Answers & Solutions JEE (Advanced)-2018 - Aakash

DATE : 20/05/2018

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472

Answers & Solutions

Time : 3 hrs.

Max. Marks: 180

for

JEE (Advanced)-2018 PAPER - 2 PART-I : PHYSICS SECTION - 1 (Maximum Marks : 24) 

This section contains SIX (06) questions.



Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is (are) correct option(s).



For each question, choose the correct option(s) to answer the question.



Answer to each question will be evaluated according to the following marking scheme: Full Marks

: +4

If only (all) the correct option(s) is (are) chosen.

Partial Marks

: +3

If all the four options are correct but ONLY three options are chosen.

Partial Marks

: +2

If three or more options are correct but ONLY two options are chosen, both of which are correct options.

Partial Marks

: +1

If two or more options are correct but ONLY one option is chosen and it is a correct option.

Zero Marks

: 0

If none of the options is chosen (i.e. the question is unanswered).

Negative Marks : – 2

In all other cases.

For Example: If first, third and fourth are the ONLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in +2 marks. Selecting only one of the three correct options (either first or third or fourth option), without selecting any incorrect option (second option in this case), will result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in –2 marks. 1

JEE (ADVANCED)-2018 (PAPER-2)

1.

A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as dK/dt = t, where  is a positive constant of appropriate dimensions. Which of the following statements is (are) true? (A) The force applied on the particle is constant (B) The speed of the particle is proportional to time (C) The distance of the particle from the origin increases linearly with time (D) The force is conservative

Answer (A, B, D) Sol. K 

1 dK dv  mv mv2 ⇒ 2 dt dt

given,



dK dv  t ⇒ mv  t dt dt

v

0



t

 v2  t 2 tdt ⇒  m 2 m 2 0

∫ vdv  ∫ v

dv    t  a m dt m

F  ma  m  constant dS   t2  t⇒ S dt m m 2

2.

Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity u0. Which of the following statements is (are) true? (A) The resistive force of liquid on the plate is inversely proportional to h (B) The resistive force of liquid on the plate is independent of the area of the plate (C) The tangential (shear) stress on the floor of the tank increases with u0 (D) The tangential (shear) stress on the plate varies linearly with the viscosity  of the liquid

Answer (A, C, D) Sol. Fv

A

u0

Plate

⎛ dv ⎞ Fv  A ⎜ ⎟ ⎝ dz ⎠ Since height h of the liquid in tank is very small ⇒ ⎛u ⎞ Fv  ()A ⎜ 0 ⎟ ⎝ h ⎠

⎛ 1⎞ Fv  ⎜ ⎟ , Fv  u0 , Fv  A, Fv   ⎝h⎠ 2

dv v ⎛ u0 ⎞   dz z ⎜⎝ h ⎟⎠


3.

An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density . It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 120° at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is 0. Which of the following statements is (are) true?

z



P R 120°

O

Q

(A) The electric flux through the shell is

3R 0

(B) The z-component of the electric field is zero at all the points on the surface of the shell (C) The electric flux through the shell is

2R 0

(D) The electric field is normal to the surface of the shell at all points

z



Answer (A, B) Sol. PQ = (2) R sin 60°

3  (2R)  2



3R

P



R

Q enclosed  



We have  

qenclosed 0

3R



120°

O

R Q



⎛ 3R ⎞ ⎜ ⎜  ⎟⎟ ⎝ 0 ⎠

Also electric filed is perpendicular to wire so Z-component will be zero. 3


4.

A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale).

f 2

45° f



(A)



(C)

 > 45°



(B) 

0 T1, the correct statement(s) is (are) (Assume H and S are independent of temperature and ratio of lnK at T1 to lnK at T2 is greater than T 2/T 1. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.) (A) H < 0, S < 0

(B)

G < 0, H > 0

(C) G < 0, S < 0

(D)

G < 0, S > 0

Answer (A, C)

Sol.

ln K1 T2  ln K2 T 1  On increasing temperature, K decreases.  H° < 0 From graph K > 1  G° < 0 – H S  ln K1 TR R T  1  2 ln K2 H S T1  T2R R

(H  T1S) T2 T2  (H  T2 S) T1 T1

–H° + T1S° > –H° + T2S°  S° < 0

SECTION 2 (Maximum Marks: 24) 

This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE.



For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.



Answer to each question will be evaluated according to the following marking scheme:

7.

Full Marks

: +3 If ONLY the correct numerical value is entered as answer.

Zero Marks

: 0

In all other cases.

