Solutions AIATS JEE(Main)-2018 Test-2 (Code-A & B) (27!08!2017)

Test - 2 (Paper-I) (Code-A) (Answers & Hints)

All In India Aakash Test Se Series fo for JE JEE (M (Main)-2018

All India Aakash Test Series for JEE (Main)-2018

TEST - 2 (Code-A) Test Date : 27/08/2017

ANSWERS PHYSICS

CHEMISTRY

MATHEMATICS

1.

(4)

31.

(3)

61.

(2)

2.

(3)

32.

(2)

62.

(4)

3.

(2)

33.

(4)

63.

(2)

4.

(4)

34.

(4)

64.

(4)

5.

(1)

35.

(1)

65.

(1)

6.

(1)

36.

(2)

66.

(1)

7.

(3)

37.

(2)

67.

(4)

8.

(4)

38.

(4)

68.

(1)

9.

(4)

39.

(4)

69.

(2)

10.

(4)

40.

(4)

70.

(4)

11.

(2)

41.

(3)

71.

(4)

12.

(4)

42.

(3)

72.

(1)

13.

(2)

43.

(1)

73.

(2)

14.

(4)

44.

(1)

74.

(4)

15.

(4)

45.

(4)

75.

(4)

16.

(2)

46.

(3)

76.

(2)

17.

(4)

47.

(4)

77.

(1)

18.

(1)

48.

(4)

78.

(4)

19.

(3)

49.

(1)

79.

(1)

20.

(3)

50.

(1)

80.

(1)

21.

(1)

51.

(3)

81.

(3)

22.

(2)

52.

(3)

82.

(1)

23.

(2)

53.

(2)

83.

(1)

24.

(4)

54.

(3)

84.

(1)

25.

(3)

55.

(2)

85.

(3)

26.

(3)

56.

(1)

86.

(3)

27.

(4)

57.

(2)

87.

(4)

28.

(2)

58.

(3)

88.

(1)

29.

(2)

59.

(3)

89.

(4)

30.

(2)

60.

(3)

90.

(4)

1/8


Test - 2 (Code-A) (Answers & Hints)

PART - A (PHYSICS) 1.

Answer (4)

6.

Answer (1)

 dB  2   R B0  dt 

2 As E × 2 l = R 

2

E



R

2l

B0 and

for equilibrium of charge

B

x

2  1  qR B0  mg  x   K  2l 

2.

B



Answer (3)



 



 

As F  q v  B

∫



 

dx

0i 2 x

  

B.dl

2

 a2

0idx cos  90 –  

a

 ∫

a

2 x

3

Force on electron will be along BO direction 3.

2

 a2

 ia 1 –1  x   0 tan   2 a a

Answer (2)

a

a

3

v y



v

30°

B v x

r 

Mv y qB

and p 

qB

v

0i     0i  – 6   24 2  4 

7.

Answer (3)

8.

Answer (4)

9.

Answer (4) As

2M

θ 9 0 –

θ

qE + mg = Kx

E B

 V speed of wave and

V



 K

E = V × B

x

3

  108  4  10–6  600 V/m

Mv y v v sin30 qB   y  2v x 2 v cos 30 p 2Mv x qB r

2

10. Answer (4) A i

r



2 3

4.

Answer (4)

5.

Answer (1) As time constant  

i

(2, 0)

L R

– t     1 – e    6  1 – R   

V



4 1

B

 4 seconds  

F

1

  3.8 A

 i  dl  B    

 2  2 

e

Power P = i 2R = 14.4 W

2/8

y  6

p

 

F

8

 

6 2 – i ˆ N

  

 

6 – i ˆ N



30  10

9

200 




All India Aakash Test Series for JEE (Main)-2018 15. Answer (4)

11. Answer (2) 2

M



L

Q ⇒

2m0



M

LQ



2m0

5

m0 R

2

Emf induced between O and A

Q 

2m0

1 5

R

2

Q

1



E

2

12. Answer (4)

final

π

α

–

α=δ

2

i

so

in loop  

 Ba2  a 2    2R  2

a

1

=π –

Total deviation = 1  2  2   –  

B  dl 

∫

 

2

qB

=

2

2m

2

2

2R

a 4

4R

 0  i  i d   

 0  i  0

d E dt

  d E    0 i  0 0  dt    

18. Answer (1)

 – 

u avg =

2

=

  

 2 –  

2m

B

Ba

17. Answer (4)

β

Correct equation is

T

R



Due to the flux change eddy currents will develop and retard the motion till a constant speed is achieved.

