Answers & Solutions. JEE (Advanced)-2018

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1 DATE : 0/05/08 Regd. ffice : Aakash Tower, 8, Pusa Road, New Delhi-0005.: Fax : Answers & Solutions Time : hrs. Max. Marks: 80 for JEE (Advanced)-08 PAPER - PART-I : PYSICS This section contains SIX (06) questions. SECTIN - (Maximum Marks : 4) Each question has FUR options for correct answer(s). NE R MRE TAN NE 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 : + If all the four options are correct but NLY three options are chosen. Partial Marks : + If three or more options are correct but NLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but NLY 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 : In all other cases. For Example: If first, third and fourth are the NLY 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 + 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 + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in marks.

2 JEE (ADVANCED)-08 (PAPER-). 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) dk dv Sol. K mv mv dt dt dk dv given, t mv t dt dt v t v t vdv tdt m m 0 0 v t m dv a dt m Fma m constant ds t t S dt m m. 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 u 0. 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 u 0 (D) The tangential (shear) stress on the plate varies linearly with the viscosity of the liquid Answer (A, C, D) Sol. F v u 0 dv Fv A dz A Plate Since height h of the liquid in tank is very small u 0 dv v dz z h u0 F v ( )A h F v,f v u 0,F v A,F h v

3 JEE (ADVANCED)-08 (PAPER-). 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 0 at the centre 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 0 Q (A) The electric flux through the shell is R 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 R (D) The electric field is normal to the surface of the shell at all points Answer (A, B) Sol. PQ = () R sin 60 z 0 (R) R P R Qenclosed R 0 q We have enclosed 0 R Q R 0 Also electric filed is perpendicular to wire so Z-component will be zero.

4 JEE (ADVANCED)-08 (PAPER-) 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 45 f (A) > 45 (B) (C) 0 < < 45 (D) Answer (D) Sol. Image of point A B f/ Q A P x C (f x) v f v f f IAB f IAB AB AB f For eight of PQ v f(f x) v (f x) f v (f x) f x IPQ f(f x) f PQ x{ (f x)} x f f (AB)x (AB)x I PQ PQ x x f PQ f I PQ = AB (Size of image is independent of x. So, final image will be of same height terminating at infinity) 4

5 JEE (ADVANCED)-08 (PAPER-) 5. In a radioactive decay chain, 90 Th nucleus decays to 8 Pb nucleus. Let N and N be the number of and particles, respectively, emitted in this decay process. Which of the following statements is (are) true? (A) (B) (C) (D) N 5 N 6 N N 4 Answer (A, C) Sol. A Decay A4 ZX ZY A Decay A ZX ZX 90Th 8Pb No. of -particles emitted 5 4 Since Z decreases by (90 8) = 8 only ence no. of decay = 6. In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 z is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 8.9 cm. Which of the following statements is (are) true? (A) The speed of sound determined from this experiment is ms (B) The end correction in this experiment is 0.9 cm (C) The wavelength of the sound wave is 66.4 cm (D) The resonance at 50.7 cm corresponds to the fundamental harmonic Answer (A, B, C) Sol. (n ) 50.7 e 4 (n ) 8.9 e cm = 66.4 cm v = f = m/s = m/s For, n =, e = 0.9 cm 5

6 SECTIN (Maximum Marks: 4) JEE (ADVANCED)-08 (PAPER-) This section contains EIGT (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. 6.5, 7.00, -0., -.0, 0.7, -7.0) 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: Full Marks : + If NLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 7. A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4 kg is at rest on this surface. An impulse of.0 Ns is applied to the block at time t = 0 so that it starts moving along the x-axis with a velocity v(t) = v 0 e t/, where v 0 is a constant and = 4 s. The displacement of the block, in metres, at t = is. Take e = 0.7. Answer (6.0) Sol. v 0 = =.5 m/s 0.4 ds =.5e t/ dt s t ds =.5e 0 0 t / dt.5 s= e t / t 0 =.5 e t / At t =, s = 0[ e ] = = 6.0 m 8. A ball is projected from the ground at an angle of 45 with the horizontal surface. It reaches a maximum height of 0 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 0 with the horizontal surface. The maximum height it reaches after the bounce, in metres, is. Answer (0.00) Sol. u sin 45 =0 g u =0 4g If speed is v after the first collision then v =60 4g as v u h max = vsin0 = g v = 0 m = 0.00 m 8g 9. A particle, of mass 0 kg and charge.0 C, is initially at rest. At time t = 0, the particle comes under the influence of an electric field E(t) E ˆ 0 sinti, where E 0 =.0 NC and = 0 rad s. Consider the effect of only the electrical force on the particle. Then the maximum speed, in ms, attained by the particle at subsequent times is. 6

7 JEE (ADVANCED)-08 (PAPER-) Answer (.00) Sol. F=qE=.0 N sin (0 t) ˆi F m a = = 0 sin (0 t) dv =0 sin (0 t) dt V t dv = 0 sin(0 t)dt v= cos(0 t) v max = m/s =.00 m/s 0. A moving coil galvanometer has 50 turns and each turn has an area 0 4 m. The magnetic field produced by the magnet inside the galvanometer is 0.0 T. The torsional constant of the suspension wire is 0 4 N m rad. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0. rad. The resistance of the coil of the galvanometer is 50. This galvanometer is to be converted into an ammeter capable of measuring current in the range 0.0 A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is. Answer (5.56) Sol. = BANi m = K 4 K 0 0. i m = = 4 BAN = = 0. A = 0.9 S 50 S= = A steel wire of diameter 0.5 mm and Young s modulus x 0 Nm carries a load of mass M. The length of the wire with the load is.0 m. A vernier scale with 0 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count.0 mm, is attached. The 0 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by. kg, the vernier scale division which coincides with a main scale division is. Take g = 0 ms and =.. Answer (.00) Sol. W L YA L (0.5) (0 ) = 0. mm 9 L.C. of vernier mm = 0. mm 0 Vernier reading = 7