The total number of compounds having at least one bridging oxo group among the molecules given below is _________. N2O3, N2O5, P4O6, P4O7, H4P2O5, H5P3O10, H2S2O3, H2S2O5

Answer (5.00) 17

JEE (ADVANCED)-2018 (PAPER-2) N2O 3 O

Sol.

N2O 5 O

O

O

O

O

8.

P OH

P

P O P

P

O

P OH

H4 P2 O5 O

O

O OO

H5 P3 O10 O O HO

O

P4O 7 O

O OO P

O N

O

P4O6 P

O

N

N—N

O

P OH

HO

O

P H

O

P H

H2 S2 O3 S OH

OH

P

O O

P

O

HO

S OH

H2 S2 O5 O O

O

HO

S O

S

OH

Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is __________. (Atomic weights in g mol–1: O = 16, S = 32, Pb = 207)

Answer (6.47) Sol.

2PbS  3O2   2PbO  2 SO2 2PbO  PbS   3Pb  SO2 3 moles of O2 produce 3 moles of lead 96 kg of oxygen produce 621 kg of lead. 1 kg of oxygen produce

9.

621  6.468  6.47 kg 96

To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction, MnCl2 + K2S2O8 + H2O  KMnO4 + H2SO4 + HCl (equation not balanced). Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2 (in mg) present in the initial solution is _________. (Atomic weights in g mol–1: Mn = 55, Cl = 35.5)

Answer (126.00) Sol. From POAC, m moles of MnCl2 = m moles of KMnO4 = x((let) and, meq of KMnO4 = meq of oxalic acid

⎛ 225 ⎞ x5  ⎜ ⎟2 ⎝ 90 ⎠ x=1  m moles of MnCl2 = 1 mg of MnCl2 = (55 + 71) = 126 mg 18


10. For the given compound X, the total number of optically active stereoisomers is _________.

HO

HO

This type of bond indicates that the configuration at the specific carbon and the geometry of the double bond is fixed

HO

This type of bond indicates that the configuration at the specific carbon and the geometry of the double bond is NOT fixed

HO

X

Answer (7.00) HO

OH

Sol.

HO

OH

Only three streocentre are present.  Total isomer = 23 = 8 But one is optically inactive. OH

OH

HO

OH

11. In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is __________. (Atomic weights in g mol–1: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis)

O NaOBr +

H3O

A

NH3, 

(60%)

B

Br2/KOH

C

(50%)

Br2(3 equiv) AcOH

(50%)

D

(100%)

Answer (495.00) O

COOH

H 3O Acetophenone 10 moles

+

 A (60%)

Yield of D in moles = 10 

NH2

NH2 Br

Br2 KOH

NH 3

NaOBr

Sol.

CONH2

Br

 B (50%)

C (50%)

60 50 50    1.5 moles 100 100 100

Amount of D = 1.5 × 330 = 495.00 19

Br

D (100%)


12. The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: 2Cu(s) + H2O(g)  Cu2O(s) + H2(g)

pH2 is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln( pH2 ) is _________. (Given: total pressure = 1 bar, R (universal gas constant) = 8 J K–1 mol–1, ln(10) = 2.3. Cu(s) and Cu2O(s) are mutually immiscible. At 1250 K: 2Cu(s) +

H2(g) +

1 O (g)  Cu2O(s); G = –78,000 J mol–1 2 2

1 O (g)  H2O(g); 2 2

G = –1,78,000 J mol–1; G is the Gibbs energy)

Answer (–14.60) Sol. (i) 2 Cu(s)  (ii) H2 (g) 

1  Cu2 O(s), G  –78 kJ / mole O2 (g)  2

1  H2 O(g), G  –178 kJ / mole O2 (g)  2

(i) – (ii) 2 Cu(s)  H2 O(g)   Cu2 O(s)  H2 (g), G  100 kJ

G  G  RT ln K  0 ⎛ pH 5 2 ⎜ 10  8  1250 ln  ⎜ pH O ⎝ 2 ⎛ pH 2 10 4 ln ⎜ ⎜ pH O ⎝ 2

⎞ ⎟0 ⎟ ⎠

⎞ ⎟  105  0 ⎟ ⎠

ln pH2 – ln pH2O  –10 Now, pH2O  XH2O  PTotal  0.01 1  10–2 

ln pH  2 ln10  –10 2

ln pH2  4.6  –10 ln pH  –14.60 2

13. Consider the following reversible reaction,

A(g)  B(g) AB(g) . The activation energy of the backward reaction exceeds that of the forward reaction by 2RT (in J mol–1). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of G (in J mol–1) for the reaction at 300 K is __________. (Given; ln(2) = 0.7, RT = 2500 J mol–1 at 300 K and G is the Gibbs energy) Answer (8500.00) 20