Here  = 

As t =

E

16. Answer (2)

β δ

2

   Bia   B  2

initial

α

Ba

1 2

0 E 2 =

0 2

 10 4

 – 

Volume of cylinder = 10 cm2 × 50 cm = 5 × 10 –4 m3

qB

U = u avg × Volume

13. Answer (2) V L

0

=

 104  5  10–4 

2

50 2

19. Answer (3) i

V R

20. Answer (3) 21. Answer (1)

V C

22. Answer (2)

At resonance V L = V C

 = 0°

Electric field between the plates of capacitor

So power factor cos  = 1

If electron moves on straight line

14. Answer (4)

F e = F m

Emf in square of side ‘a’ e1



d dt

Ba   B a 2

2

0

In small square

e2

  cos t 

Net emf = e1 – e2 = i 

i 

e1 – e2 R





2

B0 a – b

2

b2)

  cos t

4  a  b 

 a – b  B0 cos t 4

eE = evB 

 B0 b2   cos t  B0(a2 –

E 

t 

l v



lB E



 cos t

v

lB

 0



E B



0lB 

23. Answer (2) For BNet = 0

 0 i 2 R

–

 0i    0 4 R

 = 2 rad

3/8

 0


Test - 2 (Code-A) (Answers & Hints) 29. Answer (2)

VL

– V

A

V

c

F 1

F 2 F

φ V R

i

B

As after resonant frequency X L dominates

Net force per unit length on A

25. Answer (3) N

F 

e

n i s

v

v

2R

R

45°

M







v cos

2

F1

60  F22  2F1F2 cos

0 i 2  F 2 where F1  2 a F = 2 F 1 cos 30 =

0 i 2 F  2 a



2R v cos 

F 1 3

3

30. Answer (2)

B 2Rv cos 45

=



v

R

 B

C

1

e = BRv

α α

With N at high potential

– 0 9

26. Answer (3)

2

UC  UL initial  UC  UL final

⇒

1 Total initial energy  Energy in capacitor 2

1 1Q

 2 2 Q

2

   

2

C



Q

1 Q

tan 1

2 C

⇒

r

B

0H



  B  cos  H 

2

tan 2

8 4  10

–7

BV



tan  cos 

In second plane

Q

Q 

27. Answer (4)

 



2

2

2

In first plane

 2  103



BV BH sin 



tan  sin 

cos  = tancot1, sin = tan cot2

 104

cos2 + sin2 = tan2 [cot21 + cot22] cot2 = cot21 + cot22

28. Answer (2)

PART - B (CHEMISTRY) 31. Answer (3)

34. Answer (4)

In electrolytic refining impure metal act as anode. 32. Answer (2)

35. Answer (1)

33. Answer (4) CaO + SiO2 Impurity

 CaSiO3

(Slag)

In blast furnace reduction of iron oxides takes places in different temperature range.

4/8

Zone refining is used for refining of high pure metal that are used in semiconductor. Siderite  FeCO3. 36. Answer (2) 37. Answer (2) AgCl, Cu(OH)2 and Zn(OH)2 are soluble in aqueous NH3 solution.



38. Answer (4) NiSO4 + 4Py + 2NaNO 2  Na2SO4 + [Ni(Py)4](NO2)2 39. Answer (4) Bauxite contains the impurity of SiO2, iron oxide and TiO2.

Compound

E.A.N

Ni(CO)4

36

Mn(CO)5

35

Fe(CO)5

36

49. Answer (1)

40. Answer (4)

50. Answer (1)

Mn can show +7 oxidation state.

