FIITJEE ALL INDIA TEST SERIES · AITS-PT-I-PCM-JEE(Main)/205 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942

website: www.fiitjee.com

FIITJEE

ALL INDIA TEST SERIES

PART TEST – I

JEE (Main)-2019-20

TEST DATE: 9-11-2019

Time Allotted: 3 Hours Maximum Marks: 300 General Instructions:

• The test consists of total 75 questions.

• Each subject (PCM) has 25 questions.

• This question paper contains Three Parts.

• Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.

• Each part has only three sections: Section-A, Section-B and Section-C.

Section-A (01 – 20, 26 – 45, 51 – 70) contains 60 multiple choice questions which have only one

correct answer. Each question carries +4 marks for correct answer and –1 mark for wrong

answer.

Section-B (21 – 22, 46 – 47, 71 – 72) contains 6 Numerical based questions with answer as numerical value from 0 to 9 and each question carries +4 marks for correct answer. There is no negative marking.

Section-C (23 – 25, 48 – 50, 73 – 75) contains 9 Numerical answer type questions with answer

XXXXX.XX and each question carries +4 marks for correct answer. There is no negative marking.

AITS-PT-I-PCM-JEE(Main)/20



2

Physics PART – I

SECTION – A

(One Options Correct Type)

This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct.

1. For a particle projected with initial velocity ˆ ˆ û 6i 8 j 20k= + + , find angular velocity of the particle

about the point of projection, at t = 2 sec. (Given ˆg 10k= − )

(A) 1 3ˆ î j5 20

−

(B) 1 3ˆ î j5 20

− +

(C) 2 3 2ˆ î j

5 20−

(D) 2 3 2ˆ î j

5 20+

Ans. B Sol. Particle will be at Hmax at t = 2

max

20 20H 20m

2 10

= =

Hˆ ˆ ˆr 12i 16 j 20k; r 20 2= + + =

Hˆ ˆv 6i 8 j= +

y

x

z

(0, 0, 0)

H(12, 16, 20)

2

r v

r

=

( )2

ˆ ˆ160i 120 j

20 2

− += =

1 3ˆ î j5 20

− +




3

2. A solid cylinder of mass ‘m’ is kept stationary on a fixed incline of 37, by

applying a force tangentially as shown in the figure. Calculate the minimum coefficient of friction that the surface must have in order to achieve equilibrium of the cylinder by applying a force of minimum magnitude.

37

F

r

(A) 3

4

(B) 3

8

(C) 3

16

(D) none of these Ans. B Sol. Force when applied parallel to plane will have minimum magnitude. Torque about P;

F(2r) = mg sin r

min

mgsinF

2

= .…(i)

Also, F + fmax = mg sin ….(ii)

tan 3

2 8

= =

Fmin

P

mg

fmax

3. For what value of initial speed u, a projectile launched at (from the

vertical) from the top point of a hemisphere will land at 2 angular

position(from the vertical) on the hemisphere as shown in the figure.

R 2

u

(A) gR cot

(B) 2cos

gRsin

(C) gR cos

(D) Not possible Ans. C




4

Sol. Coordinates of P [(R sin 2 , – R (1 – cos 2)]

( )( )

2

2

1 xy x tan 90 g

2 ucos 90= − −

−

2 2

2 2

cos 1 R sin 2R 1 cos2 Rsin2 g

sin 2 u sin

− − = −

2 2

2 2

2 2

1 R4sin cos2sin 2cos g

2 u sin

− = −

2

2

2gRcos2 u gR cos

u

= =

R P

90 −

u

2

(0, 0)

4. A particle moving in space with a velocity 20 m/s along

positive x-axis starts experiencing a constant acceleration at t = 0 as shown in the figure. The minimum value of the ratio

min

max

R

R, where Rmin/Rmax is the minimum and maximum radius

of curvature of the particle’s trajectory for any t > 0 is

a = 5 m/s2

v = 20 m/s

x 150

(A) 1

8

(B) 1

6

(C) 1

5

(D) none of these Ans. D Sol. Particle will describe a parabolic path as shown

After long time Rmax →

min

max

R

R= 0.

a = 5 m/s2

v = 20 m/s

x

60 P O

y




5

5. An hour glass containing sand is kept on a weighing scale. At t = 0, sand starts falling from upper compartment to lower compartment at constant rate. How will the reading of scale look like when plotted against time.

(A) W

t

(B)

W

t

(C)

W

t

(D) W

t

Ans. B Sol. Till first sand particle touches lower compartment of hourglass, weight will decrease, then it

becomes constant, when last sand particle leave upper compartment it increases.




6

6. A rod of length ‘ ’ and cylinder of radius ‘r’ is kept on an incline plane as shown in the figure. The rod is pivoted, while the cylinder can roll without slipping. A light string PQ attaches top of cylinder to some point on the rod such that it is parallel to the incline. The minimum value of radius of cylinder to ensure that the string is taut when the system is released from rest is

r

P

Q

(A) 2

5

(B) 21

50

(C) 4

9

(D) none of these Ans. C

Sol. Q Pa a

3gsin r 4

gsin3

4

r9

7. A truck has its side door initially open as shown. The

dimension of the door is 2m 3m (W ) and its mass

is M = 5 kg. At t = 0, the truck starts moving with uniform acceleration a = 5 m/sec2 as shown in figure (top view).

