2020 jee-main q p k s · 2020_jee-main question paper_key & solutions sri chaitanya iit academy...

2020_Jee-Main Question Paper_Key & Solutions


Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081

www. srichaitanya.net, [email protected]

PHYSICS (SINGLE CORRECT ANSWER TYPE)

This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. The plot that depicts the behavior of the mean free time ( time between two successive

collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graphs are schematic and not drawn to scale)

1) 2) 3) 4) Ans: 1

Sol: 2

vv 8RT2 d n

M

if v is kept constant then

1T

2. Consider two solid spheres of radii 1 2R 1m,R 2m and masses 1M and 2M , respectively. The gravitational field due to sphere 1 and 2 are shown. The value of

1

2

MM

is :

1) 12

2) 23

3) 16

4) 13

Ans: 3

Sol: 223

2Gm

122

1Gm

2

1

3 12 4

mm

1

2

16

mm

3. When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy AT eV and de-Broglie wavelength A . The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is B AT T 1.5 eV . If the de-Bronglie wavelength of these photoelectrons

B A2 , then the work function of metal B is: 1) 1.5 eV 2) 4eV 3) 2eV 4) 3eV




Ans: 2 Sol: Relation between De – Broglie wavelength and K.E. is

12 e

hKEKE m

BA

B A

KEKE

1 1.52

A

A

TT

2AT eV 2 1.5 0.5BKE eV 4.5 0.5 4B eV

4. The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4 kg. (The coordinates of the same are shown in figure) are :

1) (1.25m, 1.50m) 2) (0.75m, 0.75m) 3) (0.75m, 1.75m) 4) (1m, 1.75m) Ans: 3

Sol:

51 12 2

2cm

i jj i

r

3 74 4

cmr i j

5. A thermodynamic cycle xyzx is shown on a V-T diagram




The P-V diagram that best describes this cycle is : (Diagrams are schematic and not to scale)

1) 2) 3) 4) Ans: 3 Sol : xy is isobaric expansion, yz is isochoric process, zx is isothermal compression 6. A particle of mass m is fixed to one end of a light spring having force constant k and

unstretched length l. The other end is fixed. The system is given an angular speed about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:

1) 2m

k m

l 2)

2mk m

l 3)

2

2

mk m

l 4) 2

2

mk m

l

Ans: 4

Sol:

20m l x kx 0

21l kx m

20

2

l mxk m

7. At time t=0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5s, then induced EMF in the loop is:

1) 36 V 2) 56 V 3) 48 V 4) 28 V Ans: 2

Sol: 41000 500 10 Addt 5

2 45 10 64 8 10

56 v5




8. Three charged particles A,B and C with charges – 4q, 2q and -2q are present on the circumference of a circle of radius d. The charged particles A,C and centre O of the circle formed an equilateral triangle as shown in figure, Electric field at O along x-direction is:

1) 20

2 3qd

2) 20

3qd

3) 20

3 3q4 d

4) 20

3q4 d

Ans: 2 sol:

02 2

0

4 32cos30netkq qE

d d

9. The critical angle of a medium for a specific wavelength, if the medium has relative

permittivity 3 and relative permeability 43

for this wavelength, will be:

1) 045 2) 015 3) 060 4) 030 Ans: 4

Sol: 1V

r rn 2 1sin c2

0c 30

10. Consider a uniform rod of mass M= 4m and length l pivoted about its centre. A mass m

moving with velocity making angle 4

to the rod’s long axis collides with one end

of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is:

1) 3 27

l

2) 37l

3) 37 2

l

4) 47l

Ans: 1

Sol:

oi ofL L 2 21 4

2 12 42mV mL mL

6 3 277 2

V VLL




11. The length of a potentiometer wire is 1200cm and it carries a current of 60mA. For a cell of emf 5V and internal resistance of 20 , the null point on it is found to be at 1000cm. the resistance of whole wire is:

1) 80 2) 120 3) 60 4) 100 Ans: 4

Sol:

potential gradient = 51000 1200

pV 6PV V and 3

6 10060 10

PP

VRI

12. Consider a solid sphere of radius R and mass density 2

0 2

rr 1R

, 0 r R . The

minimum density of a liquid in which it will float is:

1) 025 2) 02

3 3) 0

5 4) 0

3

Ans: 1

Sol: 2

0 2

r1 0 r RR

mg B

2 3L

44 dr R3

RR 34 5

2 30 0 L2 2

0 0

r r r 44 r dr 4 RR 3 5R 3

0 L

25

13. In finding the electric field using Gauss law the formula enc

0

qEA

is applicable. In the

formula 0 is permittivity of free space, A is the area of Gaussian surface and encq is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?

