fiitjee punjabi bagh

FIITJEE

FIITJEE Ltd., Punjabi Bagh Centre, 31-32-33, Central Market, West Avenue Road, Punjabi Bagh (West), New Delhi - 110026, Ph: 011-45634000

JEE Main- 20-07-2021-Morning PHYSICS Section-A Q1. The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas

from 0 C to 50 C when no work is done is ___________. (R is the universal gas

constant) (A) 175 R (B) 750 R (C) 250 R (D) 500 R Q2. A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a

magnetic field. Let d

r andd

r be their respective radii of circular path. The value of dr

r

is

equal to :

(A) 2 (B) 2

(C) 1

2 (D) 1

Q3. If Aand B are two vectors satisfying the relation A B A B . = Then the value of A B−

will be :

(A) 2 2A B 2AB+ − (B) 2 2A B+

(C) 2 2A B 2AB+ + (D) 2 2A B 2AB+ +

Q4. Consider a mixture of gas molecule of types A, B and C having masses

A B Cm m m . The ratio of their root mean square speeds at normal temperature and

pressure is :

(A) A B C

1 1 1

(B) A B C

0 = = =

(C) A B C

1 1 1

(D) A B C

=

Q5. A nucleus of mass M emits − ray photon of frequency ‘v’. The loss of internal energy by

the nucleus is : [Take ‘c’ as the speed of electromagnetic wave ]

(A)2

hvhv 1

2Mc

+

(B) 0

(C) 2

hvhv 1

2Mc

−

(D) hv

Q6. A radioactive material decays by simultaneous emissions of two particles with half lives of

1400 years and 700 years respectively. What will be the time after which one third of the material remains? (Take In 3 = 1.1)

(A) 340 years (B) 1110 years (C) 700 years (D) 740 years

FIITJEE


Q7. The value of current in the 6 resistance is :

(A) 6 A (B) 10 A (C) 4 A (D) 8 A

20

140V

6

5

90V

Q8. The radiation corresponding to 3 2→ transition of a hydrogen atom falls on a gold

surface to generate photoelectrons. These electrons are passed through a magnetic field

of 45 10 T.− Assume that the radius of the largest circular circular path followed by these

electrons is 7mm, the work function of the metal is : (Mass of electron = 319.1 10 kg− )

(A) 0.82 eV (B) 1.88 eV (C) 1.36 eV (D) 0.16 eV Q9. For the circuit shown below,

calculate the value of 2I :

(A) 0.15 A (B) 0.1 A (C) 25 mA (D) 0.05 A

Vi

RS=1000

100V

Iz

R

2000 VZ =50V

Q10. The arm PQ of a rectangular conductor is moving from x 0= to x = 2b outwards and then

inwards from x = 2b to x = 0 as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from x = 0 to x = b. Identify the graph the variation of different quantities with distance.

x=b x=0 x=2b

P

Q

b x=0

A

2b b

B

C

(A) A – Flux, B – EMF, C – Power dissipated (B) A – EMF, B – Power dissipated, C – Flux (C) A – Power dissipated, B – Flux, C – EMF (D) A – Flux, B – Power dissipated, C – EMF Q11. Region I and II are separated by a spherical

surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :

(A) 9.52 cm (B) 18.23 cm (C) 37.58 cm (D) 55.44 cm

I = 1.25

I

C

II

II = 1.4

O

25cm

FIITJEE


Q12. The normal reaction ‘N’ for a vehicle of 800 kg mass, negotiating a turn on a 30 banked

road at maximum possible speed without skidding is _________ 3 210 kg m / s .

[ Given 5cos30 0.87, 0.2 = = ]

(A) 6.96 (B) 10.2 (C) 12.4 (D) 7.2

Q13. The entropy of any system is given by

= +

2

2

kRS n 3

J Where and are the

constants. ,J,k and R no. of moles, mechanical equivalent of heat, Boltzmann constant

and gas constant respectively . [ TakedQ

ST

= ]

Choose the incorrect option from the following :

(A) S, , k and R have the same dimensions.

(B) and k have the same dimensions.

(C) S and have different dimensions.

(D) and J have the same dimensions.

Q14. A current of 5 A is passing through a non-linear magnesium wire of cross – section

0.04 2m . At every point the direction of current density is at an angle of 60 with unit

vector of area of cross -section. The magnitude of electric field at every point of the

conductor is : (Resistivity of magnesium 844 10 m− = )

(A) 711 10 V / m− (B) 211 10 V / m−

(C) 511 10 V / m− (D) 311 10 V / m−

Q15. The value of tension in a long thin metal wire has been changed from 1T to 2

T . The

lengths of the metal wire at two different values of tension 1

T and2

T are1and

2

respectively. The actual length of the metal wire is :

(A) 1 2 2 1

1 2

T T

T T

−

− (B)

1 2 1 2T T

(C) 1 2

2

+ (D) 1 1 2 2

1 2

T T

T T

−

−

Q16. AC voltage ( )V t 20sin t= of frequency 50Hz is applied to a parallel plate capacitor.

The separation between the plates is 2mm and the area is 21m . The amplitude of the

oscillating displacement current for the applied Ac voltage is __________.

[ Take 12

08.85 10 F / m− = ]

(A) 83.37 A (B) 55.58 A

(C) 21.14 A (D) 27.79 A

Q17. A butterfly is flying with a velocity 4 2m/ s in North – East direction. Wind is slowly

blowing at 1m / s from North to South. The resultant displacement of the butterfly in 3

seconds is :

(A) 15m (B) 20m

(C) 3 m (D) 12 2m

FIITJEE


Q18. A certain charge Q is divided into parts q and ( )Q q .− How should the charges Q and q

be divided so that q and (Q – q) placed at a certain distance apart experience maximum electrostatic repulsion ?

(A) q

Q2

= (B) Q 4q=

(C) Q 2q= (D) Q 3q=

Q19. A steel block of 10 kg rests on a horizontal

floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration 0.2 m/s2. The normal reaction R’ by the floor if mass of the

iron cylinders are equal and of 20kg each, is

_________N.

[ Take 2

sg 10m / s and 0.2= = ]

3

1 2

a=0.2m/s2

(A) 714 (B) 684 (C) 716 (D) 686 Q20. A person whose mass is 100 kg travels from Earth to Mars

in a spaceship. Neglect all other object in sky and take acceleration due to gravity on the surface of the Earth and

mars as 2 210m / s and 4m / s respectively. Identify from the

below figures, the curve that fits best for the weight of the passenger as a function of time.

(A) (c) (B) (a) (C) (b) (D) (d)

A

time

Weight

B 400N

1000N

(a)

(b)

(c)

(d) Section-B Q1. The amplitude of wave disturbance propagating in the positive x-direction is given by

( )=

+2

1y

1 x at time t 0= and

( )=

+ −2

1y

1 x 2at =t 1 s , where x and y are in metres. The

shape of wave does not change during the propagation. The velocity of the wave will be ________ m /s.