At any time t, if the door makes angle with its initial

position, then the component of force exerted by hinge on the door along its width is

Truck

Initial position of side door

W = 2m

Top view

(A) 125

sin2

(B) 250 sin

(C) 75 sin

(D) none of these Ans. A




7

Sol. MI of door about hinge : 2Mw

3

Using : I =

2w Mw

Ma cos2 3

=

3 a

cos2 w

=

hinge Ma cos

Ma

Masin

d 3 a

cosd 2 w

=

2 3asin

w =

For circular motion of CM of door;

2H

wR Masin M

2

− =

H

3 5 125R Masin Masin Masin sin

2 2 2 = + = =

8. For a particle moving in space with velocity v , which of the following is incorrect.

(A) dv d | v |

dt dt

(B)

t t

0 0

| v |dt v dt

(C)

2

1

t

t

v | a | dt

(D) d | v | dv v

dt dt v

=

Ans. C Sol. Use the definition of kinematics variables.




8

9. A stick of length L is dropped from rest. At same instant, an insect starts moving up the stick with a constant speed u relative to the stick. The maximum height attained by the insect from its initial position with respect to the ground is

L

(A) 2u

2g

(B) 2u

g

(C) 22u

g

(D) none Ans. A

Sol. insec tˆv (u gt) j= − , v 0= at

ut

g=

2

u/g u/g 2

max

0 0

uy vdt (u gt)dt

2g= = − =

10. In the arrangement shown, a bar AB is

connected through a rotating drum via a massless string which is attached to the end B of the bar. The portion of the rod near end A rests on a horizontal surface. The drum rotates

with a uniform angular speed 0. The angular

speed of bar AB at given position is

r

x

B

A

0

h

(A) 0

2 2

rx

x h

+

(B) 0

2 2

rh

(x h )

+

(C) 0

2 2

rh

x h

+

(D) 0

2 2

rx

(x h )

+




9

Ans. B

Sol. 0A/B C/B 2 2

rhv sin

CB (x h )

= = =

+

C

A

B

v cos

v sin

v

11. A car of mass m is initially at rest on a boat of mass M and is tied to

the shore as shown. The car starts accelerating from rest at t = 0 and acquires a velocity v0 in time t0. At t = t0, the car applies brake and comes to rest relative to the boat in no time. Neglect friction between the boat and the water. The time ‘t’ in which the boat strikes the wall is

m

M

L

(A) 00

1t

v+

(B) 00

L(M m)t

mv

++

(C) 0

LM

mv

(D) none of these Ans. B

Sol. 0mv (M m)v= +

0mvv

M m =

+

00

L(M m)t t

mv

+= +




10

12. Two small rings of mass ‘m’ each are contained to move on smooth wire. The rings are connected to a block of mass ‘2m’ via massless strings of length ‘ ’ each. Initially, the system is held at

rest as shown. The velocity of block when string makes 60 with

the vertical is

2m

m m

(A) 2g

(B) 3g

5

(C) g

5

(D) 3g

4

Ans. D

Sol. 2 21

1 12mg (2m)v 2 mv

2 2 2= +

…(i)

v1 cos 30 = v cos 60

…(ii)

Solving 3g

v4

=

v1 v1

v

60 30

13. A spinning drum of radius ‘R’ with its axis parallel to the ground, carries

dust particles deposited on its inner surface. Find minimum angular speed with which drum be rotated so that a dust particle separating from the drum may fall at its diametrical opposite position in space. Assume, sufficient friction present to prevent any relative slipping of dust particle on the drum.

O

(A) g

2R

(B) g

2R

(C) g

4R

(D) None of these Ans. B




11

Sol. v = R …(i)

mg sin = 2mv

R …(ii)

for motion after separation

2V

Tgcos

=

…(iii)

21 2V

2R gsin2 gcos

=

…(iv)

O

y

x

mg

v

2cos

gR Rgsin 45sin

= =

2 g gR or =

2 2R =

14. A particle moves in xy plane. The position vector of particle at any time t is 2ˆ ˆr (2t)i (2t ) j= + m.

The rate of change of at t = 2 sec, where is the angle made by the velocity vector from the

positive x-axis, is

(A) 1

14rad/sec

(B) 2

17rad/sec

(C) 4

7rad/sec

(D) 6

5rad/sec

Ans. B

Sol. ˆ ˆv(t) 2i (4t) j= +

tan = 2t

2

d 2 2rad / sec (at t 2sec)

dt 17sec

= = =




12

15. A circular groove PQM of radius R is cut in an inclined plane as shown. A small ball of mass m is released form P. The magnitude of the displacement of wedge by the

time ball reaches M is ( = 60, R = 1m, m = 2kg, M = 3kg)

y

x

M

Q

O

P

(A) 5 m

(B) 1

5m

(C) 17

5m

(D) none of these Ans. B Sol. Centre of mass of wedge and ball will not shift in x-y plane

