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JEE MAIN MODEL - HINTS AND SOLUTIONS

IIT, MEDICAL & ENGINEERING

Entrance COACHING

Now

@

MUKKAM:

BHABHA Institute of Sciences

2nd floor, Karayil Tower,

Near Bend Pipe Bridge, Orphanage Road,

Mukkam, PIN - 673602

Ph : 8943300201 , 8943300204.

Registration Started

BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY.

Course commencement: Immediate after the Plus two Board examination.

BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018

PART : A — PHYSICS

ALL THE GRAPHS/DIAGRAMS GIVEN ARE

SCHEMATIC AND NOT DRAWN TO SCALE

01. The diameter of a cylinder is measured using a

Vernier callipers with no zero error. It is found that

the zero of the Vernier scale lies between 5.10 cm

and 5.15 cm of the main scale. The Vernier scale has

50 divisions equivalent to 2.45 cm. The 24th division

of the Vernier scale exactly coincides with one of the

main scale divisions. The diameter of the cylinder is

(1) 5.112 cm (2) 5.124 cm

(3) 5.136 cm (4) 5.148 cm

Sol : Answer (2)

Reading = M.S.R + No of division of V.S

matching the main scale division

(1MSD – 1VSD)

50

45.205.02410.5

= 5.124 cm option (2) is correct.

02. The velocity - displacement graph of a particle

moving along a straight line is shown

The most suitable acceleration-displacement graph

will be

(1) (2)

(3) (4)

Sol : Answer (1)

03. At what angle the vector BA

and BA

must

act, so that the resultant is 22 BA

(1) cos-1

22

22

BA

BA (2) cos

-1

22

22

BA

BA

(3) cos-1

22

22

2 BA

BA

(4) cos-1

22

22

2 AB

BA

Sol : Answer (4)

04. A uniform cylinder of radius R is spinned about its

axis to the angular velocity 0 and then placed into a

corner (see figure). The coefficient of friction

between the corner walls and the cylinder is equal to

. How many turns does the cylinder accomplish

before it stops?





(1)

18

1 22

0

g

R (2)

14

1 22

0

g

R

(3)

12

1 22

0

g

R (4)

16

1 22

0

g

R

Sol : Answer (1)

05. A particle which is constrained to move along x-axis,

is subjected to a force in same direction which varies

with distance x of the particle from the origin as

F(x) = - kx + ax3. Here k and a are positive constant.

For x 0, the functional form of the potential energy

U(x) of the particle is:

(1) (2)

(3) (4)

Sol : Answer (4)

From the given function we can see that F = 0 at

x = 0 i.e., slope of U – x graph is zero at , x = 0.

Therefore, the most appropriate option is (4)

06. A particle of mass ‘m’ oscillates along the horizontal

diameter AB inside a smooth spherical shell of radius

R. At any instant K.E. of the particle is K. Then force

applied by particle on the shell at this instant is:

(1) R

K (2)

R

K2

(3) R

K3 (4)

R

K

2

Sol : Answer (3)

sin2

mgR

mvN

By conservation of energy





2

2

1sin mvmgR

07. The magnitudes of gravitational field at distances r1

and r2 from the centre of a uniform sphere of radius R

and mass m are Eg1 and Eg2, respectively. Then,

[a] 2

1

2

1

r

r

E

E

g

g if r1 < R and r2 < R

[b] 2

1

2

2

2

1

r

r

E

E

g

g if r1 > R and r2 > R

[c] 2

1

2

1

r

r

E

E

g

g if r1 > R and r2 > R

[d] 2

1

2

2

2

1

r

r

E

E

g

g if r1 < R and r2 < R

(1) [a] and [b] are true (2) [b] and [c] are true

(3) [b] and [d] are true (4) [a] and [d] are true

Sol : Answer (1)

08. A metallic cube as a bulk modulus ‘B ’and a density

‘ ’ . A pressure of ‘P ’ is applied uniformly from all

sides of the cube. The increase in density is

(1) PB

B

(2)

B

PB

(3) PB

B

(4)

PB

B

Sol : Answer (1)

= V

M , B =

V

PV

= VV

M

=

V

M

V

V1

1

=

B

P1

1 =

PB

B

09. A bubble is rising from bottom of a lake to the top

slowly such that its diameter gets doubled. If a

barometer placed on the bank of lake reads ‘h’, then

depth of lake is (Given is the relative density of

mercury in barometer)

(1) 5h (2) 4h

(3) 7h (4) 2h

Sol : Answer (3)

The rising of bubble is isothermal

P1V1= P2V2

VPVPgH aw 3

2 2)(

VPa 8

gH w = aP7

g

gh

g

PH

w

Hg

w

a

77

h7





10. A vessel having area of cross-section ‘A’contains a

liquid up to a height ‘h’ . At the bottom of the vessel,

there is a small hole having area of cross-section ‘a’.