Fact.

41. Answer (3) [Cr(NH3)4(CN)Cl]Br is optically inactive.

51. Answer (3)

42. Answer (3) NH3 2– Cr2O7

+ 4H2O2 +

2H+

 2CrO5 + 5H2O

43. Answer (1) 2+

Pb

Co NH3

+ 2OH  Pb(OH)2 –

Pb(OH)2 + 2OH

 [Pb(OH)4]2–

Cis (optically active). 52. Answer (3)

Soluble in excess of reagent. Cd2+ + 2OH

   

[Cr(en)3]Cl3 cannot show geometrical isomerism.

Cd(OH)2

Precipitate is insoluble in excess of reagent Ni2+ + 2OH  Ni(OH)2

Fact. 54. Answer (3)

The precipitate is insoluble in excess reagent. Mg 2+ + 2OH reagent.

53. Answer (2)

 Mg(OH2)  insoluble in excess

Hg2+ does not give precipitate with dil. HCl. 55. Answer (2) Ag2S2O3 + H2O  Ag2 S + H2SO4

44. Answer (1)

(Black) Y

56. Answer (1) 6.35

Moles of Cu =

63.5

In lanthanoids as atomic number increase size decrease.

= 0.1 mol

57. Answer (2)

Formula = [Cu(H2O)4]SO4·H2O Moles of water = 0.4

Rotten egg smell gas is H2S.

Mass of water = 0.4 × 18 = 7.2 g

Na2S + dil. H2SO4  H2S + Na 2SO4 H2S + Pb2+  PbS + 2H+

45. Answer (4)

Blackening of filter paper moistened with lead acetate solution.

46. Answer (3) Spin only magnetic moment() =

n(n  2) B.M.

n  Number of unpaired electron. Cu2+, n = 1,  = 1.73 B.M. Cu , n = 0,  = 0 +

Cr , n = 5,  = 5.9 B.M. +

Zn , n = 0,  = 0 2+

47. Answer (4) Name of positively charged ligands end with ium.

58. Answer (3) NO3

+ 3Fe2+ + 4H+  NO + 3Fe3+ + 2H2O

2+ [Fe(H2O)6]2+ + NO  [Fe(H2 O)5NO] + H2O Brown

59. Answer (3) HCl give white fumes with NH3. 60. Answer (3) [Ag(NH3)2][Ag(Cl)2]  Diamminesilver(I) dichloridoargentate(I)

5/8



PART - C (MATHEMATICS) 61. Answer (2) As

lim

0

70. Answer (4)

the function approaches to

x

3 e

 0 hence

e  R ~ {0}.

62. Answer (4) As lim f ( x )  –1, lim f ( x )  1 and f (1) = 0 1–

1

x

x

 Range = {–1, 0, 1}

 x 2 – x  5  2  – x  x  5 f ( x ) =   x 2 – x  5   9 – x

; 0

x

1

; 1

x

;

2

x

2

72. Answer (1)

 1  1   – x   –  1  x  –

y = max f

n

2

y = x

 1   –1  1   1 h0  – – h   – –  2  2 lim

  h  

) 1 , 0 (

–π

π

–3π 2

  2   2   –  1  2h   1  2h   h0    

=

0

f ( x ) is continuous everywhere but non-differentiable at x = 0, 1 and 2.

lim

=

x

71. Answer (4)

63. Answer (2)

x

;

lim

–π

π

2

2

2π

x

Always continuous.

  2  = lim  –   h0   1  2h  

73. Answer (2) y = f(x )

= –2 64. Answer (4)

 2e  e  e–  x

Simplifying lim loge  

x

x

x

   log 2  Q 

x

c

e

65. Answer (1) 66. Answer (1) 1

d 2 y

 2   As lim 3    1  3  h ⇒   3   3  

For

The required limit value is 1.

 d 2y If  2  dx

x

x

dx 2

=0

at x = a

x

67. Answer (4)

  d 2 y  2   dx x

2

lim

h .sin h

h 0

h

3

1

68. Answer (1) lim f  x 

1

 4 (from less than 4) hence G.I.F. is 3

x

69. Answer (2) The function |sin x | is not differentiable at x = n for all n being integer.

6/8

 a–

x

 0   a

Then x = a is a point of inflection. 74. Answer (4) As f ( x ) will intersect at least once in each x  (a, b) and (b, c ) and (c , d ) hence three zeros atleast. 75. Answer (4) For f ( x ) = 3ax 2 + 2bx + c  the extremum will exist if 4 b2 – 4 (3ac ) > 0 but not in all cases.



