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MOCK TEST – I (JEE MAIN)

ANSWER KEYS, HINTS & SOLUTIONS

PHYSICS

1. B

A B×

is a vector perpendicular to both A

and B

Now, ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆA B i 2 j k 3i j 2k× = − + × + −

ˆ ˆ ˆ3i 5 j 7k+ +

Now, A B

nA B

×=

×

2 2 2

ˆ ˆ ˆ3i 5 j 7k

3 5 7

ˆ ˆ ˆ3i 5 j 7k

83

+ +=

+ +

+ +=

2. B Horizontal velocity remains same

o o o30 cos60 v cos45

v = 15 2 m/s

=

⇒

3. A

( ) ( )

2 2

net

2

v 29 t

dva 29

dt

R ma 29 N

F N R

10 29 129 N

=

= =

= =

= +

= + =

4. C

p∆ after 1st impact ( )ep p= − −

( )p 1 e= +

Similarly p∆ after 2nd

impact = ( )ep 1 e+

So, ( ) 2

netp p 1 e 1 e e ........ = + + + +

( )

( )p 1 e

1 e

+=

−

5. B Force of explosion is internal and system is initially at rest

Let the velocities of the first two fragments are ˆv i & ˆvj and that of the fragment 2m be 1 2 3ˆ ˆ ˆv i v j v k+ + ,

So, 1 2 3p p p 0+ + =

( )1 2 3

1 2 3

ˆ ˆ ˆ ˆ ˆmvi mvj 2m v i v j v k 0

v vv , v and v 0

2 2

⇒ + + + + =

⇒ = − = − =

F R

N mg

N




FMT-JEE-2018-MAIN-PCM - P2

So, ( )( )

3

2 2 2 2 2

f 1 2 3

2

v vˆ ˆp 2m i j2 2

1 1 1K mv mv 2m v v v

2 2 2

3 mv

2

= − +

= + + + +

=

Energy released in explosion

f iE K K∆ = −

2 23 3mv 0 mv

2 2= − =

6. D Angular momentum about point of contact will be conserved

i fL L=

2 200

0

v2 2 vMr Mrv Mr Mrv

5 2r 5 r

6v v

7

+ = +

⇒ =

7. C

mg T ma− = and

2mgl ml aI

l2 3

2

3ga

4

= α = ×

⇒ =

So, mg

T4

=

8. C

0 0

V r

V r

0

0

K dVI

r dr

drdV K

r

rV K ln V

r

= − = −

⇒ =

⇒ = +

∫ ∫

9. D

By conservation of angular momentum, 1 2mv R mv r=

1 2 v R v r⇒ = ….(i)

By conservation of energy, 2 2

1 2

GMm 1 GMm 1mV mV

R 2 r 2− + = − +

( )1

2GMrV

R R r⇒ =

+

Now, 1

2GMrRmv R m

r R=

+

10. D By equation of continuity

Av av '=

V1

V2

R r

mg

T

a

α





2

2

A Rv ' v v

a r

v ' 400 cm/sec.

π ⇒ = = ×

π

⇒ =

11. D For cylinder A ;

p 1Q nC T∆ = ∆

For cylinder B, v 2Q nC T∆ = ∆

Hence, p 1 v 2nC T nC T∆ = ∆

( )v v 2C R 30 C T+ = ∆

For diatomic gas,

v

2

5C R

2

T 42 K

=

⇒ ∆ =

12. B

( )0

dT d

dt dt

dT K dt

θ∝ ∆θ ∝ θ − θ

⇒ = ∆θ

In first case,

odT 61 59 2 C= − = ; o30 C∆θ = , dt = 4 min.

For second case, dT = 2oC; o20 C∆θ =

dT 2 1K

dt 30 4 60

dT 2dt 6 min.

1K20

60

= = =∆θ ×

= = =∆θ

×

13. C

( )y a sin t cos t

1 1 a 2 sin t cos t

2 2

= ω + ω

= ω + ω

a 2 sin t4

π = ω +

The motion is SHM with amplitude a 2

14. B

2 2

2

aT xT 4T

x x T

ω π= = ×

24

Constt.T

π= =

15. A

Relative velocity is ( )0v v+

Opponent frequency is ( )0v v+

λ

The no. of positive crests striking per sec. is same as frequency.