8 JEE (ADVANCED)-08 (PAPER-). ne mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 00 K and the universal gas constant R = 8.0 J mol K, the decrease in its internal energy, in Joule, is. Answer (900.00) Sol. n, 5 T V = T V V T T 00 V 8 T = 5 K R U n T T 8 (5 00) = 900 J Decrease in internal energy = 900 J. In a photoelectric experiment a parallel beam of monochromatic light with power of 00 W is incident on a perfectly absorbing cathode of work function 6.5 ev. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 00%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n 0 4 N due to the impact of the electrons. The value of n is. Mass of the electron m e = 9 0 kg and.0 ev = J. Answer (4.00) Sol. P = 00 J/s 00 No. of photons per second (N) = 9 ( ) = 0 0 p m(ke) m(ev) = 0 4 F = p N = = N 4. Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = to n = transition has energy 74.8 ev higher than the photon emitted in the n = to n = transition. The ionization energy of the hydrogen atom is.6 ev. The value of Z is. Answer (.00) Sol..6 Z Z Z Z = 9 Z = 8

9 JEE (ADVANCED)-08 (PAPER-) This section contains FUR (04) questions. SECTIN (Maximum Marks: ) Each question has TW (0) matching lists: LIST-I and LIST-II. FUR options are given representing matching of elements from LIST-I and LIST-II. NLY NE 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 : + If NLY 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 : - In all other cases. 5. The electric field E is measured at a point P(0, 0, d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II. LIST-I LIST-II P. E is independent of d. A point charge Q at the origin Q. E d. A small dipole with point charges Q at (0, 0, l) and Qat (0, 0, l ). Take l d R. E d. An infinite line charge coincident with the x-axis, with uniform linear charge density S. E d 4. Two infinite wires carrying uniform linear charge density parallel to the x- axis. The one along (y = 0, z = l ) has a charge density + and the one along (y = 0, z = l ) has a charge density. Take l d. 5. Infinite plane charge coincident with the xy-plane with uniform surface charge density (A) P 5; Q, 4; R ; S (B) P 5; Q ; R, 4; S (C) P 5; Q ; R, ; S 4 (D) P 4; Q, ; R ; S 5 Answer (B) Sol. List-II Q () E 4 0 d Q( l) () Eaxis 4 d 0 E d E d () E 0d E d (4) ( l) E (d l) (d l) d E d (5) E E is independent of d 0 9

10 JEE (ADVANCED)-08 (PAPER-) 6. A planet of mass M, has two natural satellites with masses m and m. The radii of their circular orbits are R and R respectively. Ignore the gravitational force between the satellites. Define v, L, K and T to be, respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite ; and v, L, K and T to be the corresponding quantities of satellite. Given m /m = and R /R = /4, match the ratios in List-I to the numbers in List-II. LIST-I LIST-II P. v v. 8 Q. L L. R. K K. S. T T 4. 8 (A) P 4; Q ; R ; S (B) P ; Q ; R 4; S (C) P ; Q ; R ; S 4 (D) P ; Q ; R 4; S Answer (B) GMm mv GM Sol. v R R R Let R = R R = 4R m = m m = m List-I v R 4R (P) : v R R (Q) L = mvr M R v m L R(m)v () : L 4R(m)v K (m)v (R) (4) 8: K (m)v (S) T R : 8 T R 4 7. ne mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II. P R m v P 0 II P 0 I IV III V 0 V 0 V 0

11 JEE (ADVANCED)-08 (PAPER-) LIST-I LIST-II P. In process I. Work done by the gas is zero Q. In process II. Temperature of the gas remains unchanged R. In process III. No heat is exchanged between the gas and its surroundings S. In process IV 4. Work done by the gas is 6P 0 V 0 (A) P 4; Q ; R ; S (B) P ; Q ; R ; S 4 (C) P ; Q 4; R ; S (D) P ; Q 4; R ; S Answer (C) Sol. dp dp dv dv adiabatic Isothermal List-I (P) Process I Adiabatic Q = 0 (Q) Process II Isobaric W = PV = P 0 [V 0 V 0 ] = 6P 0 V 0 (R) Process III Isochoric W = 0 (S) Process (IV) Isothermal Temperature = Constant 8. In the List-I below, four different paths of a particle are given as functions of time. In these functions, and are positive constants of appropriate dimensions and. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned : p is the linear momentum, L is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path. LIST-I LIST-II P. r(t) ti ˆtj ˆ. p Q. r(t) cost ˆi sint ˆj R. r(t) (cost ˆi sint ˆj). L. K S. ˆ r(t) ti t ˆj 4. U 5. E (A) P,,, 4, 5; Q, 5; R,, 4, 5; S 5 (B) P,,, 4, 5; Q, 5; R,, 4, 5; S, 5 (C) P,, 4; Q 5; R,, 4; S, 5 (D) P,,, 5; Q, 5; R,, 4, 5; S, 5 Answer (A) Sol. When force F = 0 potential energy U = constant F 0 force is conservative Total energy E = constant