1) Only when the Gaussian surface is an equipotential surface 2) Only when the Gaussian surface is an equipotential surface and E

is constant on the

surface 3) Only when E

= constant on the surface

4) For any choice of Gaussian surface Ans: 2 Sol: Magnitude of electric field is constant & the surface is equipotential




14. Effective capacitance of parallel combination of two capacitors 1C and 2C is 10 F. When these capacitors are individually connected to a voltage source of 1 V, the energy stored in the capacitor 2C is 4 times that of 1C . If these capacitors are connected in series, their effective capacitance will be:

1) 1.6F 2) 3.2F 3) 8.4F 4) 4.2F Ans: 1

Sol: Given 1 2 10 ........( )C C F i

2 21 2

1 142 2

CV C V

1 24 ........C C ii from equation (i) & (ii) 1 2C F 2 8C F

If they are I series 1 2.

1 2

1.6eqC CC F

C C

15. A leak proof cylinder of length 1m, made of a metal which has very low coefficient of expansion is floating, vertically in water at 00 C such that its height above the water surface is 20cm. When the temperature of water is increased to 04 C , the height of the cylinder above the water surface becomes 21cm. The density of water at 0T 4 C , relative to the density at 0T 0 C is close to:

1) 1.01 2) 1.03 3) 1.04 4) 1.26 Ans: 1

Sol:

00 Cmg A 80 g 04 C

mg A79 g 0

0

4 C

0 C

80 1.01P 79




16. The magnifying power of a telescope with tube length 60cm is 5. What is the focl length of its eye piece?

1)20cm 2) 10cm 3) 30cm 4) 40cm Ans: 2

Sol: o

e

fmf

5 o

e

ff

5o ef f

60o ef f 6 60ef 10ef 17. Boolean relation at the output stage-Y for the following circuit is:

1) A.B 2) A.B 3) A+B 4) A B Ans: 1 Sol: First part of figure shown OR gate and Second part of figure shown NOT gate So pY OR NOT NOR gate Y A B A.B

18. The dimension of stopping potential 0V in photoelectric effect in units of Planck’s constant ‘h’, speed of light ‘c’ and gravitational constant ‘G’ and ampere A is:

1) 1/3 2/3 1/3 1h G ,C A 2) 5/32/3 1/3 1h C ,G A

3) 2/3 1/3 4/3 1h ,C ,G ,A 4) 2 3/2 1/3 1h G ,c A Ans: 2 Sol: a b c dV K h I G C (V is voltage)

We know 2 1h ML T [I] = A [G] = M-1L3T-2

[C] = LT-1 [V] = 2 =3 1ML T A

a c db2 =3 1 2 1 1 3 2 1ML T A ML T A M L T LT

2 =3 1 a c 2a 3c d a 2c d bML T A M L T A a c 1..................(1) b 1................. 2




19. Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 12 210 m / s by an applied magnetic field (west to east). The value of magnetic field: (Rest

mass of proton is 271.6 10 kg) 1) 0.71mT 2) 0.071mT 3) 71mT 4) 7.1mT Ans: 1

Sol:

13 27 21. . 1.6 10 1.6 102

K E V

72 10v

Bqv= ma 27 12

19 7

1.6 10 101.6 10 2 10

B

30.71 10 T So 0.71mT 20. The graph which depicts the results of Rutherford gold foil experiment with - particles is: : Scattering angle Y: Number of scattered -particles detected (Plots are schematic and not to scale)