In 3 sec. 0v v3

+

λ

16. B





From symmetry, E

due to a uniform linear charged can only be radially directed. As a Gaussian

surface, we can choose a circular cylinder of radius r and length. L, closed at each end by plane caps

normal to the axis.

o in

o in

E. ds q

E.ds E.ds q

ε =

ε + =

∫

∫ ∫

Cylindrical Plane Surface

( ) o

o

o 0

E 2 rl E.ds cos90 l

lE

2 rl 2 r

ε π + = λ

λ λ= =

ε π πε

The direction of E

is radially outwards for a line of positive charge.

17. B

( )x

y

VE 10x 5y 10 10 0

x

VE 5x 5

y

ˆE 5 j V/m.

∂= − = − + = − + =

∂

∂= − = − = −

∂

∴ = −

18. A

Work done = change in potential energy

( )

( )( )

( ) ( )

2 1

2

1

22 2

2

2 r r

2

r

U U

U 1/ 2 E C

EC1 1 E CU 1/ 2 E C

2 C' 2 C

Work done = 1/ 2 E C 1 .

= −

=

= = ε = ε

∴ ε −

19. D

The resistance of the parallel combination of 2 and 3Ω Ω resistors is given by

1 1 1 5R 1.2

R 2 3 6= + = ⇒ = Ω

This resistance is in series with 2.8Ω giving a total effective resistance 1.2 2.8 4 .= + Ω = Ω

In the steady state, charge on the capacitor C has stablised and hence no current passes through

4Ω resistor which is in series with the capacitor.

Thus the current through the circuit = 6/4 = 1.5 A,

ABV 1.5 1.2 1.8V,= × = I through 2Ω resistor = 1.8/2 = 0.9 A.

20. A

( )

( )

1 2

1 2

2

2

1 2

r 1.5 , 52cm, 40cm

r

R

rresistance R =

1.5 40

52 40

1.5 405

12

= Ω = =

−=

−

×=

−

×= = Ω

l l

l l

l

l

l l

21. D

r

l

+++++ +++++ +++++ +++++ +++++ +++ E

∞ ∞





2 2 2 2

2

22 2

31 21 2 3

1 2 3

dR

A A V m / d m

or R m

25 9 1R : R : R : : : : 125 : 15 : 1

m m m 1 3 5

ρ ρ ρ ρ ρ= = = = =

∝

= = =

l l l l l

l

l

ll l

22. A

The magnetic field at P due to the flat coil of n turns, radius r, carrying

current i is

( )( )

( )22 20 0 0 0

3/2 3 3 32 2

n r inir nirB . . d r . . .

2 2 2 2d d dd r

πµ µ µ µ µ= ≅ >> = =

π π+

23. C

0NiB

µ=

l where N = Total number of turns, l = length of the solenoid

74 10 N 100.2

0.8

−π × × ×⇒ =

44 10N

×⇒ =

π

Since N turns are made from the winding wire so length of the wire

( ) [ ]L 2 r N 2 r length of each turns= π × π =

42 34 10

L 2 3 10 2.4 10 m− ×⇒ = π × × × = ×

π

24. D

( )

( )

( )

0

1

1

22 2

t / t

2 0 0

2

3

0

t /0.2

2

bt 5t

2 2

E 12l 6A

R 2

dlE L R l

dt

E 12l l 1 e l 6A

R 2

L 400 10t 0.2

R 2

l 6 1 e

Potential drop across L = E - R l 12 2 6 1 e 12 e

−

−

−

− −

= = =

= + ×

− − ⇒ = = =

×= =

= −

= − × − =

25. D

It is mentioned in the problem that on filling the vessel with the liquid, point B is observed for the same

setting; this means that the image of point B, is observed at A, because of refraction of the ray at C. For

refraction at C,

a

sin r1.5

sin i

Now,

= µ =l

2 2

AD 10sin r =

AC 10 h=

+

Where h is the height of vessel.

N

r C

T

i h

D

5 cm

A B

Liquid

1.5µ =

R1

E

L

R2 S

r

n i

i d

P





2 2

2

2

2

2

2 2

2

BD 5sin i=

BC 5 h

10 25 h. 1.5

5100 h

25 h 9

16100 h

400 16h 900 9 h

7h 500

h 8.45 cm

=+

+∴ =

+

+∴ =

+

∴ + = +

∴ =

∴ =

26. C

In this case

( )( )

( )

( )

( )

