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PHYSICS, CHEMISTRY & MATHEMATICS

SET - A

Time Allotted: 3 Hours

Maximum Marks: 210

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

You are not allowed to leave the Examination Hall before the end of the test.

INSTRUCTIONS

Caution: Question Paper CODE as given above MUST be correctly marked in the answer OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong results.

A. General Instructions

1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.

2. This question paper contains Three Sections.

3. Section-I is Physics, Section-II is Chemistry and Section-III is Mathematics.

4. Each Section is further divided into Two Parts: Part-A & C

5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be provided for rough work.

6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices, in any form, are not allowed.

B. Filling of OMR Sheet

1. Ensure matching of OMR sheet with the Question paper before you start marking your answers on OMR sheet.

2. On the OMR sheet, darken the appropriate bubble with HB pencil for each character of your Enrolment No. and write in ink your Name, Test Centre and other details at the designated places.

3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Two Parts. (i) Part-A (01 – 10) contains 10 multiple choice questions which have one or more than one correct

answer. Each question carries +4 marks and – 2 mark for wrong answer. (ii) Part-C (01 – 10) contains 10 Numerical based questions with single digit integer as answer, ranging

from 0 to 9 (both inclusive) and each question carries +3 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate :____________________________________________

Batch :____________________ Date of Examination :___________________

Enrolment Number :_______________________________________________

BA

TC

HE

S –

17

19

FIITJEE

CPT - 1

CODE:110662

PAPER - 1

PT-1-(1719) (Paper-1) Set-A – PCM-2

FIITJEE Ltd., Bhopal Centre, 48-A, Gurukripa Plaza, Zone – 2, M.P. Nagar, Ph. 0755-42553355, 4253455 website: www.fiitjee.com

Useful Data Chemistry:

Gas Constant R = 8.314 J K1 mol1

= 0.0821 Lit atm K1 mol1

= 1.987 2 Cal K1 mol1

Avogadro's Number Na = 6.023 1023

Planck’s Constant h = 6.626 10–34 Js

= 6.25 x 10-27 erg.s

1 Faraday = 96500 Coulomb

1 calorie = 4.2 Joule

1 amu = 1.66 x 10-27 kg

1 eV = 1.6 x 10-19 J

Atomic No : H=1, D=1, Li=3, Na=11, K=19, Rb=37, Cs=55, F=9, Ca=20, He=2, O=8,

Au=79. V = 23, Cr = 24, Co = 27, Xe = 54

Atomic Masses: He=4, Mg=24, C=12, O=16, N=14, P=31, Ar=40, Cr=52, Br=80, Cu=63.5,

Fe=56, Mn=55,

Pb=207,

Au=197, Ag=108, F=19, H=1, Cl=35.5, Sn=118.6

Useful Data Physics:

Acceleration due to gravity g = 10 2m/ s

(JEE ADV PAPER – 1 SET - A) 3


SSSEEECCCTTTIIIOOONNN---111 ::: PPPHHHYYYSSSIIICCCSSS

(Multi Correct Choice Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.

1. A particle of mass m and charge q is projected in a region where an electric field is existing and given

by 0ˆE E i , with a velocity

0ˆv j from the origin at time t = 0, then choose the correct statements

(assuming 2 2

0 0 0m v 2qE mx ).

(a) radius of curvature of the particle when its x-coordinate becomes x0 is 2x0.

(b) radius of curvature of the particle when its x-coordinate becomes x0 is 4 2 x0.

(c) speed of the particle when its x-coordinate becomes x0 is 02v .

(d) speed of the particle when its x-coordinate becomes x0 is 02v .

2. Two cells of unequal emfs 1E and 2E and internal resistances 1r and 2r are joined as shown in figure.

pV and QV are the potential at P and Q respectively.

(a) The potential difference across both the cells will be equal.

(b) One of the cell, will supply energy to the other cell

(c) The potential difference across one of the cells will be greater than its emf.

(d) 1 2 2 1P Q

1 2

E r E rV v

r r

3. Three wires are carrying same constant current i in different directions. Four loops enclosing the wires

in different manners are shown. The direction of d is shown in the figure.