12 List-I (P) r(t) ti ˆtj ˆ dr vˆiˆj dt = constant p constant v constant K = constant dv a 0 F 0 U constant dt E = U + K = constant Lm(rv) 0 L constant P,,, 4, 5 (Q) r(t) costi ˆsintj ˆ dr ˆ ˆ v sin t( i) cos tjconstant p constant dt v ( sint) ( cost) constant K constant dv a r 0 dt E = constant = K + U But K constant U constant Lm(rv) m (k) ˆ constant Q, 5 (R) r(t) (costi ˆsintj) ˆ dr ˆ ˆ v [sin t( i) cos tj] constant p constant dt v constant K = constant dv a r 0 E constant,u constant dt L m(r v) m kˆ constant (S) R,, 4, 5 r(t) ti ˆ t ˆj dr v ˆi tj ˆ constant p constant dt JEE (ADVANCED)-08 (PAPER-) v ( t) constant K constant dv a ˆj 0 E constant K U dt But K constant U constant Lm(rv) t kˆ constant S 5

13 JEE (ADVANCED)-08 (PAPER-) PART-II : CEMISTRY This section contains SIX (06) questions. SECTIN - (Maximum Marks : 4) Each question has FUR options for correct answer(s). NE R MRE TAN NE 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 : + If all the four options are correct but NLY three options are chosen. Partial Marks : + If three or more options are correct but NLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but NLY 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 : In all other cases. For Example: If first, third and fourth are the NLY 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 + 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 + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in - marks.. The correct option(s) regarding the complex [Co(en)(N ) ( )] + (en = NC C N ) is (are) (A) It has two geometrical isomers (B) It will have three geometrical isomers if bidentate en is replaced by two cyanide ligands (C) It is paramagnetic (D) It absorbs light at longer wavelength as compared to [Co(en)(N ) 4 ] + Answer (A, B, D) Sol. (A) [Co(en)(N ) ( )] N C N Co N C N Co C N N C N N N N (B) [Co(CN) (N ) ( )] N CN NC N NC N Co Co Co NC N NC N N N N N CN

14 (C) Co + = [Ar]d 6 JEE (ADVANCED)-08 (PAPER-) in presence of en and N it form low spin complex. (D) [Co(en)(N ) 4 ] + has larger gap between e g and t g than [Co(en)(N ) ( )] +. So [Co(en)(N ) ( ) + absorb longer wavelength than [Co(en)(N ) 4 ] +.. The correct option(s) to distinguish nitrate salts of Mn + and Cu + taken separately is (are) (A) Mn + shows the characteristic green colour in the flame test (B) nly Cu + shows the formation of precipitate by passing S in acidic medium (C) nly Mn + shows the formation of precipitate by passing S in faintly basic medium (D) Cu + /Cu has higher reduction potential than Mn + /Mn (measured under similar conditions) Answer (B, D) Sol. (A) Manganese show pale purple colour in flame test. (B) S Cl Cu CuS Black ppt. (C) Both Cu + and Mn + form precipitate with S in basic medium. (D) E 0.4 Cu /Cu E Mn.8 V /Mn. Aniline reacts with mixed acid (conc. N and conc. S 4 ) at 88 K to give P (5 %), Q (47%) and R (%). The major product(s) of the following reaction sequence is (are) R ) Ac, pyridine ), CC + S ) 4) NaN, Cl/7-78K 5) Et, ) Sn/Cl ) / (excess) ) NaN, Cl/7-78K 4) P major product(s) (A) (B) (C) (D) Answer (D) Sol. N N N N N Conc. N Conc. S 4 + N + N 5% (P) 47% (Q) 4 % (R)

15 JEE (ADVANCED)-08 (PAPER-) N N Ac, Pyridine N N C C (R) N C C N C C + + NCl N N Et N NaN, Cl 7-78 K N (S) (S) N Sn/Cl N (excess) N NaN, Cl P 4. The Fischer presentation of D-glucose is given below. C C D-glucose The correct structure(s) of -L-glucopyranose is (are) + NCl C C (A) (B) C C (C) (D) Answer (D) 5

16 Sol. Structure of -L-Glucopyranose is JEE (ADVANCED)-08 (PAPER-) C 5. For a first order reaction A(g) B(g) + C(g) at constant volume and 00 K, the total pressure at the beginning (t = 0) and at time t are P 0 and P t, respectively. Initially, only A is present with concentration [A] 0, and t / is the time required for the partial pressure of A to reach / rd of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases) (A) In(P 0 P t) (B) t / Time [A] 0 (C) In(P 0 P t) (D) Rate constant Answer (A, D) Time [A] 0 Sol. A B + C P 0 P 0 P P P P t = P 0 + P P = P P t 0 P P Kt ln ln P P P P P0 0 0 t 0 0 t Kt ln P0 ln(p0 P t) ln (P 0 P t) Time Rate constant of reaction is independent of initial concentration. 6

17 JEE (ADVANCED)-08 (PAPER-) 6. For a reaction, A P, the plots of [A] and [P] with time at temperatures T and T are given below. )0 [A] / (mol L 5 T T )0 [P] / (mol L 5 T T Time Time If T > T, the correct statement(s) is (are) (Assume and S are independent of temperature and ratio of lnk at T to lnk at T is greater than T /T. ere, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.) (A) < 0, S < 0 (B) G < 0, > 0 (C) G < 0, S < 0 (D) G < 0, S > 0 Answer (A, C) Sol. ln K ln K T T n increasing temperature, K decreases. < 0 From graph K > G < 0 S ln K T R R T ln K S T TR R ( T S ) T T ( T S ) T T + T S > + T S S < 0 SECTIN (Maximum Marks: 4) This section contains EIGT (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. 6.5, 7.00, -0., -.0, 0.7, -7.0) 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: Full Marks : + If NLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 7. The total number of compounds having at least one bridging oxo group among the molecules given below is. N, N 5, P 4 6, P 4 7, 4 P 5, 5 P 0, S, S 5 Answer (5.00) 7