L

M

L

L

M

M

L M L L M

11 1 11

f R R

Rand F

2

11so P

f R

1 2and P

f R

and hence power of system

P=P P P 2P P

2 1 2 2P

R R R

1 RF

P 2

µ − = µ − − =

∞ −

−=

µ −= =

= − =

+ + = +

µ − µ⇒ = + =

∴ = − = −µ

i.e. the lens will be equivalent to a converging mirror of focal length ( )R / 2µ

27. A

For eyepiece,

vE = –25 cm, fE = +5 cm

E

0 E 0

0 0

u 4.17 cm 4.2 cm

L v u 12.2 cm here v 8 cm

f 1 cm u 1.1 cm

⇒ = − −

= + = =

= + ⇒ = −

≃

28. B

Wavelength associated with a particle is given by

( )o 31

o

3

h

2mE

h 150A for electron M = 9.11 10 kg

V2meV

1500.122 A

10 10

For minima

dsin n

For first minima n = 1

0.55 sin 0.122

0.122sin 0.2218

0.55

−

λ =

λ = = ×

λ = =×

θ = λ

θ =

θ = =





29. D

0

0

0

0 0

hv hv E

E hv hv

cBut v =

1 1E hc hc

= +

= −

λ

λ − λ= − =

λ λ λ λ

30. D

The resistivity of pure Si is given by

( ) ( )

( ) ( )

e e h h 1 e h

i 19

e h

16 3

1 1 1

e n n en

1 1or n

e 1.6 10 3000 0.12 0.045

1.26 10 m

−

−

ρ = = =σ µ + µ µ + µ

= =ρ µ + µ × × +

= ×

When 1019

atoms of phosphorous (donor atoms of valence five) are added per m3, the semiconductor

becomes n – type semiconductor.

19 16

e h e d hn n n N 10 n 1.26 10∴ − ≈ = = = ×∵

19 19

e e

1 1Resitivity = 5.21 m

n e 1.6 10 10 0.12−ρ = = Ω

µ × × ×

CHEMISTRY

1. A

HF has high melting point than HCl due to intermolecular hydrogen bonding while HCl < HBr < HI (melting point) due to increase in molecular weight.

2. C The room energy 3/2RT derives from kinetic (moment) energy of a gas or material. 3. A 4. A 5. B

( )ν = −a z b

Now, tan45 1 a° = =

ab = 1

ν = − =39 1 38

ν = 1444 s-1

6. A

≃b

c

T 2

T 3

ν

θ

z





7. C

Volume ( ) ( )− −= × = ×33 8 22 3a 5 10 cm 1.25 10 cm

d = 4g cm-3

Mass of unit cell = 5 x 10

-22 g

Mass of one molecule −

=× 23 1

72g

6.023 10 mole= 1.195 x 10

-22 g

No. of FeO molecules per unit cell −

−

×= =

×

22

22

5 104

1.195 10

The + −+2 24Fe 4O

8. D The decrease is ‘S’ character between bond angle 120 and 109.5 is 8.3% For 2.5

o decrease the decrease in ‘S’ character

×

= =8.3 2.5

1.9810.5

Thus S-character = 25 – 1.98 = 23.02% 9. D

Average molar mass 1 1 2 2

1 2

m x m x 28 80 32 20

x x 100

+ × + ×= =

+

= 28.8 g mol–1

A Molar volume at STP (Vm) = 24.8 L

Density 128.81.161 gL

24.8

−= =

10. A 11. D The highest rise would be recorded. If neutralization process is complete

2 4 2 4 2

2NaOH H SO Na SO H O+ → +

Let meq of NaOH x, H2SO4 = y, then x = y for complete neutralization

2 4NaOH H SO

N V N V× = ×

2 4

NaOH

H SO

V 1

V 1= then 50, 50 ml each

12. C 13. B

( )1 n

t Ao−

∴ ∝

or ( )n 1

const Ao t−

× =

or ( )1 n

t Ao const−

× =

or ( )n 1

t Ao const−

× =

or ( )−

× =n 1

2t Ao const

but from question

n 1

12

−= , n = 3

14. C Because elimination of ‘H’ is easier that of ‘D’ and this type of elimination takes place in trans manner. 15. A





NO2

CH3

NO2O2N

ExplosionCO + N

2 + H

2O + CO

2

16. D 17. C

Ag being salt is alloyed with copper. The composition of a silver alloy in expressed as its fineness. The amount of Ag in 1,000 parts of the alloy hence. 82.5% Ag and 17.5% Cu

18. C

2 3 2 2 22As S 4NaOH NaAsO 3NaAsS 2H O+ → + + soluble

19. A

Cr

O O

O

O O K

O K

OOK

20. C Follow structure of one dimensional silicates 21. B

NH2

OH2

3HNO2

O

H

N N⊕

≡ CHO

fliplicy possible

OH

NH2

HNO2

O

H

H

N N⊕

≡

O

Ring bond between C2 and C3 as well as the Hydrogen atom can migrate independently to form cyclopentane carboxaldehycle and cyclohexananone respectively.

