Solution to IIT JEE 2018 (Advanced) : Paper - II

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1 Solution to IIT JEE 08 (Advanced) : Paper - II PART I PHYSICS SECTIN (Maximum Marks:4) This section contains SIX (06) questions. Each question has FUR options for correct answer(s). NE R MRE THAN 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 : +3 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.. 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. (A), (B), (D) dk t dt K = mv dk dv mv t dt dt dv t v dt m vdv t dt m v t m 058/IITEQ8/Paper/QP&Soln/Pg.

2 () Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution v t (proportional to time) m a = dv dt m F ma m (constant). 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. (A), (C), (D) Viscous force F = A dv dh F u u 0 F A 0 h F h h F Shear stress = u0 A F A 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 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? (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 R / 0 (D) The electric field is normal to the surface of the shell at all points 3. (A), (B) Gauss law = E da Q where Q = L L = PQ = R sin 60 L = R 3 R /IITEQ8/Paper/QP&Soln/Pg. 0 Electric field lines are radially outwards, perpendicular to length of wire. Hence component of E.F. is zero along z-axis. P Q 0 Z E E

3 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (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.) (A) (B) (C) (D) 4. (D) If u = f v If u < f Image is virtual, erect and magnified. Let a point P at distance x from f u = (f x) V f x f x V f x f f (f x) f (f x) V x Magnification M = where h 0 = x hi hi v f h u x 0 f (independent of x) f P x 45 f 5. In a radioactive decay chain, 3 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) N = 5 (B) N = 6 (C) N = (D) N = 4 5. (A), (C) Z = 90 8 = 8 A = 3 = 0 90Th 3 8Pb Number of -particle N = Z = 5 = 0 Number of -particles N = 0 8 = 058/IITEQ8/Paper/QP&Soln/Pg.3

4 (4) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution 6. In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 Hz 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 83.9 cm. Which of the following statements is (are) true? (A) The speed of sound determined from this experiment is 33 m s (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 6. (A), (B), (C) For st resonance 3V 0 4 L e f.. () For nd resonance 5V 4 L e f 3V 5V 4 L e 4 L e 3 L e 5 L e 0.. () e = 3L 5L e = V 0 4 L e f V = m / s 300 V 33 = cm f SECTIN II (Maximum Marks:4) 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. 6.5, 7.00, -0.33, -.30, 30.7, -7.30) using the mouse and the onscreen 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 : +3 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 N s is applied to the block at time t = 0 so that it starts moving along the x-axis with a velocity v(t) = 058/IITEQ8/Paper/QP&Soln/Pg.4 t/ e v 0, where v 0 is a constant and = 4 s. The displacement of the block, in metres, at t = is. Take e = [6.3] t/ v v e 0 F = N.S.

5 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (5) ds t/ v v0e dt S t/ ds v0 e dt 0 0 t/ S = v 0 e 0 S = v0 e Impulse = P = mv v = s = m/ s 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 30 with the horizontal surface. The maximum height it reaches after the bounce, in metres, is. 8. [30 m] u sin 45 H = u g v sin 30 h = g v h 4 v H u u 4 9. A particle, of mass 0 3 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 0 sin t î, where E 0 =.0 N C and = 0 3 rad s. Consider the effect of only the electrical force on the particle. Then the maximum speed, in m s, attained by the particle at subsequent times is. 9. [] Velocity is maximum when acceleration is zero. F = 0 E = E 0 sin t = 0 sin t = 0 t = 0 or t v E = E 0 sin t F = qe = qe0 sin t qe qe0 a = sin t m m dv qe0 a = sin t dt m v t qe0 dv sin t m /IITEQ8/Paper/QP&Soln/Pg.5

6 (6) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution qe m cos t 0 0 v = cos t replacing t = qe0 v = m vmax t 0 qe m 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. 0. [5.56] Number of turns = 50 Area = 0 4 m 0.9 A S B = 0.0 T Torsional constant C = 0 4 G At full scale deflection = 0. rad.0 A 0. A Resistance R = 50 Torque = C = niab 4 C 0.0 Full scale current i = 4 nab i 0.amp 0.9 S = 0. R S = R S A steel wire of diameter 0.5 mm and Young's modulus 0 N m 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 m s and = 3... [3] Diameter d = 0.5 mm Y = 0 N/m Strain = Stress Y L F L AY 058/IITEQ8/Paper/QP&Soln/Pg.6

7 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (7) FL.0 L = AY L = mm L.C. of Vernier scale = 0. mm Number of division which coincide with main scale = 3.. 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 Jmol K, the decrease in its internal energy, in Joule, is.. [900 J] Adiabatic Expansion PV = constant T.V = constant V 5 T = T where V 3 /3 T T = T 5K 8 4 U = f nrt J 3. 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 = kg and.0 ev = J. 3. [4] P = 00 w KE max = h = 0 Energy of photon E = h = = 6.5 ev Number of photons incident per unit time P 00 9 n = E = 0 0 Change in momentum of photon P = mk me( v) Force F = n(p ) F = 0 0 = = = /IITEQ8/Paper/QP&Soln/Pg.7