(a) Along closed Loop-1 0B.d i

(b) Along closed Loop-2 0B.d i

(c) Along closed Loop-3 B.d 0

(d) The net work done by the magnetic force to

move a unit charge along the loop is zero.

4. In the circuit shown in the figure 1 2C C C and capacitor 1C is having initial charge Q. The switch is

closed at t 0 . Which of the following options are correct.

(a) The current at time t through the circuit will be 2t /RCQ

eRC

(b) The charge on capacitor 1C at time t will be 2t /RCQ1 e

2

(c) The charge on the capacitor 2C at time t will be 2t /RCQ1 e

2

(d) Total heat loss in the process of charge transfer will be 2Q

8C

5. Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are

ˆ ˆv 3i 4j and ˆ ˆa 2i xj . Select the correct alternative (s)

(a) x = -1.5

(b) x = 3

(c) magnetic field is along z-direction

(d) kinetic energy of the particle is constant

1 1E ,r

2 2E ,r

P Q

+

C1 C2

R

Loop-1

Loop-2

Loop-3

i i



6. In the given circuit ammeters are ideal then, which of the following statements are true?

(a) Reading of 3A will be half as shown by 2A

(b) Reading of 1A will be thrice as shown by 2A

(c) Reading of 3A will be lowest

(d) Reading of 1A will be thrice as shown by 3A

7. A metallic conductor of irregular cross-section is as shown in the figure. A constant potential difference is applied across the ends (1) and (2). Then:

(a) the current at the cross-section P equals the current at the cross-section Q

(b) the electric field intensity of P is less than that at Q

(c) the rate of heat generated per unit time at Q is greater than that at P

(d) the number of electrons crossing per unit area of cross-section at P is less then that at Q.

8. Figure shows three concentric thin spherical shells A, B and C of radii R,

2R, and 3R. Shells A and C are given charges q and 2q and shell B is earthed. Then

(a) charge on inner surface of shell C is 4

q3

(b) charge on outer surface of shell B is 4

q3

(c) charge on outer surface of shell C is 2

q3

(d) charge on outer surface of shell C is 4

q3

A

B

C

9. Four identical plates (equally spaced) and a battery are connected as shown. If the capacitance

between two consecutive plates are C then choose the correct statement.

(a) Energy supplied by the battery is 2CV .

(b) Energy linked in the space between plates 1 and 2 is 21CV

6.

(c) Potential difference between plates 2 and 4 is V.

(d) The surface charge density on plate 3 on it’s right side is

more than that of on left side.

10. A positive charge is passing through an electromagnetic field in which E & B are directed towards y-axis

& z-axis respectively. If a charge particle passes through the region undeviated, then its velocity is/are represented by (here a, b & c are constant)

(a) E ˆ ˆv i ajB

(b) E ˆ ˆv i bkB

(c) E ˆ ˆv i ciB

(d) E ˆv iB

P Q (1) (1)

1 2 3 4

V(ideal)

A3

A2

A1

R

6R

2R

R E



PART – C

(Integer Type) This section contains 10 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

1. Consider a hollow spherical shell of radius R carrying a charge Q distributed uniformly over its surface.

If the spherical shell rotates with an angular velocity about an axis passing through its centre, then

the magnetic moment of hollow sphere is given by 2Q R

Mk

. Find the value of K.

2. Consider a non conducting sphere of radius a carrying uniformly distributed charge q surrounded by a

spherical shell of radius 2a. The region between a r 2a is filled with charge density C / r, where r

is the distance from the centre. If the magnitude of electric field in the region a r 2a is constant, then

the magnitude of charge in the volume a r 2a can be written as nq. Find the value of n?

3. Consider a metallic ring of radius 1 m, mass 1 kg and carrying a current of 1A in a gravity free space in

the x-y plane with its centre O at the origin as shown in the figure. If a uniform magnetic field ˆ ˆ(3i 4 j) T

is applied, then the instantaneous acceleration of the point P (which is on the y-axis at the moment) will

be 2C m / s . Find the value of C.

4. Solid sphere A of mass M and radius R has charge Q at centre.

q

A

Q

Insulated rod of height R

X

Very Rough Insulated Surface A point charge q is at the same level with centre. The force of friction acting on sphere, immediately

after the sphere is released is 2

0

Qq

2K X then find the value of k.