22. C

23. C 24. C 25. A

N

Indole

CHCl3

C2H5ONa

N

H

Cl

Cl -HCl

N

ouinolive





Urotropine (CH2)6 N4 hexamethyl diamine = NH2(CH2)6 – NH2 26. B 27. A

(i) K2Pt Cl6 – three ions (ii) ( )3 42Pt NH Cl - one molecule

(iii) ( )3 33Pt NH Cl Cl - Two ions (iv) ( )3 35

Pt NH Cl Cl - four ions

and conductivity ∝ no. of ions 28. A At isoelectric point migration of α - amino acid not observed

29. A

( )2A

RCH OH RCHO HBr→ + +

3

2

NaOHA CHI

I(R – should be CH3 when it shows haloform)

30. B According to bent rule.

MATHEMATICS

1. C

Let 4 3 2

1 2 3 4 5n 10 x 10 x 10 x 10x x= + + + +

3 2

1 2 4 5m 10 x 10 x 10x x= + + +

∴ ( ) ( )2

4 3 5 4 510m n 10 x x 10 x x x− = − + − −

10m n− is a three digit number & 10m n

m

− is an integer where numerator is three digit & denominator

is four digit.

3 4 5 10m n 0 x x x 0∴ − = ⇒ = = =

∴ ( )4 3 3

1 2 1 2n 10 x 10 x 10 10x x= + = +

∴ 1 210x x+ is a two digit no.

∴ 1 210 10x x 99≤ + ≤

2. B

n 2 1 2 n 1a 1 a a ........a+ +− =

it follows that n 2n 1

n 1

a 1a

a 1+

+

+

−=

−

( )n 2 n 1 n 1 n 1 n 1

n 1 n 1n 1 n 1 n 2

2

1 1 1 1

a 1 a 1 a a 1 a

1 1 11 1

a a 1 a 1

11 2

a 1

+ + + + +

∞ ∞

= =+ + +

∴ = = −− − −

+ = + −

− −

= + =−

∑ ∑

3. C

NH

O

O

B

C





a . c b . c cos= = θ

Taking dot product with a, cosθ = α

Taking dot product with b, cosθ = β

Taking dot product with c

1 cos cos a b c = α θ + β θ + γ

∴

22

2 2

2

1 0 cos

a b c 0 1 cos 1 2cos

cos cos 1

So, 1 2cos 1 2cos

1 2cos

θ

= θ = − θ θ θ

= θ + γ − θ

γ = − θ

So, 2 2 2 1α + β + γ =

4. B

x 4 1 x 23 2

x 4 x 4x 2 x 2

1 x 8

x 2 x 4

5 5

5 5

1 x 8

x 4x 2

x 2 x 8

x x 6 0

Let x t

− − − − −+ +

−

+ −

=

=

−=

−+

− = −

− − =

=

( ) ( )

2

2

t t 6 0

t 3t 2t 6 0

t 3 t 2 0

t 3, 2

x 3

x 9

− − =

− − − =

− + =

= −

=

=

5. A

2 2

a b2, 1.5

R R

2RsinA2

R

sinA 1

7 c a b R

2

= =

=

⇒ =

∴ = − =

( ) ( )

( )

2 2BD AB AD

AB bcAD AC

AB AC a c

= +

= =+ +

( )

bcAE

a b

3R 7 3R 7AD , AE

142 4 7

=+

= =+

2R 7 3R 2

BD , CE7 1 7

= =+

6. B

A

B C

E D

a

b c





1st term of A.P. & G.P. = a

Let common ratio be r & common difference be d.