8 (8) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution 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 = 3 to n = transition. The ionization energy of the hydrogen atom is 3.6 ev. The value of Z is. 4. [3] E = 3.6 z n n E = 3.6z 3 3.6z 4 4 E = 3.6z 5 3.6z E E = 3.6 z z z = z = 3 SECTIN III (Maximum Marks:) This section contains FUR (04) questions. 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 : +3 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. A small dipole with point charges Q at (0, 0,l) and d Q at (0, 0, l). Take l << d. R. 3. An infinite line charge coincident with the x-axis, E d with uniform linear charge density S. 4. Two infinite wires carrying uniform linear charge E 3 d 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 3, 4; R ; S (B) P 5; Q 3; R, 4; S (C) P 5; Q 3; R, ; S 4 (D) P 4; Q, 3; R ; S 5 058/IITEQ8/Paper/QP&Soln/Pg.8

9 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (9) 5. (B) (P) (5), (Q) (3), (R) (), (4); (S) () () E.F. due to a point charge at origin. kq E E d d () E.F. at any point on axis of dipole KP 4KQL E = 3 3 d d E 3 d (3) E.F. due to an infinite long charge E = K d E d (4) E.F. due to two infinite long wires E E E E K K k (L) 4KL d L d L d L d L 4K If d >> L E = L E d d (5) E.F. due to infinite plane charge E (independent of d) 0 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 P. v v Q. L L R. K K S. T T LIST-II (A) P 4; Q ; R ; S 3 (B) P 3; Q ; R 4; S (C) P ; Q 3; R ; S 4 (D) P ; Q 3; R 4; S 6. (B) (P) (3), (Q) (), (R) (4), (S) () m m 058/IITEQ8/Paper/QP&Soln/Pg.9

10 (0) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution R R 4 (P) For orbital speed mv GMm F = R R GM v = R V R 4 V R (Q) Angular momentum L = mvr L m V R L m V R 4 (R) Kinetic energy K = mv K m V K m V (S) Time period of revolution T = R V T R V T R V 4 8 () 8 M m m 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. LIST-I 058/IITEQ8/Paper/QP&Soln/Pg.0 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 3. 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 3; R ; S (B) P ; Q 3; R ; S 4 (C) P 3; Q 4; R ; S (D) P 3; Q 4; R ; S

11 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) () 7. (C) (P) (3), (Q) (4), (R) (), (S) () I adiabatic II isobaric III isochoric IV isothermal (P) Process I is adiabatic Hence No heat is exchanged between gas and surrounding. (Q) Process II is isobaric. w = PV = 3P 0 (3V 0 V 0 ) = 6P 0 V 0 (R) Process is isochoric w = 0 (S) Process is isothermal T = 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 P. r(t) = tˆt t ˆj. p Q. r(t) = cos t î + sin t ĵ. L R. r(t) = (cos t î + sin t ĵ ) 3. K S. r(t) = ˆ ti t ˆj 4. U 5. E LIST-II (A) P,, 3, 4, 5; Q, 5; R, 3, 4, 5; S 5 (B) P,, 3, 4, 5; Q 3, 5; R, 3, 4, 5; S, 5 (C) P, 3, 4; Q 5; R,, 4; S, 5 (D) P,, 3, 5; Q, 5; R, 3, 4, 5; S, 5 8. (A) (P) (), (), (3), (4), (5); (Q) (), (5); (R) (), (3), (4), (5); (S) (5) (P) r t ˆi t ˆj dr v ˆi ˆj (constant) dt a 0 F 0 Angular momentum L mr v ˆ ˆ L m t k t k 0(conserved) 058/IITEQ8/Paper/QP&Soln/Pg.

12 () Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution (Q) r a cos t ˆi sin t ˆj Particle is moving on an elliptical path. Angular momentum about origin and total energy remains conserved. r a cos t ˆ i sin t ˆj (R) Particle moves on a circular path radius = a angular velocity = speed v = a but v constant hence P constant L m a kˆ is constant Angular momentum Kinetic energy E = mv (constant) P.E. U = constant E = constant t (S) ˆ r t i ˆj dr v ˆi t ˆj (depends on t) dt hence P mv also depends on t. dv a ˆj dt F ma m ˆj (constant) t L mr v m t k ˆ ( k) ˆ mt ˆ L k (depends on t) nly total energy remains conserved. 058/IITEQ8/Paper/QP&Soln/Pg.