Then the time taken for the liquid level to fall from

height ‘H1’ to ‘H2’ is given by

212)1( HHg (2) 21

2HH

ga

A

(3) 212

HHg

a

A

(4) gH2

Sol : Answer (2)

:

.2. 21 HH

ga

At

11. Steam at 1000

C is passed into 1.1 kg of water

contained in a calorimeter of water equivalent 0.02

kg at 25 0C till the temperature of the calorimeter

and its contents rises to 80 0C.The mass of the steam

condensed (in kg) is

(1) 0.130 (2) 0.065 (3) 0.260 (4) 0.135

Sol : Answer (1)

12. Two rods, one of aluminium and the other made of

steel, having initial lengths l1 and l2 are connected

together to form a single rod of length l1 + l2 .The

coefficients of linear expansion for aluminium and

steel are a and s , respectively . If the length of

each rod increases by the same amount when their

temperature are raised by t 0C , then find the ratio

21

1

ll

l

,

(1) a

s

(2)

s

a

(3) as

a

(4)

as

s

Sol : Answer (4)

13. When an ideal diatomic gas is heated at constant

pressure, the fraction of the heat energy supplied

which increases the internal energy of the gas is

(1) 5

2 (2)

5

3





(3) 7

3 (4)

7

5

Sol : Answer (4)

14. Which of the following graphs correctly represents

the variation of V

dPdV

with P for an ideal gas

at constant temperature?

(1) (2)

(3) (4)

Sol : Answer (1)

15. In a given forced field, the potential energy of

a particle is given as a function of its x – coordinate

as x

k

x

kxU 2

2

1 , where k1 and k2 are positive

constants. Find the period of small oscillations of the

particle about its equilibrium position in the field.

(1) 4

2

3

14

k

mk (2)

4

2

3

18

2 k

mk

(3) 4

2

3

124

k

mk (4)

4

2

3

122

k

mk

Sol : Answer (3)

16. A uniform rope having mass m hangs vertically

from a rigid support. A transverse wave pulse is

produced at the lower end. The speed (v) of the wave

pulse varies with height (h) from the lower end as





(1) (2)

(3) (4)

Sol : Answer (2)

17. A thin semi-circular ring of radius r has a positive

charge q distributed uniformly over it. The net field

E

at the centre O is

(1) jr

q ˆ2 2

0

2 (2) j

r

q ˆ4 2

0

2

(3) jr

q ˆ4 2

0

2 (4) j

r

q ˆ2 2

0

2

Sol : Answer (4)

18. Three capacitors C1, C2 and C3 are connected as

shown in the figure given below to a battery of V

volt. If the capacitor C3 breaks down electrically, the

change in total charge on the combination of

capacitors is





(1)

321

321 1

CCC

CVCC

(2)

321

2121 1

CCC

CCVCC

(3)

321

321 1

CCC

CVCC

(4)

321

221 1

CCC

CVCC

Sol : Answer (1)

19. For the given circuit,

If internal resistance of cell is 1.5 , then

(1) VP – VQ = 0 (2) VP – VQ = 4 V

(3) VP – VQ = - 4 V (4) VP – VQ = - 2.5V

Sol : Answer (4)

20. A current I flows around a closed path in the

horizontal plane of the circle as shown in the figure

given below. The path consists of eight arcs with

alternating radii r and 2r. Each segment of arc

subtends equal angle at the common centre P. The

magnetic field produced by current path at point P is

(1) r

I0.8

3 ; perpendicular to the plane of the paper

and directed inwards

(2) r

I0.8


and outwards

(3) r

I0.8


and inwards





(4) r

I0.8


and outwards

Sol : Answer (1)

21. A domain in ferromagnetic iron is in the form of a

cube of side length 1 m. Estimate the number of

iron atoms in the domain. The molecular mass of

iron is 55 g/mol and its density is 7.9 g/cm3.

(1) 8.65 10-10

atoms (2) 8 10-13

atoms

(3) 8 105 atoms (4) 8.65 10

10atoms

Sol : Answer (4)

22. An LCR circuit is equivalent to a damped pendulum.

In an LCR circuit the capacitor is charged to Q0 and

then connected to the L and R as shown below

If a student plots graphs of the square of maximum

charge (Qmax2) on the capacitor with time (t) for two

different values L1 and L2 (L1 > L2) of L then which of

the following represents this graph correctly?

(1) (2)

(3) (4)

Sol : Answer (1)

23. An infinitesimally small bar magnet of dipole

moment M is pointing and moving with a speed v in

the x direction. A small closed circular conducting

loop of radius ‘a’ and negligible self inductance lies

in the x - y plane with its centre at x = 0 and its axis

conciding with x axis. If x = 2a , the emf induced in

the loop is





2

0

16

3)1(

a

Mv

2

0

32

3)2(

a

Mv

2

0

8

1)3(

a

Mv

2

0

16

1)4(

a

Mv

Sol : Answer (2)

24. An electromagnetic radiation represented by

ttaE 0coscos1 falls over lithium surface

with work function of 2.39 eV. Maximum kinetic

energy of emitted photoelectrons will be

(Given, a = 4 N/C, 14106 rads

-1 and

15

0 106.3 rads-1

)

(1) 0 (2) 0.37 eV

(3) 0.70 eV (4) No photoemission occurs

Sol : Answer (2)

25. In the situation as shown in the figure, the focal

length of the thin plane concave lens is 20 cm. A

point object P is at distance of 20 cm from the lens.