76. Answer (2) For f ( x ) = x x (1 + loge x )

 1  f     0, as well as e

g ( x ) will not be differentiable at x

lim

x

0

1

 0,

–2 5

,

–2 3e  5

82. Answer (1)

x

A

y = f(x ) (0, 1)

(1, 1)

13

5

x

O

x



1 x = e

B

1  ⇒  ,   is longest bijective interval with e 

 1 b    e

1 e

f ( x ) = ⇒

2 ⇒ a   b       0 e 

C

12

–1

sin

tan 

e



 12 sin x  5 cos x  –1    sin  sin  x   13   5 ⇒

0

12



 4

 As x varies in real the slope of f ( x ) will be –1 or 1 only.

77. Answer (1)

83. Answer (1)

78. Answer (4)

y = f(x )

y = f(x )

f (x ) > 0

y = 1

y = x y = 0 O

x –1

y = f (x ) Always continuous

x

79. Answer (1) Let x  0

 g ( x )  g (0)

 f ( x ) > 0

 f (g ( x ))  f (g (0))

 concave upwards

 h( x )  0



But h( x )  0

R

 f (1) = 9 and f (1) = 5

1 1

x

 f ( x ) is bijective function

80. Answer (1)



dx

(f –1(d ))  0 

As f ( x ) = 6 x 2 + 3

 h( x ) is a zero function.

0

2

84. Answer (1)

 h( x ) = 0

As

2

d

2

  , x  0  2 1  x  1  x 2 1

 1 hence 

x

Therefore, f ( x ) =

x – 1 

and f ( x ) = 1 at x = 0

1

1

x  R ~ 0 2

1  x

Therefore, f –1(5) = 1 and

 (y – 1) =

1 9

 f –1  5  

1 f  1

 x – 5 

 9y – x = 4

7/8



85. Answer (3)

e – ; y =   e ; x

x

88. Answer (1) x

0

x

0

y (b, f(b))

y = e|x | is symmetric about the y -axis.

} v , u {

y

 2a  3b  2a  3 b , f    5    5

(a, f(a)) 

(, e )

x

  =

– tan

v 

2a

e

x

 3b

2f  a   3f  b  5

As evident

 Angle between tangents is  – 2tan–1e. 86. Answer (3)

v

As f ( x ) = f –1(0) = Roots of f ( x ) ⇒

b

5

2

Hence f ( x ) = x 1, x 2, x 3

 3b

Here

 = tan–1e –1

2a

5

u 



a

O

 (0, 0)

 2a  3b   5  

 f 

89. Answer (4)

9 intersection points

As f {f ( x )} < f (3 – 4 x ) is required y = f(x ) (–2,19) (–1, 14) (0,3) –3 –2 –1 1 2 x 0 x

–4

1

and f ( x ) = –2 – 6 x 2

(4,19)

 f ( x )   f ( x ) > 3 – 4 x 3 – 2 x – 2 x 3 > 3 – 4 x

3

4 x

 2 x – 2 x 3 > 0 x  ( x – 1) ( x + 1) < 0

x

3

2

(1, –8)

(3,–6)

(–4,–13)

(2, –13)

 87. Answer (4) b



2a

⇒

a –1

f  a 



ab



2

a –1

4a  a – 1 –  2a

 a – 12

0

–

1

90. Answer (4)

 f  a  , now 2





As g {f ( x )} = x 1

 g {f ( x )} =    f x

2

2a – 4 a

 a – 12

Now f (1) = 2 hence g 2 

 f (a) = 0  a = 2  f (2) is minimum 

8/8

– –1

2a

+

+





1

f  1



1 4

Test - 2 (Paper-I) (Code-B) (Answers & Hints)



TEST - 2 (Paper-I) - Code-B Test Date : 27/08/2017

ANSWERS PHYSICS

CHEMISTRY

MATHEMATICS

1.