13 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (3) PART II-CHEMISTRY SECTIN (Maximum Marks: 4) This section contains SIX (06) questions. Each question has FUR options for correct answer(s). NE R MRE THAN 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 : +3 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)(NH 3 ) 3 (H )] 3+ (en = H NCH CH NH ) 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)(NH 3 ) 4 ] 3+. (A, B, D) Complex [Co(en) (NH 3 ) 3 (H )] +3 has two geometrical isomers. NH 3 NH 3 NH 3 H en Co en Co NH 3 NH 3 H NH 3 (Fac) (Mer) [Co(NH 3 ) 3 (H ) (CN) ] + has three geometrical isomers. 058/IITEQ8/Paper/QP&Soln/Pg.3

14 (4) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution NH 3 NH 3 NH 3 NC NH 3 NC NH 3 NC H Co Co Co CN H NH 3 H NC NH 3 CN NH 3 NH 3 (Fac) (Fac) (Mer) [Co(en) (NH 3 ) 3 (H )] +3 is diamagnetic Co = [Ar] 8 4s 3d 7 Co +3 = [Ar] 8 3d 6 en is strong field ligand. t g eg 0 0 [Co(en) (NH 3 ) 3 (H )] +3 [Co (en) (NH 3 ) 4 ] +4 0 = hc 0 [Co (en) (NH 3 ) 3 (H )] +3 absorbs light at longer wavelength as compared to [Co(en) (NH 3 ) 4 ] +3. 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 H S in acidic medium (C) nly Mn + shows the formation of precipitate by passing H S in faintly basic medium (D) Cu +/ Cu has higher reduction potential than Mn + /Mn (measured under similar conditions). (B), (D) Both Cu + and Mn + gives green colour in the flame test. K sp (CuS) < K sp (MnS) nly Cu + shows the formation of precipitate by passing H S in acidic medium. E 0 0 E Cu /Cu Mn /Cu Less electro positive metals have higher reduction potential. 3. Aniline reacts with mixed acid (conc.hn 3 and conc.h S 4 ) at 88K to give P(5 %), Q (47%) and R (%). The major product(s) of the following reaction sequence is (are) ) Ac, pyridine ) Sn/HCl ) Br, CH 3 C H )Br /H (excess) R S major product(s) 3) H 3 + 3) NaN, HCl/7378 K 4) NaN, HCl/73 78 K 4) H 3 P 5) EtH, 058/IITEQ8/Paper/QP&Soln/Pg.4

15 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (5) (A) (B) (C) (D) 3. (D) NH N NH C CH 3 NH AC NH N Br (R) Br Ac, pyridine NH Br /H excess Br Br Br NaN, HCl H3P Br N Br NH Sn/HCl Br Br CH3CH Br Br Br N EtH N H 3 NaN, HCl 7378k Br 4. The Fischer presentation of D-glucose is given below. D-glucose The correct structure(s) of β-l-glucopyranose is (are) (A) (B) (C) (D) 058/IITEQ8/Paper/QP&Soln/Pg.5

16 (6) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution 4. (D) CH CH H H H H H H H H H H H H H H H H CH H CH H D-Glucose L-Glucose H H H H CH H H H C H.. H H.. CH H H H H H C H H L-Glucose H H H H H CH H H H H H H H -L-Glucopyranose 5. For a first order reaction A(g) B(g) + C(g) at constant volume and 300 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 /3 is the time required for the partial pressure of A to reach /3 rd of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases) (A) (B) (C) (D) 058/IITEQ8/Paper/QP&Soln/Pg.6

17 Rate constant IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (7) 5. (A, D) A (g) B (g) + C (g) P o k ne t 3Po Pt Kt = n(p 0) n(3po P t ) n(3po P t ) n(p o) kt n(3p P ) o t Slope = k At t = t / n3 t/3 k t /3 is independent of [A] o t t /3 [A] o For I st order Rx n, rate constant is independent of initial concentration of reaction. [A] o 6. For a reaction, A P, the plots of [A] and [P] with time at temperatures T and T are given below. If T > T, the correct statement(s) is (are) (Assume ΔH Ɵ and ΔS Ɵ are independent of temperature and ratio of lnk at T to lnk at T is greater than T /T. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equillibrium constant, respectively.) (A) ΔH Ɵ < 0, ΔS Ɵ < 0 (B) ΔG Ɵ < 0, ΔH Ɵ > 0 (C) ΔG Ɵ < 0, ΔS Ɵ < 0 (D) ΔG Ɵ < 0, ΔS Ɵ > 0 058/IITEQ8/Paper/QP&Soln/Pg.7

18 (8) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution 6. (A), (C) From the graph () and (), we get H = ve i.e. exothermic. nk T Given : nk T S h R R T T S h T R R T S (T T ) 0 R S 0 SECTIN (Maximum Marks: 4) 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. 6.5, 7.00, 0.33,.30, 30.7, 7.30) using the mouse and the onscreen 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 : +3 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 3, N 5, P 4 6, P 4 7, H 4 P 5, H 5 P 3 0, H S 3, H S 5 7. [5.00 or 6.00] N 3: N N or N N Symmetrical Unsymmetrical N 5: N N P4 6: P7 7: P P P P P P P P 058/IITEQ8/Paper/QP&Soln/Pg.8