Let wedge displacement be ˆ ˆxi yj+

rˆ ˆS ( Rcos )i Rj= − +

ˆ ˆ ˆ ˆm (x Rcos )i (y R)j M(xi yj) 0 − + + + + =

mRcos

xm M

=

+;

mRy

m M= −

+

w

1 2ˆ ˆS i j5 5

= −

16. A man is rotating a stone of mass 10 kg tied at the end of a light rope in a circle of radius 1m. To

do this, he continuously moves his hand in a circle of radius 0.6 m. Assume, both circular motions to be occurring in the same horizontal plane. What is the maximum speed with which he can throw the stone, if he can exert a pull not exceeding 1250 N on the string.

(A) 10 2 m/s

(B) 5 5 m/s

(C) 10 m/s (D) 20 m/s Ans. C




13

Sol. 2mv

TcosR

=

TRcos

v 10m

= = m/s

0.6m 1mm

T

v

17. A man of mass m while jumping up in the air applies a force F = F0 cos t on the ground. Assume,

that he remains in contact with the ground from t = 0 to t = 2

sec.

(A) Velocity of man during take off t2

=

is less than 0F

m

(B) Man is producing peak power at the instant of take off. (C) The average power produced by man is zero.

(D) The ratio of power produced at t = 8

and

3t

8

=

is more than 1.

Ans. A

Sol. a = 0Fcos t

m

0Fv sin t

m=

peak power at t = 4

.

Ratio of power is 1

Also

/2

0

(F mg)dt mv

− =

0F gv

m 2

= −

F

P

(/8)

(/4)

(3/8) t




14

18. Two beads A and B of equal mass ‘m’ are connected by a light string. The beads are positioned to move on a smooth ring in a vertical plane as shown in the figure. The tension in the string just after release is

A

B

(A) 2 mg

(B) mg

2

(C) mg

4

(D) mg

2

Ans. D

Sol. T sin = ma

…(i)

mg − T cos = ma

mg mg

Tsin cos 2

= = +

( = 45)

mg T

N1

a

Bead A

mg

T

a

N2

Bead B 19. A particle A is fixed at the origin of a fixed coordinate system. B is another particle that experiences

a force 3 2

2ˆF r

r r

= − +

due to particle A, where r is the position vector of B with respect to A.

Find the work done in moving particle B slowly from ( )1 0 0P 2r , 2r to point 0 02

r rP ,

2 2

by an

external agent, where r0 is the equilibrium position of the particle.

(A) 29

64

(B) 2

16

(C) 2

16

−

(D) 29

64

−

Ans. B




15

Sol.

f

i

r

ext ext 3 2

r

2W F dr dr

r r

= = −

0

0

r

22rrr

= − +

ext 200

3W

2r4r

= − +

Also, at equilibrium; F = 0 r0 = 2

2

2 2 2

exta

3W

4 164 4

= − =

20. A spool of mass ‘m’ has inner and outer radius R and 2R

respectively. A thread is wound on the inner disc and its free

end is pulled with a force F at an angle with the horizontal

in the direction as shown in the figure. Assume no slipping condition for thread and the spool and Icm = 2mR2, then

F

R 2R

(A) For 0 90, frictional force acts leftward only.

(B) For 0 60, it acts leftward and for 60 90 it acts rightward.

(C) For a certain between 0 and 90 friction becomes zero.

(D) none of these Ans. A

Sol. Fcos − f = maC

…(ii)

2f(2R) FR 2mR− = …(ii)

aC = 2R

…(iii)

f = ( )F

1 cos3

+ always leftward.

F

F cos

F sin

f




16

SECTION – B (Single digit integer type)

This section contains 02 questions. The answer to each question is a single Digit integer ranging from 0 to 9, both inclusive.

21. Two men A and B of mass M and 2M are standing 120 apart on a circular platform of mass 4M

and radius R. The platform is free to rotate about its centre, while men are standing at periphery.

Now men interchange their position with respect to disc by walking along periphery only. If

represents magnitude of angle rotated by platform with respect to the ground, find the ratio max

min

. Ans. 5 Sol. A and B can swipe their position by walking clockwise / anticlockwise

For max : A should turn 120 (clockwise) while B should turn

240(clockwise)

Conserving angular momentum about C:

2

2 2 4MRMR (120 ) 2MR (240 ) ( ) 0

2 − + − + − =

= 120

For min : A should turn 120 clockwise, B should turn 120

anticlockwise

120

A

B

2M

M

min

120

5

=

max

min

=

5

22. A(3 m, 4 m) and B(7 m, 1 m) are two coins on a carrom board. The

striker is placed at origin O. If the striker can be shot with a speed of 5 m/sec, then find the minimum time (in sec) taken by striker to become collinear with coin A and B.

v

y

x

A (3, 4)

B (7, 1)

(0, 0) O

Ans. 1 Sol. For tmin, striker should be shot perpendicular to line AB

2 2AB (3 7) (4 1) 5= − + − = m

1 1

h AB | OA OB |2 2

=

ˆ ˆ î j k5h 1

3 4 02 2

7 1 0

=

h = 5 m

tmin = 5/5 = 1 sec




17

SECTION – C (Numerical Answer Type)

This section contains 03 questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the second

decimal place; e.g. XXXXX.XX).

