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Aakash Educational Services Limited – Registered Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 []

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.011-47623456

REVIEW ASSIGNMENT-2 (JEE Main) (PHYSICS)

Topics covered :

Complete syllabus of JEE(Main)

Choose the correct answer :

1. If 1 m/s2 is equal to x and 1 km/min2 is equal

to y then

(1) x < y (2) x = 3.6y

(3) y = 3.6x (4) x = 360y

2. A particle moves on x-axis such that its

position is given by

x = |t – T|m (Here t is time and T is a constant)

If the average velocity of the particle for first

6 s is zero then the distance covered by the

particle in first 3 s is

(1) 2 m (2) 6 m

(3) 3 m (4) 1.5 m

3. Loudness of sound at a distance x from a point

source of sound is L. At what distance from the

point source loudness is 2

L ?

(1) 2x

(2) 2x

(3) 4x

(4) It may be any of the above values

4. a, b and c represent maximum particle

velocity, maximum particle acceleration and

wave speed respectively for a harmonic

transverse wave travelling in negative y

direction on a string. Which of these equations

may be representing the wave?

(1) 2sin

6

a a bx t y

b b ac

(2) 2sin

a a bx t y

b b ac

(3) 2

sin6

a b bx t y

b a ac

(4) 2

sina b b

x t yb a ac

5. A body is projected from a height 2

R above

surface of earth with a minimum speed so that

it does not fall on the surface of earth. If mass

and radius of earth are M and R respectively,

then the speed of the body when it is closest

to earth is

(1) 3

2

GM

R (2)

6

7

GM

R

(3) 5

6

GM

R (4)

6

5

GM

R

6. A heavy rod of mass m, length , hangs

vertically downward from a rigid support as

shown in the figure. Young's modulus of

material of rod is Y and area of cross section

is A. The elastic potential energy of rod is

(1) 2 2 3

6

M g L

AY (2)

2 2 2

6

M g L

AY

(3) 2 2

3

M g L

AY (4)

2 2

6

M g L

AY

25/03/2020

JEE Main Review Assignment-2


7. A block of mass 2.0 kg is moving on a

frictionless horizontal surface with a velocity v

m/s towards another block of equal mass kept

at rest. The spring constant of the spring fixed

at one end is 100 N/m. If maximum

compression of the spring is 10 cm, the

velocity v is

(1) 1 m/s (2) 0.5 m/s

(3) 0.25 m/s (4) 0.75 m/s

8. Two blocks of masses m1 and m2 are

connected by a light inextensible string

passing over a smooth fixed pulley of

negligible mass. Acceleration of centre of

mass of system is

(1) 1 2

1 2

m mg

m m

(2) 1 2

1 2

m mg

m m

(3) 1 2

1 2

2m mg

m m

(4) 2

1 2

1 2

m mg

m m

9. In shown figure, if mass A will go up by a

distance x, mass B will go

(1) 3x down (2) 4x down

(3) 2x down (4) 8x down

10. A cylinder whose inside diameter is 4.00 cm

contains air compressed by a piston of mass

m = 13.0 kg, which can slide freely in the

cylinder. The entire arrangement is immersed

in a water bath whose temperature can be

controlled. The system is initially in equilibrium

at temperature ti = 20°C. The initial height of

the piston above the bottom of the cylinder is

hi = 4.00 cm. The temperature of the water

bath is gradually increased to a final

temperature tf = 100°C. The height hf of the

piston is

(1) 10.18 cm (2) 5.09 cm (3) 2.03 cm (4) 20.36 cm

11. The above composite block is shown as a

2-dimensional figure below

The block is made up of materials of two

different thermal conductivities as shown. The equivalent thermal conductivity between face A and face B is given by

(1) 2 1

2

1

ln

k kk

k

(2) 2 1

2 1

k k

k k

(3) (k2 – k1) ln 2

1

k

k

(4) 2

12 1( )

k

kk k e

12. For a heat engine the temperature of the source is 127ºC. To have 60% efficiency the temperature of the sink is

(1) 160ºC (2) –200ºC (3) –113ºC (4) 113ºC 13. The two arms, of a vertical U-tube containing

a liquid, are maintained at different

temperatures, t1 and t2. The liquid columns in

the two arms have heights l1 and l2

respectively. The coefficient of volume expansion of the liquid is equal to (Initial temperature is assume to be zero)

(1) 1 2

2 1 1 2t t

l l

l l (2) 1 2

1 1 2 2t t

l l

l l

(3) 1 2

2 1 1 2t t

l l

l l (4) 1 2

1 1 2 2t t

l l

l l

Review Assignment-2 JEE Main


14. The amount of work done by the agent to

change the configuration A into B is

(1) 2

04

Q

l (2)