19 4 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (9) H P : H P P H H H HP 30: H P P H H HS 3 : H S H P H H S HS 5 : H S S H 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 = 3, Pb = 07) 8. [6.47] PbS + Pb + S n n [Equal moles] Pb wo w p w b o w pb M M 3 07 Let, pb w = kg o w P b = 07 3 = 6.47 kg 9. 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 + H KMn 4 + H S 4 + HCl (equation not balanced). Few drops of concentrated HCl 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 = 35.5) 9. [6.00] (N) 5 MnCl K S H KMn H S HCl (N) 5 (N) 4 4 KMn H C Mn C (N) 058/IITEQ8/Paper/QP&Soln/Pg.9

20 (0) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution No. of equivalents of MnCl = No. of eq. of KMn 4 = No. of eq. of H C 4 5 n MnCl = n HC4 w 5 6 = w = = 6 mg For the given compound X, the total number of optically active stereoisomers is. 0. [7.00] Total number of stereomers of compound X = 8 X has 7 optically active stereisomers X has 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 : H =, C =, N = 4, = 6, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis). [495.00] CH CNH NH NaBr H 3 + NH 3, D Br/KH 0 moles A (60%) 6 moles B (50%) 3 moles C (50%).5 moles Br (3 eq) CH 3 CH Br NH Br No. of moles of D =.5 (C 6 H 4 NBr 3 ) = = 495 gm Br D (00%).5 moles 058/IITEQ8/Paper/QP&Soln/Pg.0

21 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (). 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. However, the N gas contains mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: Cu(s) + H (g) Cu (s) + H (g) P is the minimum partial pressure of H (in bar) needed to prevent the oxidation at H 50 K. The value of ln (p ) is. H (Given: total pressure = bar, R (universal gas constant) = 8 JK mol, ln(0) =.3. Cu(s) and Cu (s) are mutually immiscible. At 50 K: Cu(s) + ½ (g) Cu (s); ΔG Ɵ = 78,000 J mol H (g) + ½ (g) H (g); ΔG Ɵ =,78,000 J mol ; G is the Gibbs energy). [4.60] Cu(s) H (g) Cu (s) (g) At 50 K Cu(s) (g) Cu (s) G J/mol = 78 kj/mol H (g) H (g) (g) G = J/mol = 78 kj/mol Given : Cu H Cu G 00 kj/mol (s) (g) (s) (g) G = G + RT nk (ph ) 0 = 00 + RT n (ph ) 8 ph 00 = 50 n 000 ph ph = % of total pressure ( bar) = 00 8 ph 00 = 50 n = nph n0 0 = nph n0 0 = nph (.3) 0 = nph 4.6 nph 4.60 bar 3. 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 ΔG Ɵ (in J mol ) for the reaction at 300 K is. (Given; ln() = 0.7, RT = 500 J mol at 300 K and G is the Gibbs energy) 3. [ ] A(g) B(g) AB(g) E b E a = RT J/mol ; RT = 500 J/mol ; A f = 4A b T = 300 K 058/IITEQ8/Paper/QP&Soln/Pg.

22 () Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution G = RT nk Kf A e K K Ae = b E a /RT f E b /RT b EbEa A F e RT A b = 4 e G = RT RT RT = 500 4e n 4e n 4e = 500 n4 n e = 500 ( n + ) (0.7) = 500 = = 8500 Absolute value of G = Consider an electrochemical cell: A(s) A n+ (aq, M) B n+ (aq, M) B(s). The value of ΔH Ɵ for the cell reaction is twice that of ΔG Ɵ at 300 K. If the emf of the cell is zero, the ΔS Ɵ (in J K mol ) of the cell reaction per mole of B formed at 300 K is. (Given: n () = 0.7, R(universal gas constant) = 8.3 J K mol. H, S and G are enthalpy, entropy and Gibbs energy, respectively.) 4. [.6] n n+ A s A aq, M B aq,m Bs xidation n A A ne. () Reduction B n + + ne B () multiply equation () by () and add both n n A B B Given E.M.F. of the cell is zero; the system is in equilibrium, G = 0 G = RT n k n [A ] G RT nk n [B ] 058/IITEQ8/Paper/QP&Soln/Pg. G RT n() G H S G G S (Given H = G) G = TS G S T RT n4 = = R n T = =.6 J k mol

