017/Paper-2 HAVE CONTROL  HAVE PATIENCE  HAVE CONFIDENCE  100% SUCCESS BEWARE OF NEGATIVE MARKING

-1 :   

–I(i) : ( :    

1.

2.

32)

     (A), (B), (C)  (D)                                                 : +4               : 0            : –2    

                   (A)  (B)  (C)  (D)  C  t    I  t 1  2   (A)  (B)  I (22 – 12)  (C)  Ct  (D) Ct = I (2 – 1)



1001CJA103517010

H-3/32


v    < 1   

//////////////////

////////////////////////////////

////////////////////////////////////////////////////////

 (B)  (C)  (D)   2m  k   (A)  4mg  (A)

4.

2m g (B)   k (C)  (D)  g/2 



H-4/32

1001CJA103517010


  (kg/m)  F(t)    a  't' bucket F(t)

chain a

(A)



t

(B)

3 8

 F(t) t  a3t 4 

3 2 2 3 3 4 a t  (D)  F(t) t  a t  4 4  x > 0, y > 0  y  0 = 1 

(C)  t 6.

3 2

 F(t)  a2 t 2   F(t) 

x-    R    max = 2.5    y

x R (A) y   Ry

(B) y =

R 2

 1 2

3 2

ˆj  iˆ  

(C)

 max  cos max = 2/5 

(D)

max  sin max = 2/5  



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H-5/32


m  b  L     M   O  O       

 

O

2MgL(sin   sin ) (A)    b mb2  ML2 sin4 



2MgL(sin   sin ) b sin  (B)    mb2  ML2 sin 4 



2MgL(sin   sin ) sin 2  (C)    mb2  ML2 sin 4  MgL(sin   sin ) sin  (D)    mb2  ML2 sin4 







H-6/32

1001CJA103517010


B  A     FA  FB  PA  PB  

B A

(A) FB = FA

(B) PB = PA

(C) PB = 2PA.

(D) FB = 2FA



1001CJA103517010

H-7/32


–I(ii) : ( : 16)                     (A), (B), (C) (D)                                           : +4               : 0            : –2     9 10    (dark oil)    100  50  20ºC  t = 0     40  30ºC    to electrical outlet thermometer light bulb

stirrer

9.

 temp.

(B) 30

10.

temp.

temp.

(A) time (s)

temp.

(C) 30 time (s)

(D) 30

time (s)

30

time (s)

 100   60  60  (A) 40ºC

(B) 50ºC

(C) 60ºC

(D) 90ºC



H-8/32

1001CJA103517010


11 12                            (wagon wheel)  1     

Figure-1

1001CJA103517010

H-9/32


12.

 (A)  (B)  (C)  (D)   

r1

direction of rotation

r2

 1  r1  2  r2   1  2    (A)

(B)

(C)

(D)



–II :  & –III :  II III  H-10/32

1001CJA103517010


–IV : ( :    

1.

32)

      0 9                                            :

+4

 :

0

              

 1kg  L  AOB,  AO = OB = 10m AOB   A  K

AB  K     (5 5  25)



O

A

B

2.

Smooth joint

 DE = 1.0 mm  C  CD = 1.5 mm  DE   n  3n 

C

D

E



1001CJA103517010

H-11/32


a  2mg/a  AB BC B    A  C  4a  (A, C  )

4h a

m  B  C h  

 4.

 M    F

0  F  F0    Mge

 F

M



H-12/32

1001CJA103517010


2R  R    = 3  ( ) 

 10

 0 0 

6.

 m  R  k  m    11p2m k

p   

 m frictionless

m k



1001CJA103517010

H-13/32


M = m   l       m   v   mv 2 T   T v m

8.

m

M

 R  2m  P   m  v   X v v/v  X

v

P

g bu



H-14/32

1001CJA103517010


-2 :   

–I(i) : ( :    

1.

32)

     (A), (B), (C)  (D)                                               : +4              : 0          : –2              (A) H2O(s)   H2O(l),

(B)

   Kp0

(C)

2SO2(g) 

(D)

N 2 O 4(g) 

= 1

          

G0    

  + O2(g)   2SO3(g)     2NO 2(g)

 25ºC  KC   K     p

 ; H

= 54 kJ

    

3.2 kJ                     57.2 kJ

2.

(A)  U = nCv,m  T

(B) W = nCv,m  T

(C) W = –P ext.(V2–V1)

(D) TV–1 =





1001CJA103517010

H-15/32


  (A)  H Hf [C(g)]   (B) ‘H’ Hcomb , Hf [H2O(l)] (C)  Hf [H2O(l)]

 

  

Hcomb , Hf [CO(g)]    

(D) 4.