Find the velocity of image formed by plane mirror

at the instant as shown in the figure.

(1) ji ˆ3ˆ2 cms-1

(2) j3 cms-1

(3) ji ˆ3ˆ2 cms-1

(4) j3 cms-1

Sol : Answer (2)





26. In an interference pattern of Young’s double still

experiment, we observe the 12th

order maxima for

wavelength 600 nm at a point on the screen. What

order will be visible at the same point, if the source

is replaced by light of wavelength 480 nm?

(1) 4th (2) 10

th

(3) 15th (4) 20

th

Sol : Answer (3)

27. An electromagnetic wave of frequency

= 3.0 MHz passes from vacuum into a dielectric

medium with permittivity = 4.0. Then

(1) wavelength is doubled and the frequency

remains uncharged

(2) wavelength is doubled and frequency becomes

half

(3) wavelength is halved and frequency remains

uncharged

(4) wavelength and frequency both remains

uncharged

Sol : Answer (3)

28. A diatomic molecules is made of two masses m1 and

m2 which are separated by a distance r. If we

calculate its rotational energy by applying Bohr’s

rule of angular momentum quantisation, its energy

will be given by (n is an integer)

(1)

22

2

2

1

222

21

2 rmm

hnmm (2)

2

21

22

2 rmm

hn

(3) 2

21

222

rmm

hn

(4)

2

21

22

21

2 rmm

hnmm

Sol : Answer (4)





29. Half-life of a neutron is about 693 s and mass of

neutron is 1.6 10-27

kg. A beam of 10000 high

speed neutrons with kinetic energy 0.08 eV is fired

in space. Number of neutrons that possibly decay in

travelling through a distance of 4 km will be

(1) 10 (2) 100

(3) 1000 (4) 10000

Sol : Answer (1)

30. For the given logic circuit

Which of these is correct?

(1) y = 0 for x1 = x3 = 0 and x2 = 1

(2) y = 0 for x1 = x2 = x3 = 0

(3) y = 1 for x1 = x2 = x3 = 1

(4) y = 1 for x1 = x2 = 1 and x3 = 0

Sol : Answer (1)

when x1 = x3 = 0 and x2 = 1, inputs to AND gate

are 1 and 1 so, y = 0 due to inversion of NOT gate

PART : B — CHEMISTRY

31. A sample of a hydrate of barium chloride weighing

of 61 g was heated until all the water of hydration is

removed. The dried sample weighed 52g. The

formula of the hydrated salt is

(atomic mass Ba = 137 amu)

(1) BaCl2. H2O (2) BaCl2.2H2O

(3) BaCl2.3H2O (4) BaCl2.4H2O

Sol : Answer (2)

Weight of hydrated BaCl2 = 61 g

Weight of anhydrous BaCl2 = 52 g

Loss in Mass = 61 – 52 = 9g

Mass of H2O removed = 9g

Moles of H2O removed = 5.018

9

Molecular mass of BaCl2 = 208

% of H2O in the hydrated BaCl2

= %75.1410061

9

10018208

1875.14

x

x

x = 2

Formula of the hydrated salt is BaCl2. 2H2O

32. At very high pressure, the compressibility

factor of one mole of a gas is given by:

(1) RT

Pb1 (2)

RT

Pb





(3) RT

Pb1 (4)

)(1

VRT

b

Sol : Answer (1)

For 1 mole of gas )(2V

aP (V – b) = RT

At very high pressure, P >2V

a so

2V

a is

negligible.

P ( V – b) = RT

PV - Pb = RT

RT

Pb

RT

PVZ 1

33. The ionisation enthalpy of hydrogen atom is

1.312 x 106 J Mol

-1. The energy required to excite

the electron in the atom from n = 1 to n = 2 is:

(1) 9.84 x 105J Mol

-1 (2) 8.51 x 10

5J Mol

-1

(3) 6.56 x 10

5J Mol

-1 (4) 7.56 x 10

5J Mol

-1.