Q2. In the reported figure, heat energy absorbed by a system

in going through a cyclic process is _________ J.

litre

P(kPa)

20

40

20 40

FIITJEE


Q3. In a spring gun having spring constant 100N / m a small

ball ‘B’ of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance ‘d’ on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground . The value of d is ___________m.

Gun

2m

ball

Q4. The frequency of a car horn encountered a change from 400 Hz to 500 Hz, when the car

approaches a vertical wall. If the speed of sound is 330 m /s. Then the speed of car is _________km /h.

Q5. A body having specific charge 8 C / g is resting on a

frictionless plane at a distance 10cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V/m is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be __________s.

100V/m

Body

Q6. A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of

mass ‘m’ travelling along the surface hits at one end of the rod with a velocity ‘u’ in a direction perpendicular to the rod. The collision is completely elastic. After collision,

particle comes to rest. The ratio of masses m

M

is 1

.x

The value of ‘x’ will be _______.

Q7. An object viewed from a near point distance of 25 cm, using a microscopic lens with

magnification ‘6’, gives an unresolved image, A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to ________cm.

Q8. In an LCR series circuit, an inductor 30 mH and a resistor 1 are connected to an Ac

source of angular frequency 300 rad /s. The value of capacitance for which, the current

leads the voltage by 45 is 3110 F.

x

− Then the value of x is ____________.

Q9. A circular disc reaches from top to bottom of an inclined plane of length ‘L’ when it slips

down the plane, it takes time 1' t ' . When it rolls down the plane, it takes time 2

t .The value

of 1

2

t

tis

3

x The value of x will be ___________.

Q10. A carrier wave ( ) ( )6

CV t 160 sin 2 10 t= volts is made to very between max

V 200 V= and

minV 120V= by a message signal ( ) ( )3

m mV t A sin 2 10 t= volts. The peak voltage m

A of

the modulating single is ________V.

FIITJEE


CHEMISTRY Section-A Q1. Chemical nature of the nitrogen oxide compound obtained from a reaction of concentrated nitric acid and P4O10 (in 4:1 ratio) is: (A) basic (B) neutral (C) amphoteric (A) acidic Q2. Compound A is converted to B on reaction with CHCl3 and KOH. The compound B is toxic and can be decomposed by C. A, B and C respectively are: (A) Primary amine, nitrile compound, conc. HCl (B) Primary amine, isonitrile compound , conc. HCl (C) Secondary amine, isonitrile compound, conc NaOH (D) Secondary amine, nitrile compound, conc. NaOH Q3. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The dihedral angles in H2O2 in gaseous phase is 90.2o and in solid phase is 111.5o. Reason R: The change in dihedral angle in solid and gaseous phase is due to the difference in the intermolecular forces. Choose the most appropriate answer from the options given below for A and R. (A) A is correct but R is not correct. (B) Both A and R are correct but R is not the correct explanation of A. (C) Both A and R are correct and R is the correct explanation of A (D) A is not correct but R is correct. Q4.

CH2 CH2

H

H

CH2

O

CH

(A) (B) (C) (D) Among the given species the Resonance stabilized carbocations are: (A) (A), (B) and (C) only (B) (C) and (D) only (C) (A), (B) and (D) only (D) (A) and (B) only Q5. A s-block element (M) reacts with oxygen to form an oxide of the formula MO2. The oxide is pale yellow in colour and paramagnetic. The element (M) is: (A) Na (B) K (C) Mg (D) Ca Q6. Green chemistry in day- to day life is in the use of: (A) Chlorine for bleaching of paper (B) Large amount of water alone for washing clothes (C) Liquified CO2 for dry cleaning of clothes (D) Tetrachloroethene for laundry Q7. Identify the incorrect statement from the following:

(A) -Glycosidic linkage makes cellulose polymer (B) Amylose is a branched chain polymer of glucose

(C) Starch is a polymer of - D glucose (D) Glycogen is called as animal starch

FIITJEE


Q8. The metal that can be purified economically by fractional distillation method is: (A) Ni (B) Cu (C) Fe (D) Zn Q9. The set in which compounds have different nature is: (A) B(OH)3 and Al (OH)3 (B) B(OH)3 and H3PO3 (C) NaOH and Ca(OH)2 (D) Be(OH)2 and Al(OH)3 Q10. The conditions given below are in the context of observing Tyndall effect in colloidal solutions: (A) The diameter of the colloidal particles is comparable to the wavelength of light used. (B) The diameter of the colloidal particles in much smaller than the wavelength of light used. (C) The diameter of the colloidal particles is much larger than the wavelength of light used. (D) The refractive indices of the dispersed phase and the dispersion medium are comparable. (E) The dispersed phase has a very different refractive index from the dispersion medium. Choose the most appropriate conditions from the options given below. (A) (A) and (D) only (B) (A) and (E) only (C) (C) and (D) only (D) (B) and (E) only Q11. According to the valence bond theory the hybridization of central metal atom is dsp2 for which one of the following compounds? (A) K2[Ni(CN)4] (B) NiCl2.6H2O (C) Na2[NiCl4] (D) [Ni(CO)4]

Q12. In the given reaction 3-Bromo-2,2-dimethyl butane C H OH

(MajorPr oduct)

2 5 ' A '⎯⎯⎯⎯→

Product A is: (A) 2-Ethoxy-2,3-dimethyl butane (B) 2-Hydroxy-3,3-dimethyl butane. (C) 2-Ethoxy-3,3-dimethyl butane (D) 1-Ethoxy-3,3-dimethyl butane. Q13. Orlon fibres are made up of: (A) Polyamide (B) Polyacrylonitrile (C) Cellulose (D) Polyesters Q14. The species given below that does NOT show disproportionation reaction is:

(A) BrO− (B) 3BrO−

(C) 2BrO− (D)

4BrO−

FIITJEE


Q15.

KMnO4

H2SO

4 ,

'A'(major product)

'B'(major product)

KMnO4

H2O ,

For above chemical reactions, identify the correct statement from the following (A) Both compound ‘A’ and compound ‘B’ are dicarboxylic acids. (B) Compound ‘A’ is dicarboxylic acid and compound ‘B’ is diol. (C) Compound ‘A’ is diol and compound ‘B’ is dicarboxylic acid. (D) Both compound ‘A’ and compound ‘B’ are diols. Q16. An inorganic compound ‘X’ on treatment with concentrated H2SO4 produces brown fumes and gives dark brown ring with FeSO4 in presence of concentrated H2SO4. Also compound ‘X’ gives precipitate ‘Y’, when its solution in dilute HCl is treated with H2S gas. The precipitate ‘Y’ on treatment with concentrated HNO3 followed by excess of NH4OH further gives deep blue coloured solution, Compound ‘X’ is: (A) Cu(NO3)2 (B) Pb(NO3)2 (C) Co(NO3)2 (D) Pb(NO2)2 Q17.