23. The resultant of two vectors Â 5i= and B is R . The direction of R is such that it makes an angle

6

with the positive x-axis in anticlockwise sense. Find the magnitude of R when B has minimum

magnitude. (Take 3 1.732= )

Ans. 00004.33 Sol. From triangle;

R = 5 cos 30 = 5 3

2

30

x

y

R

B

A = 5

24. A block of mass m = 4 kg is kept on a rough surface of = 0.25. At

t = 0, an external force F(t) that varies linearly with time starts acting on the block. The force acts for a duration of 8 sec only and F(t = 8) = 0 N. If the block starts moving at t = 2 sec and finally stops at t = 11 sec, then find the time (in sec) instant when velocity acquired by the block is maximum.

m

F(t)

Ans. 00006.80

Sol. fmax = 0.25 4 10 = 10 N

Net impulse is zero in 11 sec

If = IF

max

18 F 100

2 =

Fmax = 25N From similar triangle

−10

10

25

fk/F

2 t1

t2 8

11

t(sec)

11

25 10t 5sec

t 2= =

Block will acquire Vmax at t = t2 where

( ) 2

2

25 10t 6.8sec

3 8 t= =

−




18

25. A hemisphere of radius R = 0.5 m is rotated with a uniform angular

speed = 10 rad/sec for some time. The hemisphere is then

stopped and it was found that only 20% of its top surface area remained covered with dust particles. Find the coefficient of friction between the particles and the hemisphere.

dust

Ans. 00001.35

Sol. Given : 2

(1 cos ) 0.22

− =

cos = 0.8

From FBD of dust particle:

N = mg cos − m2r sin …(i)

N = mg sin + m2r cos …(ii)

2

2

gsin r cos

gcos r sin

+ =

− = 1.35

R

r

mg

m2r

N N




19

Chemistry PART – II

SECTION – A



26. The geometry around each iodine in dinuclear anion ( )4

2 82I OH O

−

is:

(A) Octahedral

(B) Monocapped octahedral

(C) Square pyramidal

(D) Pentagonal bipyramidal

Ans. A

Sol.

O

I

O

O

O

I

O

OO

OH

OH

O

27. Amongst Zn, Ga, Ge and As the element with the lowest first I.E. is:

(A) As (B) Zn

(C) Ga (D) Ge

Ans. C

Sol. 1Zn 906 kJ mol−=

1Ga 579 kJ mol−=

1Ge 761 kJ mol−=

1As 947 kJ mol−=




20

28. The magnetic property of which molecule changes if we assume s – p mixing is not operational for all given molecules:

(A) N2 (B) F2

(C) O2 (D) C2

Ans. D

Sol. ( ) 2 * 2 2 * 2 2 2

2 x yC 12 actual 1s 1s 2s 2s 2p 2p

Diamagnetic

→ → =

If s – p is not operational.

2 * 2 2 * 2 2 1 1

z x y1s 1s 2s 2s 2p 2p 2p

paramagnetic

=

29. Consider the following reversible reaction; ( ) ( ) ( )2 c

1A g 2B g AB g K

2+ = .

The above equilibrium is established in a 1 L flask and at equilibrium 2 moles of each A and B are present. If 2.0 moles of B are added further how many moles of AB2 should be added so that moles of A does not change?

(A) 6 (B) 8

(C) 10

(D) 12 Ans. D

Sol. ( ) ( ) ( )2A g 2B g AB g+

2

c 2

AB1K

2 A B= =

( )2

22

AB1AB 4 moles

2 2 2 = =

Now, B = 4 moles Let y moles of AB2 is added.

( )

( )2

4 y1

2 2 4

+=

y 12 =




21

30. During titration of acetic acid with aq. NaOH solution, the neutralization graph has a vertical line. This line indicates:

pH

V

(A) Neutral nature at equivalence point (B) Acidic nature at equivalence point (C) Depends on experimental proceedings (D) Alkaline nature at equivalence point Ans. D 31. Which is the poorest reducing agent?

(A) Nascent hydrogen (B) Atomic hydrogen

(C) Dihydrogen (D) All have same reducing strength

Ans. C

Sol. 2

1H H e

2

+ −⎯⎯→ +

Takes B.E. for dissociation to H and then oxidizes slow. Fact by NCERT. 32. For which of the following reactions, the degree of dissociation cannot be calculated from the V.D.

data:

I. ( ) ( ) ( )2 22HI g H g I g+

II. ( ) ( ) ( )3 2 22NH g N g 3H g+

III. ( ) ( ) ( )2 22NO g N g O g+

IV. ( ) ( ) ( )5 3 2PCl g PCl g Cl g+

(A) I and III

(B) II and IV (C) I and II

(D) III and IV




22

Ans. A

Sol. gn 0 = in I and III.

