test - 7 (code-a) (answers) all india aakash test series for jee (main)-2021 test - 7 ... · 2020....

Test - 7 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2021

All India Aakash Test Series for JEE (Main)-2021

Test Date : 20/12/2020

ANSWERS

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

TEST - 7 - Code-A

1/13

PHYSICS CHEMISTRY MATHEMATICS

1. (2)

2. (1)

3. (3)

4. (4)

5. (4)

6. (3)

7. (2)

8. (1)

9. (3)

10. (4)

11. (3)

12. (1)

13. (2)

14. (1)

15. (1)

16. (3)

17. (2)

18. (4)

19. (2)

20. (1)

21. (03.00)

22. (40.00)

23. (05.00)

24. (16.00)

25. (12.00)

26. (2)

27. (4)

28. (3)

29. (4)

30. (1)

31. (3)

32. (1)

33. (3)

34. (2)

35. (3)

36. (3)

37. (1)

38. (4)

39. (2)

40. (3)

41. (4)

42. (1)

43. (4)

44. (4)

45. (2)

46. (12.00)

47. (03.20)

48. (63.00)

49. (01.20)

50. (18.00)

51. (2)

52. (2)

53. (1)

54. (3)

55. (3)

56. (3)

57. (2)

58. (1)

59. (4)

60. (3)

61. (1)

62. (2)

63. (4)

64. (1)

65. (4)

66. (2)

67. (1)

68. (2)

69. (3)

70. (2)

71. (81.00)

72. (04.00)

73. (01.00)

74. (00.00)

75. (03.00)

All India Aakash Test Series for JEE (Main)-2021 Test - 7 (Code-A) (Hints & Solutions)


2/13

1. Answer (2)

Hint : Use law of homogeneity of dimensions

Sol.: T PabEc

[M0L0T1] [ML–1T–2]a [ML–3]b[ML2T–2]c

a + b + c = 0 ...(i)

–a – 3b + 2c = 0 ...(ii)

–2a – 2c = 1 ...(iii)

On solving equations (i), (ii) and (iii)

5

6a

1

2b

1

3c

a + b + c = 0

2. Answer (1)

Hint : Fm = ilB

Sol.: When wire AB in equilibrium,

FB = Mg, 0 1 2

2

I IMg

h

...(i)

Now wire AB is depressed by x

Restoring force F = Mg – FB

0 1 22 ( )

I IF Mg

h x

= 0 1 22 (1 / )

I IMg

h x h

1

1x

F Mg Mgh

1 1x

Ma Mgh

g

a xh

Comparing with general equation

a = – 2x

2

2h

Tg

3. Answer (3)

Hint : p

v

C

C

Sol.: p

v

C

C

TV – 1 = constant

T( – 1)V – 2dV + V – 1 dT = 0

( 1)

dV dT

V T

1

( )

dV b CTdT

V a b T

logV = 1

log( )

b T CTa b

(a – b) logV + blog T + CT = constant

Va – b Tb eCT = constant

4. Answer (4)

Hint:

Fnet = 2Kx + 2Kx cos60°

Sol.:

PART - A (PHYSICS)

Test - 7 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021


3/13

x = x cos60°

–ma = 2Kx + 2K x cos60°

–ma = 2Kx + 2Kx cos260°

=

21

2 22

Kx Kx

–ma = 2

24

KxKx

–ma = 22

KxKx

–ma = 5

2

Kx

5

2

Ka x

m

2

25

mT

K

5. Answer (4)

Hint : For insufficient length, h = l

Sol.: h = l for l less than hmax

6. Answer (3)

Hint : For frequency = No. of waves

time

Sol.: Frequency = Number of waves

time of fall of plate

21

2s ut gt

22 2 4.2 10

9.8

st

g

10

108 Hzft

7. Answer (2)

Hint : sin sinA A B B

Sol.: Applying Snell's law at points A and B

Asin lA = BsinlB

0

0 sin90 sin

1

ix

a

sin 1x

ia

2

tan 1 1 1x x

ia a

tandy

idx

2

1

1 1

x

ady dx

x

a

, solving integration

x2 + y2 – 2ax = 0

8. Answer (1)

Hint : Use concept of solid angle

Sol.: Electric flux 0

(1 cos )2

q

due to one

charge

total 2 2

0

1q d

a d

9. Answer (3)

Hint : Use Kirchhoff’s laws

Sol.: Using Kirchhoff’s loop law

V = 0 inside a loop

For upper loop nE – nir – IR = 0 ...(i)

For lower loop (N – n) E + (N – n)(I– i)r + IR = 0 ...(ii)

For entire loop NE – Nir = 0 ....(iii)

I = 0 (current in resistance R)

10. Answer (4)

Hint : Use Wheatstone balanced bridge.