18 JEE (ADVANCED)-08 (PAPER-) Sol. N N N N 5 N N P P 4 6 P P P P 4 7 P 4 5 P P P P P P 5P 0 P P P S S S S 5 8. 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 consumed is. (Atomic weights in g mol : = 6, S =, Pb = 07) Answer (6.47) Sol. PbS Pb S Pb PbS Pb S moles of produce moles of lead 96 kg of oxygen produce 6 kg of lead. S S 6 kg of oxygen produce kg To measure the quantity of MnCl dissolved in an aqueous solution, it was completely converted to KMn 4 using the reaction, MnCl + K S 8 + KMn 4 + S 4 + Cl (equation not balanced). Few drops of concentrated Cl were added to this solution and gently warmed. Further, oxalic acid (5 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl (in mg) present in the initial solution is. (Atomic weights in g mol : Mn = 55, Cl = 5.5) Answer (6.00) Sol. From PAC, m moles of MnCl = m moles of KMn 4 = x((let) and, meq of KMn 4 = meq of oxalic acid 5 x5 90 x = m moles of MnCl = mg of MnCl = (55 + 7) = 6 mg 8

19 JEE (ADVANCED)-08 (PAPER-) 0. For the given compound X, the total number of optically active stereoisomers is. This type of bond indicates that the configuration at the specific carbon and the geometry of the double bond is fixed X Answer (7.00) This type of bond indicates that the configuration at the specific carbon and the geometry of the double bond is NT fixed Sol. nly three streocentre are present. Total isomer = = 8 But one is optically inactive.. In the following reaction sequence, the amount of D (in g) formed from 0 moles of acetophenone is. (Atomic weights in g mol : =, C =, N = 4, = 6, = 80. The yield (%) corresponding to the product in each step is given in the parenthesis) Na + A (60%) N, B (50%) /K C (50%) ( equiv) Ac D (00%) Answer (495.00) Sol. Na + C N CN K N N Acetophenone 0 moles A (60%) B (50%) C (50%) D (00%) Yield of D in moles = 0.5 moles Amount of D =.5 0 =

20 JEE (ADVANCED)-08 (PAPER-). The surface of copper gets tarnished by the formation of copper oxide. N gas was passed to prevent the oxide formation during heating of copper at 50 K. owever, the N gas contains mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: Cu(s) + (g) Cu (s) + (g) p is the minimum partial pressure of (in bar) needed to prevent the oxidation at 50 K. The value of ln( p ) is. (Given: total pressure = bar, R (universal gas constant) = 8 J K mol, ln(0) =.. Cu(s) and Cu (s) are mutually immiscible. At 50 K: Cu(s) + (g) Cu (s); G = 78,000 J mol (g) + (g) (g); G =,78,000 J mol ; G is the Gibbs energy) Answer ( 4.60) Sol. (i) Cu(s) (g) Cu (s), G 78 kj/ mole (ii) (g) (g) (g), G 78 kj/ mole (i) (ii) Cu(s) (g) Cu (s) (g), G00 kj G GRT lnk0 p ln 0 p p ln 0 0 p ln p ln p 0 Now, p X P 0.00 Total ln p ln0 0 ln p ln p 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 RT (in J mol ). 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 ) for the reaction at 00 K is. (Given; ln() = 0.7, RT = 500 J mol at 00 K and G is the Gibbs energy) Answer ( ) 0

21 JEE (ADVANCED)-08 (PAPER-) Sol. A(g) + B(g) AB(g) E E RT a b a f A A f b 4 K K K f b K f A e f E a f /RT K b A e b E ab /RT K K f b Af e A b (E E )/RT ab af K = 4e RT/RT K = 4e G = RT In K = RT ( + ln 4) = 500 ( + 0.7) = 8500 J mol Absolute value is Consider an electrochemical cell: A(s) A n+ (aq, M) B n+ (aq, M) B(s). The value of for the cell reaction is twice that of G at 00 K. If the emf of the cell is zero, the S (in J K mol ) of the cell reaction per mole of B formed at 00 K is. (Given: ln() = 0.7, R (universal gas constant) = 8. J K mol., S and G are enthalpy, entropy and Gibbs energy, respectively.) Answer (.6) Sol. n AA ne n B ne B n n A B A B = G, E cell = 0 G = TS G = TS S G T n S RTln K Rln [A ] T n [B ] 8.ln S =.6 JK mol