Same moles at equilibrium as before. 33. On dissolving moderate amount of sodium metal in liquid NH3 at low temperature, which of the

following does not occur? (A) Blue coloured solution is obtained. (B) Na+ ions reformed in the solution. (C) Liquid ammonia solution becomes good conductor of electricity. (D) Liquid ammonia solution becomes diamagnetic. Ans. D Sol.

( ) ( )

( )

3 3 3x xNa Liq. NH Na NH e NH

x y

+ −

+ ⎯⎯→ +

+

Blue coloured and paramagnetic in nature. 34. Consider the following statements about carbon suboxide I. It is a linear molecule. II. It is an optically active cumulene.

III. The total number of bonds, bonds and lone pair in the compound are 12.

The correct statement is/are:

(A) I and II (B) II and III

(C) I and III (D) All of the above

Ans. C Sol. O = C = C = C = O → Linear → Optically inactive

→ 4 ,4 and 4 lone pair.




23

35. In the compound M – O – H, the M – O bond will be broken if:

(A) E.N. of M and O < E.N. of O and H

(B) E.N. of M and O = E.N. of O and H

(C) E.N. of M and O > E.N. of O and H

(D) Cannot be predicted by E.N. data

Ans. C

Sol. E.N. , more polarity. 36. Choose the INCORRECT statement: (A) Amorphous solids are short range orders (B) Amorphous solids are anisotropic in nature (C) Crystalline solids have definite heat of fusion (D) Crystalline solids are true solids Ans. B 37. The correct observed curve for a backbody at different temperature representing intensity

wavelength relation is: (T2 > T1)

(A)

T1

T2

I

(B)

T1

T2

I




24

(C)

T1

T2 I

(D)

T1

T2

I

Ans. A Sol. If T2 > T1

At high E → Intensity increases and decreases.

(Reference – NCERT) 38. The oxidation state of sulphur in the product/(s) of the reaction of S8 in basic medium is/are: (A) +2 and +6

(B) - 2 and +6

(C) - 2 and +2 (D) - 2 only.

Ans. C Sol.

( ) ( ) ( ) ( ) ( )

( ) ( )

2 2

8 2 3 2S s 12OH aq 4S aq 2S O aq 6H O

2 2

− −−

+ ⎯⎯→ + +

− +

39. The orbital overlapping in the figure results in:

(A) d – d antibonding

(B) d – d bonding




25

(C) d – d antibonding

(D) d – d bonding

Ans. B

Sol. Both d-orbitals have axial overlapping in same phase forming bonds.

40. The preparation of SO3(g) by reaction ( ) ( ) ( )2 2 3

1SO g O g SO g

2+ is an exothermic reaction

following temperature – pressure relationship for its % yield, for temperatures (T1, T2 and T3) the correct option is:

% yield

T3

T2

T1

Pressure

(A) T3 > T2 > T1

(B) T1 = T2 = T3

(C) T1 > T2 > T3

(D) Not enough information. Ans. C Sol. For exothermic reaction, increase in temperature decreases yield. 41. The existence of a solid compound in more than one modification is known as:

(A) Isomorphism (B) Allotropy

(C) Amorphism (D) Polymorphism

Ans. D 42. The INCORRECT statement amongst the following is: (A) Boron has unusually high melting point due to strong crystalline lattice. (B) In group 13, the compound with +1 O.S. are more ionic than with +3.




26

(C) Aluminium dissolves in dil. HCl to liberate H2 whereas H2 is not liberated when it reacts with NaOH.

(D) Aluminium chloride in acidified aqueous solution forms ( )3

2 6Al H O

+

ion.

Ans. C Sol. Al reacts with both acid and base to liberate H2. 43. Pick out the correct statement:

(A) r Cs

0.536r Cl

+ +

+ −= .

(B) High temperature increases coordination number. (C) Presence of excess Li+ in LiCl develops violet colouration. (D) Greater is the number of F-centres, greater is the intensity of colour developed. Ans. D

Sol. (A) r Cs

0.732r Cl

+ +

+ −= .

(B) High temperature decreases coordination temperature. (C) LiCl has pink colour. 44. The INCORRECT order is: (A) Covalent character: PbCl2 > CaCl2 > SrCl2 > BaCl2 (B) Thermal stability: PbF4 > PbCl4 > PbBr4 > PbI4 (C) Melting point: KF > KCl > KBr > KI (D) Boiling point: CHCl3 > CH3Cl > CCl4 Ans. D

45. The correct graph representing the electron gain enthalpy of chalcogens is: ( ge H is released

energy)

(A)

ge H

Se S O




27

(B)

ge H

Se S O

(C)

egH

Se S O

(D)

egH

Se S O Ans. C

Sol. ge H O 141 → → −

S 200

Se 195

S Se O

→ −

→ −



46. The radial distribution curve of the orbital with double dumbshell shape in the fourth principal shell

consists of ‘n’ nodes, n is: Ans. 1 Sol. Double dumb shell confirms the graph touches x-axis only once. 1 node.