Sol : Answer (1)

IE = 1EE

1.312 x 106 = 0 - E1

16

1 10312.1 MolJE

4

10312.1

2

10312.1 6

2

6

2

E

(Energy of electron in second orbit n = 2)

Energy required when an electron makes

transition from n = 1 to n = 2

)10312.1(4

10312.1 66

12

EEE

= 9.84 x 105 J Mol

-1

34. The hybridization of orbitals of N atom in

NO3- , NO2

+ and NH4

+ are respectively

(1) Sp, Sp2, Sp

3 (2) Sp, SP

3, SP

2

(3) SP2, Sp, Sp

3 (4) Sp

2, Sp

3, Sp

Sol : Answer (2)

35. In a fuel cell methanol is used as fuel and oxygen gas

is used as oxidizer. The reaction is:

CH3OH( )(2)()(2

3) 222 OHgCOgO

At 298 K, G of formation for CH3OH( ),

H2O( ) and CO2(g) are -166.2, -237.2 and

-394.4 kJ/Mol respectively. If standard enthalpy of

combustion of Methanol is -726 kJ/Mol. Efficiency

of the fuel cell will be

(1) 80% (2) 87%

(3) 90% (4) 97%

Sol : Answer (4)

1726 MolKJH

MolKJOHCHG f /2.166)( 3

MolKJOHG f /2.237)( 2

MolKJCOG f /4.394)( 2

)2.166()]2.237(24.394[ rG

= -702.6 KJ /Mol

% Efficiency = 100

H

G

=1726

/6.702

MolKJ

MolKJ=96.77%

36. The equilibrium constant at 298 K for a reaction

A + B C + D is 100. If the Initial

concentration of all the four species were 1M each,

then equilibrium concentration of D will be:

(1) 0.182 (2) 0.818

(3) 1.818 (4) 1.182

Sol : Answer (3)

A + B C + D

Initial conc 1 1 1 1

Equilibrium conc 1-x 1- x 1+x 1+x

Now 2

2

1

1

x

xKc

2

2

1

1100

x

x

x

x

1

110

10 – 10x = 1+ x = 9

x = 0.81

i e [D] at equilibrium = 1+0.81 = 1.818 M





37. The increasing order of the ionic radii of the given

isoelectronic species is:

(1) S2-

, Cl-, Ca

2+, K

+ (2) Ca

2+, K

+, Cl

-, S

2-

(4) K+, S

2-, Ca

2+, Cl

- (3) Cl

-, Ca

2+, K

+, S

2-

Sol : Answer (2)

38. In a face centered cubic lattice, atom A occupies the

corner positions and atom B occupies the face centre

points. If one atom of B is missing from one face

centre point, the formula of the compound is:

(1) A2B5 (2) A2B

(3) AB2 (4) A2B3

Sol : Answer (1)

Share of A from corners = 8 x 1/8 = 1

Share of B, with one face vacant = 5/2

A : B = 1: 5/2 = 2:5

Formula is A2B5

39. In 0.2 molal aqueous solution of a weak acid HX, the

degree of ionization is 0.3.Taking Kf of water is

1.85, the freezing point of the solution will be nearest

to:

(1) - 0.480oC (2) -0.360

oC

(3) -0.260oC (4) + 0.480

oC

Sol : Answer (1)

HX H+ + X

-

1 Mole 0 0

1-0.3 0.3 0.3

i = ( 1- 0.3) + 0.3 + 0.3 = 1.3

Tf = i. Kf . M = 1.3 x 1.85 x 0.2 = 0.48

Freezing point = 0oC – 0.48

oC = -0.48

0C

40. Consider the following cell reaction:

2Fe (s) + O2(g) + 4H+(aq) 2Fe

2+(aq)+ 2H2O( )

Eo = 1.67 V At [Fe

2+] = 10

-3M, Po2 = 0.1atm and

pH = 3 the cell potential at 25oC is:

(1) 1.47 V (2) 1.77 V

(3) 1.87 V (4) 1.57V

Sol : Answer (4)

2Fe Fe2+

n = 4

pH=3 [H+] = 1 x 10

-3M

2Fe (s) + O2(g) + 4H+(aq)

2Fe2+

(aq)+ 2H2O( ) Eo = 1.67 V

7

43

23

4

2

22

10)10(1.0

)101(

][

][

HPO

FeQ

Qn

EE o log059.0

VE 57.110log

4

059.067.1 7

41. A first order reaction goes as follows

K1 = 1.26 x 10-4

s-1

K2 = 3.8 x 10-5

s-1

The percentage of B in the mixture of B and C

is likely to be

(1) 80% (2) 76.83%

(3) 92% (4) 68%

Sol : Answer (2)

%83.76108.31026.1

1026.1%

54

4

BOf

42. Coagulation of 90 ml of a negative sol requires 10ml

of 0.5 M NaCl. The coagulation value of NaCl is

(1) 25 (2) 50

(3) 75 (4) 100

Sol : Answer (2)

10ml of 0.5 M NaCl = 10 x 0.5 millimoles of NaCl =

5 millimoles of NaCl.

100 ml of total sol requires

NaCl = 5 millimoles

1000 ml (ie 1 litre) sol requires

NaCl = 50 millimoles

43. In the context of Hall Heroult process for the

extraction of Al, which of the following statements is

false?