4/13

Sol.: Using principle of meter bridge

111 99.8

1 100 99.8

x

y

22.2 + 111y = 99.8 + x ...(i)

again, 1 0.3

111 100 0.3

x

y

99.7 + y = 33.3 + 111x ...(ii)

x = 0.59 cm y = 0.70 cm

11. Answer (3)

Hint : F = ilB and iR

Sol.: Current in circuit = R

Bvl

iR

Heat Q = i F = Bil

F

Q BvlBl

F

iBl

Q

Fv

12. Answer (1)

Hint: Using momentum transfer and energy

conservation.

Sol.: Using momentum transfer and energy

conservation.

U U

mvc c

2U

vmc

...(i)

Applying energy-conservation

21

(1 cos )2

mv mg

2

21 sin22 2

mv mgl

sin2

U

mc gl

( 0)

2U

mc gl

13. Answer (2)

Hint : Use formulae of collision

Sol.: Velocity of centre of mass after collision

= 1 1 2 2

1 2( )

M v M v

M M

( 2 ) 2 3

22 2

e e gh ev gh

where 1 2v e gh 2 2 2v e gh

2 2 2v u as

23

0 2 22

egh gh

294

hh e

14. Answer (1)

Hint : Radius of curvature 2v

a

Sol.: ˆ ˆx yv v i v j

ˆ ˆ10 (0 10 3)v i j

| | 1000v

Radius of curvature of path 2v

ra

2 2

10cos

10 30

1000

cosr

g

5/210r m

15. Answer (1)

Hint : Wfriction = F × S

Sol.: cos .frw Mg s



5/13

. cos

frwMgs

...(i)

21

.2

frw mv Mg h

21

sin2 cos

frwmv ss

tan

fr

Ew

16. Answer (3)

Hint : Use F = ma

Sol.: T + 16 = 80

T = 64 N

17. Answer (2)

Hint : More B.E./A means more stability

Sol.: More B.E./A means more stability

18. Answer (4)

Hint : 0A

cd

Sol.: 0A

cd

On increasing temperature

0 0 1

2

(1 2 )

(1 )

A A Tc

d d T

0 1 2(1 2 ) (1 )A T T

cd

20 1 2 1 2(1 (2 ) 2 )A

T Td

c = constant with temperature

21 – 2 = 0

21 = 2

19. Answer (2)

Hint : ( 1)A

Sol.:

= ( – 1)A – ( – 1) A

= ( – )A

= 0.3°

20. Answer (1)

Hint : Diode is unidirectional

Sol.: Given circuit works as half wave rectifier.

21. Answer (03.00)

Hint : 2

0 2 21 2

1 1

E E Zn n

Sol.: a – b = 593 Å

22 2

1 1 1

2 3

BRZ

22 2

1 1 1

1 2

LRZ

After putting values in equation Z = 3

22. Answer (40.00)

Hint : Diode is unidirectional

Sol.: Simplifying the circuit

Voltage across capacitor 2C = 2

Q

C

= 2

eqC V

C

= 2 120

3 2

C

C

= 40 V



6/13

23. Answer (05.00)

Hint : 0

and2

dvE E

r dr

Sol.: dv E dr

02

2

x

x

v drr

0

2ln

2

x

x

9 69 10 40 10 0.6931 2

= 5 × 105 V

24. Answer (16.00)

Hint : s vdt

Sol.: Speed of point P at time t

2 2 20 2 cos(180 )v v v v

2 sin2

v

distance covered in one revolution

s = 00

T

v dt

where t

v

tR

s = 8R = 16 m

25. Answer (12.00)

Hint : Bandwidth fmax – fmin

Sol. : Signal will have frequency range from

to c m c mf f f f

26. Answer (2)

Hint: Thermal stability of carbonates of group-2

elements increases down the group.

Sol.: Solubility order : BeSO4 > MgSO4 > CaSO4

> BaSO4

Thermal decomposition temperature order :

BeCO3 < MgCO3 < CaCO3 < BaCO3

Density order : Li < K < Na < Rb < Cs

Electronegativity order ; B > Tl > In > Ga > Al

27. Answer (4)

Hint: Chlorine reacts with excess of NH3 to give

N2.