(2)

31.

(3)

61.

(4)

2.

(2)

32.

(3)

62.

(4)

3.

(2)

33.

(3)

63.

(1)

4.

(4)

34.

(2)

64.

(4)

5.

(3)

35.

(1)

65.

(3)

6.

(3)

36.

(2)

66.

(3)

7.

(4)

37.

(3)

67.

(1)

8.

(2)

38.

(2)

68.

(1)

9.

(2)

39.

(3)

69.

(1)

10.

(1)

40.

(3)

70.

(3)

11.

(3)

41.

(1)

71.

(1)

12.

(3)

42.

(1)

72.

(1)

13.

(1)

43.

(4)

73.

(4)

14.

(4)

44.

(4)

74.

(1)

15.

(2)

45.

(3)

75.

(2)

16.

(4)

46.

(4)

76.

(4)

17.

(4)

47.

(1)

77.

(4)

18.

(2)

48.

(1)

78.

(2)

19.

(4)

49.

(3)

79.

(1)

20.

(2)

50.

(3)

80.

(4)

21.

(4)

51.

(4)

81.

(4)

22.

(4)

52.

(4)

82.

(2)

23.

(4)

53.

(4)

83.

(1)

24.

(3)

54.

(2)

84.

(4)

25.

(1)

55.

(2)

85.

(1)

26.

(1)

56.

(1)

86.

(1)

27.

(4)

57.

(4)

87.

(4)

28.

(2)

58.

(4)

88.

(2)

29.

(3)

59.

(2)

89.

(4)

30.

(4)

60.

(3)

90.

(2)

1/8



PART - A (PHYSICS) 1.

1 1 Q

Answer (2)

 2 2

1 α α

– 0 9

2 Q 

C Q

1 Q

2

2 C

2

⇒

2

2

6.

   

2

Q

Q 

2

Answer (3) N 

In first plane



  cos B   H 

tan 1

tan 

BV



tan 2



BV BH sin 

tan 



e

sin 

2.



2R v cos 

B 2Rv cos 45

Answer (4)

A

VL

F 1

– V

V

c

F 2

φ

F

V R

B

C

2



F1

Answer (2) For BNet = 0

60  F22  2F1F2 cos

 0i 2 R

0 i 2   F 2 F where 1 2 a F = 2 F 1 cos 30 =

i

As after resonant frequency X L dominates 8.

Net force per unit length on A F

v

With N at high potential 7.

Answer (2)



v cos

e = BRv

cos2 + sin2 = tan2 [cot21 + cot22] cot2 = cot21 + cot22





R

 B =

cos  = tancot1, sin = tan cot2

R

45°

M

In second plane

v

v

2R

cos 

n i s

–

 0 i    0 4 R

 = 2 rad

F 1 3

9.

Answer (2)

2

F 

0 i 2 a

Electric field between the plates of capacitor E 

3

3.

Answer (2)

If electron moves on straight line

4.

Answer (4)

F e = F m

  r

5.

B

0H



8 4  10

–7

 2  10

3

 104

Answer (3) UC  UL initial  UC  UL final ⇒

1 2

2/8

Total initial energy  Energy in capacitor

eE = evB  t 

l v



lB E



10. Answer (1)

v

lB

 0

 

E B

0lB 

 0



11. Answer (3)

At resonance V L = V C

12. Answer (3)

 = 0°

13. Answer (1)

So power factor cos  = 1

1

u avg =

2

0

0E 2 =

2

19. Answer (4)

 10 4

Volume of cylinder = 10 cm2 × 50 cm = 5 × 10 –4 m3

final

U = u avg × Volume

=

0

 104  5  10 –4 

2

50

∫ B  dl   i  i 

2

=π –

d E dt

  d E    0 i  0 0  dt    

As t =

T

 2 –  

2

2m

15. Answer (2) Due to the flux change eddy currents will develop and retard the motion till a constant speed is achieved.