CHO

OH

OH

O

OH(I) (II) (III) (IV)

Which among the above compound/s does / do not form silver mirror when treated with Tollen’s reagent? (A) Only (IV) (B) (III) and (IV) only (C) Only (II) (D) (I), (III) and (IV) only Q18. Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Sharp glass edge becomes smooth on heating it upto its melting point. Reason R: The viscosity of glass decreases on melting. Choose the most appropriate answer from the options given below. (A) A is false but R is true. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false (D) Both A and R are true and R is the correct explanation of A. Q19. The correct structure of Rhumann’s Purple, the compound formed in the reaction of ninhydrin with proteins is:

N

O

OO

O

(A)

FIITJEE


(B) N

O

OO

O

(C) N

O

OO

O

(D) N N N

O

OO

O

Q20. The correct order of intensity of colors of the compounds is:

(A) ( ) ( )2 22

4 2 6 4NiCl Ni H O Ni CN

+ −−

(B) ( ) ( )222

4 24 6NiCl Ni CN Ni H O

+−−

(C) ( ) ( )2 22

2 46 4Ni H O NiCl Ni CN

+ −−

(D) ( ) ( )22 2

4 24 6Ni CN NiCl Ni H O

+− −

Section-B

Q1. The number of lone pairs of electrons on the central I atom in 3I− is___________.

Q2. An average person needs abut 10000 KJ energy per day.The amount of glucose (molar

mass = 180.0 g mol−1) needed to meet this energy requirement is_________ g, (Nearest integer)

(Use : cH (glucose) = − 2700kJ mol−1)

Q3. To synthesise 1.0 mole of 2-methylpropan-2-ol from Ethylethanoate _________ equivalents of CH3MgBr reagent will be required. (Integer value)

Q4. The spin only magnetic moment value for the complex [Co(CN)6]4− is______ BM. [At. No. of Co= 27] Q5. At 20oC, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20oC above an equimolar mixture

of benzene and methyl benzene is ____________ 10−2. (Nearest integer)

FIITJEE


Q6. 2 2 32SO (g) O (g) 2SO (g)+

In an equilibrium mixture, the partial pressures are

SO3P 43kPa= ; O

2P 530= Pa and

SO2P 45kPa= . The equilibrium constant pK = _____________ 210− (Nearest integer)

Q7. 250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl

molecules in the solution after complete reaction is__________ 2110 (Nearest integer)

( )23AN 6.022 10=

Q8. The number of nitrogen atoms in a semicarbazone molecule of acetone is__________. Q9. The inactivation rate of a viral preparation in proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral

inactivation is ___________ 3 110 min− − . (Nearest integer)

[ Use: n10 = 2.303; log10 3=0.477; property of logarithm: log xy = y log x]

Q10. The Azimuthal quantum number for the valence electrons of Ga+ ion is____________. (Atomic number of Ga= 31)

FIITJEE


MATHEMATICS Section-A

Q1. Let ‘a’ be a real number such that the function ( ) 2f x ax 6x 15, x R= + − is increasing in

3,4

−

and decreasing in

3,

4

. Then the function ( ) 2g x ax 6x 15, x R= − + has a :

(A) local minimum at 3

x4

= (B) local maximum at 3

x4

=

(C) local minimum at 3

x4

= − (D) local maximum at 3

x4

= −

Q2. Let ijA a = be a 3 x 3 matrix, where

ij

1 , if i j

a x , if i j 1

2x 1 , otherwise

=

= − − = +

Let a function f :R R→ be defined as ( ) ( )f x det A= . Then the sum of maximum and

minimum values of f on R is equal to :

(A) 88

27− (B)

20

27

(C) 88

27 (D)

20

27−

Q3. If z and are two complex numbers such that ( ) ( )3

z 1and arg z arg w2

= − = , then

1 2zarg

1 3z

−

+ is :

(Here arg(z) denotes the principal argument of complex number z)

(A) 4

(B)

3

4

−

(C) 4

− (D)

3

4

Q4. If in a triangle ABC, AB = 5 units, 1 3B cos

5

− =

and radius of circumcircle of ABC is 5

units, then the area (in sq. units) of ABC is :

(A) 6 8 3+ (B) 8 2 2+

(C) 10 6 2+ (D) 4 2 3+

Q5. Words with or without meaning are to be formed using all the letters of the word

EXAMINATION. The probability that the letter M appears at the fourth position in any such word is :

(A) 1

9 (B)

1

11

(C) 2

11 (D)

1

66

FIITJEE


Q6. The coefficient of 256x in the expansion of ( ) ( )100101 21 x x x 1− + + is :

(A) 100

15C− (B) 100

16C

(C) 100

16C− (D) 100

15C

Q7. The probability of selecting integers a 5, 30 − such that ( )2x 2 a 4 x 5a 64 0+ + − + ,

for all x R is :

(A) 2

9 (B)

1

6

(C) 7

36 (D)

1

4

Q8. The Boolean expression ( ) ( )p ~ q q ~ p is equivalent to :

(A) q p (B) ~ q p

(C) p q (D) p ~ q

Q9. Let a be a positive real numbers such that

a x x

0e dx 10e 9

− = −

where [x] is the greatest integer less than or equal to x. Then a is equal to :

(A) e10 log 2+ (B) ( )e10 log 1 e− +

(C) ( )e10 log 1 e+ + (D) e10 log 3+

Q10. Let 2 3

A , a Ra 0

=

be written as P + Q where P is a symmetric matrix and Q is skew

symmetric matrix. If ( )det Q 9= , then the modulus of the sum of all possible values of

determinant of P is equal to : (A) 18 (B) 36 (C) 24 (D) 45

Q11. If and are the distinct roots of the equation ( )1 1

2 4 2x 3 x 3 0,+ + = then the value of

( ) ( )96 12 96 121 1 − + − is equal to :

(A) 2556 3 (B) 2528 3

(C) 2452 3 (D) 2456 3

Q12. The number of real roots of the equation

( )1 1 2tan x x 1 sin x x 14

− − + + + + = is :

(A) 4 (B) 1 (C) 0 (D) 2

Q13. The value of the integral ( )1

e

1

log 1 x 1 x dx−

− + + is equal to :

(A) e2log 2 1

4

+ − (B)

elog 2 12

+ −

(C) e

1 3log 2

2 4 2

+ − (D)

e

12log 2

2 2

+ −

FIITJEE


Q14. The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6

observations are 2, 4, 5 and 7, then the remaining two observations are : (A) 10, 11 (B) 3, 18 (C) 1, 20 (D) 8, 13

Q15. Let the tangent to the parabola 2S : y 2x= at the point P(2, 2) meet the x-axis at Q and

normal at it meet the parabola S at the point R. Then the area (in sq. units) of the triangle PQR is equal to :

(A) 35

2 (B)

15

2

(C) 25 (D) 25

2

Q16. Let a function f :R R→ be defined as

( )

xsinx e if x 0

f x a x if 0 x 1

2x b if x 1

−

= + − −

where [x] is the greatest integer less than or equal to x. If f is continuous on R, then

( )a b+ is equal to :