23 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (3) SECTIN 3 (Maximum Marks: ) This section contains FUR (04) questions. 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 : +3 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 3. [Ti(H ) 3 Cl 3 ] R sp 3 d 3. [Cr(NH 3 ) 6 ] 3+ S d sp 3 4. [FeCl 4 ] The correct option is (A) P 5; Q 4, 6; R, 3; S (B) P 5, 6; Q 4; R 3; S, (C) P 6; Q 4, 5; R ; S, 3 (D) P 4, 6; Q 5, 6; R,; S 3 5. (C) P : dsp (6) [Ni (CN) 4] Ni 4S 3d 8 5. Ni(C) 4 6. [Ni(CN) 4 ] Ni + 3d 4s 4p CN is a strong ligand Pairing takes place. dsp hybridization square planar Q : sp 3 (4), (5) (4) [FeCl 4 ] Fe 4s 3d 6 Fe 3d 4s 4p Cl Weak ligand No pairing takes place. sp 3 hybridization Tetrahedral structure. 058/IITEQ8/Paper/QP&Soln/Pg.3

24 (4) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution (5) Ni(C) 4 Ni 4s 3d 8 3d 4s 3p C strong ligand pairing from electrons of s orbital. sp 3 hybridization Tetrahedral R ; () () [Fe F 6 ] 4 Fe 4s 3d 6 ; Fe + 3d 6 3d 4s 4p 4d F weak field ligand No pairing sp 3 d hybridization octahedral S; d sp 3, 3 () [Ti (H ) 3 Cl 3 ] Ti 4s 3d Ti 3+ 3d 3d 4s 4p Weak ligands H, Cl d sp 3 octahedral (3) Cr(NH ) 3 Cr : 4s 3d 5 ; Cr 3+ 4s 0 3d 3 3d 4s 4p d sp 3 octahedral soln. (C) 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) 058/IITEQ8/Paper/QP&Soln/Pg.4

25 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (5) LIST I LIST II P.. I, NaH Q.. [Ag(NH 3 ) ]H R. 3. Fehling solution S. 4. HCH, NaH 5. NaBr The correct option is (A) P ; Q,3; R,4; S,4 (B) P,5; Q 3,4; R 4,5; S 3 (C) P,5; Q 3,4; R 5; S,4 (D) P,5; Q,3; R,5; S,3 6. (D) (P) : Ph H Ph Iodo form reaction P (), (5) Ph (Q) : H N Ph Me HS4 I, NaH X H Me NaBr H X H Me Gives Tollens & Fehlings Test Q (), (3) HN N Ph Ph X H H Me () Ag (NH 3 ) H (3) Fehlings solution 058/IITEQ8/Paper/QP&Soln/Pg.5

26 (6) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution (R) Ph Ph H Me Ph + CH H S 3 4 Me Me H Ph (S) Me Ph Haloform reaction Ph CH 3 () [Ag(NH 3 ) ]H + AgN 3 () Fehlings X X () I, NaH () NaBr P, 5 ; Q, 3 ; R, 5 ; S, 3 7. LIST-I contains reactions and LIST-II contains major products. LIST I LIST II P.. Q.. R. 3. S Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option. (A) P,5; Q ; R 3; S 4 (B) P,4; Q ; R 4; S 3 (C) P,4; Q,; R 3,4; S 4 (D) P 4,5; Q 4; R 4; S 3,4 7. (P) (), (4) Na Elimination + Br + H (Q) Me Substitution + + H + H Br Br + MeH (R) Me () Br base NaMe R (4) 058/IITEQ8/Paper/QP&Soln/Pg.6 (4)

27 (S) IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (7) Substitution Na + MeBr -NaBr Me Soln. (B) P, 4 ; Q ; R 4 ; S 3 (3) 8. 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 << ; degree of hydrolysis of salt <<; [H + ] represents the concentration of H + ions) LIST I P. (0 ml of 0. M NaH + 0 ml of 0.M acetic acid) diluted to 60 ml Q. (0 ml of 0. M NaH + 0 ml of 0. M acetic acid) diluted to 80 ml R. (0 ml of 0. M HCl + 0 ml of 0. M ammonia solution) diluted to 80 ml S. 0 ml saturated solution of Ni(H) in equilibrium with excess solid Ni(H) is diluted to 0 ml (solid Ni(H) is still present after dilution). LIST II. the value of [H + ] does not change on dilution. the value of [H + ] changes to half of its initial value on dilution 3. the value of [H + ] changes to two times of its initial value on dilution 4. the value of [H + ] changes to / times of its initial value on dilution 5. the value of [H + ] 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 3; S (B) P 4; Q 3; R ; S 3 (C) P ; Q 4; R 5; S 3 (D) P ; Q 5; R 4; S 8. [D] P : CH3CH NaH CH3CNa H 0ml of 0.m 0ml of 0.m The value of H + does not change on dilution for buffer solution P(). Q : NaH CH3CH CH3CNa H 0mlof 0.M 0ml of 0.M C = 0.05 M C = 0.05 M on dilution to 80 ml M V = M V K H H 058/IITEQ8/Paper/QP&Soln/Pg.7