22 JEE (ADVANCED)-08 (PAPER-) This section contains FUR (04) questions. SECTIN (Maximum Marks: ) Each question has TW (0) matching lists: LIST-I and LIST-II. FUR options are given representing matching of elements from LIST-I and LIST-II. NLY NE 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 : + If NLY 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 : In all other cases. 5. Match each set of hybrid orbitals from LIST-I with complex(es) given in LIST-II. List-I List-II P. dsp. [FeF 6 ] 4 Q. sp. [Ti( ) Cl ] R. sp d. [Cr(N ) 6 ] + S. d sp 4. [FeCl 4 ] The correct option is (A) P 5; Q 4,6; R,; S (B) P 5,6; Q 4; R ; S, (C) P 6; Q 4,5; R ; S, (D) P 4,6; Q 5,6; R,; S Answer (C) Sol.. [FeF 6 ] 4 d 4s 4p Fe + 5. Ni(C) 4 6. [Ni(CN) 4 ] igh spin complex because F is weak field ligand. sp d. [Ti( ) Cl ] Ti + dsp dsp. [Cr(N ) 6 ] + Cr + dsp dsp 4. [FeCl 4 ] Fe + : d 6, Cl is weak field ligand. d 4s 4p sp

23 Ni + dsp JEE (ADVANCED)-08 (PAPER-) 5. Ni(C) 4 Ni 0 d 8 4s, C is strong field ligand Ni 6. [Ni(CN) 4 ] sp CN is strong field ligand. 6. 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) P. Q. R. S. List-I N S 4 N S 4 AgN The correct option is (A) P ; Q,; R,4; S,4 (B) P,5; Q,4; R 4,5; S (C) P,5; Q,4; R 5; S,4 (D) P,5; Q,; R,5; S, Answer (D) X List-II. I, Na. [Ag(N ) ]. Fehling solution 4. C, Na 5. Na

24 JEE (ADVANCED)-08 (PAPER-) Sol. (P) + S 4 I/Na C (Q) P, 5 N Na + N C [Ag(N ) ] C (R) Q, + S 4 Fehling Solution C Na I /Na R, 5 C C (S) + AgN C Fehling solution [Ag(N ) ] C C S, 7. LIST-I contains reactions and LIST-II contains major products. List-I List-II P. Na +. Q. +. R. + Na. 4

25 JEE (ADVANCED)-08 (PAPER-) S. Na Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option. (A) P,5; Q ; R ; S 4 (B) P,4; Q ; R 4; S (C) P,4; Q,; R,4; S 4 (D) P 4,5; Q 4; R 4; S,4 Answer (B) Sol. (P) Na + + with halide, elimination predominates. (Q) + + (R) + Na (S) Na + P, 4; Q ; R 4; S. 8. Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of the solutions on [ + ] are given in LIST-II. (Note: Degree of dissociation () of weak acid and weak base is << ; degree of hydrolysis of salt <<; [ + ] represents the concentration of + ions) List-I List-II P. (0 ml of 0. M Na + 0 ml of. the value of [ + ] does not change on dilution 0. M acetic acid) diluted to 60 ml Q. (0 ml of 0. M Na + 0 ml of 0. M. the value of [ + ] changes to half of its initial acetic acid) diluted to 80 ml value on dilution R. (0 ml of 0. M Cl + 0 ml of 0. M. the value of [ + ] changes to two times of its ammonia solution) diluted to 80 ml initial value on dilution S. 0 ml saturated solution of Ni() in 4. the value of [ + ] changes to times of its equilibrium with excess solid Ni() is initial value on dilution diluted to 0 ml (solid Ni() is still present after dilution). 5. the value of [ + ] changes to times of its initial value on dilution Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option is (A) P 4; Q ; R ; S (B) P 4; Q ; R ; S (C) P ; Q 4; R 5; S (D) P ; Q 5; R 4; S Answer (D) 5

26 JEE (ADVANCED)-08 (PAPER-) Sol. (P) C C old 0 0 C C old 0 Buffer with [Salt] = [Acid] p does not change on dilution (P) () (Q) [ C C ] old = [ CC ] = new 80 CC + c C C x x [ ] old [ ] new Kh c /40 /80 [ ] = new [ ] [ ] old old new [ ] [ ] [ ] new (Q) (5) old (R) 00. [N 4] old [N ] 4 new 80 N 4 N4 c y y y [ ] old [ new ] Kh c /40 /80 new [ ] [ old] [ ] old [ new ] (R) (4) (S) For a saturated solution, x [ ] K sp irrespective of volume of solution. [ + ] remains constant. (S) () END F CEMISTRY 6

27 JEE (ADVANCED)-08 (PAPER-) PART-III : MATEMATICS SECTIN (Maximum Marks : 4) This section contains SIX (06) questions. Each question has FUR options for correct answer(s). NE R MRE TAN NE 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 : + If all the four options are correct but NLY three options are chosen. Partial Marks : + If three or more options are correct but NLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but NLY 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 : In all other cases. For Example: If first, third and fourth are the NLY 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 + 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 + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in - marks.. For any positive integer n, define f n : (0, ) R as n f(x) n jtan (x j)(x j ) for all x (0, ) (ere, the inverse trigonometric function tan x assumes values in Then, which of the following statement(s) is (are) TRUE?,.) 5 j (A) tan (f (0)) 55 0 j j j j (B) (f (0))sec (f (0)) 0 (C) For any fixed positive integer n, (D) For any fixed positive integer n, lim tan(f n(x)) n x lim sec (f (x)) x n Answer (D) n Sol. f n(x) tan (n x) tan (x) tan (nx)x f(x) (nx) x n 7