2

08

Q

l

(3) 2

02

Q

l (4)

2

0

3

4

Q

l

15. A charged particle of specific charge

(charge/mass) is released from origin at time

t = 0, with velocity 0ˆ ˆ( )v v i j

in uniform

magnetic field 0ˆB B i

. Co-ordinates of the

particle at time 0

tB

are

(1) 0 0 0

0 0 0

2, ,

2

v v v

B B B

(2) 0

0

, 0, 02

v

B

(3) 0 0

0 0

20, ,

2

v v

B B

(4) 0 0

0 0

2, 0,

v v

B B

16. Consider a conducting ring of radius r placed

in a plane containing two parallel infinite wires

as shown. The wires carry equal and opposite

current i. Distance of the wires from centre of

the ring are also shown. For magnetic field at

the centre of the ring to be zero, the ring

should be carrying current

(1) 2

i

, anticlockwise (2)

2

i

, clockwise

(3) 6

i

, anticlockwise (4)

6

i

, clockwise

17. Two identical rings of identical radii and

resistance of 36 each are placed in such a

way that they cross each others centre C1 and

C2 as shown. Conducting joints are made at

intersection points A and B of the rings. An

ideal cell of emf 20 V is connected across AB.

The power delivered by cell is

(1) 80 W

(2) 100 W

(3) 120 W

(4) 200 W

18. The value of current passing through battery

immediately after (K) is closed, If initially all

capacitors are uncharged, will be

(1) 5

E

R

(2) 3

E

R

(3) 6

5

E

R

(4) Zero

19. Resistance of a wire at temperature °C is

given by relation 0 1b

R R a

. Here

R0 is the temperature at 0°C. The temperature

coefficient of the material of the wire at °C is

(1) 2

2

( )a b

(2) 2

2

( )

( )

a b

a b

(3) 2

( )

( )

a b

a a b

(4) 2

a b

a b



20. A source of red light is placed at the bottom of two transparent media having refractive index

= 2 and 2 as shown in the figure. The

maximum area in which red is seen when observed from above is nearly

(1) 50.1 cm2 (2) 40 cm2

(3) 100 cm2 (4) 20 cm2

21. A ray of light falls on a prism of refractive index

2 at an angle i with the normal. What

should be the minimum value of i for the total internal reflection will take place at side AC is

(1) 30° (2) 45°

(3) 60° (4) 90°

22. If in the given diagram OO is the principal axis.

P is the point object and P is the image, then the mirror used and location of object (P) w.r.t. mirror is

(1) Concave mirror with P beyond centre of curvature

(2) Concave mirror with P between focus and pole

(3) Concave mirror with P between focus and centre of curvature

(4) Convex mirror with P at 2.5f on right,

where f is the focal length of the mirror

23. Select the correct alternative.

(1) In Young's double slit experiment (YDSE),

if one of the slits is covered, the intensity

at the centre becomes half

(2) If YDSE is performed under water instead

of air, fringe width will become 1

times of

the value in air

(3) If one of the slits is covered with a

transparent sheet, the interference pattern

shifts to the opposite side in which

transparent sheet is introduced

(4) All of these

24. A graph is plotted between rate of

disintegration and number of active nuclei.

What is half life of the radioactive substance?

(Here time is in second)

(1) 3 s (2) 1

s3

(3) 3 ln2s (4) ln2

s3

25. A nucleus of mass number A originally at rest

emits -particle with kinetic energy E. The

kinetic energy of the daughter nucleus is equal

to

(1) 4E

A (2)

4

( – 4)

E

A

(3) ( – 4)

4

A E (4)

( – 4)

AE

A




REVIEW ASSIGNMENT-2 (JEE Main) (PHYSICS)

[ANSWERS]

1. (2)

2. (3)

3. (4)

4. (3)

5. (4)

6. (4)

7. (1)

8. (4)

9. (1)

10. (2)

11. (1)

12. (3)

13. (1)

14. (1)

15. (4)

16. (1)

17. (2)

18. (3)

19. (2)

20. (1)

21. (2)

22. (3)

23. (2)

24. (4)

25. (2)

25/03/2020



[ANSWERS]

1. Answer (2)

2

2

1km

10001m/s1

minute3600

x = 3.6 km/minute2

= 3.6y

2. Answer (3)

Above curve shows the position versus time curve.