28

47. After balancing the given reaction:

( )5 4 2 4 2 33Ca PO F SiO C aP bCaF cCaSiO dCO+ + ⎯⎯→ + + +

The value of [d – (a + b + c)] is: Ans. 7

Sol. ( )5 4 2 4 2 334Ca PO F 18SiO 30C 3P 2CaF 18CaSiO 30CO+ + ⎯⎯→ + + +

SECTION – C (Numerical Answer Type)

This section contains 03 questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the second

decimal place; e.g. XXXXX.XX).

48. Equal volumes of 0.02 M AgNO3 and 0.02 M HCN were mixed. If [Ag+] at equilibrium is x × 10-5,

calculate x. Take Ka(HCN) = 9 × 10-10 and Ksp(AgCN) = 4 × 10-16. Ans. 00006.67 Sol. After mixing [AgCN] = 0.01 M, [HCN] = 0.01 M

aHCN H CN K+ −+

( )Ag CN AgCN s+ −+ sp1/ K

( ) 6a

sp

KAg HCN H AgCN s K 2.25 10

K

0.01 0.01

x x 0.01 x 0.01 x

+ ++ + = =

− −

− −

2

0.01 xK

x

−= since x is small w.r.t. 0.01

2

0.01K

x=

5x 6.6 10 M Ag− + = =

49. The ratio of distances of nearest neighbours in BCC lattice by next to next nearest in FCC lattice

is: Ans. 00000.71

Sol. Nearest in BCC 3a

;2

Next to next nearest 3

a2

= in FCC

Ratio 3a 2 1

0.712 3a 2

= = =




29

50. The concentration of oxalic acid is ‘X’ mol L– 1. 40 ml of this solution react with 16 mL of 0.05 M acidified KMnO4. What is the pH of ‘X’ M solution? Assume oxalic acid dissociates completely: (log2 = 0.3010).

Ans. 00001.30 Sol. (Oxalic) N1V1 = N2V2 (KMnO4) M × 40 × 2 = 0.05 × 16 × 5

M 0.05 16 5

0.0540 2

= =

pH = - log (0.05) =1.3




30

Mathematics PART – III

SECTION – A



51. Let

1

30

dx

1 x =

+ ,

( )

1n n

3 3

r 1

3nn

n r

limn

=

→

+

=

, then log is equal to

(A) log 2 – 1 +

(B) log 2 – 3 + 3

(C) 2 log 2 –

(D) log 4 – 3 + 3

Ans. B

Sol. ( )1 1

3

30 0

3log log 1 x dx log2 3 dx

1 x = + = − +

+ = log 2 – 3 + 3

52.

( )3

2 2

xcos 1dx

x 2xcos 1

+

+ +

is equal to

(A) 2

xc

x 2xcos 1+

+ +

(B) 2

xcosc

x 2xcos 1

− +

+ +

(C) 2

2xc

x 2xcos 1+

+ +

(D) 2

2xcosc

x 2xcos 1

+

+ +

Ans. A

Sol. Put x + cos = sin tan

( )3

2 2

xcos 1dx

x 2xcos 1

+

+ +

= ( )cot sin cos d + = –cot cos + sin + c




31

= 2

xc

x 2xcos 1+

+ +

53. ( )

x 1f x

x 1

+=

+ for

5 1f : 0, , 3

2 2

→

(where [.] denotes the greatest integer function and {.}

represents fractional part of x), then which of the following is true? (A) f(x) is many-one discontinuous function

(B) ( ) ( )( ) ( )x 1 x 1

min lim f x , lim f x f 1− +→ →

=

(C) f(x) is surjective differentiable function every where (D) f(x) is injective function Ans. D

Sol. ( )

1; 0 x 1

x 1

2f x ; 1 x 2

x

3 5; 2 x

x 1 2

+

=

−

f(x) is discontinuous and bijective function

( )x 1

1lim f x

2−→= ; ( )

x 1lim f x 2

+→= and f(1) = 2

54. Let f(x) be a decreasing function defined on (0, ). If f(2a2 + a + 1) < f(3a2 – 4a + 1), then the range

of a is

(A) a > 1 or 1

a3

(B) a (0, 5)

(C) 1

a3

(D) ( )1

a 0, 1, 53

Ans. D Sol. f(2a2 + a + 1) < f(3a2 – 4a + 1)

2a2 + a + 1 > 3a2 – 4a + 1

a (0, 5)

Also, 2a2 + a + 1 > 0 and 3a2 – 4a + 1 > 0, we get 1

a3

or a > 1

So, ( )1

a 0, 1, 53




32

55. Let f: R → R be a function such that f(0) = 1 and for x, y R, f(xy + 1) = f(x)·f(y) – f(y) – x + 2 holds

(where [.] denotes the greatest integer function). Then which is the correct option?