28 (A) x = 0 is not in domain (B) x = 0 is not in domain JEE (ADVANCED)-08 (PAPER-) (C) n lim tan(f n(x)) lim 0 x (nx)x x (D) lim (tan (f (x)) 0 x n. Let T be the line passing through the points P(, 7) and Q(, 5). Let F be the set of all pairs of circles (S, S ) such that T is tangent to S at P and tangent to S at Q, and also such that S and S touch each other at a point, say, M. Let E be the set representing the locus of M as the pair (S, S ) varies in F. Let the set of all straight line segments joining a pair of distinct points of E and passing through the point R(, ) be F. Let E be the set of the mid-points of the line segments in the set F. Then, which of the following statement(s) is (are) TRUE? (A) The point (, 7) lies in E (B) The point 4 7, 5 5 does NT lie in E (C) The point, lies in E (D) The point 0, does NT lie in E Answer (B, D) Sol. M P Q PMQ = 90 T 5 7 Locus of M x + y y 9 = 0...() Equation of chord of circle whose midpoint is (, ) is S = T x + y (y + ) 9 = + 9 It passes through (, ) + + = 0 Locus : x + y x y + = 0...() (, ) R(, ) x + y y 9 = 0 ption (A) is incorrect although it satisfies eq. () otherwise the line T would touch the second circle on two points. Also (4/5, 7/5) satisfies eq. () but again in this case one end of the chord would be (, 7) which is not included in E. Therefore 4 7, 5 5 does not lie in E. (/, ) does not satisfy equation (), therefore it does not lie in E. (0, /) does not satisfy (), so does not lie in E. 8

29 JEE (ADVANCED)-08 (PAPER-). Let S be the set of all column matrices real variables) x + y + 5z = b x 4y + z = b b b b such that b, b, b R and the system of equations (in x y + z = b has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each b b b S? (A) x + y + z = b, 4y + 5z = b and x + y + 6z = b (B) x + y + z = b, 5x + y + 6z = b and x y z = b (C) x + y 5z = b, x 4y + 0z = b and x y + 5z = b (D) x + y + 5z = b, x + z = b and x + 4y 5z = b Answer (A, D) Sol. Given system of equations can be written as 5 x b 4 y b AX B z b b 5 b 5 b 5 A B 4 b 0 0 6b b b b b b 7 7 b b b b b For solution of this system, b b b b + 7b b = 0 (b, b, b ) = ( 7K + K, K, K ) where K, K R...(i) (A) 0 any possible set of (b, b, b ) would give a solution. every set of values of (i) provides atleast a solution of system of option (A). b b b (B ) 5 6b 0 0 bb 0 0 0bb b b 0 0 b b 0 0 b b For solution of this system, b + b + b = 0 (b, b, b ) = ( K K, K, K ), K, K R...(ii) bviously, the set represented by (i) is not contained in (ii). This system doesn t have solution for every set of (b, b, b ) of (i). (C) Equations are of parallel planes or identical planes. For solution, these planes should be identical for that, (b, b, b ) = ( K, K, K )...(iii) Every such value of (b, b, b ) is contained in (i) but not every value of (b, b, b ) of (i) is present in (iii). Not every (b, b, b ) of (i) gives solution to the system of equations of option (C). (D) 0 every set of values of (i) provides atleast a solution of system of option (D). 9

30 JEE (ADVANCED)-08 (PAPER-) 4. Consider two straight lines, each of which is tangent to both the circle x y and the parabola y = 4x. Let these lines intersect at the point Q. Consider the ellipse whose center is at the origin (0,0) and whose semi-major axis is Q. If the length of the minor axis of this ellipse is, then which of the following statement(s) is (are) TRUE? (A) For the ellipse, the eccentricity is and the length of the latus rectum is (B) For the ellipse, the eccentricity is and the length of the latus rectum is (C) The area of the region bounded by the ellipse between the lines x (D) The area of the region bounded by the ellipse between the lines x and x = is ( ) 4 and x = is ( ) 6 Answer (A, C) Sol. y mx is also tangent to x y m m m m Common tangents are y = x + and y = x Q(, 0) x y Equation of ellipse is b = a ( e ) y ( e ) e Length of latus rectum = b a y x Q(, 0) (0, 0) x x = x A x dx x x sin x 4 sq. units 0

31 JEE (ADVANCED)-08 (PAPER-) 5. Let s, t, r be non-zero complex numbers and L be the set of solutions z = x + iy (x, y R,i ) of the equation sz tz r 0, where z xiy. Then, which of the following statement(s) is (are) TRUE? (A) If L has exactly one element, then s t (B) If s = t, then L has infinitely many elements (C) The number of elements in L {z : z + i = 5} is at most (D) If L has more than one element, then L has infinitely many elements Answer (A, C, D) Sol. sztzr 0...(i) Taking conjugate of (i) sztzr 0...(ii) Eliminating z from (i) and (ii) (ssztszsr) (tszttztr) 0 z( s t ) tr rs (A) If (B) If s t, then z has unique value s t and rtrs0, then z has infinitely many values If s t and rtrs0, then z has no value L may be empty set or infinite set (C) Locus of z is null set or singleton set or a line in all cases. It will intersect given circle at most two points. (D) If L has more than one element, then L has infinite elements. 6. Let f : (0, ) R be a twice differentiable function such that f(x)sint f(t)sinx lim sin x tx t x for x (0, ). If f, then which of the following statement(s) is (are) TRUE? 6 (A) f x (B) f(x) x 6 for all x (0, ) (C) There exists (0, ) such that f() = 0 (D) f f 0 Answer (B, C, D) f(x)sint f(t)sinx Sol. lim sin x tx t x Using L.. rule and N.L. theorem, f(x) cosx f(x) sinx = sin x f(x) xc sinx...(i)