2a 2d ar+ = ……(1)

& a d ar 0.25+ = + ……(2)

From (1) & (2), 2ar 2ar a 0.5 0− + − =

( )

( ) ( )

( )

2 4a 4a a 0.5 0

2a 0

2a0 2 a R

2a

a a 0.5 0 & a 4a 4a a 0.5 0

1a , 0 ,

2

∴ − − ≥

≥

< < ⇒ ∈

− > − + − >

∈ −∞ ∪ ∞

7. C

Let the subsets be n words from the alphabet 0, 1. Let an be the no. of binary words with no two

successive ones. The words can start either with 0 and may continue in n 1a − ways or they start with 10

& may continue in n 2a − ways

n n 1 n 2a a a− −= + 1 2a 2, a 3= =

3 4 5 6 7 8 9a 5 a 8 a 13 a 21 a 34 a 55, a 89= = = = = = =

8. A

( )

35

24

1f x

x

3 1

xxI dx1

1x

=

−=

−∫

3

7 3

2

2 6

3 1

x x dx1

1 1

x x

−=

−

∫

Let

2 6

1z

1 1

x x

=

−

[ ]

80

729

15

64

80/729

15/64

10

7

1 dz I

2 z

1 logz

2

1 2 log

2 3

10, 7

∴ =

=

=

= α = β =

∫

9. D





( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

( )

2 2 2

o

cos C cos B sin A cos B A cos B A

cosC . cos A B cos C cos B A

cosC cos A B cosCcos B A

cosC cos A B cos B A 0

cosC 0 or 2sinBsinA 0 Not possible

C 90

== − = + −

⇒ π − + = π − −

⇒ − + = − −

⇒ − + + − =

⇒ = =

⇒ =

( )P sinA sin 90 A 0 0

1 1 sin2A 0 P

2 2

= − + +

= ⇒ < ≤

10. B

2x2 e 1− < (if x 0→ )

2

2 2

2 2

x

x x

x x

e 1 2

2 e 2 e1 0 1 if x 0

e 1 e 1

x1

tanx

+ >

− −< ⇒ < < →

+ +

<

11. A

Differentiating with respect to x

( ) ( ) ( )

( )

( )

2 2

2

1 1

20 0

1 xf x xf ' x f x 1 1

x 1 x x 1

1xf ' x 1

x 1

1g x dx 1 dx

x 1

+ − = + −

+ − +

= −+

= −

+ ∫ ∫

12. C

( )

( )

4

1 2 3 4 1 2 3 4

12 3 4 4

1 2 3 4

1 2 3 4

a a a a 256 a a a a

a a a aa a a a

4

A.M. G.M.

But A.M. G.M. A.M. G.M.

a a a a

+ + + ≤

+ + +⇒ ≤

⇒ ≤

≥ ⇒ =

∴ = = =

13. B

( ) ( )( )

( )

32 sin 4x0

b b

a a0

2

0

cos6xcos7xcos8xcos9xI 10 dx

1 e

2I 10 cos6xcos7xcos8xcos9x dx f x dx f a b x dx

I 5 cos6xcos7x cos16x cos2x dx

π

π

π

=+

− = + −

= +

∫

∫ ∫ ∫

∫

∵

2

0

5 cos6xcos7xcos2x dx 0

π

= +∫





4

0

I 10 cos6xcos7xcos2x dx

π

= ∫

14. D

Let A

cot x,2

= B C

cot y, cot z2 2

= =

( )

( ) ( ) ( )

( )

( ) ( ) ( )

22 22

2 2 2

2 2 2

3s a b c sx y z

r r

6 x 2y 3z x y z

7

13x 160y 405z 72 xy yz zx 0

3x 12y 4y 9z 18z 2x 0

x y z 9 1 4

y z x 4 9

− + ++ + = =

∴ + + = + +

⇒ + + − + + =

⇒ − + − + − =

∴ + + = + +

15. B

r

671 671

r 1 r 1r

1 1 1

t r 2 r 3

1 1 1 1 1 1 1 1 1......

t r 2 r 3 3 4 4 5 670 671

1 1 668

3 671 2013

2014

= =

= −+ +

∴ = − = − + − + + − + +

= − =

⇒ λ =

∑ ∑

16. C

( ) ( )

( )( )

3

2

2

z 1 2z z 1 0

z 1 z z 1 2z 0

z 1, ,

+ + + =

+ − + + =

⇒ = − ω ω

( )

( ) ( ) ( )

2014 1007

2

f z z z 1

f 1 0 but f 0 f

= + +

− ≠ ω = = ω

2 and ∴ ω ω are two common roots.