Sol.:

3 3 2 2 2Cu HNO (conc.) Cu(NO ) NO H O

2 2 2 2 2CaC H O C H Ca(OH)

2 2Br KOH(cold & dil) KBr KBrO H O

3 2 4 2NH (excess) Cl NH Cl N

28. Answer (3)

Hint: Coagulation value of the ion causing

coagulation is inversely proportional to its

coagulation power which is directly proportional

to its valency.

Sol.: Fe2O3. 2H2O is positive colloid and will be

coagulated by the anion of an electrolyte. Since

the coagulation value is inversely proportional to

coagulation power of the ion causing

coagulation. The Cl– ion of NaCl will have the

maximum coagulation value

29. Answer (4)

Hint: For a first order reaction, the time for x%

conversion is given by, x n100

t t100 x

Sol.: t90%; t99%; t99.9% = 1 : 2 : 3

90%

99.9%

t 1

t 3

99.9%90%

t 10t 3.33 years

3 3

PART - B (CHEMISTRY)



7/13

30. Answer (1)

Hint: Both the monomers of PHBV have a chiral

cente each.

Sol.: PHBV monomers

3 2

OH|

CH — CH — CH — COOH;*

3 2 2

OH|

CH — CH — CH — CH — COOH*

Bakelite monomers are HCHO and phenol

Nylon 6,6 monomers are Adipic acid

(CH2)4(COOH)2 and Hexamethylene diamine

(CH2)6(NH2)2

Natural rubber monomer is Isoprene

3

2 2

CH|

CH C — CH CH

31. Answer (3)

Hint: Hoff mann bromamide degradation

reaction

Sol.: 6 5 2

O| |

C H — C — NH

2Br

KOH C6H5 — NH2

(x-Aromatic-1°-amine)

I. C6H5NH2 3CHCl

KOH C6H5NC (pungent smell)

II. C6H5 — NH2 (Aromatic amine) not prepared

by Gabriel phthalimide synthesis.

III. 6 5 2C H SO Cl

6 5 21 amine

C H — NH PPT

IV. 20 5 C

NaNO

6 5 2 6 5 2HClC H — NH C H N Cl

32. Answer (1)

Hint: 2 moles of H2O2 on complete dissociation

give 1 mole of O2.

Sol.: 2 2 2 22H O 2H O O (g)

2 × 34 g 2 × 18 g 22.4 L at STP

? 5.6 L

17 g H2O2 decomposed

50 g of 68% w/w H2O2 contain 34 g H2O2 and

16 g H2O

After decomposition, remaining H2O2 = 34 – 17

= 17 g

On decomposition of 17 g H2O2, water formed is

9 g.

Final water = 16 + 9 = 25 g

Molality = 17 1000

20 m34 25

33. Answer (3)

Hint: A buffer should either have weak acid or

weak base along with its salt with SB or SA

Sol.:

(1) NH4Cl + NaOH NaCl + NH4OH

mmol 20 10 0 0

10 0 10 10

Mixture of NH4OH and NH4Cl will function as

basic buffer.

(2) CH3COOH + NH4OH CH3COONH4 + H2O

mmol 1 1 0 0

0 0 1 1

CH3COONH4 is a salt of weak acid and weak

base it will function as a salt buffer

(3) NaHCO3 + HCl H2CO3 + NaCl

Initial 1 1 — —

Remains 0 0 1 1

WA H2CO3 and salt NaCl last (Not a buffer)

(4) Na3PO4 + NaH2PO4 gives 2

4HPO and

remaining 3

4PO

which acts as buffer

34. Answer (2)

Hint: Solid PCl5 exists as 4 6PCl PCl

Sol.: SCl4 sp3d AB4E see saw

Solid PCl5 exists as ionic compound form 4PCl

and 6PCl

with sp3 and sp3d2 hybridisation.

XeOF4 sp3d2 AB5E square pyramidal.

4ICI

sp3d2 AB4E2 square planar.

35. Answer (3)

Hint: H2O2 has extensive H-bonding and

reduces acidified KMnO4 to Mn2+ ion.



8/13

Sol.: H — H and D — D bond lengths are equal.

In ortho hydrogen, nuclear spin is in same

direction ½ + ½ = 1

Extent of hydrogen bonding in H2O2 is more than

that in H2O

H2O2 reduces KMnO4 solution in acidic medium.

36. Answer (3)

Hint: Organic compounds having N-atom not as

diazonium salt or part of a cyclic ring can be

detected by Lassaigne’s test.

Sol.: In Lassaigne’s extract nitrogens detected

in the form of NaCN.