(A) 2 (B) 5 (C) 3 (D) 4

Q17. Let ˆ ˆ ˆ ˆ ˆa 2i j 2k and b i j= + − = + . If c is a vector such that a.c c , c a 2 2= − = and the

angle between ( )a b and c is6

, then the value of ( )a b c is :

(A) 3 (B) 2

3

(C) 3

2 (D) 4

Q18. Let ( )y y x= be the solution of the differential equation

( )x 2 ye 1 y dx dy 0, y 1 1

x

− + = = −

Then the value of ( )( )2

y 3 is equal to :

(A) 61 4e+ (B) 31 4e−

(C) 31 4e+ (D) 61 4e−

Q19. Let [x] denote the greatest integer x , where x R . If the domain of the real valued

function ( )x 2

f xx 3

− =

−

is ( ) ) ), a b, c 4, , a b c,− then the value of a b c+ +

is:

(A) 3− (B) 8

(C) 2− (D) 1

Q20. Let ( )y y x= be the solution of the differential equation

FIITJEE


y y 1

x tan dy y tan x dx, 1 x 1, yx x 2 6

= − − =

Then the area of the region bounded by the curves ( )1

x 0, x and y y x2

= = = in the

upper half plane is :

(A) ( )1

16

− (B) ( )1

18

−

(C) ( )1

24

− (D) ( )1

312

−

Section-B

Q1. Let

1 1 0

A 0 1 1

0 0 1

−

= −

and 20 7B 7A 20A 2I= − + , where I is an identity matrix of order 3 x 3.

If ijB b = , then 13b is equal to………

Q2. Let y mx c, m 0= + be the focal chord of 2y 64x= − , which is tangent to

( )2 2x 10 y 4+ + = . Then, the value of ( )4 2 m c+ is equal to …….

Q3. Let a, b, c, d be in arithmetic progression with common difference .

If

x a c x b x a

x 1 x c x b 2

x b d x d x c

+ − + +

− + + =

− + + +

, then the value of 2 is equal to………

Q4. Let a,b,c be three mutually perpendicular vectors of the same magnitude and equally

inclined at an angle , with the vector a b c+ + . Then 236cos 2 is equal to…….

Q5. Let T be the tangent to the ellipse 2 2E : x 4y 5+ = at the point ( )P 1,1 . If the area of the

region bounded by the tangent T, ellipse E, lines x = 1 and

1 1x 5 is 5 cos ,

5

− = + +

then + + is equal to…..

Q6. There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2

are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ……….

Q7. The number of rational terms in the binomial expansion of

12011

644 5

+

is…..

Q8. Let P be a plane passing through the points ( ) ( ) ( )1,0,1 1, 2,1 and 0,1, 2− − . Let a vector

ˆ ˆ ˆa i j k= + + be such that a is parallel to the plane P, perpendicular to

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆi 2j 3k and a. i j 2k 2+ + + + = , then ( )2

− + equals…….

FIITJEE


Q9. If the value of ( ) 2

x 2

x

x 0lim 2 cosx cos2x

+

→− is equal to ae , then a is equal to…….

Q10. If the shortest distance between the lines ( )1ˆ ˆ ˆ ˆ ˆ ˆr i 2j 2k i 2j 2k , R= + + + − + ,

( )2ˆ ˆ ˆ ˆ ˆ0 and r 4i k 3i 2j 2k , R = − − + − − is 9, then is equal to…….

FIITJEE


ANSWER: JEE Main- 20-07-2021-Morning

PHYSICS CHEMISTRY MATHEMATICS

Section-A Ans1. D Ans2. A Ans3. A Ans4. C Ans5. A Ans6. D Ans7. B Ans8. A Ans9. C Ans10. A Ans11. C Ans12. B Ans13. B Ans14. C Ans15. A Ans16. D Ans17. A Ans18. C Ans19. D Ans20. A Section-B Ans1. 2 Ans2. 100 Ans3. 1 Ans4. 132 Ans5. 1 Ans6. 4 Ans7. 25 Ans8. 3 Ans9. 2 Ans10. 40

Section-A Ans1. D Ans2. B Ans3. D Ans4. D Ans5. B Ans6. C Ans7. B Ans8. D Ans9. A Ans10. B Ans11. A Ans12. A Ans13. B Ans14. D Ans15. B Ans16. A Ans17. C Ans18. B Ans19. B Ans20. A Section-B Ans1. 3 Ans2. 667 Ans3. 2 Ans4. 2 Ans5. 78 Ans6. 172 Ans7. 226 Ans8. 3 Ans9. 106 Ans10. 0

Section-A Ans1. D Ans2. A Ans3. B Ans4. A Ans5. B Ans6. D Ans7. A Ans8. C Ans9. A Ans10. B Ans11. C Ans12. C Ans13. B Ans14. A Ans15. D Ans16. C Ans17. C Ans18. D Ans19. C Ans20. B Section-B Ans1. 910 Ans2. 34 Ans3. 1 Ans4. 4 Ans5. 1.25 (As per NTA 1) Ans6. 777 Ans7. 21 Ans8. 81 Ans9. 3 Ans10. 6

FIITJEE


SOLUTION: JEE Main- 20-07-2021-Morning PHYSICS Section-A Sol1. Since process is isochoric, so

VQ nC T= f

n R T2

=

( )

54 R 50 500R

2

= =

Sol2. As we know that radius of circular path in magnetic field is given as

mv 2mK

rqB qB

= = , so

Charged particle Charge Mass

Deuteron e 2m

Alpha particle 2e 4m

d d

d

r m q 12 2

r m q 2

= = =

Sol3. = A B A B = ABcos ABsin = 045

− = + − 2 2A B A B 2ABcos = + − 2 2 1A B 2AB

2= + −2 2A B 2AB

Sol4. As we know that root mean square speed is given as

3RT

vM

= , so

A B C

v v v A B C

1 1 1

v v v as A B C

m m m .

Sol5. Using conservation linear momentum, we can write

= =

h hMv

c

=

hv

Mc

JBC

-ray M M

v

JBC

Rest

Loss of internal energy= Gain in kinetic energy+ energy of gamma ray

= + = +

2

21 1 hMv h M h

2 2 Mc

= +

2

hh 1

2Mc

Difficulty Level: Difficult and Very good Question Sol6. As we know that equivalent half life of radioactive material decays by simultaneous

emissions, is given as

= +1 2

1 1 1

T T T= +

1 1

700 1400=

3

1400 =

1400T years

3

−= t

0N N e − = t0

0

NN e

3

− = t1e

3( ) =t n 3

( )

=

n 3t

( )

( )= =

n 3 1.1 1400T 740.74 years

n 2 0.693 3

P D

t =0 N0 0

t =t N

FIITJEE


Sol7. Method-I: Using Kirchhoff’s Law for

loopBDABB , we can write

1140 20I 6I 0− + + = 110I 3I 70 + = …....(i)

Using Kirchhoff’s Law for loop ACBBA ,

we can write

( )1 15 I I 90 6I 0− + − =

15I 11I 90 − + = ……..(ii)