(A) ( ) x 1lim f x 1

+→=

(B) ( ) x 2lim f x 2

+→=

(C) ( ) x 0lim f x 0

−→=

(D) ( ) x 0lim f x 1

−→=

Ans. C Sol. f(xy + 1) = f(x) f(y) – f(y) – x + 2 ..... (1) f(xy + 1) = f(x) f(y) – f(x) y + 2 ..... (2) Thus, f(x) + y = f(y) + x Put y = 0, we get f(x) = x + 1 56. Tangent are drawn to the curve y = cos x from the origin. Then the point of contact lies on the

curve x2 – y2 = ax2y2. This value of ‘a’ is equal to (A) –1

(B) 1

2

(C) 1 (D) 2 Ans. C

Sol. Let the point of contact be (, cos ), then y cos

sinx

− = −

−

–cot = , then, we get a = 1

57. If the curve y = x2 + ax + b and y = cx – x2 touch each other at (1, 0). Then the area bounded by

the curve y = x2 + ax + b with x-axis is

(A) 2

3

(B) 5

6

(C) 1

3




33

(D) 1

6

Ans. D Sol. 1 + a + b = 0

c – 1 = 0 c = 1

Also, x 1

dyc 2x

dx =

= − = –1

x 1

dy2x a 1

dx =

= + = −

a = –3; b = 2

Area bounded by y = x2 + ax + b with x-axis

( )2

2 3 22

11

x 3x 1x 3x 2 dx 2x

3 2 6

− + = − + =

58. Number of solutions of

x

2 30

dtx

4 t t=

− − in (0, 1) is/are

(A) 1 (B) 2 (C) 3 (D) none of these Ans. D

Sol. Let ( )x

2 30

dtf x x

4 t t= −

− −

( )2 3

1f x 1 0

4 x x = −

− − x (0, 1)

f(x) is decreasing in (0, 1)

Also, f(0) = 0

f(x) < 0 x (0, 1)

No solution

59. It is given that there are two sets of real numbers A = {a1, a2, ....., a100} and B = {b1, b2, ....., b50}. Total number of onto functions ‘f’ from A to B such that

f(a1) f(a2) f(a3) ..... f(a100) is/are

(A) 100C50 (B) 99C50 (C) 100C49 (D) 99C51 Ans. B




34

Sol. Let first x1 elements of set A maps with b1 and next x2 elements maps with b2 and so on last x50 elements maps with b50

x1 + x2 + ..... + x50 = 100

Number of positive integral solutions = 99C49

60. ( )( )xx x

x 0lim x x

+→− is equal to

(A) 0 (B) 1 (C) –1 (D) does not exist Ans. C

Sol. x

x 0lim x 1

+→=

61. If for all x, y the function f is defined by f(x) + f(y) + f(x)·f(y) = 1 and f(x) > 0, f(x) is differential

everywhere, then

(A) f(0) > f(1)

(B) f(x) = 0 for all x

(C) f(0) < f(1)

(D) none of these Ans. B Sol. Put y = x, we get 2f(x) + (f(x))2 = 1

Differentiate w.r.t to x, we get 2f(x) + 2f(x)f(x) = 0

f(x) (1 + f(x)) = 0

f(x) = 0, because f(x) > 0

62. If ( ) ( )n

n

n 1

xf x 1 loga

n!

=

= + , then at x = 0, f(x)

(A) has no limit (B) is discontinuous (C) is continuous but not differentiable (D) is differentiable Ans. D




35

Sol. ( )( )

n

x loga x

n 0

xlogaf x e a

n!

=

= = =

( ) xf x a= always continuous and differentiable

63. ( )2

nlim cos n n→

+ , n Z is equal to

(A) 0 (B) 1 (C) –1 (D) does not exist Ans. A

Sol. ( ) ( )n2 2cos n n 1 cos n n n + = − − +

2

nlim n n n→

− + 1

2−

( ) ( )n2

nlim cos n n 1 cos 0

2→

+ = − =

64. n/m means that n is a factor of m, then the relation ‘ / ’ is (A) reflexive and symmetric (B) transitive and reflexive (C) transitive and symmetric (D) equivalence Ans. B Sol. ‘ / ‘ is reflexive since every natural number is a factor of itself ‘ / ‘ is transitive. If n is a factor of m and m is a factor of P, then surely n is a factor of P ‘ / ‘ is not symmetric, for example 2 is a factor of 4 but 4 is not a factor of 2

65. If f: R → R is a continuous function satisfying f(0) = 1 and f(3x) – f(x) = x x R and

( ) ( )nn

xlim f x f P x

3→

− =

, then which of the following is correct?

(A) ( )P =

(B) ( )3e

P e2

=

(C) P(5) = 10 (D) P(11) = 5.5 Ans. D




36

Sol. f(3x) – f(x) = x

Replace x by x

3, we get

( )x x

f x f3 3

− =

x x x

f f3 9 9

− =

.......................... ..........................

n 1 n n

x x xf f

3 3 3−

− =

Adding, we get ( )n

x 1 1 1f x f x .....