C6H5N2Cl decomposes to give N2 and does not

form NaCN. C6H5NH2 gives NaCN in

Lassaigne’s solution.

37. Answer (1)

Hint: Magnitude of heat of hydrogenation is

inversely proportional to the stability of alkene.

Sol.: More the -H atoms, more the

hyperconjugate structures lesser the heat of

hydrogenation.

38. Answer (4)

Hint: Method of preparation of acrylic acid from

ethyne

Sol.: CaC2 2H O

C2H2 1eq HCl

CH2 == CHCl Mg

Dry ether CH2 == CHMgCl

2

2

(i) CO|(ii) H O / H

(or) CH2 == CH — COOH

39. Answer (2)

Hint: Various methods used for concentration of

oxide and sulphide ores

Sol.: Conceptual

40. Answer (3)

Hint: Glucose is an aldose which is oxidised by

Br2 water whereas fructose is a ketose which is

not oxidised by Br, water

Sol.: Both glucose and fructose reacts with

Phenyl hydrazine, Fehling's, solution and

Tollen’s reagent. Bromine water is weak

oxidising agent. It oxidises glucose of gluconic

acid and does not affect fructose.

41. Answer (4)

Hint: Benzaldehyde cannot be synthesised by

Reimer Tiemann reaction

Sol.: C6H5CHO 32

CH MgBr

H O

Cu

6 5 32 alcohol

OH|

C H — CH — CH

6 5 3

O||

C H — C — CH

C6H5CHO cannot be prepared by Reimer-

Tiemann reaction.

42. Answer (1)

Hint: Elevation in boiling point of an ideal

solution depends on molality and van't Hoff

factor (i)

Sol.: b bT k m i

1

2

b 1 1

b 2 2

T m i

T m i

111.6 10 i 11.7 10 2

158 58.5

1i 2

i for Mg(OH)2 = 1 + (n – 1)

2 = 1 + (3 – 1)

2 = 1 = 0.5

43. Answer (4)

Hint: EMF of an electrochemical cell is given by

the difference between reduction potentials of

cathode and anode half cells

Sol.: o

cellE = O.P of anode – O.P. of cathode

= 0.4 – 0.1 = 0.3 V

At Anode : (A A+3 + 3e–) × 2

At cathode : (B+2 + 2e– B) × 3

Cell reaction : 2A + 3B+2 2A+3 + 3B



9/13

3 2

o

2 3

0.059 [A ]E E log

n [B ]

2

3

0.059 (0.1)0.3 log

6 (0.01)

40.3 0.01 log10

0.3 0.04 0.26 V

44. Answer (4)

Hint: Element having oxidation state in between

its highest and lowest oxidation states can act as

both oxidising as well as reducing agent.

Sol.: Conceptual

45. Answer (2)

Hint: Atoms having half-filled and fully filled

d-subshell have stable configuration

Sol.: Chromium (Z = 24) 1s2 2s2 2p6 3s2 3p6 3d5

4s1

(n + l) = 3 i.e. 3s and 2p, total 2 + 6 = 8

electrons

m = –1 in 2p, 3p and 3d (2 + 2 + 1 = 5e–)

n = 3 are 3s2 3p6 3d5 (2 + 6 + 5 = 13)

l = 0 are 1s2 2s2 3s2 4s1 (2 + 2 + 2 + 1 = 7)

46. Answer (12.00)

Hint: Molecular formula of a compound is

integral multiple of its empirical formula

Sol.: C = 40%, H = 6.66%, O = 53.34%

EF = CH2O

V.D. = 60, M.wt = 2 × 60 = 120.

120

n 430

M.F = CH2O × 4 = C4H8O4 = CxH2yOz

So, x = 4, y = 4 and z = 4

x + y + z = 4 + 4 + 4 =12

47. Answer (03.20)

Hint: CFSE of a complex depends on no. of

electrons in t2g and eg.

Sol.:

[FeF6]–3, F– is W.F.L

Fe+3(d5), 3 22g gt e , CFSE :

(–0.4 × 3 + 0.6 × 2) O = 0

[Fe(CN)6]–4, CN– is S.F.L

Fe+2(d6), 6 02g gt e , CFSE :

(–0.4 × 6 + 0.6 × 0) O = –2.4 O

[NiCl4]–2, Cl– is W.F.L

Ni+2(d8), 4 4g 2ge t , CFSE :

(–0.6 × 4 + 0.4 × 4) t = –0.8 t

So, a + b + c = 0 – 2.4 – 0.8 = –3.2 = –0.1 k

k = 32

48. Answer (63.00)

Hint: Crystalline solids having ABC ABC

packing have fcc unit cell

Sol.: ABC, ABC, ABC … is CCP lattice

% packing efficiency = 74

Free space = 100 – 74 = 26

If 50% of atoms are removed then occupied

space = 74

37%2

Total free space = (74 – 37) + 26 = 37 + 26 = 63%.