) – 2  Fe(3 (aq.)  Cl (aq .)  FeCl (aq.)

5.

   (A) Fe(NO3)3   (B) AgNO3   (C)   (D)  HOMO              (A) N2

6.

(B) O2

(C) F2

(D) N 22

             (A) Mn2(CO)10 (C)



(B) [PtCl3(CH2=CH2)]  (D) [Fe(CN)6]4–



H-16/32

1001CJA103517010


 CH3 H

OH

H

H

H2SO4

P (Major)

Br2 CCl4

Q (Major)

CH3

8.

(A)

 (Q) ,  -2,3- (B)  (P) ,  -2-

(C)

 (Q) ,  -2,3- (D)  (P) ,  -2-

 F > Cl > Br > I (A) H2O (C)





(B)



(D) DMSO 



1001CJA103517010

H-17/32


–I(ii) : ( : 16)                     (A), (B), (C) (D)                                         : +4              : 0          : –2       9 10   

 t2g  eg  0 UV- 0   9.

        2

10.

(A)  TiCl 6 

(B) [Fe(CN)6]3–

(C) [Zn(H2O)4]2+

(D) [Fe(H2O)5NO]SO4

               

(A) CO > CN 

(C) CN > NO2



(B) NO2 > ON O 

(D) NH3 > CN



H-18/32

1001CJA103517010


11 12    CH3 Cl Cl

11.

12.

CH3

CH3 Br Br

Br

Cl Br

Cl

Br

CH3

CH3

Cl

(P)

(Q)

CH 3

CH3

Br

Cl

Br

Br

Cl CH3

Cl (R)

(S)

 (A) P

Q   

(B) Q

R   

(C) Q

 S  

(D) P

 R  

 (A)

 Q 

(B) P, Q, R

S  

(C) R  S

(fold) (C2)  

(D) P ,

 

–II :  & –III :  II III  1001CJA103517010

H-19/32


–IV    

1.

: ( : 32)

               0 9                                   :

+4

 :

0

              

(Hg)   ?     : fH0 (Hg, g) = 67 kJ/mol S0 (H,g,l) = 77.4 ; S0 (Hg,g) = 177.4 J/K/mol

2.

( )                       25 ml 0.1M NaOH 25 mL 0.08 M HCl          500 ml        pH   ( )                     



H-20/32

1001CJA103517010


               - x         p-p x –y       y  

4.

                [Fe(CN)6]3–, [Co(H2O)6]3, [NiF6]2–, [Cr(H2O)6]3, [Fe(CN)6]4–, [Co(C2O4)3]3–, [Ni(en)3]2+, [PtCl4]2–

5.

                            [Pt(en)3]4+, [Cr(gly)3]3, [PtBrCl I(Py)], [ZnBrClI(Py)], [PtCl2Br2]2–, [Co(H2O)4Br2]

6.

 CH3–CH–CH–CH–CH–CH3 OH OH OH OH



1001CJA103517010

H-21/32


SN1    CH3–CH2–Br           CH2–Br

(i) CH 3–Br

(ii) CH3–O–CH2–Br (iii)

(iv) CH2=CH–CH2–Br O

H3C

(v) H3C ——C–Br H3C

(vi) CH3–CH2–I

(vii) H–C–CH2–Br

(viii) CH3–NH–CH2–Br

Alc. KOH

8. H3C

Br

  

H-22/32

1001CJA103517010


-3 :  

–I(i) : ( :

32)



    



(A),      



              



              

1.

2.

3.

  :

+4

  :

0

(B), (C)

(D)          

                    

  : –2     36x3 – 189x2 + 320x – 175 = 0 , ,   || < || < ||   ?   (A) , ,   (B) ,     (C) ( – )( – )  363 – 812 – 58 + 145  (D) ( – )( – )  363 – 812 – 58 + 145 7

x  0 < a < b  2, 2  , 2x – 5  a, b  2   (A) x  2  (B) x  1  2 2 2 2 (C) a + b  63  (D) a + b  3  cos – 4sin = 1,   (0, 2)  sin + 4cos 

(A)

15

(B) 4

(C) –4

(D)  15



1001CJA103517010

H-23/32


ƒ(x) = cosx  g(x) = n|x| fog(x)  gof(x)  R1  R2     ? (A) R1 = {cos: 0    }

 3   (B) R1  sin  :     2 2 

(C) R2 = {–e : – <  < }

1 (D) R1  R 2  n :    1 e   n 1

5.

n 2 2 k 0 k  n  (n  2)k  n  1

An    n = 1, 2, 3, ..... 