5. A miliammeter of range 10mA and resistance 9 is joined in a circuit as shown. The meter gives full scale deflection for current I when A and B are used as its terminals, i.e. current enters at A and leaves at B (C is left isolated). The value of current I in ampere

9 ,10mA

0.1 0.9

A B C

a 2a

P

x

y

O



6. A uniform, 500 g metal bar 100 cm long carries a current I in a uniform, horizontal, 0.5 T magnetic field as shown in figure. The bar is hinged at b but rests unattached at a. What is the largest current, in ampere, that can flow from a to b without breaking the electrical contact at a ? Assume that the wire be inclined to the horizontal at an angle of 60º as shown. (Take g = 10 ms-2).

7. A small current carrying loop having current I0 is placed in the plane of paper as shown. Another

semicircular loop having current I0 is placed concentrically in the same plane as that of small loop, the radius of semicircular loop is R(R >> a). Find the force applied by the smaller ring on bigger ring in

newton. (Given R = 1 m, I = I0 =

0

40

A, a = 0.1 m)

8. The figure shows a network of five resistances and two batteries

Ratio of current (in ampere) through the 30V battery and 15V battery is X. Find X

9. A current I flows in a circuit shaped like on isosceles trapezium ABCD. Given AB = 2CD where CD= . Magnetic field at point P located in the

plane of the trapezium ABCD at a distance a from the mid-point of CD is

found to be 22

o

naan

I

, where value of n is

10. Figure shows one quarter of a simple circular loop of wire that carries

a current of 10A. Its radius is a = 5cm. A uniform magnetic field, B =

4T, is directed in the +Xdirection, Then the torque on the entire

loop is found to be 4n 10n5 N-m where value of n is

Z

Y

X

60o

I

B

a

B

A

D

C

P



SSSEEECCCTTTIIIOOONNN---222 ::: CCCHHHEEEMMMIIISSSTTTRRRYYY

PART– A


1. In which of the following product is formed according to Hofmann’s rule ?

(a)

Br

3Me CO K

(b)

Br

EtO K

(c)

N+ OH /

(d)

F

EtO K

2. Which reagent(s) can be used in the conversion of 2-butanone to propanoic acid. (a) NaOH, NaI/H+ (b) Fehling solution (c) NaOH, I2/H+ (d) Tollen’s reagent 3.

O

3 3 2

4

CH C C Na CH I H

Pd BaSO(A) (B) (C)

(a)

(A) is

CH2 C C CH3

Na+O

-

(b)

(B) is CH3O

CH2 C C CH3

(c)

(C) is

C C

CH

H H

CH3CH3

OCH3

(d)

(C) is

C C

H

CH H

CH3

CH3O

CH3

4. Which of the following will give Cannizzaro reaction ?

(a) CHO

(b)

CHO

(c)

Cl

Cl

Cl

CHO

(d) HCHO

5. Pyrimidine bases present in DNA are (a) Cytosine (b) guanine (c) thymine (d) adenine 6. Acetaldehyde can be obtained from which of the following reactions ?

(a) Cu/

3 2CH CH OH (b)

CCH3 Cl

O

4

2

Pd BaSO

H ,

(c) 3 2 2(CH COO) Ca (HCOO) Ca

(d) CH3 CH C Ph

CH3

3

2

O

Zn H O



7. Which of the following compounds will give iodoform test ?

(a)

CH3 C

O

H

(b)

CH3 C

O

CH3

(c)

CH3 C

O

Cl

(d)

CH3 C

O

NH2 8.

O CH2 CH CH2*

?

(a) OH

CH2 CH CH2*

(b) OH

CH2 CH CH2*

(c) OH

CH2 CH CH2*

(d) OH

CH2 CH CH2*

9. The phenomenon of mutarotation is shown by (a) Glucose (b) Fructose (c) Cellulose (d) Starch 10. Which of the following compounds give positive Tollen’s test ?

(a)

H C

O

OH

(b) O

OH

(c) CH3 CH OC2H5

OH

(d)

CH3 C

O

H

PART – C


1. How many of the following are reducing sugars ? Glyceraldehyde, Glucose, Fructose, Sucrose, Starch, Maltose 2. The presence/absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA ?

3.