Sol. A(g) + B(g) AB(g)

Eab – Eaf  2RT Af 4 Ab K

Kf Kb

Kf  A f e

–E a f /RT

Kb  Ab e –Eab /RT

Kf A f (Eab –Eaf )/RT  e Kb Ab 

K = 4e2RT/RT K = 4e2

G° = – RT In K = – RT (2 + ln 4) = – 2500 (2 + 2 × 0.7) = – 8500 J mol–1 Absolute value is 8500.00 14. Consider an electrochemical cell: A(s) | An+ (aq, 2 M) || B2n+ (aq, 1 M) | B(s). The value of H for the cell reaction is twice that of G at 300 K. If the emf of the cell is zero, the S the cell reaction per mole of B formed at 300 K is .

(in J K–1 mol–1) of

(Given: ln(2) = 0.7, R (universal gas constant) = 8.3 J K–1 mol–1. H, S and G are enthalpy, entropy and Gibbs energy, respectively.) Answer (–11.62) Sol.

A   An  ne – B2n  2ne –  B 2A  B2n   2An  B

H° = 2G°, Ecell = 0 G° = H° – TS° 

G° = TS°  S 

S 

– RT ln K [An ]2  – R ln 2n T [B ]

 – 8.3  ln



G T

22 1

S° = – 11.62 JK–1 mol–1 21


SECTION 3 (Maximum Marks: 12) 

This section contains FOUR (04) questions.



Each question has TWO (02) matching lists: LIST-I and LIST-II.



FOUR options are given representing matching of elements from LIST-I and LIST-II. ONLY ONE of these four options corresponds to a correct matching.



For each question, choose the option corresponding to the correct matching.



For each question, marks will be awarded according to the following marking scheme: Full Marks

: +3 If ONLY the option corresponding to the correct matching is chosen.

Zero Marks

: 0

If none of the options is chosen (i.e. the question is unanswered).

Negative Marks : –1 In all other cases. 15. Match each set of hybrid orbitals from LIST-I with complex(es) given in LIST-II. List-I P.

dsp2

Q.

sp3

R.

3

sp d

S.

d2sp3

List-II 1. [FeF6]4– 2. [Ti(H2O)3Cl3]

2

3. [Cr(NH3)6]3+ 4. [FeCl4]2– 5. Ni(CO)4 6. [Ni(CN)4]2–

The correct option is (A) P  5; Q  4,6; R  2,3; S  1 (B) P  5,6; Q  4; R  3; S  1,2 (C) P  6; Q  4,5; R  1; S  2,3 (D) P  4,6; Q  5,6; R  1,2; S  3 Answer (C) Sol.

1. [FeF6]4– 3d

4s

4p

Fe+2

High spin complex because F– is weak field ligand. sp3d2 2. [Ti(H2O)3Cl3] Ti

3+

2

3

2

3

d sp d2 sp3

3. [Cr(NH3)6]3+ 3+

d sp

Cr

d 2sp 3

4. [FeCl4]2– Fe+2 : 3d6, Cl– is weak field ligand. 3d

4s

4p

sp 3

22


5. Ni(CO)4 Ni0 – 3d84s2, CO is strong field ligand Ni

sp

3

6. [Ni(CN)4]2– Ni+2 dsp

2

CN– is strong field ligand. 16. The desired product X can be prepared by reacting the major product of the reactions in LIST-I with one or more appropriate reagents in LIST-II. (given, order of migratory aptitude: aryl > alkyl > hydrogen)

O Ph

OH Ph

Me

X List-I

HO P.

List-II

Ph Me

Ph Me H 2N

Q.

OH

H Me

R.

Ph Me

S.

OH

+ HNO2

2. [Ag(NH3)2]OH

Ph

Me

Br

1. I2, NaOH

Ph

Ph

HO

+ H2SO4

OH

+ H2SO4

3. Fehling solution

+ AgNO3

4. HCHO, NaOH

Ph H

Ph Me

OH

5. NaOBr The correct option is (A) P  1; Q  2,3; R  1,4; S  2,4 (B) P  1,5; Q  3,4; R  4,5; S  3 (C) P  1,5; Q  3,4; R  5; S  2,4 (D) P  1,5; Q  2,3; R  1,5; S  2,3 Answer (D) 23


Ph

Me

Sol. (P) HO Ph

Ph + H2SO 4

O

Me Ph

HO Me

Me

Ph

H OH + HNO2

(Q) H2N

Me

Ph

O

Me Ph

[Ag(NH3)2]OH

H

Fehling Solution

Ph

(R) HO Me

Ph

Ph OH + H2SO4

Ph

O

Ph Me

Me

Me

NaOBr

Ph Me Ph

R  1, 5

Ph (S) Ph

COOH

Me

Q  2, 3

Ph

C – OH

Me Ph

P  1, 5

I2/NaOH

COOH

Ph Me Ph

COOH

Ph

Ph OH + AgNO3

CHO

Ph

Br Me

Me Fehling solution

[Ag(NH3)2]OH

Ph

Ph COOH

Ph

COOH

Ph

Me

Me

S  2, 3 17. LIST-I contains reactions and LIST-II contains major products. List-I

P.