28 (8) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution K H KnC C K a K K H a C H C H C 0.05 H C 0.05 H 0.05 H 0.05 H H R : NH4H HCl NH4Cl H 0ml of0.m 0ml of0.m C = 0.05 M C = 0.05 M H KnC H C H C H C H H R (4) S : The value of H H C 0.05 H C 0.05 does not change on dilution. 058/IITEQ8/Paper/QP&Soln/Pg.8

29 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (9) PART- III MATHEMATICS SECTIN (Maximum Marks: 4) This section contains SIX (06) questions. Each question has FUR options for correct answer(s). NE R MRE THAN 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 : +3 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, ) as n f n(x) tan j (x j)(x j) for all x (0, ) (Here, the inverse trigonometric function tan x assumes values in, ) Then, which of the following statement(s) is (are) TRUE? 5 (A) j j (B) j tan f (0) 55 0 j f (0) sec (f (0)) 0 (C) For any fixed positive integer n, (D) For any fixed positive integer n,. (A, B, D) f (x) n n tan j (x j) (x j ) = n j j (x j) (x j ) lim tan(f n (x)) n x x tan (x j) tan (xj ) lim sec (f (x)) = tan (xn) tan (x) tan x(xn) n tan(f (x)) x nx n n n 058/IITEQ8/Paper/QP&Soln/Pg.9

30 (30) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution Also, limtan(f (x)) 0 n x limsec (f n (x)) x tan (f (0)) j j 5 5 j j j tan (f (0)) j 55 Also (B) option is correct. and (A), (D) are correct.. 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? 4 7 (A) The point (, 7) lies in E (B) The point, 5 5 does NT lie in E (C) The point, lies in E 3 (D) The point 0, does NT lie in E. (D) Let M (h, k). PMQ = 90 (prove yourself) k 5 k 7 M h h E = Locus of M is x + y y 39 = 0 P Q Equation of chord of circle whose midpoint (h, k) is (,7) (,5) T = S xh + yk (y + k) 39 = h + k k 39 Since this locus passes through (, ). E h + k h k + = 0 ptions (B) & (C) are incorrect. (A) is also incorrect as in that case one circle will be a point option (D) is correct. b 3. Let S be the set of all column matrices b b 3 such that b, b, b 3 and the system of equations (in real variables) x + y + 5z = b x 4y + 3z = b x y + z = b 3 has at least one solution. Then, which of the following system(s) (in real variables) has b (have) at least one solution for each b S? b 3 (A) x + y + 3z = b, 4y + 5z = b and x + y + 6z = b 3 (B) x + y + 3z = b, 5x + y + 6z = b and x y 3z = b 3 (C) x + y 5z = b, x 4y + 0z = b and x y + 5z = b 3 (D) x + y + 5z = b, x + 3z = b and x + 4y 5z = b 3 058/IITEQ8/Paper/QP&Soln/Pg.30

31 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (3) 3. (A),(C), (D) For atleast one solution, either 0 or = = = 3 = 0. 5 = = = b 5 b 4 3 b 3 Also, 3 = 0 b 5 3 = 0 b + 7b 3b 3 = 0 b 3 0 b + 7b 3b 3 = 0 b For option (A), = For option (B), = So no solution. For option (C), = , so unique solution , = 0, = Also, = = 3 = 0. So, infinitely many solution. 5 For option (D), = Hence (A), (C), (D) are correct. b 3 5 b 6 b 3 = 54 0, so unique solution. 3 = 3 (b + b + 3b 3 ) 0 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 semimajor 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 ( ) 4 (D) The area of the region bounded by the ellipse between the lines x and x = is x and x = is ( ) 6 058/IITEQ8/Paper/QP&Soln/Pg.3

32 (3) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution 4. (A, C) Eq. of tangent to y = 4x is y = mx + m () Eq. of tangent to x + y = is m y = mx () Comparing () and (), we get m = m = Q (, 0) Eq. of ellipse is x y = Eccentricity = = Length of latus rectum = = Q Area = = x dx x x sin (x) = ( ) 4 5 Let s, t, r be non-zero complex numbers and L be the set of solutions z = x + iy (x, y, i = ) of the equation sz tz r 0 and z x iy. 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 5. (A), (B), (C) & (D) Let, s = s is t = t it (s, s, t, t, r, r R) r = r + ir sz tz r 0 (s t )x (t s )y r i (s t )x s t y r 0 s t x t s y r 0 And s t x s t y r 0 s t t s i) If, then locus is two coincident lines. s t s t 058/IITEQ8/Paper/QP&Soln/Pg.3 s t t s ii) If, then liens are intersecting and hence locus is a point. s t s t If s t, L has exactly one point. And if s \ t, L has infinitely many elements as it s a straight line.