OCH3conc. HI (a) mole of HI is consumed

OCH3

conc. HI (b) mole of HI is consumed



a + b = ? 4. How many of the following are addition polymers ? Polythene, PVC, natural rubber, Bakelite, nylon-6,6, Teflon. 5. How many of the given compounds would give aldol condensation ?

CH3 C

O

CH3

C CH3

O

CHO CHO

CHO

, , , ,

CH3

CH3

CHO

CH3

,

O

CHO

CHO,

6.

O

COOH

COOH

HOOC COOHHOOC

HOOC

HOOC

‘n’ moles of CO2 are evolved in this reaction. Value of ‘n’ is _____ 7.

O

O

OH OH3

2

' xmole ' CH MgBr

H /H O

Find out the value of ‘x’. 8. Of the following reactions, how many reactions are used for the preparation of amines

(a) 4LiAlHR C N (b)

4LiAlH

2R C NH

O

(c) O

2Br OH

2R C NH

(d) O

NaOH

3R C CH

(e)

N-K

+

O

O

2

R X

H /H O

(f)

R CH3

O

2 4N H ,OH

(g)

2 5P O /

2R C NH

O

(h) 2H ,Ni

2 2R CH NO

9. How many moles of HI reacts with glycerol to give 2-iodopropane ? 10. Lassaigne’s test is used for the detection of N, O, S, Cl, P, As, K, Ca

PT-1-(1719) (Paper-1) Set-A – PCM-10


SSSEEECCCTTTIIIOOONNN---333 ::: MMMAAATTTHHHEEEMMMAAATTTIIICCCSSS

PART – A


1. Let f, g : R R be defined by f(x) 3x 1 | 2x 1| and 1

g(x) (3x 5) | 2x 5 |5

, then

(a) fog = gof (b) (fog)-1 = gog (c) y = min (fog(x), (fog(x))2, (fog(x))3, … (fog(x))101) is not differentiable at exactly three distinct values of x.

(d)

100 times

fogofogofog...fog (5) 3

2. If 2

1

2

1 x 1f(x) sin

2 1 x

, then which of the following is(are) correct ?

(a) 1

f '( 1)4

(b) Range of f(x) is 0,

2

(c) f '(x) is an odd function (d) x 0

f(x) 1lim

x 2

3. Identify which of the following function(s) is (are) bijective ?

(a) 1 xf : ( ,0] 0, ,f(x) sin (e )2

(b) 1 1f : [ 1,1] { 1,01}, f(x) sgn(sin | x | cos | x |)

(c) f : [ 3,0] [cos3,1],f(x) cosx

(d) f : R I R,f(x) ln{x}

Note : sgn(y) and {y} denote signum function of y and fractional part function of y respectively

4. Let

2

3

n(1 x) x px; 0 x 1

xg(x) q ; x 0

xtan

3; 1 x 0

x

then

(a) g(x) is continuous at x = 0, if 1 1

p , q2 3

(b) g(x) is discontinuous at x = 0, if 1

q3

(c) g(x) has irremovable discontinuity at x = 0 if 1

p2

(d) g(x) has removable discontinuity at x = 0 if 1 1

p ,q2 3

5. Let

1 axn

1 bxf(x) , x 0x

1, x 0

. If f '(0) 1 , then

(a) a + b = 1 (b) 1

a b2

(c) 1 3

a ,b2 2

(d) a – b = 1



6. Let f(x) be a non constant twice derivable function defined on R such that f(2 + x) = f(2 - x) and

1f ' 0 f '(1)

2

. Then which of the following is/are correct ?

(a) f(-4) = f(8)

(b) Minimum number of roots of the equation f ''(x) 0 in (0, 4) are 4.