49. Answer (01.20)

Hint: Root mean square velocity = 3p

d

Sol.: Crms = 53p 3 2.4 10

d 8

5 40.9 10 9 10 300

Crms = 300

300

V 1.2250

50. Answer (18.00)

Hint: At equilibrium G RTlnK

Sol.: 2 2H I 2HI

[H2] : [I2] : [HI] at equilibrium = 1 : 2 : 1

2 2

c

2 2

[HI] 1k 0.5

[H ][I ] 1 2

G° = –2.303 RTlogK.

= –2.303 × 2 × 300 × log0.5

= –2.303 × 2 × 300 × (–0.3)

= 23.03 × 18 = 23.03 × G

G = 18



10/13

51. Answer (2)

Hint: Here P(1) and P(2) are roots of P(x) = 0

Sol.: Given that P(P(1)) = P(P(2)) = 0

i.e. P(x) has two roots P(1) and P(2)

1 + a + b, 4 + 2a + b are roots P(x)

Sum of roots = –a

4a + 2b + 5 = 0

52. Answer (2)

Hint: 22 1 ( 1) 2t t

2 2 1t t

Sol.: 2 22 1 ( 1) 2 2 1t t t t

Now 4 3 23 3 9 6t t t t

2 2( 4)( 2 1) 2t t t t

= 0 + 2 = 2

53. Answer (1)

Hint: Use

1 1 1

1 1 1 1 1 1 1 1 1

r q ppq qr rp pqr

p q q r r p p q r r q p

Sol.: 1 1 1

1p q r

ip q r

2 2 2

2 2 21 1 1 1 1 11 1 1

2 2p q r pq qr rp

ip q q r r pp q r

1 1 1

1 1 1

22

r q ppqri

p q r r q p

= 2i – 0 = 2i

54. Answer (3)

Hint: Here 1 211

4 5,2 5

N N

Sol.: Here N1 = 4! × 5! and N2 = 11!

2! 5!

2

1

433

7

N

N

55. Answer (3)

Hint: Use concept of restricted permutation

Sol.: Required probability = 7

2

5! 2! 1

7! C

56. Answer (3)

Hint: Use operations on rows

Sol.:

0

p q y r z

p x q r z

p x q y r

Apply 1 1 3 R R R and 2 2 3, R R R we get

0

0 0

x z

y z

p x q y r

[ ( )] [0 ( )] 0 x yr z q y z y p x

[Expansion along first row]

0 xyr xzq xzy yzp zyx

2 2 p q r

xyr zxq yzp xyzx y z

57. Answer (2)

Hint: Here

4 2 2

8 8

4 1 (2 1 2 )(2 1 2 )n

n nT

n n n n n

Sol.:

4 2 2

8 8

4 1 (2 1 2 )(2 1 2 )n

n nT

n n n n n

2

12 1

(2 1 2 )nS

n n

by telescopic

sum

Hence 201680

.841

S

58. Answer (1)

Hint: 1 2 1 2

1 2 1 2

G G A A

H H H H

Sol.: In this case 1 2 1 2

1 2 1 2

G G A A

H H H H

PART - C (MATHEMATICS)



11/13

59. Answer (4)

Hint: Mean

1

1 n

rr

x xn

Sol.: Let x1, x2, x3, … xn be n observations

1

1 n

rr

x xn

Let 10iix

yk

then 1

1 110

n

rr

y y xn k

10x k

yk

60. Answer (3)

Hint: Use [ ( ) ( )] ( )x xe f x f x dx e f x c

Sol.: 2sec tan sin2tan

2

x x x xe x x dx

2

2 2sec tan 2cos cossec tan

2 2

x x x x xe x x x dx

2

2sec tan cos

2

x x x xe c

61. Answer (1)

Hint: Here 21

1n nI I

n

Sol.: Clearly 21

1n nI I

n

2

11

n n

nI I

The common difference is ‘1’

62. Answer (2)

Hint: Use 0 1 2... 2n

nC C C C

and

11 2

0

2 1...