(1) Na3AlF6 serves as the electrolyte

(2) CO and CO2 are produced in this process





(3) Al2O3 is mixed with CaF2 which lowers the M.P.

of the mixture and brings conductivity.

(4) A13+

is reduced at the cathode to form Al

Sol : Answer (1)

44. Which is thermodynamically the most stable

allotropic form of phosphorus?

(1) Red (2) White

(3) Black (4) Yellow

Sol : Answer (3)

45. How many moles of KMnO4 are required to oxidise

one mole of ferrous oxalate FeC2O4 in acidic

medium?

(1) 5

2 (2)

5

1

(3) 4

5 (4)

5

3

Sol : Answer (4)

3MnO4- + 24 H

+ + 5FeC2O4 3Mn

2+ +

12 H2O + 5Fe3 + 10 CO2

5 moles of FeC2O4 = 3 Moles of KMnO4

1 mole of FeC2O4 = 5

3moles of KMnO4

46. Low spin complex of d6 cation in an octahedral field

will have the following energy in total.

(1) Po

5

12 (2) Po 3

5

12

(3) Po 25

2

(4) Po

5

2

Sol : Answer (2)

For low spin complex d6 is t2g

222 eg

o.

Total energy will be o + energy of 3 pair of

electrons

Net o + 3p = -6 x 0.4 o + 3p

= Po 35

12

47. In the reaction

Ag2O + H2O22Ag + H2O + O2

H2O2 act as

(1) reducing agent (2) Oxidising agent

(3) Bleaching agent (4) None of these

Sol : Answer (1)

48. The least stable carbonate of alkali metal is:

(1) Na2CO3 (2) Li2CO3

(3) K2CO3 (4) Cs2CO3

Sol : Answer (2)

Li2CO3 is unstable to heat, Li+ being very small in

size polarises a large CO32-

ion leading to the

formation of more stable Ci2O and CO2

49. Silica is soluble in:

(1) HCl (2) HNO3

(3) H2SO4 (4) HF

Sol : Answer (4)

HF is the only acid in which silica (SiO2) is soluble

due to the formation of Silicon tetra fluoride as

SiO2 + 4HFSiF4 + 2H2O

50. A hydrocarbon contains 20% hydrogen and 80% of

carbon. The empirical formula is:

(1) CH4 (2) CH3

(3) CH2 (4) CH

Sol : Answer (2)

Elemen

t

Percentag

e

Atomi

c mass

Relativ

e

number

of

atoms

Simples

t ratio

H 20 1 20 366.6

20

C 80 12 6.66 166.6

66.6

Empirical formula = CH3





51. CaC2 +H2O BA HgSOSOH 442 /

(1) C2H2 and CH3CHO (2) CH4 and HCOOH

(3) C2H4 and CH3COOH (4) C2H2 and CH3COOH

Sol : Answer (1)

CaC2 + 2H2O CH CH + Ca(OH)2

CHOCH

OHCHCHOHCHCH mTautomeris

HgSO

SOH

3

224

42

52. BABrCHCH HKMnOKOHaq /.

234

DCKOH

BrNH

23 D is:

(1) CH3Br (2) CH3CONH2

(3) CH3NH2 (4) CHBr2

Sol : Answer (3)

23233

/

23

.

23

23

4

NHCHCONHCHCOOHCH

OHCHCHBrCHCH

KOH

BrNH

HKMnOKOHaq

53. ZYXPhenolOHKMnO

AlClanhy

ClCHdustZn /

.

4

3

3

The product ‘z’ is:

(1) Toluene (2) Benzaldehyde

(3) Benzoic acid (4) Benzene

Sol : Answer (3)

54. In the following compounds

(1) iii > iv> i > ii (2) i > iv > iii > ii

(3) ii > i > iii > iv (4) iv > iii > i > ii

Sol : Answer (4)

Electron withdrawing group stabilizes the phenoxid

anion.

Electron withdrawing effect is maximum at ortho and

para position than meta position.

55.

What is the major product ‘A’

(1) (2)

(3) (4)

Sol : Answer (3)

Ring expansion

56. Which one is most reactive towards nuceloplilic

addition reaction?

Sol : Answer (4)

Electron withdrawing groups increase the reactivity

towards nuclephilic addition.

57. When CH3CONH2 is heated with P2O5, the product

is:

(1) CH3COOH (2) CH3COONH4

(3) CH3CN (4) CH3COCl’





Sol : Answer (3)

OHNCCHNHCCHD

OP

Oll

232352

58. CBANHHCNHPClHNO 332

252

recognise the compound ‘C’ from the following

(1) Propane nitrile (2) Methyl amine

(3) Ethyl amine (4) Acetamide

Sol : Answer (3)

22323

23223

3

32

NHCHCHClCHCH

OHCHCHNHCHCH

NH

PClHNO

59. Polysaccharides have which of the following

linkage?