Adding equation (1) with twice the equation (2), we have

125I 250=1

250I 10A

25 = =

Method-II: Using concept of equivalent cell, we can write

AB

140 0 902100 0 540020 6 5V

1 1 1 15 50 60

20 6 5

+ ++ +

= =+ +

+ +

7500

60 Volt125

= =

1

60I 10A

6 = =

20

140V

6

5

90V

A

B B B

C D

I I1 I-I1

Sol8. h 3.4 1.51 1.89 eV = − =

As we know that radius of circular path in magnetic field is given as

2mK

rqB

=2 2 2r q B

K2m

=

n=3

n=2

n=1

-1.51eV

-3.4eV

-13.6eV

h

( ) ( ) ( )

2 23 19 4

31

7 10 1.6 10 5 10K eV

2 9.1 10

− − −

−

=

2107.7 10 eV−= 1.08eV=

h K 1.89 1.08 0.81eV = − = − =

Difficulty Level: Difficult and Very good Question

Sol9. 1

50I 25mA

2000= = , and i ZV V 100 50

I 50mA1000 1000

− −= = =

z 1I I I 25mA = − =

Vi

RS=1000

100V

Iz

R

2000 VZ =50V

Vi

RS=1000

100V

Iz

R

2000 VZ =50V

A A

B B B

C

I1

I

FIITJEE


Sol10. When arm PQ of a rectangular

conductor is moving from x 0= to x =

b, the flux( =ax) linked with loop increases and while moving from

=x b to x = 2b and from =x 2b to x =

b, flux ( =ab) remains constant and

then flux ( =ax) deceases to zero as it

moves from =x b to x = 0 .

in

d dxe a av

dt dt

= − = − = − Induced emf

( )

2

ineP

R= Power dissipated

x=b x=0 x=2b

P

Q

a

x

Sol11. Since direction of incident ray (DIR) is from left to write , so considering refraction

at point M , we can write

2 1 2 1

v u R

− − =

1.4 1.25 1.4 1.25

v 40 25

− − =

− −

1.4 0.15 1.25 1.2 6.25 7.45

v 25 40 200 200

+ − = + = =

200 1.4

v 37.58cm7.45

= − = −

I = 1.25 I

C

II II = 1.4

O

25cm

DIR

P I

Sol12. In the frame of vehicle, vehicle is in equilibrium under the influence of pseudo force FP

2

P

mvF

R=

0 0

PN mgcos30 F sin30= +

2

0 0mvN mgcos30 sin30

R = + .(i)

, and

2

0 0

S

mvf cos30 mgsin30

R= −

FBD in the Frame of vehicle

fS

FP

N

mg

300

2

0 0mvN cos30 mgsin30

R = − ….(ii)

By doing 0 0(1) cos30 (2) sin30 − , we have

( )0 0N cos30 sin30 mg− =0 0

mgN

cos30 sin30 =

−

3 2800 10 800 10N 10.2 10 kgm / s

0.87 0.2 0.5 0.77

= = =

−

Difficulty Level: Difficult and Very good Question Sol13.

Physical Quantity Formula Dimension

Entropy ( S) dQS

T=

− − 2 2 1ML T K

Number of mole () mol

FIITJEE


Mechanical equivalent of heat (J) =

WJ

Q

Dimensionless

Boltzmann constant (k) =

Energyk

Temperature

− − 2 2 1ML T K

Gas constant (R) =a

R kN − − 2 2 1ML T K

a

N Avogadro’s Number

2J kR 2 kR − − 2 2 1k ML T K , and

2S − − − − 2 2 1 2 2 2 1ML T K ML T K

0 2 0 0M L T K Dimensionless

Sol14. I JAcos= 05 J 0.04 cos60 = 25J 250A / m

0.02 = =

E

J =

8 5E J 44 10 250 11 10 V / m− − = = =

Sol15. Assuming Tension in metal wire suspended from roof varies linearly and 0 and T0 (

developed due its own weight ) are tensions are ends of of wire of length . So

( ) −1 1

T , and ( ) −2 2

T

−

=−

1 1

2 2

T

T

− =

−

1 2 2 1

1 2

T T

T T

Sol16.

= 0A

Cd

= =

C

0

1 dX

C A

= = =0 0 0 0 0

0

C

V V A 2 fV AI

X d d

−

−

−

= = =

126

0 3

2 3.14 50 20 8.85 10 1I 27.79 10 A 27.79 A

2 10

Sol17. ( )= + = +0 0

BWˆ ˆ ˆ ˆv 4 2 cos45 i sin45 j 4 i 4 j , and = −

Wgˆ ˆv 0 i j

( ) ( )= + = + + − = +Bg BW Wg

ˆ ˆ ˆ ˆ ˆ ˆv v v 4 i 4 j 0 i j 4 i 3 j

= Bg

v 5 m / s Speed of Butterfly

Magnitude of displacement of Butterfly= = =Bg

v t 5 3 15 m

Sol18. Here r is fixed

( )−

= 2

0

q Q qF

4 r( ) = −

2

0

dF 1Q 2q

dq 4 r = −

2

0

dF 1

dq 2 r

q Q-q

r For maxima or minima of force , its first derivative should be zero.

( )= − = 2

0

dF 1Q 2q 0

dq 4 r =

Qq

2

Since Second Derivative is always negative so maxima will occur at this value of q.

FIITJEE


Sol19. − = = =70g N 70a 70 0.2 14

= − =N 700 14 686N

3

1 2 a=0.2m/s

2

70g N

Sol20. = 1

W 1000N Weight of person at surface of

earth = 2

W 400N Weight of person at surface of

Mars. There will be a neutral point –N where weight will be zero because at this point net gravitational field due to earth and mars is zero.

a

N

Earth

Mars

Section-B

Sol1. The equation of wave at any time t will be( )

=+ −

2

1y

1 x vt, so =v 1 2 =v 2m / s

Sol2. Heat absorbed in cyclic process = Work done = 100 Joule Sol3. Let ball starts its motion with horizontal velocity v0 , so with the help of conservation of

mechanical energy , we can write

=2 2

0

1 1mv mx

2 2 = = =

0

k 100v x 0.05 0.5 10 m / s

m 0.1

t = Time required to fall the ball = = 2h 1

2 sg 10

= = =0

1d v t 0.5 10 2 1m

10

Sol4. Initially wall will act as observer, so with the help of Doppler’s effect, we can write

=−

0

0

cf ' f

c v

Now wall will act as source of sound of frequency f’, so With the help of Doppler’s effect, we can write

+ +

= =−

0 0

0

0

c v c vf '' f ' f

c c v

wall

Car

v0

c

+

= −

0

0

c v500 400

c v − = +

0 05c 5v 4c 4v

= = = =0

c 330 330 18v m / s km / h 132 km / h

9 9 9 5

Sol5. = F qE Electric force acting on the body

−

−

= = = =

62

3

F q 8 10a E 100 0.80 m / s

m m 10Acceleration of body

FIITJEE


= = = 2 2 0.1

T 2 2 1 sa 0.8

Time period of Motion

Sol6. Using conservation of linear momentum, we can write

Rest

A B

v1

JAC

u

A B

JBC m

M

m

M

=1

mu Mv =1

muv

M………..( 1)