1) 2) 3) 4) Ans: 2

Sol: 4

1Nsin

2




(NUMERICAL VALUE TYPE) This section contains 5 questions. Each question is numerical value. For each question, enter the correct numerical value(in decimal notation, truncated/rounded-off to second decimal place.(e.g. 6.25, 7.00, ‐0.33, ‐.30, 30.27, ‐127.30). Marking scheme: +4 for correct answer , 0 if not attempted and 0 in all other cases. 21. A point object in air is in front of the curved surface of a plano-convex lens. The radius

of curvature of the curved surface is 30cm and the refractive index of the lens material is 1.5, then the focal length of the lens (in cm) is ________

Ans: 60.00

Sol: 1 2

1 1 11f R R

1R 2R 30cm 1 1 11.5 1f 30

1 0.5

f 30 f = 60cm

22. A body A, of mass m= 0.1kg has an initial velocity of 13i ms

. it collides elastically with

another body, B of the same mass which has an initial velocity of 15 jms

. After

collision, A moves with a velocity v 4 i j

. The energy of B after collision is written

as x J10

. the value of x is _________

Ans: 1 Since collision is elastic , initial K.C.= final K.E

B1 1 10.1 9 0.1 25 0.1 32 K.E2 2 2

B1K.E 0.1 34 322

X 0.110

x 1

23. Four resistances of 15 , 12 , 4 and 10 respectively in cyclic order to form Wheatstone’s network. The resistance that is to be connected in parallel with the resistance of 10 to balance the network is __________ .

Ans: 10.00

Sol:

10R 12 15 410 R

on solving R= 10




24. A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is _______Hz

Ans: 106.00

Sol: BV

pipe

air

BV 2 1V B 2

airpipe

VV2

pipe

n

n 1 Vf

2l

pipe1 0

V 300f f 105.75Hz If 2 1.412 2 2l

= 106.05Hz If 2 1.414

25. A particle is moving along the x-axis with its co-ordinate with time ‘t’ given by 2x t 10 8t 3t . Another particle is moving along the y-axis with its coordinate as a

function of time given by y(t)= 35 8t . At t=1 s, the speed of the second particle as measured in the frame of the first particle is given as . Then in m/s) is _______

Ans: 580 Sol: 2

AX 3t 8t c

Av 6t 8 i

=2 i

3BY 10 8t 2

Bv 24t j

B Av v v 24 j 2 i

2 2v 24 2 v= 580




CHEMISTRY (SINGLE CORRECT ANSWER TYPE)

This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. The number of bonds between sulphur and oxygen atoms in 2

2 8S O and the number of

bonds between sulphur and sulphur atoms in rhombic sulphur, respectively, are: 1) 8 & 8 2) 4 & 6 3) 8 & 6 4) 4 & 8 Ans: 1

Sol:

O S O O S O

O

O O

O

12 8 While counting number of bonds we should count -bonds also. But as the options do

not contain 12, the better option is 8 and 8 2. A graph of vapour pressure and temperature for three different liquids X, Y and Z is

shown below:

The following inferences are made: A) X has higher intermolecular interactions compared to Y B) X has lower intermolecular interaction compared to Y C) Z has lower intermolecular interactions compared to Y The correct inference(s) is/are: 1) C 2) A 3) B 4) A & C Ans: 3 Sol: ‘X’ is more volatile than Y and Y is more volatile than ‘Z’. Therefore intermolecular

forces decrease in the order Z, Y, X. 3. The complex that can show fac- and mer- isomers is: 1) [CoCl2(en)2] 2) [Co(NH3)4Cl2]+ 3) [Co(NH3)3(NO2)3] 4) [Pt(NH3)2Cl2] Ans: 3 Sol: The complex 3 3Ma b type will exhibit facial and meridional isomerism.