3. Answer (4)

L = 10 log 0

I

I

If distance becomes n times, as I 2

1

r for point

source I becomes 2

1

n times

L= 10 log 2

0

I

n I

= 10 log 0

I

I – 10 log n2

= 10 log 0

I

I – 20 log n

= 2

L

20 log n = 2

L

n may be any value depending on L

4. Answer (3)

vmax = a = A, amax = b = A2, v = c

= ,b

a A =

2

,a b

kb v ac

x = Asin (t + ky + )

5. Answer (4)

The particle will revolve around earth in elliptical path as shown, when projected at A

By conservation of angular momentum,

11 2 2

33

2 2

vRv v R v

By conservation of energy,

2 21 2

1 2 1–

2 3 2

GMm GMmmv mv

R R

Solving v2 = 6

5

GM

R

6. Answer (4)

Taking small element of length dy at distance y

from lower end. Stress = 1

,2

M g dUY

L A dV stress

× strain.

21 (Stress)

Volume2

dUY

21(Stress)

2A dy

Y

2 2

6

M g LU

YA

7. Answer (1)

At maximum compression, velocity of both the block will same

2v = 2v1 + 2v1

1 2

vv

From conservation of energy

2 2

2 21 1 1 1

2 2 2 2 2 2

v vmv m m kx

v = 1 m/s

8. Answer (4)

Acceleration of both blocks

2 1

1 2

m ma g

m m

... (1)

21 2 1 2

cm

1 2 1 2

ˆ ˆ( ) ( )m a m a m a j m a ja

m m m m

1 2 1 2 2 1

1 2 1 2 1 2

ˆ ˆ( )m m m m m m

a j gjm m m m m m

2

1 2

1 2

ˆ( )m m

g jm m



9. Answer (1)

Point P will go up, pulley y will go down by x will

make 2x string free and length x due to

movement of Q upward.

10. Answer (2)

i f

i f

v v

T T i f

i f

h h

T T

ff i

i

Th h

T

373

4.00 5.09 cm293fh

11. Answer (1)

Thermal resistance of small element is

dR = 2

2 1 1

1

( )

ldx

k k x k lb

R = 2 1 12

12 1

( )ln

( )

k k l k lldR

k lk k b

k =l

AR = 2 1

2

1

ln

k kk

k

12. Answer (3)

2 2

1

601 1

100 (273 127)

T T

T

2 20.6 1 0.4400 400

T T

T2 = 160 K = –113°C

13. Answer (1)

2 1

2 11 1t t

l l

14. Answer (1)

2 2

2 – 22

kQ kQW U

l l

= 2 2

04

kQ Q

l l

15. Answer (4)

Time period = 2 m

Bq

2

B

0 2

Tt

B

path = helical

axis of helix = x-axis

x-coordinate = 0

0

v

B

z-coordinate = –zr 0

0

2v

B

So, 0 0

0 0

2, 0,

v v

B B

16. Answer (1)

Magnetic field due to wires at the centre of the

ring

= 0 0 0

2 (3 ) 2 (6 ) 4

i i i

r r r

directed down the plane of ring

Ring should carry anticlockwise current,

say i, such that

| |B

due to ring = | |B

due to wires

0 0

2 4

i i

r r

2

ii

(anticlockwise)

17. Answer (2)

AC1 = AC2 = C1C2 = radius

AC1B = 120



1 1 1 1 1

24 12 12 24R

R = 4

2 2(20)

100 W4

VP

R

18. Answer (3)

net 3 2

R RR

5

6

R

6

5

Ei

R

19. Answer (2)

0 2

dR bR a

d

dR

Rd

0 22

2

0

( )

( )1

bR a

a bb a b

R a

20. Answer (1)

2 sin 2 sini r and 2 sin 1 sin2

r

1

sin 30 and 452

i i r

Radius of the circle on the upper surface is

3.46 tan30 2 tan 45 = 4 cm

Area = R2 = (4 × 10–2)2 = 16 × ×

10–4 cm2

= 50.1 cm2

21. Answer (2)

r1 aC

11

1sinr

r1 45° r1 = 45°

and r1 + r2 = 75°

r2 = 30°,

Also, 12

11 sin sin sin 2

2i r i

i = 45°

22. Answer (3)

A concave mirror with object placed between

focus and centre of curvature gives an inverted

and enlarged image.

23. Answer (2)

In any medium wavelength becomes 0

= fringe width = D

d

will become

1

times.

24. Answer (4)

–dN

Ndt

so slope = 1

tan30º

3

Half life ln2

3

25. Answer (2)

1 2

2 1

E m

E m

2

– 4

4

E A

E

E2 = 4

( – 4)E

A

Aakash Educational Services Limited – Registered Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 [Page 1]


REVIEW ASSIGNMENT-2 (JEE Main) (CHEMISTRY)

Topics covered :

Complete syllabus of JEE (Main)


1. 0.250 g of a diacidic organic base required

15 ml of N/5 HCl for complete neutralisation.