Using conservation of angular momentum about centre of mass of Rod, we can write

= +2

2mvmu M

2 12 2 = 6mu M =

6mu

M………..( 2)

Using definition of e, we can write

+ =1

v u2

+ =mu 3mu

uM M

=m 1

M 4

Sol7. For microscopic lens

F P F B

A

B’

A’

= +D

m 1f

= +25

6 1f

=f 5cm

Now in second case, this lens acts as objective of compound microscope, so according to question

= =O e

o e

LDm m m

f f

= = =

e

o

LD 60 25f 25 cm

mf 12 5

Sol8. Here = 045 and current leads voltage , so C L

X X

= = =0Xtan tan45 1

R = − =

C LX X X R

= + C

X R L − = + =3

CX 1 30 10 300 10

=

110

C

− = = =

31 1 1C 10 F

10 10 300 3

R C L

Sol9. Case-I: when disk slides down

FIITJEE


=

1

2Lt

gsin………..(i)

Case-II: when disk rolls down

= = = = =

+ ++ +

2

2 22

2L 2L 2L 2L 4Lt

gsin gsin gsin gsin 3gsin

11 k R / 2 11 1 2R R

…. (ii)

= =

1

2

t 2L 3gsin 3

t gsin 4L 2

Sol10. m 1 2

c 1 2

A V V 200 120 1

A V V 200 120 4

− − = = = =

+ + Modulation index

Here ( ) ( )6

CV t 160 sin 2 10 t= V and ( ) ( )3

m mV t A sin 2 10 t= V , so

c

m

A 160A 40V

4 4= = =

FIITJEE


CHEMISTRY Section-A Sol1. Anhydride of HNO3 produced by the treatment of P4O10 (anhydring reagent ) in ratio of

4:1 is N2O5 which is acidic in nature. ( )3 4 10 2 5 3 4

Acidic nature

4HNO P O 2N O HPO+ ⎯⎯→ +

Sol2. 1o − Amine produced isocyanide (iso nitrile) compound with CHCl3 and KOH

2 3 2R NH CHCl 3KOH RNC 3KCl 3H O− + + ⎯⎯→ + +

(A) (B)

Isonitrile ( toxic nature)

H3O+ / HCl

(C)

R NH2 + HCOOH

(A) Sol3. Following are the dihedral angles of H2O2 in gas phase and solid (Reverse with respect

to question).

O

OH

H

111.5o

gas phase

O

OH

H

90.2o

Solid phase Change of angle is the changing of intermolecular interaction in different phase

Sol4. Only (A) and (B) structures have resonance effect. In (A) benzyllic resonance and in (B)

allylic resonance of carbocation. Sol5. An alkali metal oxide of formula MO2 is super oxide which is formed by K and Rb. KO2

has pale yellow colour and paramagnetic nature, while alkaline earth metal oxides formula MO2 are peroxide and diamagnetic nature.

Sol6. Nowdays liquid CO2 used as washing clothes, because Cl2 and C2H2Cl4 are toxic

Sol7. Amylose is polysaccharide made of -D- glucose unit of linear chain polymer not branched

300-600

O

CH2OH

OH

OH

OH

H H

O

O

OH

CH2OH

OH

H H

O

O

OH

CH2OH

OH

OH

H H

FIITJEE


Sol8. Due to volatile nature , Zn – metal can be economically purified by fractional distillation refining process.

Sol9. B(OH)3 is orthoboric acid having acidic nature, while Al(OH)3 has amphoteric nature that

is acidic and as well as basic Sol10. Tyndall effect is the optical characteristics of light in which comparable particle size of

colloids and bombarded light wavelength can produce this type of effect. For scattering effect of light in Tyndall effect refractive index of dispersed phase and dispersion medium should be greatly different.

Sol11. ( ) 2 4

K Ni CN

2 8 0Ni 3d 4s+ →

3d 4s 4p

dsp2 hybridisation

Pairing developed due to strong ligand effect.

2 2NiCl .6H O

2 8 0Ni 3d 4s+ →

3d 4s 4p

sp3d

2 hybridisation

4d

2 4Na NiCl

+ →2 8 0Ni 3d 4s

3d 4s 4p

sp3 hybridisation

4d

( )4

Ni CO

8 2Ni 3d 4s→

3d 4s 4p

sp3 hybridisation

4d

Pairing developed due to strong ligand (CO).

FIITJEE


Sol12.

Br C2H

5OH

-Br CH3 Shifting

C2H

5OH / -H+

OC2H5

( 2- Ethoxy-2,3- dimethyl butane)

Sol13. Orlon fibres are the polymer of acrylonitrile

CH2 CH CN H2C CH

CN

nAcrylonitrile Polymerisation

Polyacrylonitrile (PAN)

Sol14. In −

4BrO ,Br has +7 O.S, therefore it only acts as a oxidizing reagent , does not undergo

disproportionation reaction Sol15.

KMnO4

H2SO

4 /

COOH

COOH

(A) Major Product

(B) Major Product

OH

OH

KMnO4

H2O , 273 K

Sol16. Conc

H SO2 4X ⎯⎯⎯⎯→ Brown fumes + Brown ring test with FeSO4 / H2SO4

⎯⎯⎯→H S

HCl2X Y(ppt) Conc

HNO3

⎯⎯⎯→dissolved NH OH4

⎯⎯⎯⎯→ Deep blue colour solution

In cation analysis of Cu++ ions, precipitate formed is CuS on treating with H2S and HCl

which dissolved in HNO3 and produced blue colour complex solution ( )++

3 4Cu NH in

NH4OH. Brown fumes and brown ring performance given by nitrate sample. Hence salt is

( )3 2Cu NO .

FIITJEE


Sol17. All structure can produce –CHO functional group which gives +ve test of Tollen’s reagent except structure (II), because if forms ketonic group after tautomerisation

OH OTautomerisation

Sol18. Both statement are correct for the glass body heating , but reason is not correct

explanation during heating process of glass, constituents unit rupture of glass body and gives the edge smoothness.

Sol19.

OH

OH

O

O

+ HCR

COOHNH2

OH- /

O

O

N

O

O

Ninhyidrin Amino Acid Rhumann's purple

Sol20. For Ni2+, crystal field CFSE magnitude shows the magnitude of absorbed light.

Following are the energy absorbed order ( ) ( )222

4 2 4NiCl Ni H O Ni CN

−+−

Hence order of colour of compounds are.

( ) ( )2 22

4 2 6 4NiCl Ni H O Ni CN

+ −−

Section-B

Sol1. 3I− anion has three lone pair

I I I

Sol2. 2700 kJ energy released from 180gm (1 mole) of glucose

1kJ energy released from 180

2700gm of glucose

10000 kJ energy released from 180 10000

gm2700

of glucose

Amount of glucose = 18000

666.66 667gm27

=

FIITJEE


Sol3.