32 JEE (ADVANCED)-08 (PAPER-) (A) f 6 c 0 f(x) xsinx f 4 4 x (B) As sinx x 6 4 x xsinx x 6 4 x f(x) x 6 (C) f(x) is continuous in [0, ] and differentiable in (0, ) and f(0) = f() = 0 f() = 0, (0, ) (D) f(x) = x cosx sinx, f(x) = x sinx cosx f f 0 SECTIN (Maximum Marks : 4) This section contains EIGT (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. 6.5, 7.00, 0.,.0, 0.7, 7.0) 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: Full Marks : + If NLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 7. The value of the integral (x ) (x) dx is. Answer (.00) Sol. I / 0 6 (x ) (x) / 4 dx / 0 8 x (x ) x 6 / 4 dx / x ( x) x 0 / dx Put x dx t dt x ( x) / / t I ( ) ( )( ) / t /

33 JEE (ADVANCED)-08 (PAPER-) 8. Let P be a matrix of order such that all the entries in P are from the set {, 0, }. Then, the maximum possible value of the determinant of P is. Answer (4.00) a a a Sol. P a a a a a a The maximum possibility of P can be 6 if a a a a a a a a a a a a but, the moment zero. a a a a is set as (say) and a a a a as or, automatically a a a a takes ence, P 6. Next possibility is 4. P is one such possibility. 9. Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If is the number of one-one functions from X to Y and is the number of onto functions from Y to X, then the value of ( ) is. 5! Answer (9.00) Sol. = Number of one-one functions from X to Y = Number of onto functions from Y to X = Make 5 groups out of 7 elements of Y and permutate these groups to 5 elements of X = 7 [5 + 5 ] = 9 0. Let f: R R be a differentiable function with f(0) = 0. If y = f(x) satisfies the differential equation dy ( 5y)(5y ), dx Answer (0.40) dy Sol. ( 5y)(5y ) dx dy (5y )(5y ) then the value of lim f(x) is. x dx dy 4 5y 5y 5y ln x C 0 5y dx

34 JEE (ADVANCED)-08 (PAPER-) 5y f(0) 0 ln x 5y x 5y ln 5y 5y 0 y 5y 5 lim f(x) x. Let f: R R be a differentiable function with f(0) = and satisfying the equation f(x + y) = f(x)f(y) + f(x)f(y) for all x, y R. Then, the value of log e (f(4)) is. Answer (.00) Solution : f(0) =, f : R R f(x + y) = f(x) f(y) + f(x) f(y) x, y R Put x = y = 0 f(0) = f(0) f(0) f(0) = / Put y = 0, f(x) = f(x) f(0) + f(x) f(0) f(x) = f(x) + f(x) f (x) = f(x) ln(f(x)) = x +C f(0) = C = 0 ln(f(x)) = x ln(f(4)) = 4 =. Let P be a point in the first octant, whose image Q in the plane x + y = (that is, the line segment PQ is perpendicular to the plane x + y = and the mid-point of PQ lies in the plane x + y = ) lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length of PR is. Answer (8.00) Sol. P = (,, ), Q = (0, 0, K) as it lies on z-axis K mid point of PQ, i.e.,,, satisfies x + y = + = 6...() d.r. of PQ = ( 0, 0, K) = (p, p, 0) = & = K...() () & () = = P = (,, K) = (,, K) distance of P from x-axis = 5 + = 5 + K = 5 + K = 5 K = 4 Length of PR = K = 8 4

35 JEE (ADVANCED)-08 (PAPER-). Consider the cube in the first octant with sides P, Q and R of length, along the x-axis, y-axis and z-axis, respectively, where (0, 0, 0) is the origin. Let S,, be the centre of the cube and T be the vertex of the cube opposite to the origin such that S lies on the diagonal T. If p SP, q SQ, r SR and t ST, then the value of (pq) (rt) is. Answer (0.50) Sol. p SP ˆi ˆj k ˆ = (i ˆ ˆjk) ˆ q SQ ( ˆi + ˆjk) ˆ r SR ( ˆi ˆj + k) ˆ Y (0,,0) Q R(0, 0, ) Z S,, T (,,) P(,0,0) X t ST (i ˆ+ ˆj + k) ˆ ˆi ˆj k ˆ pq (i ˆ + ˆ j) 4 ˆi ˆj k ˆ ˆi ˆj k ˆ r t ( ˆ i + ˆ j) (p q) (r t) (i ˆ+ ˆj) ( ˆi + ˆj) kˆ kˆ 4 4 (p q) (rt) Let X ( C ) ( C ) ( C )... 0( C ), where 0 C r, r {,,..., 0} denote binomial coefficients. Then, the value of X is. 40 Answer (646.00) Sol. ( + x) 0 = 0 C C x + 0 C x C 0 x 0 differentiate both sides w.r.t x, 0( + x) 9 = 0 C + 0 C x + 0 C x C 0 x 9...() ( + x) 0 = 0 C 0 x C x C 0...() Coefficient of x 9 in 0( + x) 9 ( + x) 0 is same as ( 0 C ) + ( 0 C ) + ( 0 C ) ( 0 C 0 ) = Coeff. of x 9 in 0( + x) 9 = 0 9 C 9 = X X 40 = = 646 5