(c) 4

4

f(2 x)sinx dx 0

(d) 2 4

cos t cos t

0 2

f(t)5 dt f(4 t)5 dt x

7. If

2x

2

x

dtf(x) , x 0

(log t) , then f(x) is

(A) Monotonically increasing in (2, ) (B) monotonically increasing in (1, 2)

(C) Monotonically decreasing in (2, ) (D) monotonically decreasing in (0, 1) 8. If the tangent at the point (p, q) to the curve x3 + y3 = k meets the curve again at the point (a, b), then

(A) 2

2

q b p

p a q

(B)

2 2

2 2

q b p ap a

p a q bq b

(C) a b

1p q (D)

a b1

p q

9. If I =

2 2

sin x cos x

sin x cos x sin x cos x sin x cos x

dx = cosec–1(g(x)) + c x R, then

(A) g(x) = 1 + sin2x (B) g(x) = 1 – sin2x

(C) g(x) 0 (D) – 1 g(x) 1

10. If

1

0324 xx

dxI then

(A) 6

I (B)

6

I

(C) 24

I (D)

4

I

PART – C


1. If 2 2t 2y

xt 00

tx x (t 1) 1f(y) lim dx

e .t

Then find the number of solution(s) of the equation 2f( | x |) 1 0

2. Let

3 x

2

33

(x 2) , 3 x 1f(x) and g(x) f(t)dt, 3 x 2

x , 1 x 2

. Find the number of extremum points of

g’(x).

PT-1-(1719) (Paper-1) Set-A – PCM-12


3. A polynomial y = f(x) of degree 4 increases in the interval ( ,1) (2,3) and decreases in the interval

(1, 2) (3, ) and satisfy f(0) = 1 and f '(0) 6 . Find the value of f(2).

4. If x

4 3 2

2

f(x) (t bt (b 1)t bt b) dt

increases x R then find the number of integers in the range

of b.

5. The value of

4

0

sinx cosxsinx dx

2 2

is _______ [where [x] represents greatest integer

function]

6. If f(x) + f(– x) = 2 x R, where f(x) is continuous, differentiable and invertible function, then the value

of 1 x

1

1 x

f t dt

is equal to _________

7.

n 1

n

1lim 1 ln 1

n

is equal to _______

8. The number of points in the interval [0, ] at which the function f(x) = max{|sinx|, |cosx|} is non–differentiable are ______

9. The function ‘f’ satisfies the functional equation 3f (x) + 2f x 59

10x 30x 1

for all real x 1. The

value of f (7) is ______

10. Let f be a positive function. Also let I1 = k k

2

1 k 1 k

xf x 1 x dx, I f x 1 x dx

, when

2k 1 > 0. Then 2

1

I

I is equal to ________



ANSWERS (ADV) PAPER – 1 (SET – A)

SR. NO PHYSICS C.CODE CHEMISTRY C. CODE MATHS C.CODE

1. BC P120305 ACD C124606 AC M120311

2. ABCD P120204 C C121403 AC M121502

3. ABD P120303 ABC C121807 AC M120310

4. ABC P120217 ABD C121405 ABCD M120405

5. ACD P120305 AC C121705 AC M120411

6. ACD P120212 ABCD C121401 ABCD M120813

7. ABCD P120201,202 AB C121402 AD M120604

8. ABC P120111 B C121806 ABC M120601

9. ACD P120118 AB C122901 AC M120712

10. BD P120308 ABCD C121403 BC M122810

1. 3 P120309 4 C122903 4 M1220813

2 3 P120109 2 C121705 2 M120608

3. 8 P120307 3 C122607 3 M120608

4. 7 P120104 4 C121903 5 M120616

5. 1 P120212 4 C121405 8 M120807

6. 5 P120308 5 C122811 0 M120813

7. 8 P120309 2 C121504 0 M110610

8. 3 P120109 5 C124506 2 M113626

9. 4 P120307 5 C121804 4 M120323

10. 3 P120104 3 C122001 2 M120813

PT-1-(1719) (Paper-1) Set-A – PCM-14


HINTS & SOLUTIONS : PHYSICS : MORE THEN ONE CORRECT

1. 2 2 00 0

qEv v 2 x

m

0v 2v

0 0n

2 2

x 0

qE va

m v v

2

n

vR

a =

32 2 2

0 0 002

0 0

[m v 2qE mx ]4 2 x

qE v m

(b) and (c)

2. (A) p.d. across each cell P QV V

(B) If I is clockwise then 2E is supplying and for reverse case reverse is the answer.

(C) P.D. E ir (when battery supplies energy)

E ir (when battery consumes energy).