CH3 O CH3

O

CH3MgBr

CH3 CH3

OH

CH3

+ C2H5OH

Ethylethanoate (2eq) / H2O 2- Methylpropan-2-ol

Sol4. In ( )4

6Co CN

− , Co has +2 oxidation state

3d

4s 4p 4d3d

4s 4p 4d

Co+2

In pressureof strong Ligand

Hence, electronic configuration of 2Co + is 7 0Ar 3d 4s . In complex ( )4

6Co CN

− , given

ligand CN− is strong hence, after pairing in d- subshell, total number of unpaired electron

= 1

spin magnetic moment ( )1 1 2 3= + = = 1.73 BM

= 1.73BM 2.0 BM

Sol5. oBP = vapour pressure of benzene at o20 C 70torr=

oMP = vapour pressure of toluene at o20 C 20torr=

Mixture is equimolar, B MX 0.5andX 0.5= =

Total vapour pressure ( )TP 70 0.5 20 0.5 45torr= + =

Mole fraction of benzene in vapour phase ( )o

' B BB

T

P X 70 0.5X torr

P 45

= =

20.777 77.7 10 torr−= =

Ans = 78 (nearest integer)

Sol6. Kp = ( )

( ) ( )

( )

( ) ( )( )

−= =

22

SO 1

2 2

SO O

3

2 2

P 431.72 kPa

45 0.53P P

( )12

Pk 172 10 kPa−−=

Sol7. NaOH + HCl ⎯⎯→ NaCl + 2H O

(Milimole) t= 0 250 0.5 500 1.0 0 0

t = − 375 125 125 millimole

unreacted HCl mole = 375

1000

number of HCl molecule un reacted 23375

6.023 101000

=

21 21225.75 10 226 10=

FIITJEE


Sol8. Reaction of acetone and semicarbazide

OCH3

CH3

+ NH2 NH C NH2

OH+

-H2O

NCH3

CH3

NH C

O

NH2

Semicarbazide Semicarbazone

Sol9. Kt = 2.303 log

0

t

A

A

Here, [A0] = 100 , [At] = 90 and t = 1 min

= 10

100K 1 2.303log

90

K 2.303 log10 log9 2.303 1 2log3 = − = −

3 1K 2.303 1 2 0.477 0.105938 105.938 10 min− − = − = =

Rounded = 106

Sol10. Electronic configuration of Ga+ ion = [Ar] 103d 24s o4p

Last electron goes into S- orbital, hence

Azimuthal quantum number ( ) for last electron = 0

FIITJEE


MATHEMATICS Section-A

Sol1. ( ) 2f x ax 6x 15= + −

( )3 3

f ' x 2ax 6 0 x a 4a 4

−= + = = = = −

( ) 2g x 4x 6x 15= − − +

( )3

g' x 0 8x 6 0 x4

= − − = = −

( )3

g" x 8 0 x4

−= − = is a point of local maxima for g(x).

Sol2. ( )

1 x 2x 1

det A x 1 x

2x 1 x 1

− +

= − −

+ −

( ) ( ) ( )( )2 2 21 1 x x 2x x x 2x 1 x 2x 1= − − − − + + + − −

2 3 3 2 21 x 2x 2x x 4x 2x 2x 1= − + + + − − − −

( ) 3 2f x 4x 4x 4x= − −

( ) ( )2 1f ' x 4 3x 2x 1 0 x 1,

3

−= − − = =

( )1 4 4 4

f 1 f 43 27 9 3

− + − = − + − +

4 12 36 20 88

4 427 27 27

− − + −= − + = − + =

Sol3. Let iw r.e =

( )arg w =

( )1

z sin icosr

= −

i iizw e . re i

r

− = =

1 2zw 1 2i

arg arg1 3zw 1 3i

− − =

+ +

( )( )

( )1 2i 1 3i1 2i 5 5i 1

1 i1 3i 1 9 10 2

− −− − −= = = − +

+ +

( )1 3

arg 1 i2 4

− − + =

iie

r

−

z

w

( )ire

Sol4. 3

cosB , R 55

= =

o510 C 30

sinC= =

( )( )14 3 4 3 6 8 3

2 = + = +

30o

A

B C 3

5 4

4 3 Sol5. EXAMINATION, required probability

FIITJEE


10! / 2! 2! 2! 1

11! / 2! 2! 2! 11= =

Sol6. Coefficient of ( ) ( ) 100 100

256 3 3x in 1 x x 1 x− − −

= -Coefficient of 255x in ( )100

31 x−

( )85 100 100

85 151 . C C= − − =

Sol7. D<0

( ) ( )2

4 a 4 64 5a 0 + − −

2a 13a 48 0 + −

( )( )a 16 a 3 0 + −

16 a 3 −

but 5 a 30−

5 a 3 −

a 5, 4, .....,1,2 = − − (integers)

Probability 8 2

36 9= =

Sol8. ( ) ( )P ~ q q ~ p →

( ) ( )~ P ~ q P ~ p

~ p q

p q →

+ +

-16 3

-

Sol9. ( )a x

0e dx 10 e 9= −

a ax x

0 ae dx e dx

= +

1 a xx

0 0a e dx e dx

= +

( ) aa e 1 e 1= − + −

( )a

a e e a 1= + − +

For a

a 10 & e 11 9= − = −

i.e. a

e 2=

a 10 ln2= +

Sol10. T TP P, Q Q= = −

( ) ( )T TA A A A

A2

+ + −=

TA A

P2

+=

2 3 2 a1

a 0 3 02

= +

FIITJEE


a 32

4 3 a1 2

a 3 0 a 320

2

+ +

= = + +

T

3 a0

0 3 aA A 1 2Q

a 3 0 a 32 20

2

− − −

= = = − −

( )( )

2a 3

det Q 9 9 a 3 6 a 9, 34

−= = − = = −

For a = 9

( )2 6

det P 366 0

= = −

For a = -3

( )2 0

det P 00 0

= =

Sol11. 1 1

2 4 2x 3 x 3 0+ + =

Put 1

43 a=

Equation : 2 2x ax a 0+ + =

2

2x x1 x aw, aw

a a

+ + = =

2aw, aw = =

96 96 96 24a w 3 = =

12 12 12 31 a w 1 3 1 26 − = − = − =

( ) ( )96 12 96 12 241 1 52 3 − + − =

Sol12. Domain of definition

1, 0= −

For x 0= LHS = 2

x 1= − LHS = 2

No solution.