O2N

Co

NH3

NH3

NH3

O2N

NO2

facial

Co

NH3

NH3

O2N

NO2Meridional

NO2

H3N

4. Arrange the following compounds in increasing order of C – OH bond length: Methanol, Phenol, p-ethoxyphenol 1) methanol < phenol < p-ethoxyphenol 2) Phenol < methanol < p-ethoxyphenol 3) methanol < p-ethoxyphenol < phenol 4) Phenol < p-ethoxyphenol< methanol Ans: 4 Sol:

Methanol

Pure Single bond

p-ethoxyphenol

Partial double bond but relatively less than phenol

Partial double bond due to delocalisationof lone pair of oxygen

Phenol

3CH OH+M-group....O Et

..

..H O ....H O

5. For the Balmer series in the spectrum of H atom, 2 2

1 2

1 1Hv R

n n

, the correct

statements among (I) to (IV) are: (I) As wavelength decreases, the lines in the series converge (II) The integer n1 is equal to 2 (III) The lines of longest wavelength corresponds to n2=3 (IV) The ionization energy of hydrogen can be calculated from wave number of these

lines 1) (II), (III), (IV) 2) (I), (II), (III) 3) (I), (II), (IV) 4) (I), (III), (IV) Ans: 2 Sol: Decrease in wavelength corresponds to increase in energy gap but energy gap between

successive energy levels decreases. For Blamer series n1=2 and n2 = 3,4,5,……. (2) 3 2 has highest wave length in the series.




6. The major product of the following reaction is:

1) 2)

3) 4) Ans: 3 Sol:

OH

ProtonationH

OH2

H2O

-H2O

OH2

CH2

Resonance Stabilized Carbocation-H+

OH

7. The decreasing order of reactivity towards dehydrohalogenation (E1) reaction of the

following compounds is:

(A) (B) (C) (D) 1) B>A>D>C 2) B>D>A>C 3) D>B>C>A 4) B>D>C>A




Ans: 3 Sol:

Cl

Cl

Most stable carbocation Least stable carbocation 8. The predominant intermolecular forces present in ethyl acetate, a liquid, are: 1) London dispersion, dipole-dipole and hydrogen bonding 2) Dipole-dipole and hydrogen bonding 3) Hydrogen bonding and London dispersion 4) London dispersion and dipole-dipole Ans: 4 Sol: Relative order of intermolecular forces: Dipole-dipole > dipole –induced dipole > London dispersion forces Dipole – dipoleprimary forces and London dispersion forces secondary forces. 9. Which of the following statement is not true for glucose? 1) Glucose reacts with hydroxylamine to form oxime 2) Glucose gives Schiff’s test for aldehyde 3) Glucose exists in two crystalline forms and 4) The pentaacetate of glucose does not react with hydroxylamine to give oxime Ans: 2 Sol: Glucose gives negative Schiff’s test (Informative, NCERT based). 10. When gypsum is heated to 393 K, it forms: 1) Dead burnt plaster 2) CaSO4.0.5 H2O 3) Anhydrous CaSO4 4) CaSO4. 5H2O Ans: 2 Sol: When gypsum is heated to 393k i.e. 1200C converts into plaster of paris 4 2CaSO .0.5H O 11. The strength of an aqueous NaOH solution is most accurately determined by titrating:

(Note: consider that an appropriate indicator is used) 1) Aq. NaOH in a burette and aq. Oxalic acid in a conical flask 2) Aq. NaOH in a volumetric flask and conc. H2SO4 in a conical flask 3) Aq. NaOH in a pipette and aqueous oxalic acid in a burette 4) Aq. NaOH in a burette and conc. H2SO4 in a conical flask




Ans: 1 Sol: NaOH should not be taken in burette because it reacts with glass and the joints will be

cemented. But in NCERT, it was given NaOH is taken in burette. So answer is 1. 12. The most suitable reagent for the given conversion is:

1) LiAlH4 2) B2H6 3) H2/Pd 4) NaBH4 Ans: 2 Sol: Carboxylic acids are reduced to corresponding alcohol readily by B2H6. 13. The first ionization energy (in kJ/mol) of Na, Mg, Al and Si respectively, are: 1) 496, 577, 737, 786 2) 786, 737, 577, 496 3) 496, 737, 577, 786 4) 496, 577, 786, 737 Ans: 3 Sol: Ionisation energy of Al is less than both Mg and Si. So the order is 496,737,577,786. 14. The rate of a certain biochemical reaction at physiological temperature (T) occurs 106

times faster with enzyme than without. The change in the activation energy upon adding enzyme is:

1) +6(2.303)RT 2) +6RT 3) -6(2.303)RT 4) -6RT Ans: 3

Sol:

a a21E E

RT610 e

i.e. Energy of activation decreases by 6 2.303RT . 15. Among the gases (a) – (e), the gases that causes greenhouse effect are: (a) CO2 (b) H2O (c) CFCs (d) O2 (e) O3 1) a & d 2) a, b, c & e 3) a, c, d & e 4) a, b, c & d Ans: 2 Sol: Environmental chemistry.




16. A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at 630C while the other boils at 600C. What is the best way to separate the two liquids and which one will be distilled out first?

1) Simple distillation, 3-methylpentane 2) Fractional distillation, isohexane 3) Fractional distillation, 3-methylpentane 4) Simple distillation, isohexane Ans: 2 Sol: Mixture of liquids having less difference in boiling points are separated by fractional

distillation, in which liquid with lower boiling point comes out first. In this case isohexane (B.Pt. 600C).

17. The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

1) 9 3

2 ,1 10 MXY 2) 6 3,2 10 MXY 3) 9 3

2 ,2 10 MX Y 4) 9 3

2 ,4 10 MXY Ans: 4 Sol: From the unit M3

3 3

210 2 10

2s aq aq

XY X Y

23 3 9 310 2 10 4 10 MspK

18. As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:

1) 3 2 4 3 36K Fe CN K CrO AlCl KBr KNO

2) 3 3 2 4 36K Fe CN AlCl K CrO KBr KNO

3) 3 3 2 4 36AlCl K Fe CN K CrO KBr KNO

4) 3 2 4 3 36K Fe CN K CrO KBr KNO AlCl

Ans: 4 Sol: Ferric Hydroxide is positive sol. Therefore negative ion is effective for coagulation. More

is the –ve charge, greater is the flocculation power and lower is the flocculation value.




19. The major products A and B in the following reactions are:

1)

2)

3)

4) Ans: 4 Sol:

CN PeroxideHeat

CN

Stable radical

[A]

CN + Proceeds via

formation of secondary radical

CN

20. The third ionization enthalpy is minimum for: 1) Fe 2) Mn 3) Ni 4) Co Ans: 1 Sol: Iron by loosing 3rd electron get stable d5 config. So it will have lesser 3rd IP.

(NUMERICAL VALUE TYPE) This section contains 5 questions. Each question is numerical value. For each question, enter the correct numerical value(in decimal notation, truncated/rounded-off to second decimal place.(e.g. 6.25, 7.00, ‐0.33, ‐.30, 30.27, ‐127.30). Marking scheme: +4 for correct answer , 0 if not attempted and 0 in all other cases. 21. Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams)

of the salt required to achieve 10 ppm of iron in 100 kg of wheat is_____ Atomic weight: Fe = 55.85; S = 32.00; O = 16.00 Ans: 4.95 to 4.97

Sol: 4 2FeSO .7H O 63

x10 10100 10

i.e. x= 1gm No. of moles of 1Fe55.85

No. of moles of 4 21FeSO .7H O

55.85

Amount of 4 2 161FeSO H O 55.85 0 18 7

55.85 =4.97gms




22. What would be the electrode potential for the given half cell reaction at pH=5? 0

2 2 red2 4 4 ; 1.23VH O O H e E

(R=8.314 J mol-1 K-1; Temp = 298 K; oxygen under std. atm. Pressure of 1 bar) Ans: 1.52 to 1.53

Sol: 40.059 1E 1.23 log

4 H

45

0.059 11.23 log4 10

0.0591.23 204

=1.23+0.295 =1.53 V

23. The number of chiral centres in pencillin is_____ Ans: 3.00 Sol:

N

S

COOH

NH

O

R

O

24. The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride

ions in 0.3 g of [Co(NH3)6]Cl3 is______.