 ? (A) An < Bn 6.

(B) An > Bn

(C) A n 

n 1

n 2 k 0 n  kn  k

Bn   

 3

2

(D) Bn 



 9

 dy    ƒ: [0, )  R, y = ƒ(x)  (x2  x)    y  1 , ƒ(1) = 0   dx   (A) ƒ(2) 

1 3

(B) ƒ(3) 

5 2 2

(C) y = ƒ(x), x = 0, x = 2  x   ƒ(x) dx  0

1

2 (D) y = ƒ(x), x = 0, x = 2  x   t2  3t  2 dt 0





H-24/32

1001CJA103517010


8.

x  [1, 5] ƒ(x)  |ƒ'(x)| < 2  x  [1, 5]  ƒ(2) = 4   ?  (A) ƒ(5) – ƒ(1) = –6

(B) ƒ(5) + ƒ(1) = 3

(C) ƒ2(5) – ƒ2(1) = –2

(D) ƒ2(5) – ƒ2(1) = –8

  1 <  < 2 <  < 3  ƒ(x) = (x + )(x – ) + 2  (A) ƒ(x) = 0  (B)  5|ƒ(x)| = 1 4  (C)  3|ƒ(x)| = 25 4  (D)  ƒ(|x|) = 1

4  

1001CJA103517010

H-25/32


–I(ii) : ( :  

16)

               



    (A), (B), (C) (D)                         



              



  :

+4

  :   :

0 –2

                         9 10    36

ƒ(x)  ƒ(x   ƒ(x)dx  12 

+ 4) = ƒ(x + 2) – ƒ(x)  x  R



0

1

I   (x  x 2 )30 [x 40  (1  x)40 ]dx



0

120

9.

 ƒ(x)dx  0

10.

(A) 24 I  (A)

2 101. 100C30

(B) 40

(B)

(C) 36

2 71. 101C30

(C)

(D) 48

2 101

101.

C30

(D)

2 71.

100

C30



H-26/32

1001CJA103517010


11 12    (2  x)3 ,  2/3 x ,  ƒ(x)   x 1 2  e , x  e, 

3  x  1 1  x  1 1x2 2x3

x

g(x)   ƒ(t)dt , x  (1, 3) 



0

11.

 (A) ƒ(x), x = 1 (B) ƒ(x) 2  (C)  (1, 2) ƒ(x)  (D)  (–1, 0)  ƒ(x) 

12.

g(x) (A) x = 1 + n2 (B) x = e  (C)  (D)  (2, 3)  g(x) 



–II : 

&

–III : 

II III  1001CJA103517010

H-27/32


–IV : ( :

32)



     



0 9               



            



              

1.

  :

+4

  :

0

   

P(x, y) 0  x  3, y = |kx – 1| + |kx – 2| + kx  k  [0, 1]    R   P  A            k, k  [0, 1]  [A] 

x–y

[ [x] 2.

          



] 

e5x  2e3x  ex dx , x > 0 = g(x) e6x  1

g(2) 

– g(1) = –An(e2 – e + 1) +

1 n(e4 – e2 + 1), B

A  B  A + B   

H-28/32

1001CJA103517010


 ƒ : R  R, ƒ(x) = xn|x|  n  ƒ(x)  

4.

 (x, y)  log x (xy)  log y x 2 y

5.

2log y x

 4y  log 2 8



ƒ, n n  4   ƒ() = ƒ() = 0, ƒ() = ƒ() = 1,  ƒ() = ƒ(µ) = –2   <  <  <  < µ <   g(x) = ƒ'(x)[ƒ"'(x) + ƒ""(x)] + ƒ"(x)[ƒ"(x) + ƒ"'(x)] 



1001CJA103517010

H-29/32


cos–1x +

7.

 y = sin(x 2x + y = 

8.

cos–1x2 = /4



– y), –5  x  6





ƒ(x)  lim  Ry, y  y = ƒ(x), x > 0  n 

x n  x n x n  x n

 Ry

 

H-30/32

1001CJA103517010




1001CJA103517010

H-31/32


 16.  17.  

  

I(i)

I(ii)

IV

 

    +4

 8   

   4     (0-9)

  

8

—

+4

 —    +4

  

0

–2

0

–2

  

 

    32      16      0

—

  — 

32

  ................................................................................................    .............................................

            ____________________________

____________________________





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