33 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (33) 6. Let f : (0, ) be a twice differentiable function such that f (x) sin t f (t)sin x lim sin x for all x (0, ) tx t x If f, then which of the following statement(s) is (are) TRUE? 6 (A) f x (B) f (x) x for all x (0, ) 6 (C) There exists (0, ) such that f() = 0 (D) f f 0 6. (B, C, D) f (x) sin(t) f (t) sin(x) lim sin (x) tx tx f (x) cos(t) f (t) sin(x) lim sin (x) tx f (x) cos xf (x)sin(x) sin (x) d f (x) dx sin x f(x) = x sin(x) ( C 0) For f(x) = x sin(x), option (B), (C), (D) are correct. SECTIN (Maximum Marks: 4) 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. 6.5, 7.00, 0.33, 30, 30.7, 7.30) 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: Full Marks : +3 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) is. 7. [.00] / ( 3 ) dx (x ) ( x) 0 = / /4 6/4 dx ( 3) dx Let x / t 0 x x ( x) x 058/IITEQ8/Paper/QP&Soln/Pg.33

34 (34) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution = 3 ( 3 ) dt dx t = ( 3) ( 3) dt ( x) 8. Let P be a matrix of order 3 3 such that all the entries in P are from the set {, 0, }. Then, the maximum possible value of the determinant of P is. 8. (4.00) a b c = a b c a b c = (a b c 3 + a 3 c b + a c b 3 ) (a b 3 c + a c 3 b + a 3 c b ) {a i, b i, c i } {, 0, } i =,, 3 Maximum value can be 6, but that s not possible. (For = 6, 3 entries should be and 6 entries should be in first bracket and 3 entries should be and 6 entries should be in second bracket. That s a contradiction!) 5. (Even if one element is 0) max = 4 which is possible. 0 Ex : 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! 9. (9.00) = 7 C 5 5! = 5! = 7 C C 3! + 7 C 3 5 4! = ! 0. Let f : be a differentiable function with f(0) = 0. If y = f(x) satisfies the differential equation dy ( 5y)(5y ) dx then the value of lim f (x) is. x 0. [0.40] dy ( 5y) (5y ) dx ( 5y) (5y ) dy dx 4 ( 5y) (5y ) 5y log e x c 54 5y 5y e 5y 5y 0x e 5y 0xc ( f (0) 0) 058/IITEQ8/Paper/QP&Soln/Pg.34

35 y x IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (35) 0x ( e ) 0x 5 ( e ) lim f (x) 5. Let f : 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 Then, the value of log e (f(4)) is.. (.00) f(0) = Put x = 0, y = 0 f(0) = f(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) f '(x) dx = f (x) dx log e (f(x)) = x + c x f(x) = e (C = 0 since f(0) = ) log e f(4) = log e (e ) =. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) 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.. (8.00) Let P (x 0, y 0, z 0 ) x x0 y y0 z z0 (x0 y0 3) = 0 x = 3 y 0 y = 3 x 0 Since, mirror images lies on zaxis 3 y 0 = 0, 3 x 0 = 0 x 0 = 3, y 0 = 3 Also, z 0 y0 = 5 z 0 = 4 PR = 4 (4) = 8 3. Consider the cube in the first octant with sides P, Q and R of length, along the x-axis, yaxis 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 058/IITEQ8/Paper/QP&Soln/Pg.35

36 (36) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution that S lies on the diagonal T. If p SP, q SQ, r SRandt ST, then the value of (p q) r t) is 3. (0.5) SP = p = î ˆ j kˆ SQ = q = î + ˆ j kˆ SR = r = î ˆ j kˆ ST = t = î + ˆ j kˆ ˆi ˆj kˆ p q = ˆ i ˆ j ˆi ˆj kˆ ˆi ˆj r t = kˆ (p q) (r t ) x z R T S,, y Q(0,,0) P(0,,0) Let X C C 3 C C0 where 0 C r, r {,,., 0} denote binomial coefficients. Then, the value of X is (646) 0 0 X = r Cr r r C0 r r Cr r0 X = X = X = 0 5. C 0 X /IITEQ8/Paper/QP&Soln/Pg.36

37 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (37) SECTIN 3 (Maximum Marks: ) This section contains FUR (04) questions. 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 : +3 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 E {x x : xand 0} x and x E {x E :sin loge is areal number} x. (Here, the inverse trigonometric function sin x assumes values in, ) Let f : E x be the function defined by f(x) = log e x and g : E be the function defined by g(x) = sin x loge x LIST I LIST II P The range of f is. e,, e e Q The range of g contains. (0, ) R The domain of f contains 3., S The domain of g is 4. (, 0) (0, ) 5. e, e 6. e (, 0), e The correct option is: (A) P 4; Q ; R ; S (B) P 3; Q 3; R 6; S 5 (C) P 4; Q ; R ; S 6 (D) P 4; Q 3; R 6; S 5 5. (A) P 4, Q, R, S E : x x 0 x (, 0) (, ) E : x x 0 x and log x 058/IITEQ8/Paper/QP&Soln/Pg.37