3 36

3

267.46 g/mol

169.87 g/mol

M

M

Co NH Cl

AgNO

Ans: 26.80 to 27.00

Sol: 3v 0.125 0.3 10

3 26 46

V=27.89 ml 25. The magnitude of work done by a gas that undergoes a reversible expansion along the

path ABC shown in the figure is_______.

Ans: 48.00

Sol: 31Work 36 6 4 48pa.m2




MATHEMATICS (SINGLE CORRECT ANSWER TYPE)

This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. If the equation, 2x bx 45 0, b R has conjugate complex roots and they satisfy

z 1 2 10 , then:

1) 2b b 12 2) 2b b 30 3) 2b b 72 4) 2b b 42 Ans: 2 Sol: Let z i be roots of the equation So 2 b and 2 2 45

Now, 1 2 10z 2 21 40

2 21 5 2 1 5 2 6

Hence 6b b2 – b = 30 2. If a, b and c are the greatest values of 19

pC , 20qC and 21

rC respectively, then:

1) 1) 11 22 21a b c 2)

11 22 42a b c 3)

10 11 42a b c 4)

10 11 21a b c

Ans: 2 Sol: We know n

rC is greatest at middle term(s)

19 19 1910 9pa C C C 20 20

10qb C C 21 21 216 10 11C C Cc

Now, using 11

nr

nr

C nrC

, we get 19 19 199 9 920 21 2010 11 10

a b cC . C . C

1 2 42 11a b c

/

11 22 42a b c

3. Let the line y mx and the ellipse 2 22 1x y intersect at a point P in the first quadrant. If

the normal to this ellipse at P meets the co–ordinate axes at 1 03 2

,

and 0, , then is

equal to:

1) 2 23

2) 23

3) 23

4) 23

Ans: 3 Sol: Let P be (x1, y1)

Equation of normal at P is 1 1

12 2x yx y




It passes through 11

1 1 1 1023 2 6 2 3 2

, xx

So 12 2

3y (as P lies in 1st quadrant) , So 1 2

2 3y

4. The locus of a point which divides the line segment joining the point 0 1, and a point on the parabola, 2 4x y , internally in the ratio 1:2 is:

1) 24 3 2x y 2) 2 3 2x y 3) 29 3 2x y 4) 29 12 8x y Ans: 4 Sol: Let P(0, -1) , Q (2t, t2) and R (h, k)

Given that R divides PQ in 1:2, 22 2

3 3t th ,k

Hence locus is 2

233 2 9 12 82hk x y

5. The inverse function of 2 2

2 2

8 8 1 18 8

x x

x xf x ,x ,

, is

1) 1 14 1e

xlogx

2) 1 14 1e

xlogx

3) 81 14 1e

xlog e logx

4) 81 14 1e

xlog e logx

Ans: 4

Sol: 2 2

2 28 88 8

x x

x xy

2

21 81 8

x

xyy

4 181

x yy

8141

yx logy

8

1 14 1

ylogy

x

18

1 14 1

xf x logx

6. Let the volume of a parallelopiped whose coterminous edges are given by

3 u i j k , v i j k ,

and 2w i j k ,

be 1cu.unit. If be the angle between the edges

u and

w , the cos can be:

1) 76 3

2) 76 6

3) 57

4) 53 3

Ans: 1




Sol: Volume of parallelepiped formed by coterminous vectors a,b,c

is a b c

1 11 1 1 3 3 1 2

2 1 1

or 4

For 2 ,2 1 2 5

66 6cos

For 4 ,

32 1 4 7

66 18cos

7. Let, then 1f x x cos sin x , x ,2 2

which of the following is true?