36 JEE (ADVANCED)-08 (PAPER-) This section contains FUR (04) questions. SECTIN (Maximum Marks : ) Each question has TW (0) matching lists: LIST-I and LIST-II. FUR options are given representing matching of elements from LIST-I and LIST-II. NLY NE 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 : + If NLY 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 : In all other cases. 5. Let x E x R :x and 0 x and x E xe :sin log e is a real number. x ere, the inverse trigonometric function sin x assumes value in,. Let f:e R be the function defined by x f(x) loge x and g:e R be the function defined by LIST-I x g(x) sin loge. x LIST-II P. The range of f is. e,, e e Q. The range of g contains. (0, ) R. The domain of f contains., S. The domain of g is 4. (, 0) (0, ) The correct option is: (A) P 4; Q ; R ; S (B) P ; Q ; R 6; S 5 (C) P 4; Q ; R ; S 6 (D) P 4; Q ; R 6; S 5 Answer (A) Sol.For domain of f(x); x 0 x0 or x x e, e e (, 0), e

37 JEE (ADVANCED)-08 (PAPER-) e D f :(,0) (, ),, e e For range of f(x); x e Let y loge x y x e Now x < 0 or x > y y e 0 y e e or y e y 0 < e y < or 0 y e < y < 0 or e y > 0 Range of f(x) is (,0) (0, ) For domain of g(x); x loge x and x 0 x x e e x and x < 0 or x > x(e ) 0 x and (e )x e 0 x and x < 0 or x > e x,, e e For range of g(x); x loge 0 x or x 0log x g(x) 0 or 0 g(x) Range of g(x) =,0 0, R, P 4, S, Q 6. In a high school, a committee has to be formed from a group of 6 boys M, M, M, M 4, M 5, M 6 and 5 girls G, G, G, G 4, G 5. (i) Let be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly boys and girls. (ii) Let be the total number of ways in which the committee can be formed such that the committee has at least members, and having an equal number of boys and girls. (iii)let be the total number of ways in which the committee can be formed such that the committee has 5 members, at least of them being girls. 7

38 JEE (ADVANCED)-08 (PAPER-) (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 girls and such that both M and G are NT in the committee together. LIST-I LIST-II P. The value of is. 6 Q. The value of is. 89 R. The value of is. 9 S. The value of 4 is The correct option is: (A) P 4; Q 6; R ; S (B) P ; Q 4; R ; S (C) P 4; Q 6; R 5; S (D) P 4; Q ; R ; S Answer (C) Solution : (i) = 6 C 6 C = 0 0 = 00 (ii) = 6 C 5 C + 6 C 5 C + 6 C 5 C + 6 C 4 5 C C 5 5 C 5 = = 46 (iii) = 5 C 6 C + 5 C 6 C + 5 C 4 6 C + 5 C 5 = = 8 (iv) 4 = ( 5 C 6 C 4 C 5 C ) + ( 5 C 6 C 4 C 5 C 0 ) = (50 0) + (60 6) + 5 = = 90 = x y Let : a b, 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. LIST-I LIST-II P. The length of the conjugate axis of is. 8 Q. The eccentricity of is. 4 R. The distance between the foci of is. S. The length of the latus rectum of is 4. 4 The correct option is: (A) P 4; Q ; R ; S (B) P 4; Q ; R ; S (C) P 4; Q ; R ; S (D) P ; Q 4; R ; S 8

39 JEE (ADVANCED)-08 (PAPER-) Answer (B) Sol. y L b 0 a N x M b tan0 a b a...(i) Now area of LN = ab ab ab 4...(ii) From (i) and (ii) we have a, b = Now b 4 a e Distance between foci = ae 8 The length of latus rectum = b 4 4 a 8. Let f : R R, f :, R, f : x (i) f (x) = sin e,,e R and f 4 : R R be functions defined by (ii) f (x ) = sinx if x 0 tan x if x 0, where the inverse trigonometric function tan x assumes values in, (iii) f (x) = [sin(log e (x + ))], where, for t R, [t ] denotes the greatest integer less than or equal to t, x sin if x 0 (iv) f 4 (x) = x 0 if x 0 9

40 40 JEE (ADVANCED)-08 (PAPER-) List-I List-II P. The function f is. NT continuous at x = 0 Q. The function f is. continuous at x = 0 and NT differentiable at x = 0 R. The function f is. differentiable at x = 0 and its derivative is NT continuous at x = 0 S. The function f 4 is 4. differentiable at x = 0 and its derivative is continuous at x = 0 The correct option is : (A) P ; Q ; R ; S 4 (B) P 4; Q ; R ; S (C) P 4; Q ; R ; S (D) P ; Q ; R 4; S Answer (D) Sol.(i) f : R R x f(x) sin e = sin x e f (x) is continuous everywhere, but f (0) does not exist, therefore f (x) is not differentiable function at x = 0. (ii) f :, R sinx ; x 0 tan x f (x) ; x0 sinx ; x 0 tan x LL at x = 0 is RL at x = 0 is f (x) is discontinuous at x = 0, (iii) f :,e R f(x) sin(log (x) e x e x e 0log e(x) 0sin(log e(x )) [sin(log e (x + )] = 0 f (x) = 0 f (x) is continuous and differentiable at x = 0

41 JEE (ADVANCED)-08 (PAPER-) (iv) f 4 : R R x sin ; x0 f 4(x) x 0 ; x 0 lim f (x) lim x sin 0 x 4 x0 x0 = Value of f 4 (x) at x = 0 f 4 (x) is continuous at x = 0 sin f(0h) f(0) h Now, f (0 ) lim lim 0 h0 h h0 h sin f(0h) f(0) h f (0 ) lim lim 0 h0 h h0 h f(x) is differentiable at x = 0 Now, f 4 (x) sin x. cos x x x = sin cos x x is oscillating Function which discontinuous at x = 0 END F TE QUESTIN PAPER 4