By KVL 1 2

1 2

E Ei

r r

(Anticlockwise)

1 2P Q 1 1

1 2

E r E rV V E i r

r r

3. Work done by magnetic force on a charge 0 in any part of its motion.

‘S’ is matching for all parts (i), (ii), (iii), (iv)

For loop 1 inI i i i i 0B.d ( i)

For loop 2 inI i i i i 0B.d (i)

For loop 3 inI i i i i 0B.d ( i)

(Note : That current will be taken as positive which produces lines of magnetic field in the same sense

in which d is taken)

4. dq 2q Q

Rdt C C

2t/RCQ

q (1 e )2

2t/RCdq Qi (e )

dt RC

The charge on the first capacitor

2t/RCQq' Q q (1 e )

2

(a), (b) and (c)

5. v a v.a 0 6 4x 0 x = –1.5

Further magnetic field is perpendicular to the plane of velocity.

So, magnetic field is along z-direction

Also work done by a magnetic force is zero i.e., kinetic energy of a particle remains constant if only magnetic force is acting on it.

(a),(c) and (d)



6. From Kirchhoff’s law

16RI IR E 0 .....(1)

1 12R(I I ) (I I )R IR E 0

On solving 1

II

3

1

2II I

3

Reading of 1A I

Reading of 2

2IA

3

Reading of 3

1A

3

(a),(c) and (d)

8. Whole charge q given to shell A will appear on its outer surface. If a Gaussian surface is drawn within

material or shell B, net charge enclosed by it should be zero. Therefore, charge on its inner surface will be –q. Let q be the charge on its outer surface then charge on inner surface of C will be –q and its outer surface will be 2q + q as total charge on outer shell must be 2q.

Since shell B is earthed, its potential should be zero. VB = 0

0

1 q q q q 2q q0

4 2R 2R 2R 3R 3R

Solving we get 4

q q3

So choice (a) and (b) is correct. Charge on outer surface of outer shell is 2q + q

4q 2q

2q3 3

2 + q q1

A

C

B

q–q

–q1

q

So choice (c) is correct. Choice (d) is wrong obvious from above explanation.

(a), (b) and (c)

10. In both case (b) and (d), netF 0 so it passes the region undeviated.

(b,d)

NUMERICALS :

1.

2

/2 0 20

3

0

Q(2 Rsin d ) (Rsin )

Q4 RB 2

6 R4 R

or

A3

A2

A1

R

6R

2R

R E

I1

I

PT-1-(1719) (Paper-1) Set-A – PCM-16


q

M (L)2m

2Q R

M3

and

0

3

2MB

4 r

0QB

6 R

2. r

2 2 2

a

Cq' (4 r )dr 2 C(r a )

r

2 2

2

0

q 2 C(r a )E

4 r

constant

2

qC

2 a

2 2

2

qq' 2 [(2a ) a ] 3q

2 a

3. MB 5 N m

210 rad / sI

24a R 8 m / s

5

4.

A

f N

Mg

2

0

Qq 1.

4 X

2

0

Qq 1. f Ma

4 x

22

f R MR5

2MR.(a)

5

Solving 2

0

Qqf

14 x

5. 3 310 10 9.9 0.1 i 10 10

or I = 1A

9 10mA

0.1 0.9

A B C

6. gravity = B

mgl

2

cos(60º) = BIl sin(90º) l

2. This gives

mgcos 60º

IlBsin90º



7. 0

3

MB

4 R

where M = I a2

22

0 0

3 3

IaI aB

4 R 4R

dF = BI0dl = BI0 (Rd) = 2

00 3

IaI Rd

4R

dFx = dFsin

dFx = 2

0 0

2

II a sin d

4R

Fx = 2

0 0

2

II a

2R

and Fx = 0

Fnet = Fx = 2

0 0

2

II a

2R

= 8 newton

8. 9. Hints/ Solution :

CDAB BBB

10. Hints/ Solution :

sinIAB

CHEMISTRY : (Paper - 1) 2. Iodoform reaction

2NaOH I H

3 2 3 3 2 3 2CH C CH CH CH CH C O Na CH CH C OH

O O O

3.