1) f 02

'

2) f is not differentiable at x = 0

3) f ' is increasing in ,02

and decreasing in 0,2

4) f ' is decreasing in ,02

and increasing in 0,2

Ans: 4

Sol: 1f x x cos sin x 1

2sin sin xx

2

x x

2 0

02

x xxf xx

x x

22 0

022

xxf ' xx

x

xf ' is increasing in 02

,

and decreasing in 02

,

8. Let f : R R be such that for all 1 x 1 x x xx R 2 2 , f x and 3 3 are in A.P., then the

minimum value of f x is: 1) 2 2) 3 3) 4 4) 0 Ans: 2

Sol: 1 12 2 3 3

2x x x x

f x

Using AM GM, we get 3f x




9. Let 21 1f x sin tan x sin cot x 1 , x 1 .If 1dy 1 d sin f x

dx 2 dx and y 3

6

then

y 3 is equal to :

1) 6 2) 5

6 3)

3 4) 2

3

Ans: Given 2 by NTA, But should be BONUS

Sol: Let 1 xtan 2 4 4 2

, ,

as 1x

2

221 2

1xf x sin cos sinx

Now 1 12

2 42

24 2

, ,sin f x sin sin

, ,

12

1 11

d sin f x , xdx x

2

11

dydx x

1

11

2

11

C tan x , xy

C tan x , x

13

6 2y C

But we cannot find C2 from given information This question must be BONUS 10. The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively.

Each of these 10 observations is multiplied by p and then reduced by q, where p 0 and q 0 . If the new mean and new s.d. become half of their original values, then q is equal to :

1) 10 2) –10 3)–20 4) –5 Ans: 3 Sol: If each observation is multiplied with p & then q is subtracted

New mean 1x px q 10 20 1p q ...... and new standard deviations

2 11 11 22 2

p p p p

If 12

p then q = 0 (from equation(1))

If 12

p , then q = - 20




11. 2

12 x

2x 0

3x 2lim7x 2

is equal to:

1) 1e

2) 2e 3) 2

1e

4) e

Ans: 3

Sol: Let 22

220

142

2220

22 20

1 47 2

1 3 2 17 23 2

7 2xx

x

xxlim .x x

xlimx xxL lim e e e e

x

12. For a > 0, let the curves 21C : y ax and 2

2C : x ay intersect at origin O and a point P. Let the line x=b (0 < b < a) intersect the chord OP and the x–axis at points Q and R, respectively. If the line x= b bisects the area bounded by the curves, 1 2C and C and the area

of 1OQR2

, then ‘a’ satisfies the equation:

1) 6 3x 6x 4 0 2) 6 3x 12x 4 0 3) 6 3x 12x 4 0 4) 6 3x 6x 4 0 Ans: 2 Sol:

2

0

16 4 412 3

ba a

xax dxa

3 3 22

3 3 6a b b a

a

Also 2 1 1

2 2b b

22 1

3 3 6a a

a

2 24 a aa

42

416 4a a aa

a6 – 12a3 + 4 = 0




13. The shortest distance between the lines

x 3 y 8 z 3

3 1 1

and x 3 y 7 z 6

3 2 4

is:

1) 7 302

2) 3 3) 2 30 4) 3 30

Ans: 4 Sol: 6 15 3AB i j k

4 22p i j k

7q i j k

1 4 22 6 15 31 1 7

i j kp q i j k

S.D = 36 225 93 30

36 225 9

AB. p q

p q

14. If c is a point at which Rolle’s theorem holds for the function, 2

exf x log

7x

in

the interval 3, 4 , where R , then f c" is equal to:

1) 37

2) 112

3) 124

4) 112

Ans: 4

Sol: 3 4 12f f 2

21212

xxx x

f '

0

121

12

f ' cc

f " c

15. If

1/6

2/33 6

cos xdx f x 1 sin x csin x 1 sin x

where c is a constant of integration,

then f3

is equal to:

1) 98

2) –2 3) 2 4) 98

Ans: 2




Sol: sinx = t cosx dx = dt

2 23 6 3 37

611 1

dt dtIt t t

t

Put 3 276

1 112

dtr r drtt

1

12 6 36 3

2 6 21 1 1 1 1 12 2 2 2

r dr sin xr c c sin x cr sin x sin x

21 cosec2

f x x and 3 23

f

16. Which one of the following is a tautology? 1) P P Q 2) Q P P Q 3) P P Q 4) P P Q Q Ans: 4 Sol: p q p q p p q p p q q q p p q p q p p q p q p p q