2 3 1

nC CC

n

Sol.:

100100 100 100

100 100

0 0 0

( 1 1)

1 ( 1)

rr r

r r r

CrC C

r r

101 100

100 (2 1) 99(2 ) 12101 101

99, 1, 101a b c

2 9999ab c

63. Answer (4)

Hint: Here cos = 0 OR

1sin 0

Sol.: 1

cos sin 0

1cos 0 OR sin 0

1OR +

2n

Infinite number of solutions are in N.

64. Answer (1)

Hint: Set B = Rf

Sol.: f(x) = x6 – 3x2 – 10

f(x) = 6x5 – 6x = 0

= 6x(x4 – 1) = 0

So f(–1) = –2, f(1) = –12, f(0) = –10, f(2) = 42,

f(–2) = 42

So Rf = [–12, 42] = B

65. Answer (4)

Hint: Intersection of pair of lines will be centre

= 3

3,2

Sol.: Lines intersect at 3

3,2

which is centre

of the required circle.

Given circle is x2 + y2 – 4x – 3y = 0 whose centre

is 3

2,2

; 5

2r

–2 –1 0 1 2

– + +–

Min. Max. Min.



12/13

For sufficient circle d = |r1 – r2|

5

52

r

15 9 225

92 4 4

r k

45k

66. Answer (2)

Hint: Equation hyperbola is 2 2

1.4 1

x y

Sol.: It passes through 1

3, .2

7

0 or4

Length of transverse axis = 3.

67. Answer (1)

Hint: Use Cramer's Rule

If D = 0 Infinite solutions

Sol.: 2( 1) ( 2)D

2

1 ( 1) ( 2)D

2 3 0D D

For = 1, –2 infinite solution

1, –2 unique solution

N1 = 11; N2 = 0; N3 = 2

2 2 21 2 3 125N N N

68. Answer (2)

Hint: Use concept of cross and dot product

Sol.: ( ) ( )x y x y

ˆ ˆ ˆ ˆ ˆ3 15 13 (5 )i j k i j

ˆ ˆ ˆ13 65 78i j k

ˆ ˆ ˆ13 5 6i j k

69. Answer (3)

Hint: Since lines are perpendicular then

a1a2 + b1b2 + c1c2 = 0

Sol.: Equation of line ‘L’ is

ˆ ˆ ˆ ˆ ˆ( ) ( 10 16 )r j k i j k

Point B is (–10, 16 + 1, – 1)

AB2 = 357 2 = 1 = ±1

B = (–10, 17, 0) or (10, –15, –2) then

a + 2b + 3c = 24 OR – 26

70. Answer (2)

Hint: Use concept of negation

Sol.: ~ [~ (~ )]q p q

~ (~ ) ~ (~ )q p q (De. Morgan law)

( ~ )q p q

( ) ( ~ )q p q q (Distributive law)

( )q p F (Complement law)

( )q p (Identity law)

71. Answer (81.00)

Hint: Here

21 1 ( ( 1))cot

2 1n

n nT

n

Sol.:

21 1 ( ( 1))cot

2 1n

n nT

n

1

2 2

2 1tan

1 ( 1)

n

n n

2 21

2 2

( 1)tan

1 ( 1)

n n

n n

So 1 2 1 28 tan (9) tan 1S

2 21

2 2

9 1tan

1 9 1

1 180 40tan tan82 41

81m n

72. Answer (04.00)

Hint: Put x = rcos; y = rsin

Sol.: Put x = rcos; y = rsin



13/13

Then (r2sincos)r2(cos2 – sin2) = r2

2 4 4cosec 4 4sin4

r

73. Answer (01.00)

Hint: Use Newton-Leibniz rule

Sol.: 2

2cos cos 2 cosx

x

dt dt x x x

dx

at x = 0 the value is ‘1’

74. Answer (00.00)

Hint:

1

.dy dx

dx dy

Now differentiate w.r.t. x

Sol.:

1dy dx

dx dy

22

21

d y dx d dx dy

dy dy dy dxdx

22 2

2 2

d y dx d x dy

dy dxdx dy

32 2

2 20

d y dy d x

dxdx dy

75. Answer (03.00)

Hint: Use concept of GIF.

[x] is discontinuous at integers.

Sol.: f() = [tan[cot]]

If cot , [cot ] cotI

f() = 0

If cot I, f() = 1

Points of discontinuity occurs if cot I and

cot 3.7312

i.e. cot = 1, 2, 3

3 points of discontinuity

test - 7 (code-a) (answers) all india aakash test series for jee (main)-2021 test - 7 ... · 2020....

Documents