3 9 273

− = + +

( )nn

x xlim f x f

23→

− =

( )x

P x2

=

66. The value of ( )( )3 2

0

dx

1 x 1 x

+ + is equal to

(A) 3

(B) 2

(C) 6

(D) 4

Ans. D

Sol. ( )( )3 2

0

dxI

1 x 1 x

=+ +

..... (1)

Put x = 1

t

( )( )

3

3 20

xI dx

1 x 1 x

=+ +

..... (2)

Adding equation (1) and (2), we get

2

0

dx2I

21 x

= =+




37

I4

=

67. Let f(x) be a real valued differentiable function defined for all x 1 such that ( )( )42

1f x

x f x =

+ and

f(1) = 1, then

(A) ( )xlim f x 1

4→

+

(B) ( )lim f x 14→

+

(C) ( )x 0lim f x 1

4→

−

(D) we can’t say Ans. A

Sol. ( )( )4 22

1 1f x

x 1x f x =

++

( )x x

21 1

dxf x

1 x

+

( ) ( )1f x tan x f 14

− − +

( ) 1f x tan x 14

− + −

68. If and the roots of the equation x2 + t2x – 2t = 0, ( )2

2 21

1 1 1I t x x dx

−

= + + +

, then

(A) ( )2

2

3t 3I t 3

8 4t= + +

(B) ( )2

2

3t 3I t 3

8 4t= − +

(C) ( )2

2

3t 3I t 3

8 4t= − + −

(D) ( )2

2

3t 3I t 3

8 4t= − + +

Ans. A

Sol. ( )2 2 2

2

2 21

t 1 1 1 3t 3I t x x dx 3

4 t 2t 84t 4t−

= + + + − = + +




38

69. Let ( )x 0lim f x 24

+→= and ( )

x 0lim f x 3

−→= . If ( ) ( )2 3 6 2

x 0 x 0lim f 2x x lim f x x

→ →− = − , then is equal to

(A) 27 (B) 21 (C) 12 (D) none of these Ans. D

Sol. ( ) ( )3 2

x 0 x 0lim f 2x x lim f x 24

+→ →− = =

( ) ( )6 2

x 0 x 0lim f x x lim f x 3

−→ →− = =

= 8

70. Let f: (0, ) → (0, ) be a differentiable surjective function and F(x) is the anti-derivative of f(x) such

that 2(F(x) – f(x)) = f2(x) for x R+, then

(A) ( )

x

f xlim 2

x→=

(B) ( )

x

f xlim 1

x→=

(C) f(x) is strictly decreasing function (D) can’t say Ans. B Sol. 2(F(x) – f(x)) = f2(x)

and ( )dF

f xdx

=

( ) ( ) ( )( )F x f x 1 f x = +

( )( )

( )

f xf x

1 f x =

+

( )xlim f x→

→

As f(x) is increasing surjective function

( )

x

f xlim 1

x→=




39



71. Let Mn = {0.a1a2 ..... an, such that ai = 0, 1, for 1 i n – 1, an = 1}, be a set of decimal fractions, Tn

and Sn be the number and sum of the elements of Mn respectively, then n

nn

54Slim

T→ is equal to

Ans. 3

Sol. The frequency of ai = 1 is the same as that of ai = 0 for 1 i n – 1 and an = 1, then

n 1 n 1n 2 n 1 n

1 1 1 1 1S 2 ..... 2

2 10 10 10 10

− −

−

= + + + +

n

n 1 nn nn

S 1 1 1 1lim lim 1

T 18 1810 10−→ →

= − + =

72. Let area bounded by the parabola y2 + 4y = 4x and y = mx + m – 2 is 9 sq. units, then value of |9m|

is equal to Ans. 6

Sol. Area bounded by y2 = 4x and y = mx is 3

8

3m

3

89

3m=

2m

3=

SECTION – C

(Numerical Answer Type) This section contains 03 questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the second

decimal place; e.g. xxxxx.xx).

73. The number of non-zero integral values of ‘a’ for which the function ( )2

4 3 27xf x x ax 1

2= + + + is

concave upward for x R

Ans. 00012.00

Sol. f(x) = 12x2 + 6ax + 27 0

a [–6, 6]

74. Let f(x) = ax + cos 2x + sin x + cos x is defined for x R and a R, f(x) is strictly increasing

function the range of a is ,2

, then is

Ans. 00004.25




40

Sol. f(x) = a – 2 sin 2x + cos x – sin x = a + 2(cos x – sin x)2 + (cos x – sin x) – 2

Minimum values of 2(cos x – sin x)2 + (cos x – sin x) – 2 is 17

8−

So, 17

a 08

− for increasing function

17

a8

75. Let ( )( ) ( )1

2

0

12x 3f x f x dx

15− = and ( )

1

0

1f x dx

k= , then k is equal to

Ans. 00009.00

Sol. ( )

21 4 2

0

x x 1f x dx

9 3 45

− − =

( )2x

f x3

= ( )1

0

1f x dx

9=

Documents

FIITJEE ALL INDIA TEST SERIES · AITS-PT-I-PCM-JEE(Main)/205 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942