The molecular mass of the base is

(1) 41.66 (2) 83.33

(3) 166.66 (4) 124.99

2. Thermite mixture contains

(1) 3 parts Fe2O3 and 2 parts Al by mass

(2) 3 parts Al2O3 and 4 parts Al by mass

(3) 1 part Fe2O3 and 12 parts Al by mass

(4) 3 parts Fe2O3 and 1 part Al by mass

3. Alkaline earth metal having minimum density is

(1) Mg (2) Be

(3) Ba (4) Ca

4.

(1)

(2)

(3)

(4)

5. The reactivity order of the following w.r.t.

electrophile is

I.

II.

III.

IV.

(1) I > II > III > IV

(2) II > I > III > IV

(3) II > III > I > IV

(4) I > III > II > IV

6. 15 mL of a gaseous sample believed to

contain C2H4 and C2H2 requires 45 mL of O2

for complete combustion. Determine

percentage composition of C2H2 in the

mixture.

(1) 15 mL (2) 7 mL

(3) 5 mL (4) Zero

7. Kinetic energy of the electron in 3rd shell of

Li+2 is

(1) –13.6 eV (2) 13.6 eV

(3) –27.2 eV (4) 27.2 eV

8. If x-axis is the internuclear axis then which of

the following is the correct combination for p-

d bond?

(1) px-dxy

(2) py-dyz

(3) 2z zp -d

(4) pz-dzx

25/03/2020



9. Out of the following mentioned reactions, the

one which has highest rate is

(I)

(II)

(III)

(IV)

(1) I (2) II

(3) III (4) IV

10. Identify the non-narcotic analgesic among the

following.

(1) Morphine (2) Paracetamol

(3) Codeine (4) Heroin

11. A sample of ideal gas is compressed from

initial volume of 2V0 to V0 using three different

processes

(A) Reversible isothermal

(B) Reversible adiabatic

(C) Irreversible adiabatic under a constant

external pressure

Then

(1) Final temperature of gas will be highest at

the end of process (B).

(2) Magnitude of enthalpy change of sample

will be highest in isothermal process

(3) Final temperature of gas will be highest at

the end of process (C).

(4) Final pressure of gas will be highest at the

end of process (B).

12. FeC2O4 solution of some concentration is

titrated with KMnO4 in presence of HCl and in

presence of H2SO4 separately. If VHCI and

2 4H SOV represent volume of oxidising agent

consumed at the two cases respectively, then

which of the options is correct ? (Ignore the

reaction between FeC2O4 and Cl2)

(1) VHCI = 2 4H SOV

(2) VHCI > 2 4H SOV

(3) VHCI < 2 4H SOV but VHCI 2 4H SO

1V

2

(4) VHCI = 2 4H SO

1V

2

13. Vapour pressure of C6H6 and C7H8 mixture at

50°C is given by P (mm Hg) =180 XB+ 90,

where XB is the mole fraction of C6H6. A

solution is prepared by mixing 936 g benzene

and 736 g toluene and if the vapours over this

solution are removed and condensed into

liquid and again brought to the temperature of

50°C, what would be the new mole fraction of

C6H6 in the vapour state ?

(1) 0.93 (2) 1.93

(3) 1.00 (4) 2.93

14. 20 ml of KOH solution was titrated with 0.20 M

H2SO4 solution in a conductivity cell. The data

obtained were plotted to give the graph shown

below :

The concentration of the KOH solution was :

(1) 0.30 mol L–1 (2) 0.15 mol L–1

(3) 0.12 mol L–1 (4) 0.075 mol L–1

15. 84Po210 decays with particle to 82Pb206 with a

half-life of 138.4 days. If 1.0 g of 84Po210 is

placed in a sealed tube, how much helium will

accumulate in 69.2 days. Express the answer

in cm3 at STP.

(1) 11.25 cm3

(2) 21.25 cm3

(3) 31.25 cm3

(4) 41.25 cm3



16.

The products of ozonolysis are (considering

benzene does not undergo ozondysis)

(1)

(2)

(3)

(4)

17. Identify the final product

3 4

2 43 2 6 4 3

HNO MnO /HH SO

PhCH p O N C H CH (X)

2 52

3

P O(i) SOCl(ii) NH ,

(Y) (Z)

(1) p-O2N—C6H4—CO2H

(2) p-O2N—C6H4—CN

(3) p-O2N—C6H4—CONH2

(4) p-O2N—C6H4—CH2CN

18.