(1) Glycosidic linkage (2) H-bond

(3) Peptide linkage (4) No linkage

Sol : Answer (1)

In polysaccharides different monosaccharides units

are linked to one another by glycosidic linkage

60. Salol is

(1) Acetyl salicylic acid (2) Phenyl salicylate

(3) Methyl salicylate (4) None of these

Sol : Answer (2)

Phenyl salicylate is known as salol. It is used as an

antiseptic. It is prepared by the reaction of salicylic

acid with phenol.

PART : C — MATHEMATICS

61. The principal value of tan-1

(cot 4

43) is:

(1) 4

3 (2)

4

3

(3) 4

(4)

4

Sol : Answer (4)

4

43cottan 1

=

44

44cottan 1

=

411cottan 1

=

4222cottan 1

=

4cottan 1

= 1tan 1

= 4

62. The distance of the point (1, -2, 4) from the plane

passing through the point (1, 2, 2) and perpendicular

to the planes x – y + 2z = 3 and 2x – 2y + z = 0 is:

(1) 2 (2) 2

(3) 2 2 (4) 2

1

Sol : Answer (3)

Normal vector =

122

211

ˆˆˆ

kji

=

22ˆ41ˆ41ˆ kji

= kji ˆ0ˆ3ˆ3

Equation of the plane is 02313 yx

0933 yx

Distance = 222 033

92313

= 18

963

= 222

4

23

12

23

12

63. If the vector a kji ˆˆˆ , kjbi ˆˆˆ and kcji ˆˆˆ ,

are coplanar, then the value of

cba 1

1

1

1

1

1





(1) -1 (2) 2

1

(3) 1/2 (4) 1

Sol : Answer (4)

0

11

11

11

c

b

a

Taking

C2 R2 – R1,

C3 R3 – R2

0

101

111

01

c

bb

aa

0111]011[ bcacba

0111111 bacacba

0111111 bacacba

Dividing by cba 111

01

1

1

1

1

cba

a

011

1

1

1

11

cba

a

11

1

1

1

1

1

cba

64. Let a

and b

be two unit vector such that

3ba

. If )(32 babac

then 2 c

=

(1) 55 (2) 51

(3) 43 (4) 37

Sol : Answer (1)

3ba

32

ba

3ˆˆ.ˆ2ˆ

22 bbaa

31cosˆˆ21 22 ba

01cos1121

3cos22

1cos2

2

1cos

3

babac

32

baababbababac ˆˆ.ˆ6ˆˆ.12ˆ.ˆ4ˆˆ9ˆ4ˆˆ2

222

= 002

14

3sin11941 2

24

395

24

275

= 4

277

= 4

2728

= 4

55

2

55ˆ c

55ˆ2 c

65. If A =

13

14then the determinant of the matrix

A2019

– 2A2018

- A2017

is:

(1) 25 (2) -25

(3) 75 (4) -75

Sol : Answer (1)

13

14A

201720182019 2 AAA

IAAA 222017

IAAA 222017

IAAA 222017

515

5201

2017





25251

66. The domain of the function f(x) =

2

1sin 1

xis

(1) [-3, 3] (2) [-1,1]

(3) ),1[]1,( (4) ),3(]3,(

Sol : Answer (2)

22

1sin0 1

x

1

2

10

x

210 x

11 x

11 x

11 x

Domain = 1,1

67. If 4)1

1(

2

bax

x

xxlt

x, then

(1) a = 1, b = 4 (2) a = 1, b = - 4

(3) a = 2, b = -3 (4) a = 2, b = 3

Sol : Answer (2)

41

12

bax

x

xxlt

x

41

112

x

xbaxxxlt

x

41

11 22

x

bxaxaxxxlt

x

41

11 2

x

xbaxalt

x

101 aa

41

1

x

xbalt

x

41

1

ba

4

4

31

3

41

b

b

b

ba

ba

68. The value of )2sin(

2/

2

x

dttlt

x

x

=

(1) (2) 2/

(3) 4/ (4) 8/

Sol : Answer (3)

0

02sin

2

2

x

tdt

lt

x

x

RHLx

xlt

x

..22cos

2

420cos

2

69. For xxfRx sin2log)(, and g(x) = f(f(x))

then

(1) g1(0) = cos (log 2)

(2) g1(0) = - cos (log 2)

(3) g is differentiable at x = 0 and g1(0) = -sin (log 2)

(4) g is not differentiable at x = 0

Sol : Answer (1)

xxf

xfxffxg

xffxg

sin2log

'''

xx

xxf cos

sin2log

sin2log'

0cos0sin2log

0sin2log0'

f

= - 1

2logcos2logsin2log

2logsin2log2log1

f

2logcos1

2logcos





0'0'0' fffg

0'2log' ff

2logcos12logcos

70. For ),2

5,0(

x define f(x) = dtttx

sin0 then f

has

(1) Local maximum at and 2

(2) Local minimum at and 2

(3) Local minimum at and Local maximum at 2

(4) Local maximum at and Local minimum at 2

Sol : Answer (4)

dtttxf

x

sin0

x

xxxxf

xxxf

2

1sincos''

sin'