17. A

Put x = y = 0 ( )f 0 0⇒ =

diff. w.r.t. x keeping y constant

( ) ( ) ( )x y x y x xf ' x y f ' x e x y e xe e 2y+ ++ = + + + − − +

Replace y by x & x by 0, then

( ) x xf ' x 1 e xe 1 2x= + + − +

x xe xe 2x= + +

( ) ( ) ( ) ( ) ( )11

f x f x f 1 f 0

0 0

f ' x e dx e e e ∴ = = − ∫

18. B

BD = 2r

BC = 2R

∴ 2r + r + R = 2R

R = 3r

r

E

r B C

R

R

A

30o 60

o





R

r3

⇒ =

19. B

On simplifying ( )f x 3=

( ) 3 g x x 3x 1∴ = − +

( )( )g g x 0= when ( )g x , , = α β γ

Where , , α β γ are roots of ( )g x 0=

( )g x = α has 1 solution

( )g x = β has 2 solution

( )g x = γ has 3 solution

20. A

Let any point of xy 4= 2

be R 2t, t

Let circumcentre of CPQ∆ by (h, k) which is midpoint of CR

∴

2t 0 2h & k

2 2t

1t h =k

t

hk=1

xy=1

+= =

=

⇒

⇒

21. A

( ) ( )( )( ) ( ) ( ) ( )( )( )1 3 5 7 2 4 6 8P x x a x a x a x a x a x a x a x a= − − − − + − − − −

( )

( )

( )

( )

( )

( )

( )

( )

1

2

3

4

5

6

7

8

P a ve

P a ve

P a ve

P a ve

P a ve

P a ve

P a ve

P a ve

= +

= −

= −

= +

= +

= −

= −

= +

Hence, ( )P x 0= has all positive real roots.

22. D

Let x + 1 = t in 1st integral

2 2 2

2 2 2t 2 t 2 t 2e

e et 2 t 2 e 22 2 2e e2 2 2

1 11 11

e e edt t ln t . e dt dt ln te dt e

t t t

− − −− − −

+ = + − =

∫ ∫ ∫ ∫

23. C

12 3 14 5G ,

3 3

19 5,

3

+ + =

H(3, 5) O

(6, 7)

A (3, k)

D (6, 4) B C Ey = 4

x

y

R P

C Q

α

β 1 3

x

1 1− 1−

O

y

γ





Equation of AD ( )

194

3y 4 x 65 6

−− = −

−

( )7

y 4 x 63

−⇒ − = −

point A(3, k) satisfies this

( )

7k 4 3

3

k 7 4 11

−− = −

= + =

∴ Circumradius ( ) ( )2 2

6 3 7 1 9 16 5= − + − = + =

∴ Area of circumcircle 25= π

24. B

( ) ( )

2

T

n 1

AA I A 1

adj adj A A 1−

= ⇒ = ±

= = ±

25. C

( ) nlim x

f x e cos 1 nn

→∞

= −

n

2

lim

1 x2sin

2 ne

1

n

→∞−

=

( ) n

x

2

1

n

limx

e

x e 1

g x e e1

n

→∞

−

=

−

= =

( )

( )

( )( )

( )

1

1

1 1

g x lnx

1f x 2ln ; 0 x 1

x

1g f x ln 2ln for 0 x 1

x

Domain of h x is (0, 1)

−

−

− −

=

= < ≤

= < <

∴

26. A

3 2cos xcos x cos x

4+ ≥

cos A, cosB & cosC are nonnegative

2 2 2

1 3 2x x cos A cos B cos C 2cos A cosBcosC 2x+ ≥ + + + =

1 2 3 2

2 2 2

3x x x 3x

2

cos A cos B cos C 2cos A cosBcosC 1

⇒ + + ≥ =

+ + + =∵

27. C

( )

12 8 4

4 4 4 3

12!C C C

4!× × =

28. D





S is not symmetric & T is reflexive, symmetric & transitive.

29. B

Let ˆ ˆ ˆr xi yj zk= + +

where x, y, z are positive integers

x y z 12∴ + + ≤

The no. of values of r is no. of positive integral solutions of 12

n 1 12

2 9n 3

x y z 12 C C−

=

+ + ≤ = =∑

30. D

( )1f 1 & ( )1f 3 are of opposite sign so one root lies in

between 1 & 3. By graph ( )1f x has min. 2 roots

By graph ( )2f x has min. 3 roots so ( )f ' x has 2

roots by Rolle’s theorem.

Let ( ) ( ) ( )'

1 2g x f x . f x= has min. 4 roots

⇒ ( )g' x has min. 3 roots

0 1 2 3 4

( )1f x

( )2f x

0 1 2 3 4

0 2 4

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