(1) HCHO & HCOCH(OCH3)2

(2) (CH3O)2CH – COOH & HCHO

(3) (CH3O)2CH – CHO + HCOOH + HCHO

(4) HCOOH & HCOCH(OCH3)2

19. When tin is treated with concentrated nitric

acid :

(1) It is converted into stannous nitrate

(2) It is converted into stannic nitrate

(3) It is converted into metastannic acid

(4) It becomes passive

20. Match column-I with column-II

Column-I (Chemical reaction)

I. 800 C/Pt3 2 24NH 5O 4NO 6H O

II. 2400 450 C/CuCI2 2 24HCI O 2CI 2H O

III. 2 5V O2 2 3450 500 C

2SO O 2SO

IV. Fe Mo2 2 3500 C

2N 3H 2NH

Column-II (Name of process)

(a) Contact process

(b) Ostwald’s process

(c) Deacon’s process

(d) Haber’s process

(1) I-a, II-b, III-d, IV-c

(2) I-b, II-c, III-a, IV-d

(3) I-a, II-d, III-c, IV-b

(4) I-a, II-c, III-b, IV-d



Integer Value Correct Type Questions

21. The ratio of Fe+3 and Fe+2 ion per unit cell of

Fe3O4 in octahedral voids is _________.

22. 24.95 g of CuSO4·5H2O is dissolved in 171 g

of H2O. If the percentage lowering of vapour

pressure at 100°C is x, x is closest to integer

________ .

23. Which among the following species exhibit

geometrical isomerism?

[Co(NH3)4(H2O)2]Cl2, [Ni(H2O)6]Cl2,

[Co(H2O)5Cl]Cl, [Pt(NH3)2Cl2],

[Pt(en)Cl2], [Pt(en)2Cl2],

[PdCl2BrI], [Co(NH3)4Cl2],

[Co(en)2Br2]

24. Find the pH at isoelectric point of the following

-amino acid

Given : 1apK = 2,

2apK = 4 and 3apK = 8

25. Initially only PCl5 (g) is present at 6 atm. If at

equilibrium it is partially converted into PCl3

and Cl2 gases and total pressure becomes 8

atm, the value of Kp in atm will be ________.




REVIEW ASSIGNMENT-2 (JEE Main) (CHEMISTRY)

[ANSWERS]

1. (3)

2. (4)

3. (4)

4. (3)

5. (1)

6. (4)

7. (2)

8. (4)

9. (4)

10. (2)

11. (3)

12. (2)

13. (1)

14. (1)

15. (3)

16. (2)

17. (2)

18. (3)

19. (3)

20. (2)

21. (1)

22. (2)

23. (6)

24. (3)

25. (1)

25/03/2020



[SOLUTIONS]

1. Answer (3)

At the completion of reaction,

Molecular equivalent of acid = Molecular equivalent of base

0.25 1

2 1000 15Molecular weight 5

Molecular weight = 0.25 2 1000 5

15

= 166.66 g

2. Answer (4)

Fact.

3. Answer (4)

Density of Ca = 1.55 g cm3–, Be > Mg > Ca < Sr < Ba

Due to very large volume.

4. Answer (3)

5. Answer (1)

Activating power

–OCH3 > –CH3 > –Cl > –NO2

6. Answer (4)

2 4 2 2 2

x mL 3x mL

C H 3O 2CO 2H O

2 2 2 2 2

2.5(15 x) mL(15 x) m

5C H O 2CO H O

2

Total volume of O2

2.5(15 – x) + 3x = 45

x = 15

Volume of C2H2 in the sample = 0

7. Answer (2)

Kinetic energy of e–

– E = 2

2

13.6 313.6 eV

3

8. Answer (4)

9. Answer (4)

Toluene is more reactive than benzene and

formed carbocation as electrophile is best

favoured in so, reaction IV has most rate of

reaction.

10. Answer (2)

Fact.

11. Answer (3)

Work done on the system in an irreversible

adiabatic process is maximum. So, rise in

temperature of the system in such a process will

also be maximum.