qB

=

2

 –  =

2

Emf induced between O and A E



2

Ba

   Bia 

2m

 –  qB

20. Answer (2)

16. Answer (4)

1

β

Total deviation = 1  2  2   –  

d

 

1

Here  = 

  

 0  i  0

α=δ

2

δ

0

–

β

Correct equation is 

π

α

14. Answer (4)

 

initial

α

2

i

so

in loop  

 Ba 2  a 2  B    R 2 2  

a 2

E R B

2

 2

Ba

2

M



L

2R

Q ⇒

2m0

M



LQ 2m0



5

m0R

2

Q 

2m0

1 5

R

2

Q

21. Answer (4)

a 4

A

4R

i

17. Answer (4)

y  6

Emf in square of side ‘a’ e1



d dt



Ba

2

  B0a2  cos t  e2

In small square

(2, 0)

 B0 b2   cos t 

Net emf = e1 – e2 = B0(a2 – b2)  cos t i 

i 

e1 – e2 R





2

B0 a – b

2

  cos t

B  

F

 2  2 

4  a  b 

 a – b  B0 cos t

 i  dl  B    

 

F

4

8

 

6 2 – i ˆ N

  

 

6 – i ˆ N

22. Answer (4)

18. Answer (2) V L

As

E B

 V speed of wave and

V



 K



30  10

9



200 

E = V × B i V C

V R

3

  108  4  10–6  600 V/m 2

3/8


Test - 2 (Paper-I) (Code-B) (Answers & Hints) 28. Answer (2) v y


v

30°

B v x

B

θ 9 0 –

θ

x

B



2

a

∫ B.dl  ∫  

  

qB

dx

and p 

0i 2 x

Mv y

r 

 a2

2 x

3

2

a

 ia 1 –1  x   0 tan   2 a a

r p



2

r 

a

x

v v sin30 qB  y  2Mv x 2 v x 2 v cos 30 qB p 2 3

29. Answer (3)  

As F

0i     0i  – 6   24 2  4 

 q v  B  

 

Force on electron will be along BO direction

26. Answer (1)

30. Answer (4)

As time constant  

i

v

a

3



qB

Mv y

0idx cos 90 –  

a

2M

L R



4 1

 4 seconds

 dB  2   R B0 dt  

2 As E × 2 l = R 

2

– t   1   1 – e    6  1 –   3.8 A R e   

E

V



R

2l

B0 and

for equilibrium of charge

qE + mg = Kx

2

Power P = i R = 14.4 W

 qR 2B0   mg  x   K  2l  1

27. Answer (4)

PART - B (CHEMISTRY) 31. Answer (3)

34. Answer (2)

[Ag(NH3)2][Ag(Cl)2]  Diamminesilver(I) dichloridoargentate(I) HCl give white fumes with NH3. 33. Answer (3) + 3Fe2+ + 4H+  NO + 3Fe3+ + 2H2O

2+ [Fe(H2O)6]2+ + NO  [Fe(H2O)5NO] + H2O Brown

4/8

Na2S + dil. H2SO4  H2S + Na 2SO4 H2S + Pb2+  PbS + 2H+

32. Answer (3)

NO3

Rotten egg smell gas is H2S.

Blackening of filter paper moistened with lead acetate solution. 35. Answer (1) In lanthanoids as atomic number increase size decrease.

Test - 2 (Paper-I) (Code-B) (Answers & Hints) 36. Answer (2)


Ag2S2O3 + H2O  Ag2 S + H2SO4 (Black) Y

Pb2+ + 2OH–  Pb(OH)2 Pb(OH)2 + 2OH

37. Answer (3) Hg2+ does not give precipitate with dil. HCl.

Soluble in excess of reagent. Cd2+ + 2OH

38. Answer (2) Fact.

 [Pb(OH)4]2–

   

Cd(OH)2

Precipitate is insoluble in excess of reagent

39. Answer (3) [Cr(en)3]Cl3 cannot show geometrical isomerism. 40. Answer (3)

Ni2+ + 2OH  Ni(OH)2 The precipitate is insoluble in excess reagent. Mg 2+ + 2OH reagent.