25,0,sin2,

,00sin

0sin0'

xx

xx

xxxf

02

1sincos11

f

0222

12sin2cos2211

f

71. The integral

dx

xx

xx335

912

)1(

52

(1) cxx

x

235

10

)1(2

(2) cxx

x

235

5

)1(2

(3) cxx

x

235

10

)1(2

(4) cxx

x

235

5

)1(

Sol : Answer (1)

dx

xx

xx

335

912

1

52

= dx

xxx

xx

3525

912

1

52

=

dxxxx

xx

35215

912

1

52

=

dxxx

xx

352

93

1

52

[put , txx 521

dtdxxx

dx

dtxx

63

63

52

52

]

=

3t

dt =

c

xxc

t

252

2

12

1

2

c

xxc

t

252212

1

2

1

= c

xx

2

52

1112

1

= c

x

xx

2

5

35 12

1

=

cxx

x

235

10

12

72. The value of

3log

2log 2

2

)6sin(logsin

sindx

xx

xx2

(1) )2/3log(2

1 (2) )2/3log(

4

1

(3) log(3/2) (4) 1/6 log(3/2)

Sol : Answer (2)

3log

2log 22

2

6logsinsin

sindx

xx

xx





[Put tx 2

3log3log

2log2log

2

2

tx

tx

dtdxx

dx

dtx

]

3log

2log 26logsinsin

sin dt

tt

t

3log

2log 6logsinsin

sin

2

1dt

tt

t

2

2log3log

2

1

b

a

ab

xbafxf

dxxf

2

2

3log

4

1

3. The area of the region {(x, y): y2

}142 xyandx

is:

(1) 7/32 (2) 5/64

(3) 9/32 (4) 15/64

Sol : Answer (3)

142, 2 xyandxyyx

xy 22 (1)

14 xy (2)

Solving (1) and (2)

8

1

2

1

01812

012128

012816

011016

021816

214

2

2

2

2

xorx

xx

xxx

xxx

xx

xxx

xx

2

1,

8

1,3,

2

1

74. If a curve y = f(x) passes through the point (1, -1) and

satisfies the differential equations y (1 + xy)dx = x dy

then f(-1/2) =

(1) -4/5 (2) 2/5

(3) 4/5 (4) -2/5

Sol : Answer (3)

xdydxxyy 1

1) - , 1 ( through pass This

2

2

2

2

2

cx

y

x

xdxy

xd

xdxy

xd

xdxy

xdyydx

dxxyxdyydx

xdydxxyydx





5

4

5

82

8

5

2

1

2

1

8

1

2

1

2

1

2

1

2

2

1

2

11

2

1

1

1

2

y

y

y

yx

x

y

x

c

c

75. The probability of a man hitting a target is 2/5. He

fires at the target k times (k, a given number). Then

the minimum k so that the probability of hitting the

target at least once is more than 7/10 is:

(1) 3 (2) 5

(3) 2 (4) 4

Sol : Answer (1)

P [hitting at least 1 time ] = 1 –P [ hitting no time]

=

k

5

31

10

3

5

3

10

7

5

31

k

k

From options k =3

76. The value of

4/3

4/ cos1

x

dx

(1) 2 (2) 4

(3) -2 (4) 0

Sol : Answer (1)

4

3

4cos1

x

dxI

= 4

3

4 2

2cos2

x

dx

= 4

3

4

2

2sec

2

1

dxx

=

4

3

42

12

tan

2

1

x

4

3

4

2tan

x

8

tan8

3tan

21212

77. The integral

dxxxxxxx

xx2523235

22

)coscossinsincos(sin

cossin

(1) cx

3cot1

1 (2) c

x

3cot1

1

(3) cx

)tan1(3

13

(4) cx

)tan1(3

13

Sol : Answer (4)

dx

xxxxxx

xx2523235

22

)coscossinsincos(sin

cossin

=

dxxx

xx233

22

)cos(sin

cossin

Dividing by cos6 x

Then = dx

x

xx

)1(tan

sectan3

22

[ Put

dx

dtxx

tx

22

3

sectan3

1tan

]

dx

t

dt

)(

32

c

t

t

3

1

13

1 1

c

x

)1(tan3

13





78. If 5(tan2x – cos

2 x) = 2 cos 2x + 9, then the value of

cos 4x is:

(1) 1/3 (2) 2/9

(3) -7/9 (4) -3/5

Sol : Answer (3)