12. Answer (2)

Part of KMnO4 will be used to oxidise HCl in

addition to oxidise FeC2O4. Therefore, vol of

KMnO4 used at the end point in presence of HCl

(VHCl) will be more than that used in presence of

H2SO4 2 4H SOV

13. Answer (1)

Vapour pressure of pure C6H6 B(P ) = 270 mm

Vapour pressure of pure toluene T(P ) 90 mm

B T

936 736n 12; n 8

78 92

XB = 0.6 and XT = 0.4

1 1B TP 0.6 270;P 0.4 90;

1 11 B TP P P 162 36 198 mm

1 1B T

162 36Y 0.82;Y 0.18

198 198

2BP 0.82 270;

2 2 2T 2 B TP 0.18 90; P P P

= 221.4 + 16.2

= 237.6 mm

2B

221.4Y 0.93

237.6



14. Answer (1)

Let the molarity of KOH be x mol/L

20 × x = (0.2 × 2) × 15; x = 0.30 mol/L

15. Answer (3)

No= initial number of moles of 0

1P

210

1t

2 half life of P0 = 138.4 days

n= no. of half lives in 69.2 days = 69.2 1

138.4 2

Nt, number of moles of p0 left at t = 69.2 days

12

t 0

1N N

2

No. of moles of He formed

= N0 – Nt

= 0 0

1 2 1N 1 N

2 2

= 0.293 N0

Volume of He gas formed at STP

= 0.293 22400

31.25cc210

16. Answer (2)

17. Answer (2)

18. Answer (3)

19. Answer (3)

20. Answer (2)

21. Answer (1)

Fe3O4

Fe+2 in 1

th4

of octahedral voids

Fe+3 in 1

th4

of octahedral voids + 1

th8

of

tetrahedral voids



22. Answer (2)

Mass of CuSO4 in 25 g CuSO4·5H2O = 15.98 g

Mass of water = 196 – 15.98 = 180 g

o

15.982P 2 0.1 0.2159.5 0.019615.98 180 2 0.1 10 10.2P 2159.5 10

% = 0.0196 × 100 = 1.96 2

23. Answer (6)

[Ni(H2O)6]Cl2, [Co(H2O)5Cl]Cl, [Pt(en)Cl2] does

not show geometrical isomerism.

24. Answer (3)

The pH at isoelectric point of the given amino

acid is given by

p1 = 1 2a apK pK

2

= 2 4

2

= 3

25. Answer (1)

6 – p + 2p = 8

p = 2

Kp = 2 2

16 2



REVIEW ASSIGNMENT-2 (JEE Main) (MATHEMATICS)

Topics covered :

Complete syllabus of JEE (Main)


1. The number of distinct rational numbers of the

form p/q, where , {1, 2, 3, 4, 5, 6}p q is

(1) 23 (2) 32

(3) 36 (4) 28

2. The point (1, 2) is one extremity of focal chord

of parabola 2 4y x . The length of this focal

chord is

(1) 2 (2) 4

(3) 6 (4) None of these

3. If [x] denotes the integral part of x, then domain

of the function

13 3 2( ) sin

( 1)( 2)( 3) 2

x xf x

x x x

is

(1) [0, 2) (2) [0, 2) – {1}

(3) (– , 3) – {1, 2} (4) None of these

4. If the function 3 2( ) 11 6f x ax bx x

satisfies conditions of Rolle’s theorem in [1, 3]

and

12 0,

3f then value of a and b are

respectively

(1) 1, –6 (2) –1, 6

(3) –2, 1 (4) –1, 1

2

5. If n arithmetic means are inserted between two

sets of numbers a, 2b and 2a, b, where

a, b R. Suppose that mth arithmetic mean

between these two sets of numbers is same,

then the ratio a : b equals

(1) 1:n m m (2) 1:n m n

(3) : 1m n m (4) : 1n n m

6. If 1 be a cube root of unity and

7(1 ) l m , then the value of l m

(1) 0 (2) 1

(3) 2 (4) –1

7. If a and b are chosen randomly from the set

consisting of numbers 1, 2, 3, 4, 5, 6 with

replacement. Then probability that

2/

0lim

2

xx x

x

a b= 6 is

(1) 1

3 (2)

1

4

(3) 1

9 (4)

2

9

8. From origin, chords are drawn to the circle

2 2 2 0x y y . The locus of the middle

points of these chords is

(1) 2 2 0x y y

(2) 2 2 0x y x

(3) 2 2 2 0x y x

(4) 2 2 0x y x y

25/03/2020



9. Identify the pair of functions which are not

identical.

