# IIT JEE-Advanced 2013 Solved papers by Triumph Academy

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IIT JEE-Advanced 2013 Solved papers by Triumph Academy

IIT JEE ADVANCED – 2013 / Paper-1 JEE ADVANCED -2013 [1] Paper - I PHYSICS Part I: Physics Section – I (Single Correct Answer Type) This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is (A) 5.112 cm (B) 5.124 cm (C) 5.136 cm (D) 5.148 cm Ans. [B] Value of one Vernier division Value of main scale division V 2.45 cm = 0.049 cm 50 , S 0.05 cm Least count, V C S V 0.00 1 cm b g . 24 0.001 5124 . cm Hence, diameter 510 2. A ray of light travelling in the direction along the direction (A) 30o d d i 1 i 3 j is incident on a plane mirror. After reflection, it travels 2 i 1 i 3 j . The angle of incidence is 2 (B) 45o (C) 60o Ans. [A] Angle between the incident and reflected ray will be FG 1 IJ FG 1 IJ FG 3 IJ FG 3 IJ H 2K H 2K H 2 K H 2 K 1 cos 2 b1gb1g 60 0 Hence, angle made by incident ray with normal, 30 0 (D) 75o [2] IIT JEE ADVANCED – 2013 / Paper-1 PHYSICS In the Young's double slit experiment using a monochromatic light of wavelength , the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is 3. g 2 b b (A) 2n 1 Ans. [B] Intensity at a point, I I m cos2 So, for I b b FG IJ H K FG IJ H 2K Im 1 or , cos2 2 2 2 g 8 (C) 2n 1 b b g 16 (D) 2n 1 g 2 2 L 2n 1 g 4 L 2n 1 or 4. g 4 (B) 2n 1 Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is (A) 1 : 4 (B) 1 : 2 (C) 6 : 9 (D) 8 : 9 Ans. [D] From pM 0 RT , p (as T = constant) M0 p1 1 M 02 8 p2 2 M 01 9 5. Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity k and the other 2k . The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuraiton II is (A) 2.0 s Ans. [A] RI (B) 3.0 L L 3L kA 2 kA 2 kA 1 kA 2 kA 3kA L RII RII L L L 3kA Q Ht so T tt R R t1 RI 9 t2 2s t2 RII 2 (C) 4.5 s (D) 6.0 s [3] IIT JEE ADVANCED – 2013 / Paper-1 PHYSICS 6. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 × 108 m s–1. The final momentum of the object is (A) 0.3 1017 kg ms1 (B) 10 . 10 17 kg ms1 (C) 3.0 1017 kg ms1 (D) 9.0 1017 kg ms1 Ans. [B] The total energy absorbed, E Pt 30 10 3 100 10 9 3 10 9 J Hence momentum delivered i.e. the final momentum of object, p 7. E 3 10 9 10 . 10 17 kg m s 1 c 3 108 A particle of mass m is projected from the ground with an initial speed u0 at an angle with the horizontal. At the highest point of its trajectory, it makes a completelyinelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u0 . The angle that the composite system makes with the horizontal immediately after the collision is (A) (B) (C) (D) 4 4 4 2 Ans. [A] At the highest point, v2 v1 u cosi L eu 2 ghj j MMu N 2 2 Fu 2 gG H 2 sin 2 2g I OP JK PQ 1 2 j u cos j Applying conservation of linear momentum, v v v v 1 2 so tan 2 1 450 2 v1 8. OP y h c x y h PQ stant of appropriate dimensions), when the particle is taken from the point ba, 0g to the point b0, a g along The work done on a particle of mass m by a force, K LM MN c x x 2 2 i 3/ 2 y 2 2 3/ 2 j (K being a con- a circular path of radius a about the origin in the x y plane is (A) 2K a (B) Ans. [D] Here F K LM xi yj MM e x y j N 2 2 3/ 2 K a OP Kr PP r Q 3 (C) K 2a (D) 0 i.e., at all points F is acting radially outwards from origin while particle is moving tengentially. Thus F v and work done is zero. [4] IIT JEE ADVANCED – 2013 / Paper-1 PHYSICS 9. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by appyling forces at two ends, the ratio of the elongation in the thin wire to the that in the thick wire is (A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00 Ans. [C] Both rod are joined end to end, hence tension is same in both elongation, F , so AY 1 1 A2 1 0.50 2 2 A1 2 10. 2 2 1 The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one third the size of the object. The wavelength of light inside the lens is 2 times the wavelength in free 3 space. The radius of the curved surface of the lens is (A) 1 m (B) 2 m (C) 3 m Ans. [C] Here, v 8, n 1 3 and m . So u 24 cm, applying lens equation 3 0 2 b g FGH R1 R1 IJK 1 1 1 n 1 8 24 f b g 1 F 1I F 1 I G J G J R 3m 6 H 2K H RK 1 2 (D) 6 m [5] IIT JEE ADVANCED – 2013 / Paper-1 PHYSICS Section–II Multiple Correct Answer Type This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE or MORE THAN ONE is / are correct. 11. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, b g b g c c h h . , the correct statement(s) is y x , t 0.01 m sin 62.8 m 1 x cos 628 s1 t . Assuming 314 (are) (A) The number of nodes is 5. (B) The length of the string is 0.25 m. (C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m. (D) The fundamental frequency is 100 Hz. Ans. [B, C] For fifth harmonic number of nodes are 6. f 5v 2 1 L 0.25 m 2 L 628 4 midpoint of the string is antinode. So maximum displacement is 0.01 m f0 f 20 Hz. 5 x=0 x=L x=L/2 A solid sphere of radius R and density is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3. The complete arrangement is placed in a liquid of density2 and is allowed to reach equilibrium. The correct statement(s) is (are) 12. (A) the net elongation of the spring is 4R 3 g . 3k (C) the light sphere is partially submerged. (B) the net elongation of the spring is 8R 3 g 3k (D) the light sphere is completely submerged. Ans. [A, D] The average density of complete system is equal to that of liquid so both the spheres will be completely submerged. For equilibrium of system shown in figure, b g b g Fsp 2 Vg 3 Vg i.e. Fsp Vg and hence extension x Vg 4R 3 g k 3k [6] IIT JEE ADVANCED – 2013 / Paper-1 PHYSICS A particle of mass M and positive charge Q, moving with a constant velocity u1 4i ms1 , enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side 1 after 10 milliseconds with a velocity u2 2 3i j ms . The correct statement(s) is (are) 13. d i (A) The direction of the magnetic field is -z direction. (B) The direction of the magnetic field is +z direction. (C) The magnitude of the magnetic field 50M units. 3Q (D) The magnitude of the magnetic field is 100M units. 3Q Ans. [A, C] From the given situation it is apparent that the magnetic field is in - z direction. v2 T Particle deflects by 30° so duration is . 12 FG 2M IJ 1 10 10 H qB K 12 B or o 30 3 y 50M units. 3Q x v1 Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 and 2 respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, 14. 1 along the line joining the centres of the spheres, is zero. The ratio can be 2 (B) (A) – 4 32 25 (C) 32 25 (D) 4 Ans. [B, D] The situation described can correspond to two situations as shown in figure. For first, for E p1 0, field of both must be equal and opposite i.e. both must have same polarity of P1 charge i.e. LM N 1 4 R 3 3 b g 4 0 2 R OP Q R 2 3 0 2 2R 1 4 2 Similarly in second case, for fields to cancel, polarity of charge must be opposite on them. LM N 1 4 R 3 3 b g 4 0 2 R 1 32 2 25 OP LM QN 2 2 b g OPQ b5Rg 4 2R 3 4 0 2 3 P2 2R 5R R [7] IIT JEE ADVANCED – 2013 / Paper-1 PHYSICS 15. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to charge the capacitor C2. After some time, S2 is released and then S3 is pressed. After some time, S1 S2 C1 S3 C2 2V0 V0 (A) the charge on the upper plate of C1 is 2CV0. (B) the charge on the upper plate of C1 is CV0. (C) the charge on the upper plate of C2 is 0. (D) the charge on the upper plate of C2 is – CV0. Ans. [B, D] On pressing S1, C1 gets fully charged with potential 2V0. Now on releasing S1 and pressing S2, the charge of C1 gets redistributed to get same potential on C2 and C1 given by V b g V C1 2V0 C1 C2 0 Thus charge on upper plates of C1 and C2 = + CV0 and on lower is – CV0. Now on releasing S2 and pressing S3, C1 becomes isolated and charge on its plates gets unchanged while due to battery C2 now gets charged by potential V0 in a manner that its upper plate now has – CV0 charge [8] IIT JEE ADVANCED – 2013 / Paper-1 PHYSICS Section – III: Integer Answer Type This section contains 5 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. (both enclusive) 16. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is ______ Ans. [1] hv eVo or Vo h v e e slope of graph for both metals is same FG h IJ H eK Ratio of slopes is 1. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is______ 17. Ans. [4] T1/ 2 1386 t 80 s T1/ 2 , so the activity remains nearly constant for the given duration. Percentage of nuclei decayed FG H IJ K N At ln 2 100 t 100 4 N0 A/ T 18. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in m s–1) of the particle is zero, the speed (in m s–1) after 5 s is_______ 1 2 Ans. [5] Pt K mv 2 or or 19. 0.5 5 b g 1 0.2 v 2 2 v = 5 m / s. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s–1 [9] IIT JEE ADVANCED – 2013 / Paper-1 PHYSICS about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s–1) of the system is_________ Ans. [8] Applying angular momentum conservation o M m F MR I F MR GH 2 JK GH 2 2 0 2 e I jJK 2 2mr 2 ' or = 8 rad / s. 20. A bob of mass m, suspended by a string of length l1 is given minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l2 , which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical l1 plane, the ratio l is________. 