92cos2costan5 22 xxx

91cos22coscos

sin5 22

2

2

xx

x

x

7cos4

cos

coscos15 2

2

42

x

x

xx

xxxx 2442 cos7cos4cos5cos55

05cos12cos9 24 xx

18

18014412cos 2

x

= 18

32412

18

1812

18

30

18

6 or

3

5

3

1 or

9

71

9

2

13

12

12cos24cos

3

11

3

121cos22cos

3

1cos

2

2

2

2

xx

xx

x

79. If , C are the distinct roots of the equation

x2 – x + 1 = 0, then 107101

(1) 1 (2) 2

(3) -1 (4) 0

Sol : Answer (1)

2,

112

2

214101

1072101107101

80. Let R = { (3, 3), (6, 6), (9, 9), (12, 12), (6, 12),

(3, 9), (3, 12), (3, 6) } be a relation on the set

A = { 3, 6, 9 12}. The relation is:

(1) An equivalence relation

(2) Reflexive and symmetric only

(3) Reflexive and transitive only

(4) Reflexive only

Sol : Answer (3)

81. Let z, w be complex numbers such that 0 wiz

and arg (zw) = , then arg (z) =

(1) 4

5 (2)

2

(3) 4

3 (4)

4

Sol : Answer (3)

0 wiz

iwz

wiz

wiz

argarg

4

3arg

2

3arg2

arg2

argarg

z

z

z

wi

82. If twice the 11th term of an AP is equal to 7 times of

its 21st term, then its 25

th term is equal to:

(1) 24 (2) 120

(3) 0 (4) 48

Sol : Answer (3)

Let a is the first term of AP and d is common difference

∴ 11th term will be a + 10d

21st term will be a + 20d

We need to find 25th term which is a + 24d

Here given that 2 times of 11th term is equals to

7 times the 21st term





2( a + 10d) = 7( a + 20d)

2a + 20d = 7a + 140d

5a + 120d = 0 or

a + 24d =0 (which is 25th term)

∴ 25th term of given AP = 0

83. The number of arrangements of the letters of the

word ‘BANANA’ in which two N’s do not appear

adjacently is:

(1) 40 (2) 60

(3) 80 (4) 100

Sol : Answer (1)

1804

720

!2!2

!6

two N’s come together 602

120

!2

!5

!2

!2

required 12060180

84. If x4 occurs in the r

th term in the expression of

(x4 + 1/x

3)

15 then r =

(1) 7 (2) 8

(3) 9 (4) 10

Sol : Answer (3)

r

r

rx

xCt

3

154

41

115

rxC 760

415

8

756

4760

r

r

thr 9181

85. In ,ABC if (a + b + c) ( a – b + c) = 3ac, then

(1) 060B (2)

030B

(3) 060C (4)

090 CA

Sol : Answer (1)

accbacba 3

0

222

222

222

22

60

2

1cos

2

1

2

32

3

B

B

ac

bca

acbca

acbcaca

acbca

86. The number of integral values of m, for which the x

co-ordinate of the point of intersection of the lines

3x + 4y = 9 and y =mx + 1 is also an integer is:

(1) 2 (2) 0

(3) 4 (4) 1

Sol : Answer (1)

Solving, 1,943 mxyyx we get m

x43

5

x is an integer if 3 + 4 m = 1, -1, 5, -5

4

8,

4

2,

4

4,

4

2 m

so m has two integral values.

87. The number of common tangents to the circles

x2 + y

2 – 4x – 6y–12 = 0 and

x2 + y

2 + 6x + 18y + 26= 0 is:

(1) 1 (2) 2

(3) 3 (4) 4

Sol : Answer (3)

88. The point of intersection of the normals to the

parabola y2 = 4x at the ends of its latus rectum is:

(1) (0,2) (2) (3, 0)

(3) (0,3) (4) (2,0)

Sol : Answer (2)

End of the latus rectum are

12,21,1 2 ttt

Equation of the normal at tt 2,2 is

22 tttxy

Equation of the normal for 1t are

3xy and 3 xy

They intersect at (3, 0) Or





89. Let the equation of two ellipses be E1 : 123

22

yx

and E2 : 116 2

22

b

yx, if the product of their

eccentricities is 1/2, then the length of the minor axis

of E2 is:

(1) 8 (2) 9

(3) 4 (4) 12

Sol : Answer (3)

e1 e2 = 2

1

2

1

2

2

1

1 a

c

a

c

2c1 c2 = a1a2 …………………….(1)

2

1

1

16

123

bc

c

2 216 b = 43

4(16 - b2) = 48

64 - 48 = 4b2

16 = 4 b2

b = 2

2b = 4

90. If the standard deviation of the number 2, 3 a and 11

is 3.5, then which of the following is true

(1) 3a2 – 32 a + 84 = 0 (2) 3a

2 – 34a + 91 = 0

(3) 3a2 – 23a + 44 = 0 (4) 3a

2 - 26a + 55 = 0

Sol : Answer (1)

As we learnt in ,

Standard Deviation -

If x1, x2...xn are n observations then square root of the

arithmetic mean of

Where x is mean

iit, medical & engineering entrance c oaching now mukkam · 2019-01-02 · jee main model -...

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