(1) 21 1

tan cos ; x

y x yx

(2) 1 1tan cot ; y x y

x

(3) 2

sin arc tan ; 1

xy x y

x

(4) 1 1cos tan ; sin tany x y x

10. If z is a complex number satisfying

4 3 22 1 0,z z z z then the set of

possible values of | z | is

(1) {1, 2}

(2) {1}

(3) {1, 2, 3} (4) {1, 2, 3, 4}

11. Integral of 1 2cot (cot cosec )x x x with

respect to x is:

(1) 2lncos2

xc

(2) 2lnsin2

xc

(3) 1

lncos2 2

xc

(4) ln sin x ln(cosec x cot x) + c

12. 2

cos

0

cos sine d

(1) (2) 2

(3) 0 (4) 3

13. For each real number x such that – 1 < x < 1,

let A(x) be the matrix 1 1(1 )

1

xx

x

and

.1

x yz

xy

Then,

(1) A(z) = A(x) + A(y)

(2) A(z) = A(x) [A(y)]–1

(3) A(z) = A(x) A(y)

(4) A(z) = A(x) – A(y)

14. The length of the perpendicular from the origin

to the plane passing through the point a and

containing the line r b c is

(1) [ ]

| |

a b c

a b b c c a

(2) [ ]

| |

a b c

a b b c

(3) [ ]

| |

a b c

b c c a

(4) [ ]

| |

a b c

c a a b

15. If the standard deviation of 1 2, , ..... nx x x is

3.5, then the standard deviation of

1 22 3, 2 3,..., 2 3nx x x is

(1) –7 (2) –4

(3) 7 (4) 1.75

16. If 1( )y x is a solution of the differential

equation / ( ) 0,dy dx f x y then a solution of

the differential equation ( ) ( )dy

f x y r xdx

(1) 11

1( ) ( )

( )r x y x dx

y x

(2) 11

( )( )

( )

r xy x dx

y x

(3) 1( ) ( )r x y x dx

(4) None of these

17. The distance between the directrices of the

ellipse 22 24 8 16 3 10x y x y is K

then 2

K is

18. The number of distinct real roots of the

equation

cos sin sin

sin cos sin 0

sin sin cos

x x x

x x x

x x x

which lies

in the interval 4 4

x

is______



19. Given /2

0

1lnsin ln

2 2xdx

and

2/2

0

.ln2

sin 2

x kdx

x

then k = ______

20. The number of ordered pairs (x, y) where

, [0, 10]x y satisfying

22 sec1

sin sin .2 12

yx x

is 2K then K =

21. If (a, b] is the domain of the function

0.52 8

log2

xf x

x

then value of b – a

is___

22. In a triangle ABC, 5, 2,6

a b A

and

c1, c2 are the possible values of the third side.

Then |c1 – c2| is

23. 4 341 2

nn n n no n nC C C C C

4 6 4 93 ...n n n

n nC C C to n terms is equal

to (1 )nk then the value of k is

24. Let PN be the ordinate of a point P on the

hyperbola 2 2

2 21

97 79

x y and the tangent at P

meets the transverse axis in T. Then the value

of 2020

ON OT

is

(O being origin and [] denotes the greatest

integer function)

25. Find natural number n so that area bounded

by y = nx2, 21

2y nx and 2 4 3 0y y is

greatest.




REVIEW ASSIGNMENT-2 (JEE Main) (MATHEMATICS)

[ANSWERS]

1. (1)

2. (2)

3. (2)

4. (1)

5. (3)

6. (3)

7. (3)

8. (1)

9. (4)

10. (2)

11. (2)

12. (2)

13. (3)

14. (3)

15. (3)

16. (1)

17. (8)

18. (2)

19. (2)

20. (8)

21. (4)

22. (04)

23. (2)

24. (4)

25. (1)

25/03/2020



[ANSWERS]

1. Answer (1)

1, 1, 2, 3, 4, 5, 6 6p q

2, 1, 3, 4, 5, 6 3 [ (2, 4), (2, 6)]p q

3, 1, 2, 4, 5, 6 4 [ (3, 6)]p q

4, 1, 3, 5, 6 3 [ (4, 6)]p q

5, 1, 2, 3, 4, 6 5p q

6, 1, 5 2p q

2. Answer (2)

The parabola 2 4 ,y x here 1a and focus is

(1, 0).

The focal chord is ASB. This is clearly latus

rectum of parabola, its value = 4

3. Answer (2)

For f(x) to be defined

(i) 3 2

1, 0, 12

x

3 2

1 2 2 3 2 42

xx

0 3 6 0 2x x ...(A)

(ii) 3 0x and 1, 2, 3x

x < 3 and 1, 2x ...(B)

From (A) and (B), we have

0 1x or 1 < x < 2

Domain 0,1 1,2 0,2 1

4. Answer (1)

f(1) = f(3)

a + b + 11 – 6 = 27a + 9b + 33 – 6

13a + 4b = –11

and 23 2 11f x ax bx ...(i)

2

1 12 3 2

3 3f a

1

2 2 11 03

b

1 4

3 43 3

a

1

2 2 11 03

b

...(ii)

From eqs. (i) and (ii), we get

a = 1, b = –6.