Sol13. ( )1

0I 2 ln 1 x 1 x dx= − + +

Put x cos=

then ( )0

2

I 2 ln 2 sin cos . sin d2 2

= + −

( )2

2

00

cos sin12 22ln 2 2 ln sin cos . cos . cos d

2 2 2sin cos

2 2

−

= + + − + +

FIITJEE


( )2

0I 2ln 2 1 sin d

= + −

20

ln2 cos

= + +

( )ln2 12

= + −

ln2 12

= + −

Sol14. 2 4 5 7 x y

6.56

+ + + + += …………(i)

( )2 2 2 2 2 2

22 4 5 7 x y6.5 10.25

6

+ + + + +− = ………….(ii)

(i) x y 21 + =

2 2(ii) x y 221 + =

( ) ( )(i)& (ii) x,y 10,11 =

Sol15. Tangent at ( )P 2,2 to ( ) ( )2y 2x is y 2 x 2= = +

x

y 12

= +

Normal at P is y 2x 6= − +

9

R , 32

−

( )Q 2, 0−

( ) ( ) ( )1 9 1 25

PQR 2 3 2 3 2 2 0 6 10 92 2 2 2

= − − − + − = + + =

Sol16. ( ) ( ) ( )f 0 f 0 f 0− += =

1 a 1 1 a 0 − = − = − =

( ) ( ) ( )f 1 f 1 f 1− += =

a 1 2 b 2 b a b 3 − = − = − + =

Sol17. 2 2

a.c c , c a 2c.a 8, a 3, a b 3= + − = = =

2

c 9 2 c 8 c 1 + − = =

( ) 3a b c a b c sin

6 2

= =

Sol18. x

2

ydyxe dx 0

1 y+ =

−

( ) x 2x 1 e 1 y c − − − =

Point ( )1, 1 c 0− =

3 2x 3 2e 1 y= = −

FIITJEE


2 6y 1 4e= −

Sol19. x 2

0x 3

−

−

x 2 or x 3

2 x 2, − x 3 or x 3 −

2 x 3 − , x 3 or x 4 −

( ) ) )x , 3 2, 3 4, , a b c 3 2 3 2 − − − + + = − − + = −

+ +

2 3

-

Sol20. y vx=

( )dv

D.E. tanv. v x v tanV 1dx

+ = −

dv

tanvx 1 0dx

+ =

1 y

tanvdv dx 0 x ccosx x

+ = =

Point 1

,2 6

C 1 =

y

x cosx

=

1y xcos x− =

( )1

1 42

02

A xcos xdx cos . . sin d

−

= = −

4

2

1sin2 d

2 8

−= − =

Section-B

Sol1.

1 1 0

A 0 1 1

0 0 1

−

= −

20 7B 7A 20A 2I= − +

2

1 1 0 1 1 0 1 2 1

A 0 1 1 0 1 1 0 1 2

0 0 1 0 0 1 0 0 1

− − −

= − − = −

3

1 2 1 1 1 0 1 3 3

A 0 1 2 0 1 1 0 1 3

0 0 1 0 0 1 0 0 1

− − −

= − − = −

4

1 3 3 1 1 0 1 4 6

A 0 1 3 0 1 1 0 1 4

0 0 1 0 0 1 0 0 1

− − −

= − − = −

2 3

13a of A, A , A ,.....

are 0, 1, 3, 6,……….

FIITJEE


For 0, 1, 3, 6, ……..

n nS 0 1 3 6 ..... t= + + + + +

n n 1 nS 0 1 3 ..... t t−

= + + + + +

( )n n n 1t 1 2 3 .... t t−

= + + + + −

( )n 1 . n

2

−=

( ) ( )13b 7 190 20 21 0= − +

1330 420 910= − =

Sol2. 0 16m c= − + ……..(i)

( )

2

m 10 C2

m 1

− +=

+………..(ii)

(i) & (ii) ( )2 236m 4 m 1 = +

1 8

m , C2 2 2

= =

( )4 2 m c 34+ =

( )16, 0−

( )10, 0−

2

F

Y mx 6, m 0= +

Sol3. 1 1 2 2 2 3R R R , R R R→ − → −

a c 1 b c a b

b d 1 c d b c

x b d x d x c

− + − −

= − − − −

− + + +

1 1 2 2 2 3C C C & C C C→ − → −

1 1 2

a 1 b 2b c a a b 1 0

b 1 c 2c b d b c 1 0 , R R R

b d c x c b x c

+ − − − − − + −

= − − − − − = − − − → −

− − + − +

2 2

2 0 0

1 0 2 2 1

b x c

= − − − = = =

− +

Sol4. a b c= = =

a . b b . c c . a 0= = =

2

2a b c 3+ + =

( )a . a b c a a b c cos+ + = + +

2

a 0 0 a a b c cos + + = + +

2 2 1

3 cos cos3

= =

2

2 236cos 2 36 1 4

3

= − =

FIITJEE


Sol5. Area = ( )5

2

1

1 9 5 15 1 5 x dx

2 4 2

−− − −

( ) ( )1

0

2

1cos

5

1 114 10 5 sin d

8 2−

= − + − −

114 5 5 1 1A 5 cos

8 4 4 25

− −= + − −

15 5 5 15 cos

4 4 4 5

−= − −

5 5 5

, ,4 4 4

− − = = =

5

4 + + =

x=1 x 5=

x 4y 5+ =

Sol6.

Bowlers Batsmen Wicket Keepers

(6) (7) (2)

4 5 1

54

6 2

6 7 Required number of ways

6 7 2 6 7 2 6 7 2

4 6 1 5 5 1 4 5 2C C C C C C C C C= + +

15 7 2 6 21 2 15 21 1= + +

210 252 315 777= + + =

Sol7. 120 r

120 r /64r 1 rT C 4 . 5

−

+=

For r 1T+ to be rational

r should be multiple of 6 Number of such terms = 21 Sol8. For the plane P,

( ) ( )1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆn 2j i j 3k j i 3j k k 3i2

= − − + − = + = − +

a . n 0 3 0= − = ……….(i)

( )a . 1,2,3 0 2 3 0= + + = ……….(ii)

( )a . 1,1,2 2 2 2= + + = ………(iii)

(i), (ii) & (iii) 1, 5, 3 = = − =

( )2

81 − + =

Sol9. ( ) 2

x 2

x

x 0lim 2 cosx. cos2x

+

→− ( )1

( ) 2

x 21 cosx. cos2x .

x

x 0lime

+−

→=

FIITJEE


( )

( )2

1 cosx . cos2xx 2

x

x 0lime

−+

→=

( )1 2sin2xsinx cos2x cosx.

2 cos2x2.

2x

x 0lime

− −

→=

12. 1

3 a2

x 0lime e e

+

→= = =

a 3 =

Sol10. dist. =( )1 2 1 2

1 2

A A . b b

b b

( ) ( )4 , 2, 3 . 2,2,1

3

− − − −=

8 2 4 3 2

53 3

− − − − = = − −

2 2

5 9 5 93 3

− − = − − =

2

4, 143

= − , but 0

6 =

( )1b 1, 2,2= −

( ) 23, 2, 2 b− − =

( )1A ,2,2

( )2A 4,0, 1− −

( )1 2

1 2

2,2,1b b

3b b

=

fiitjee punjabi bagh

Documents