NH3

 Mg(OH 2)  insoluble in excess

49. Answer (3) Co 2–

Cr2O7

NH3

+ 4H2O2 + 2H+  2CrO5 + 5H2O

50. Answer (3) Cis (optically active).

[Cr(NH3)4(CN)Cl]Br is optically inactive.

41. Answer (1)

51. Answer (4)

Fact.

Mn can show +7 oxidation state.

42. Answer (1)

52. Answer (4)

43. Answer (4) Compound

E.A.N

Ni(CO)4

36

Mn(CO)5

35

Fe(CO)5

36

Bauxite contains the impurity of SiO2, iron oxide and TiO2. 53. Answer (4) NiSO4 + 4Py + 2NaNO 2  Na2SO4 + [Ni(Py)4](NO2)2 54. Answer (2)

44. Answer (4) Name of positively charged ligands end with ium. 45. Answer (3)

55. Answer (2)

Spin only magnetic moment() = n  Number of unpaired electron. Cu2+, n = 1,  = 1.73 B.M. Cu+,

n = 0,  = 0

Cr +, n = 5,  = 5.9 B.M. Zn2+, n = 0,  = 0

n(n  2) B.M.

56. Answer (1) Siderite  FeCO3. 57. Answer (4) Zone refining is used for refining of high pure metal that are used in semiconductor. 58. Answer (4)

46. Answer (4)

CaO + SiO2

47. Answer (1)

Impurity

6.35

Moles of Cu =

AgCl, Cu(OH)2 and Zn(OH)2 are soluble in aqueous NH3 solution.

63.5

= 0.1 mol

(Slag)

In blast furnace reduction of iron oxides takes places in different temperature range.

Formula = [Cu(H2O)4]SO4·H2O

59. Answer (2)

Moles of water = 0.4

60. Answer (3)

Mass of water = 0.4 × 18 = 7.2 g

 CaSiO3

In electrolytic refining impure metal act as anode.

5/8



PART - C (MATHEMATICS) f  a 

61. Answer (4) As g {f ( x )} = x



4a  a – 1 –  2a

2





 a – 12

2

2a – 4 a

 a – 12

 f (a) = 0  a = 2  f (2) is minimum

1

 g {f ( x )} =    f x

65. Answer (3)

Now f (1) = 2 hence g  2 

1

f  1



1 4

As f ( x ) = f –1(0) = Roots of f ( x ) Hence f ( x ) = x 1, x 2, x 3

9 intersection points

⇒

y = f(x )

62. Answer (4) As f {f ( x )} < f (3 – 4 x ) is required

(4,19)

(–2,19) (–1, 14)

and f ( x ) = –2 – 6 x 2

 f ( x )   f ( x ) > 3 – 4 x –4

3

3 – 2 x – 2 x > 3 – 4 x

(0,3) –3 –2 –1 1 2 x 0 x 1

3

2

(1, –8)

 2 x – 2 x 3 > 0 x  ( x – 1) ( x + 1) < 0

(–4,–13)

4

x

x

3

(3,–6) (2, –13)

66. Answer (3) +

+



– –1

–

0

1

63. Answer (1)

e – ; y =   e ; x

x

x

0

x

0

y = e|x | is symmetric about the y -axis.

y

y (b, f(b)) 

(, e ) } v , u {

(a, f(a))

a

O

2a

 3b

b

 = tan–1e    = – tan–1 e

Here

v 

2a  3b

2

 Angle between tangents is  – 2tan–1e.