38 (38) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution x x (, 0) (, ) e x e x 0 x x e x e 0 (e ) x x(e ) e 0 0 e (x) x x, e (, ) x (, ) e, e Intersection x, e e, e e So, domain of g is x,, e e x Range of x is R+ {} Range of f is R {0} or (, 0) (0, ) Range of g is, {0} 6. In a high school, a committee has to be formed from a group of 6 boys M, M, M 3, M 4, M 5, M 6 and 5 girls G, G, G 3, 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 3 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 3 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. (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. 36 Q. The value of is. 89 R. The value of 3 is 3. 9 R. The value of 4 is The correct option is: (A) P 4; Q 6; R ; S (B) P ; Q 4; R ; S 3 (C) P 4; Q 6; R 5; S (D) P 4; Q ; R 3; S 6. (C) (i) 6 C 3 5 C = 00 (P 4) (ii) 6 C 5 C + 6 C 5 C + 6 C 3 5 C C 4 5 C C 5 5 C 5 = 46 (Q 6) (iii) 5 C 6 C C 3 6 C + 5 C 4 6 C + 5 C 5 6 C 0 = 38 (R 5) (iv)( 5 C 6 C 4 C 5 C ) + ( 5 C 3 6 C 4 C 5 C 0 ) + 5 C 4 6 C 0 = 89 (S ) 058/IITEQ8/Paper/QP&Soln/Pg.38

39 IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (39) x y 7. Let H :, where a > b > 0, be a hyperbola in the xy-plane whose conjugate a b axis LM subtends an angle of 60 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. 8 Q The eccentricity of H is. 4 3 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 ; R ; S 3 (B) P 4; Q 3; R ; S (C) P 4; Q ; R 3; S (D) P 3; Q 4; R ; S 7. (B) LMN = 4 3 y ab 4 3 ab = 4 3 L(0, b) a a a a 3 b tan30 a o b = 3 3 Length of conjugate axis b = 4. 4 e 3 3 SS = ae = (0, 0) 30 a N(a, 0) M(0, b) b 4 4 L.R. = a Let f : f :,, f 3:, e andf 4 : be functions defined by x (i) f (x) sin e sin x ifx 0 (ii) f (x) tan x, where the inverse trigonometric function of tan x if x 0 3 x 058/IITEQ8/Paper/QP&Soln/Pg.39

40 (40) Vidyalankar : IIT JEE 08 Advanced : Question Paper & Solution assume values in,, (iii) f 3 (x) = [sin(log e (x + ))], where, for t than or equal to t, x sin if x 0 (iv) f 4(x) x 0 if x 0, [t] denotes the greatest integer less 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 3 is 3. 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 3; R ; S 4 (B) P 4; Q ; R ; S 3 (C) P 4; Q ; R ; S 3 (D) P ; Q ; R 4; S 3 8. (D) (P) f (x) = x x e.cos e e x f (0) does not exists. sin x (Q) f (x) =, x 0 tan x, x 0 sin x, x 0 tan x sin x =, x 0 tan x, x 0 f (0 + ) = f (0 ) = Not continuous at x = 0 (R) f 3 (x) = sin log (x x e, f 3 : (, Given < x + < e / 0 < log e(x ) f 3 (x) = 0 Always continuous and differentiate. e ) R 058/IITEQ8/Paper/QP&Soln/Pg.40

41 (S) f 4(x) = IIT JEE 08 Advanced : Question Paper & Solution (Paper II) (4) x sin, x 0 x 0, x 0 bviously, f 4 (x) continuous and differentiable at x = 0. f 4 (x) = x sin x cos, x 0 x x x 0, x 0 = x sin cos, x 0 x x 0, x 0 h sin cos 0 f h h 4 (0 ) = lim h0 h = cos lim sin h h 0 h h = does not exists. f (x) is not differentiate at x = 0. 4 For continuity f 4 (x) f 4 (0 ) lim h sin cos h0 h h does not exists. f (0 ) not continuous at x = 0. 4 Either n(a) S S 3 S 5 S S 4 S S 4 S S 5 S 3 S S 4 S S 3 S 5 S S 4 S S 5 S 3 S S 5 S 3 S S 4 S S 5 S 4 S S 3 S 3 S 5 S S 4 S S 3 S 5 S S 4 S S 3 S S 4 S S 5 S 3 S S 5 S S 4 S 4 S S 3 S S 5 S 4 S S 5 S S 3 S 5 S S 3 S S 4 S 5 S S 4 S S 3 END F THE QUESTIN PAPER 058/IITEQ8/Paper/QP&Soln/Pg.4