T T T T T T T T T T T F F F T T F T T T F T T F T F F F T F F F T F T T F F F F 17. For which of the following ordered pairs , , the system of linear equations

x 2y 3z 1 3x 4y 5z 4x 4y 4z is inconsistent? 1) 4,6 2) 1,0 3) 3, 4 4) 4,3 Ans: 4

Sol: Note 3 3 1 2

3 4 51 2 3 00 0 0

3 4 51 2 3 2 34 4 4

D R R R R

Now let P3 = 4x + 4y + 4z - = 0. If the system has solutions it will have infinite solution, so 3 1 2P P P

Hence 3 4 4 2 4 2 2& & So for infinite solution 2 2 for 2 2 system inconsistent




18. Let two points be A 1, 1 and B 0, 2 . If a point ' 'P x , y be such that the area of

PAB 5sq.units and it lies on the line, 3x y 4 0 , then a value of is : 1) 4 2) 3 3) –3 4) 1 Ans: 2

Sol: 0 2 1

1 1 1 12 1

Dx' y'

2 1 10x' y' x' 2 2 10x' y' x' 3 12x' y' or 3 8x' y' 3 2,

19. Let A and B be two independent events such that 1P A3

and 1P B6

. Then, which of

the following is TRUE?

1) 1P A / B3

2) 2P A / B3

3) 1P A / B3

' ' 4) 1P A / A B4

Ans: 1 Sol: A & B are independent events so

13

P AAB

p

20. Let y y x be a solution of the differential equation, 2 2dy1 x 1 y 0, x 1dx

. If

1 3y2 2

, then 1y2

is equal to:

1) 12

2) 32

3) 32

4) 12

Ans: 4

Sol: 1 12 2

01 1dy dx sin y sin x c

y x

At 1 11 32 2 2

x ,y c sin y cos x

Hence 11 1 12 2 2

y sin cos




(NUMERICAL VALUE TYPE) This section contains 5 questions. Each question is numerical value. For each question, enter the correct numerical value(in decimal notation, truncated/rounded-off to second decimal place.(e.g. 6.25, 7.00, ‐0.33, ‐.30, 30.27, ‐127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases.

21. The least positive value of ‘a’ for which the equation, 2 332x a 10 x 2a2

has real

roots is ____________ Ans: 8 Sol: 0D

2 3310 4 2 2 0

2a a

210 4 33 4 0a a

2 4 32 0 4 8a a , ,a

22. The number of all 3 3 matrices A, with entries from the set {–1, 0, 1} such that the sum of the diagonal elements of TAA is 3, is ___________

Ans: 672 Sol: tr (AAT) = 3 2 2 2 2 2

11 12 13 21 33 3a a a a .......... a Possible cases

96

0 0 0 0 0 0 111 10 0 0 0 0 0 1 1 1 1 8 84 8 6720 0 0 0 0 011 1 3

0 0 0 0 0 0 11 1 3

, , , , , , , ,, , , , , , , , C, , , , , , ,

, , , , , , , ,

23. An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is _

Ans: 490 Sol: 0 Red, 1Red, 2Red, 3Red Numebr of ways 7 5 7 5 7 5 7

4 1 3 2 2 3 1 35 175 210 70 490C C . C C . C C . C

24. The 20

k 1sum 1 2 3 ..... k

is

Ans: 1540

Sol: 20

1

12k

k k

20

2

1

12 k

k k




20 21 41 20 2112 6 2

1 420 41 20 212 6 2

1 2870 2102

= 1540 25. Let the normal at a point P on the curve 2 2y 3x y 10 0 intersect the y–axis at

30,2

. If m is the slope of the tangent at P to the curve, then m is equal to

Ans: 4 Sol: 1 1x , yP

1

1

62 6 01 2

xyy' x y' y'y

1

1

1

1

1 26

32 y

x

yx

1 1 19 6 1 2 1y y y 1 2x

Slope of tangent = 123

= 4 4n

Prepared by Sri Chaitanya Faculty

2020 jee-main q p k s · 2020_jee-main question paper_key & solutions sri chaitanya iit academy...

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