Q.

R.

List-II + ONa

1.

Br +

HBr

2.

OMe +

NaOMe

3.

Br

24

C – OH

Me Ph

O

Ph

NaOBr

Ph

I2/NaOH

O

Ph

OH

Br

OMe

Ph Me Ph

COOH


+

S.

MeBr

4.

ONa

O

5. Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option. (A) P  1,5; Q  2; R  3; S  4 (B) P  1,4; Q  2; R  4; S  3 (C) P  1,4; Q  1,2; R  3,4; S  4 (D) P  4,5; Q  4; R  4; S  3,4 Answer (B) Sol. (P)

+ ONa + Br OH with 3° halide, elimination predominates.

(Q)

OMe + HBr

(R)

Br + NaOMe

(S)

ONa + MeBr

Br + MeOH

OMe

P  1, 4; Q  2; R  4; S  3. 18. Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of the solutions on [H+] are given in LIST-II. (Note: Degree of dissociation () of weak acid and weak base is 1



x(e  1)  1 0 x 1

and

(e  1)x  e  0 and x < 0 or x > 1 x 1



1 ⎤ ⎡ e ⎛ ⎞ x  ⎜ ,  ⎥  ⎢ e  1,  ⎟  e 1 ⎝ ⎦ ⎣ ⎠

For range of g(x);

⎛ x ⎞ ⎛ x ⎞ 1  loge ⎜  0 or 0  log ⎜ ⎟ ⎟1 ⎝ x  1⎠ ⎝ x  1⎠    g(x)  0 or 0  g(x)  2 2



⎡  ⎞ ⎛ ⎤ Range of g(x) = ⎢  , 0 ⎟  ⎜ 0, ⎥ ⎣ 2 ⎠ ⎝ 2⎦

R  1, P  4, S  1, Q  2 16. In a high school, a committee has to be formed from a group of 6 boys M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5. (i) Let 1 be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls. (ii) Let 2 be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls. (iii) Let 3 be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls. 37


(iv) Let 4 be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls and such that both M1 and G1 are NOT in the committee together. LIST-I

LIST-II

P. The value of 1 is

1. 136

Q. The value of 2 is

2. 189

R. The value of 3 is

3. 192

S. The value of 4 is

4. 200 5. 381 6. 461

The correct option is: (A) P  4; Q  6; R  2; S  1 (B) P  1; Q  4; R  2; S  3 (C) P  4; Q  6; R  5; S  2 (D) P  4; Q  2; R  3; S  1 Answer (C) Solution : (i)

1 = 6C3 × 6C2 = 20 × 10 = 200

(ii) 2 = 6C1 ×

5

C1 +

6

C2 ×

5

C2 +

6

C3 ×

5

C3 +

6

C4 × 5C4 +

6

C5 × 5C5

= 30 + 150 + 200 + 75 + 6 = 461 (iii) 3 = 5C2 × 6C3 + 5C3 × 6C2 + 5C4 × 6C1 + 5C5 = 200 + 150 + 30 + 1 = 381 (iv) 4 = (5C2 × 6C3 – 4C1 × 5C1) + (5C3 × 6C1 – 4C2 × 5C0) = (150 – 20) + (60 – 6) + 5 = 130 + 60 – 1 = 190 – 1 = 189 17. Let H :

x2 a2



y2 b2

 1 , where a > b > 0, be a hyperbola in the xy-plane whose conjugate axis LM subtends

an angle of 60º at one of its vertices N. Let the area of the triangle LMN be 4 3 . LIST-I

LIST-II

P. The length of the conjugate axis of H is

1. 8

Q. The eccentricity of H is

2.

R. The distance between the foci of H is

3.

S. The length of the latus rectum of H is

4. 4

The correct option is: (A) P  4; Q  2; R  1; S  3 (B) P  4; Q  3; R  1; S  2 (C) P  4; Q  1; R  3; S  2 (D) P  3; Q  4; R  2; S  1 38

4 3 2 3


Answer ( B )

y

Sol.