O

3

N2

CH C C Na

S

O-Na

+

C C CH3

(A)

N2

3

S

CH I

OCH3

C C CH3

(B)

2 4H ,P d BaSO

H H

CH3 CH3O

(C)

4. -

3CH C Cl gives substitution reaction with OH

O

8. Claisen rearrangement. 10. Formic acid and Hemi-acetals give positive Tollen’s test. Numerical : 3.

PT-1-(1719) (Paper-1) Set-A – PCM-18


OCH3conc. HI

(1 mol)

OH

3CH I

OCH3 I

3CH Iconc. HI

(2 mol)

7.

O

O

3

2

(1) CH MgBr

(2) H /H O

OH

OCH3

3

2

(1) CH MgBr

(2) H /H O OHCH3

CH3OH

9.

OH OH OH I I I I I

2 2I I3mol HI 1molHI 1mol HI

2 2 2 2 2 2 2 3 2 3 3 3CH CH CH CH CH CH CH CH CH CH CH CH CH CH CH CH CH CH

I I

MATHS :

HINTS AND SOLUTIONS :

1.

f(x) and g(x) are inverse of each other.

2 3 2011y min{fog(x), f(fog(x)) (fog(x)) ,...(fog(x)) }

2 3 2011y min{x,x ,x ,...x }

Not differentiable at x = 0, 1 .

2. Put x tan

1

1

1

sin sin ; 02 2

f(x) sin sin2

sin sin ; 02 2

2

2

1; x 0

2(1 x )f(x)

1; 0 x

2(1 x )

1 1

f '(1) ; f '( 1)4 4

(A) is correct

Range of f(x) 0,4

. Hence (B) is incorrect

Also f '(x) is an odd function. (C) is correct

x 0

f(x) 1 1 1lim As, f(0 ) andf(0 ) (D)

x 2 2 2

is incorrect

3. (A) x ( ,0]

x 1 xe (0,1] sin (e ) 0,2

1 xf(x)sin (e ) is one-one and onto both.

(B) 1 1f(x) sgn(sin | x | cos | x |)

f( x) f(x) f(x) is many one

(c) f(x) cosx



From the graph it is clear that f(x) is one-one and onto both.

(D) f(x) ln{x}

Range of f is ( ,0)

f is many one into

4.

32

3h 0

1 hh p ....

12 3g(0 ) lim ; g(0 ) ,g(0) q

3h

Now check the options. 5. Since, f(x) is continuous at x = 0, so

x 0 x 0

1 ax 1 axn n 1 1

1 bx 1 bxLim 1 Lim 1

x x

x 0

1 ax 1 bxLim 1 a b 1

(1 bx)x

Now, f '(0) 1

h 0

1 ahn

1 bh1

hf '(0) Lim 1h

By solving we get

2 22

2h 0

b a(a b 1)h .h ...

2Lim 1h

2 2b a

a b 1 0 and 12

2 2a b 1and b a 2

b a 2

1 3

a ,b2 2

Ans

6. We have f(2 – x) = f(2 + x) Replacing x by 2 – x, we get

f '(x) f(4 x) …(1)

Put x = -4 in (1), we get f(-4) = f(8) on differentiating (1) w.r.t. x, we get f’(x) = -f’(4 – x) …(2)

Put 1

x ,1,2 in (2)2

, we get

1 7

f ' 0 f '(1) f '(2) f ' f '(3)2 2

Now, consider a function y = f’(x)

As f’(x) satisfy Rolle’s theorem in 1 7 7

,1 ,[1,2], 2, , ,32 2 2

respectively.

So, by Rolle’s theorem, the equation of f’’(x) = 0 has minimum 4 roots in (0, 4)

Now, consider

/ 4

1

/ 4

I f(2 x)sinxdx

…(3)

PT-1-(1719) (Paper-1) Set-A – PCM-20


Applying king property, we get

/ 4 / 4

1

/ 4 / 4

I f(2 x)sin( x)dx f(2 x)sin(x)dx

1 1I I

Hence 1I 0

Again, Consider 2

cos t

2

0

I f(t)5 dt

Put 4 t y dt dy

So, 2

cos (4 y)

2

4

I f(4 y)5 ( dy)

4 4

cos y cos t

2 2

f(4 y)5 dy f(4 t)5 dt

7. 2 2 2 2

2x 1 x 2f (x)

(logx ) (logx) 2(logx)

f(x) > 0 for x > 2

and f(x) < 0 for x < 2 Hence (A), (D) are correct answers.