5. Answer (3)

Let 1 2, ..... nA A A be airthmetric means between

a and 2b , then 2

1mb a

A a mn

Again , let 1 2, ........ nB B B be arithmetic means

Between 2a and b then 2

21m

b aB a m

n

Now ,

2 2

21 1m m

b a b aA B a m a m

n n

1 1

b a a mm a

n b n m

6. Answer (3)

is cube root of unity.

21 0

21

Now if 7(1 ) l m

2 7( ) l m

14 l m

12 2 l m

3 4 2( ) l m

2 l m

1 l m

can comparison l = 1, m = 1

X

Y

B

A(1, 2)

S(1, 0)



7. Answer (3)

2/

0lim 6

2

xx x

x

a b

0

1 1lim

6

x x

x

a b

x xe

log log 6a be

ab = 6

(a, b) = (1, 6), (6, 1), (2, 3), (3, 2)

Required probability = 4 1

6 6 9

8. Answer (1)

1T S

i.e., 2 21 1 1 1 1 1( ) 2xx yy y y x y y

Passes through (0, 0)

2 2 0x y y

9. Answer (4)

Conceptual

10. Answer (2)

The given equation is 2 2( 1)( 1) 0z z z .

2, , ,z i w w w being an imaginary cube root

of unity. Thus | | 1z .

11. Answer (2)

I = 21 2cosec cot 2cotx x x

= 2 2cosec 2cosec cot cotx x x x dx

= (cosec cot )x x dx

12. Answer (2)

Real part of 2

0

iee d

put ie z

Real part of 2

0

12

zedz

i z

.

13. Answer (3)

1

11( )

1 (1 )(1 )1

1

x y

xyx y xyA z A

xy x y x y

xy

( ). ( ) ( )A x A y A z

14. Answer (3)

Given plane passes through and a b

containing the line is [ ] 0AP AB c

( ) (( ) ) 0r a b a c

( ) [ ]r b c c a a b c

length of r from the origin [ ]

| |

a b c

b c c a

15. Answer (3)

We know that if ii

x Ad

h

then | |x dh .

In this case 3 / 2

2 31/ 2

ii

xx

.

So 1

2h .

Thus 1

2 3.5 7| |d xh

.

16. Answer (1)

( ). 0dy

f x ydx

( )dy

f x dxy

ln ( )y f x dx

( )

1( )f x dx

y x e Then for given equation I.F =

( )f x dxe

Hence Solution 1 1. ( ) ( ). ( )y y x r x y x dx

11

1( ). ( )

( )y r x y x dx

y x

17. Answer (8)

2 2

2 2 1 ( 3 10)( 2)

2 4

x yx y

1, , 0 ,

2h k z e

Perpendicular distance from (2, 0) to

3 10 0x y is a

aee

2 6 42

aa a

Distance between directrics = 2

16a

Ke



18. Answer (2)

The given equation can be identified as

2(cos sin ) (2sin cos ) 0x x x x

4

x

and 1 1tan

2x

19. Answer (2)

/2 /2/22 2 2

00 0

cos cot 2 cotI x ec xdx x x x xdx

/22

/2

00 0

0 2 lnsin 2 lnsintanx

xLt x x xdx

x

0

10 2 lnsin 2 ln ln2

2 2xLt x x

0 0

2

1.coslnsin sin 011/x x

xx xLt Ltx

x

20. Answer (8)

2

2 1 1 1 1sin sin sin

2 2 4 2x x x

and

22 sec(sec ) 1, 2 2yy

It is possible only when 21sin , sec 1

2x y

5 13 17, , ,

6 6 6 6x

0, , 2 , 3y

No. of ordered pairs = 16

21. Answer (4)

0.52 8

log 02

x

x

2 8 1

2

x

x

(2, 6] 4x b a

22. Answer (04)

cos 2 2 2

2 2 3 1 02

b c aA C C

bc

Now C1 + C2 = 2 3, C1C2 = – 1

So |C1 – C2| = 4

23. Answer (2)

Coefficient xn in 4 4 311 1

n nn noC x C x

4 62 1 ... 1 1

n n nn nnC x C x

Coefficient xn in 31 1 1 3

nn nx x

24. Answer (4)

ON OT = 97 cos 97sec = 972

97

42020

25. Answer (1)

3

1

2Area

y ydy

n n

is greatest when n is least

y = 1

y nx = 2

y nx = 1/2 2

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