5 2f  a   3f  b 

67. Answer (1)

5

As f ( x ) = 6 x 2 + 3

As evident v

 f ( x ) is bijective function

 2a  3b    5 

 f (1) = 9 and f (1) = 5

 f 

Therefore, f –1(5) = 1 and

64. Answer (4) b



6/8

2a a –1

x

x

5

u 

 (0, 0)

 2a  3b  2a  3 b  , f    5    5

⇒

ab



2a

2

a –1

 f  a  , now

 (y – 1) =

1 9

 9y – x = 4

 x – 5 

 f –1 5  

1 f  1



68. Answer (1)

72. Answer (1) Let x  0

y = f(x )

 g ( x )  g (0)

f (x ) > 0 y = x

 f (g ( x ))  f (g (0))  h( x )  0 But h( x )  0

–1

y = f (x )

 h( x ) = 0 x

 h( x ) is a zero function. 73. Answer (4) y = f(x ) y = 1

 f ( x ) > 0  concave upwards 2

d



2

dx

(f –1(d ))  0 

y = 0 x

R

O

x

Always continuous

69. Answer (1)

74. Answer (1)

A

75. Answer (2) For f ( x ) = x x (1 + loge x )

13

5

 1 e

 f     0, as well as

 B

⇒

–1

sin

tan 

x

x

1

y = f(x )

 12 sin x  5 cos x  –1    sin  sin x    13   5



0

C

12

f ( x ) =

lim x

⇒

12

0



(0, 1)

(1, 1)

 4

O

 As x varies in real the slope of f ( x ) will be –1 or 1

x 1 x = e

only.

1  ⇒  ,   is longest bijective interval with e 

70. Answer (3) g ( x ) will not be differentiable at

x

 0,

–2 5

,

As

0



1

2

x

3e  5

1  1  , x  0  1 hence   2 1  x  1  x 2

Therefore, f ( x ) =

x – 1 

and f ( x ) = 1 at x = 0

1 2

1  x

1 e

–2

71. Answer (1) 1

 1 b    e

x  R ~ 0

2 ⇒ a   b       0 e  e

76. Answer (4) For f ( x ) = 3ax 2 + 2bx + c  the extremum will exist if 4 b2 – 4 (3ac ) > 0 but not in all cases. 77. Answer (4) As f ( x ) will intersect at least once in each x  (a, b) and (b, c ) and (c , d ) hence three zeros atleast.

7/8



78. Answer (2)

83. Answer (1)

y = f( x )

lim f  x 

1

 4 (from less than 4) hence G.I.F. is 3

x

84. Answer (4) 2

x

h .sin h

lim h 0

h

1

3

85. Answer (1) 1

 2   As lim 3    1  3  h ⇒   3   3   x

For

d 2 y 2

=0

dx

x

at x = a

The required limit value is 1.

 d2y If  2  dx

  d 2 y  2   dx x

 a–

x

86. Answer (1)

 0 

87. Answer (4)

 a



Simplifying lim loge  

2e x

e e



x

Then x = a is a point of inflection.

x

– x

   log 2  Q 

c

e

88. Answer (2)

79. Answer (1) y = max f

n

lim

y = x

x

) 1 0, (

–π

π

–3π

–π

π

2

2

2

=

2π

x

 x 2 – x  5  2  – x  x  5 f ( x ) =   x 2 – x  5   9 – x

;

x

; 0

0 x x

;

2

x

2

 1   –1  1   1 h 0  – – h   – –  2  2 lim

  h  

=

  2   –  1  2h   h 0   

lim

lim

= –2

1

; 1

1

 1  1   x  – x    

  2   2  –  1  2h   1  2h   h 0    


–

–

= Always continuous.

89. Answer (4)

2

As lim f ( x )  –1, lim f ( x )  1 and f (1) = 0 1–

x

90. Answer (2)

82. Answer (2)

As

The function |sin x | is not differentiable at x = n for all n being integer.



1

x

 Range = {–1, 0, 1}

f ( x ) is continuous everywhere but non-differentiable at x = 0, 1 and 2.

8/8

x

lim

0

the function approaches to

x

e  R ~ {0}.





3 e

 0 hence

Solutions AIATS JEE(Main)-2018 Test-2 (Code-A & B) (27!08!2017)

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