8. 22

2 2

dy p3x

dx 3y q (Slope of tangent)

Since this is the line joining (p, q) and (a, b)

2

2

q b p

p a q

… (1)

As these two points line on the curve. p3 + q3 = a3 + b3 = k.

Hence p3 a3 + q3 b3 = 0 (p a) (p2 + ap + a2) + (q b) (q2 + bq + b2) = 0

2 2

2 2

q b p ap a

p a q bq b

… (2)

From (1) and (2), 2 2 2

2 2 2

p p ap a

q q bq b

p2 (bq + b2) = q2 (ap + a2) pq (pb aq) = (p2b2 q2a2) pq = (pb + aq)

b a

1q p … (3)

9. We have I =

2 2

sin x cos xdx

sin x cos x sin x cos x sin x cos x

= cosec–1(g(x)) + c

I =

2 2

cos2xdx

1 sin2x sinxcosx sin xcos x

Let 1 + sin2x = y

I = 2

dy

y y 1

I = cosec–1(y) + c

g(x) = y = 1 + sin2x.

10. 04 32 xx 1,0x

And 32 xx 1,0x

22322 444 xxxxx



2322 2444 xxxx

2322 24

1

4

1

4

1

xxxx

1

02

1

032

1

02 2444 x

dx

xx

dx

x

dx

2446

1

032

xx

dx

INTEGER TYPE :

1. 2 2t 2x

xt 00

tx x (t 1) 1f(x) lim dx

e .t

x 2t 2

x 2

t 00

(t 1) 1f(x) e limx x dx

t

Now 2t 2

t 0

(t 1) 1lim

t

(2t 2) n(t 1)

t 0

e 1lim 2

t

x

x 2 x 2

0

f(x) e (x 2x)dx e x

Now 2f( | x |) 1 ...(1)

2

|x|

x 1

2e

Let 2

x

xf(x) ; x 0

e

2

x

(2x x )f '(x)

e

x

x(2 x)

e

Number of solution of the equation (1) is 4

2. Clearly,

3

2

3

(x 2) , 3 x 1g'(x) f(x)

x , 1 x 2

g'(x) is a continuous function

PT-1-(1719) (Paper-1) Set-A – PCM-22


sign scheme of g''(x)

g'(x) has a local maximum at x = -1 and local minimum at x = 0

Hence, g'(x) has two extremum points.

3. As y = f(x) is polynomial hence it is differentiable.

f '(x) k(x 1)(x 2)(x 3)

f '(0) 6k

6 6k k 1

Hence f '(x) (x 1)(x 2)(x 3)

3 2(x 6x 11x 6)

4 3 2x 6x 11x

f(x) 6x C4 3 2

f(0) = 1 hence C = 1

4 3 2x 6x 11x

f(x) 1 6x4 3 2

Hence, f(2) = 1 – 4 + 16 – 22 + 12 = 3

4. 4 3 2f '(x) x bx (b 1)x bx b

4 3 2 2(x bx bx ) (x bx b)

2 2(x 1)(x bx b)

Now f '(x) 0 x R

2x bx b 0 x R

20 b 4b 0

b(b 4) 0 0 b 4 .

5. Since sin x 1

02 2

x R sin x

02

similarly cos x

02

and

4

0 0

sin x dx 4 sin x dx 8

.

6. f() =

f(– ) = 2 –

f–1() =

f–1(2 – ) = –

f–1(x) + f–1(2 – x) = 0 value of integration is zero.



7.

n 1

n

1lim 1 log 1

n

=

n

n

11

nlim 1 log

11

n

=

n

n

1 1lim 1 log 1 log 1

n n

=

n

n n

1 1lim 1 log 1 lim log 1

n n

= 1 + 0 – loge = 1 –1 = 0 .

9. 3f (x) + 2fx 59

10x 30x 1

For x = 7, 3f (7) + 2f (11) = 100 For x = 11, 3f (11) + 2f (7) = 140 Solving, we get f (7) = 4.

10. I1 = k

1 k

xf x 1 x dx

= k

1 k

1 k k x f 1 k k x 1 1 k k x dx

b b

a a

using f x dx f a b x dx

= k

1 k

1 x f x 1 x dx

= I2 I1 2I1 = I2 2

1

I2

I .

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