L b O

30°

a

x

N

M

tan30 

 b

b a

a

...(i)

3

Now area of OLN = 

1 ab 2

1 ab  2 3 2

 ab  4 3

...(ii)

From (i) and (ii) we have a  2 3, b = 2 Now e  1 

b2 a2

 1

4 2  12 3

Distance between foci = 2ae

 22 3 

2 3

8

2b2 2  4 4   The length of latus rectum = a 2 3 3  ⎛ ⎞ ⎛  ⎞ 18. Let f1 : »  » , f2: ⎜ – 2 , 2 ⎟  » , f3 : ⎜⎜ – 1, e 2 – 2 ⎟⎟  » and f4 : »  » be functions defined by ⎝ ⎠ ⎝ ⎠

(i) f1(x ) = sin

(ii) f 2 (x ) =



2



1 – e–x ,

⎧ sinx ⎪ ⎪ tan –1 x if ⎨ ⎪ ⎪⎩ 1 if

x0

, where the inverse trigonometric function tan –1 x assumes values

x0

⎛  ⎞ in ⎜ – , ⎟ ⎝ 2 2⎠ (iii) f3(x) = [sin(loge(x + 2))], where, for t  » , [t ] denotes the greatest integer less than or equal to t,

(iv) f4(x) =

⎧ 2 ⎛ 1⎞ ⎪ x sin ⎜ ⎟ if ⎝x⎠ ⎨ ⎪ if 0 ⎩

x0 x0 39


List-I P.

List-II

The function f1 is

1. NOT continuous at x = 0

Q. The function f2 is

2. continuous at x = 0 and NOT differentiable at x = 0

R. The function f3 is

3. differentiable at x = 0 and its derivative is NOT continuous at x = 0

S. The function f4 is

4. differentiable at x = 0 and its derivative is continuous at x = 0

The correct option is : (A) P  2; Q  3; R  1; S  4 (B) P  4; Q  1; R  2; S  3 (C) P  4; Q  2; R  1; S  3 (D) P  2; Q  1; R  4; S  3 Answer ( D ) S o l .( i ) f1 : R  R 2 ⎞ ⎛ ⎛ 1 f1(x)  sin ⎜ 1  e x ⎟ = sin ⎜ 1  2 ⎜ ⎝ ⎠ ex ⎝

⎞ ⎟⎟ ⎠

f1(x) is continuous everywhere, but f1(0) does not exist, therefore f1(x) is not differentiable function at x = 0.

⎛  ⎞ (ii) f2 : ⎜  , ⎟  R ⎝ 2 2⎠ ⎧ sinx ⎪ tan1 x ; x  0 ⎪⎪ ; x0 f2 (x)  ⎨1 ⎪ sinx ⎪ ; x0 ⎪⎩ tan1 x

LHL at x = 0 is –1 RHL at x = 0 is 1 f2(x) is discontinuous at x = 0,  ⎛ ⎞ (iii) f3 : ⎜ 1, e 2  2 ⎟  R ⎜ ⎟ ⎝ ⎠

f3 (x)  sin(loge (x  2) 

∵ 1  x  e 2  2 

1  x  2  e2  0  loge (x  2) 

 2

 0  sin(loge (x  2))  1 [sin(loge(x + 2)] = 0 f3(x) = 0 f3(x) is continuous and differentiable at x = 0 40


(iv) f4 : R  R

⎧ 2 ⎛ 1⎞ ⎪x sin ⎜ ⎟ ; x  0 f4 (x)  ⎨ ⎝x⎠ ⎪0 ; x0 ⎩

⎛ 1⎞ lim f4 (x)  lim x2 sin ⎜ ⎟  0 x 0 x 0 ⎝x⎠ = Value of f4(x) at x = 0  f4(x) is continuous at x = 0 ⎛ 1⎞ sin ⎜ ⎟ f(0  h)  f(0) ⎝h⎠ 0 Now, f (0 )  lim  lim 1 h0 h0 h h ⎛ 1⎞ sin ⎜ ⎟ f(0  h)  f(0) ⎝h⎠ 0  lim f (0 )  lim 1 h0 h0 h h  f(x) is differentiable at x = 0

1 ⎛ 1⎞ ⎛ 1⎞ Now, f4 (x)  2  sin ⎜ ⎟  x2 . 2 cos ⎜ ⎟ x x ⎝ ⎠ ⎝x⎠

⎛ 1⎞ ⎛ 1⎞ = 2  sin ⎜ ⎟  cos ⎜ ⎟ is oscillating x ⎝ ⎠ ⎝x⎠ Function which discontinuous at x = 0 END OF THE QUESTION PAPER

41