2 Ans. [5] At the highest point, speed of bob 1 is v1 g1 . For 1 D, elastic collision between objects of equal masses, velocities get interchanged. v2 gl1 5gl2 (for complete circle), or l1 5. l2 [10] IIT JEE – 2013 / Paper-I CHEMISTRY Part II Section–I : Straight Correct Answer Type This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 21. The compound that does NOT liberate CO 2 , on treatment with aqueous sodium bicarbonate solution, is (A) Benzoic acid (C) Salicylic acid Ans: (D) 22. (B) Benzenesulphonic acid (D) Carbolic acid (Phenol) Concentrated nitric acid upon long standing, turns yellow-brown due to the formation of (A) NO (B) NO2 (C) N 2O (D) N 2 O4 Ans: (D) 23. Methylene blue, form its aqueous solution, is adsorbed on activated charocal at 25C . For this process, the correct statement is (A) The adsorption requires activation at 25C (B) The adsorption is accompanied by a decrease in enthalpy. (C) The adsorption increases with increase of temperature. (D) The adsorption is irreversible. Sol: (B) 24. Sulfide ores are common for the metals (A) Ag , Cu and Pb (B) Ag, Cu and Sn Sol: (A) 25. (C) Ag, Mg and Pb (D) Al, Cu and Pb The arrangement of X ions around A ion in solid AX is given in the figure (not drawn to scale). If the radius of X is 250 pm, the radius of A is (A) 104 pm Sol: (A) r e (B) 125 pm j 2 1 r 104 pm (C) 183 pm (D) 57 pm [11] IIT JEE – 2013 / Paper-I CHEMISTRY 26. Upon treatment with ammoniacal H 2S , the metal ion that precipitates as a sulfide is b g b g (A) Fe III bg (B) Al III (C) Mg II bg (D) Zn II Sol: (D) 27. The standard enthalpies of formation of CO 2 (g) H 2O ( ) and glucose (s) at 25C are 400 kJ / mol,300 kJ / mol and 1300 kJ / mol , respectively. The standard enthalpy of combustion per gram of glucose at 25C is (A) 2900 kJ Sol: (C) (B) 2900 kJ (C) 1611 . kJ (D) 16.11 kJ C6 H12 O 6 (s) 6O 2 (g) 6CO 2 (g) 6H 2 O ( ) b g b g b g H oC 6 400 6 300 1300 2900 kJ / mol 2900 kJ / g 180 1611 . kJ / g 28. Consider the following complex ions, P, Q and R. P FeF6 3 b g , Q V H 2O 6 2 b g and R Fe H 2O 2 6 The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is (A) R Q P Sol: (B) (B) Q R P (C) R P Q (D) Q P R [12] IIT JEE – 2013 / Paper-I CHEMISTRY 29. In the reaction P Q R S the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is (A) 2 Sol: (D) 30. (B) 3 (C) 0 (D) 1 is independent of intial concentration of P. First order with respect to P and zero order with respect to Q. (From given figure) t 1/ 2 KI in acetone, undergoes SN 2 reaction with each of P, Q R and S. The rates of the reaction vary as H 3C Cl P Cl Q Cl R (A) P Q R S (B) S P R Q (C) P R Q S (D) R P S Q Sol: (B) S [13] IIT JEE – 2013 / Paper-I CHEMISTRY Section–II Multiple Correct Answer(s) Type This section contains 5 multiple choice questions. Each question four 4 choices (A), (B), (C) and (D), out of which ONE or MORE are correct. 31. The pairs (s) of coordination complexes/ions exhibiting the same kind of isomerism is (are) b g Cob NH g Cl b g Ptb NH g b H OgCl (A) Cr NH 3 5 Cl Cl 2 and Cr NH 3 4 Cl 2 Cl (B) 3 4 (C) CoBr2Cl 2 2 2 and 3 2 and PtBr2Cl 2 2 2 b g bNO g Cl and PtbNH g Cl Br (D) Pt NH 3 3 3 3 3 Sol: (B, D) 32. Among P, Q, R and S, the aromatic compound(s) is/are AlCl 3 P NaH Q b NH g CO (A) P Sol: (A, B, C, D) 4 2 3 100 115C R HCl S (B) Q (C) R (D) S [14] IIT JEE – 2013 / Paper-I CHEMISTRY 33. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to (A) p (empty) and * electron delocalisations. (B) * and electron delocalisations (C) p (filled) and electron delocalisations. (D) p (filled) * and * electron delocalisations Sol: (A) 34. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are) (A) G is positive (B) S system is positive (C) Ssurroundings 0 (D) H 0 Sol: (B, C, D) 35. b g b g b g The initial rate of hydrolysis of methyl acetate 1M by a weak acid HA , 1M is HA , 1M of that of b g a strong acid HX, 1M , at 250C. The Ka of HA is (B) 1 10 5 (A) 1 104 Sol: (A) H H HA (C) 1 106 1 100 HX H Ka HA 10 . 10 2 M H A HA e1.0 10 je1.0 10 j 1 10 2 ~ 2 1 4 (D) 1 10 3 [15] IIT JEE – 2013 / Paper-I CHEMISTRY Section–III Integer Answer Type This section contains 5 questions. The answer to each is a single-digit integer, ranging from 0 to 9 (both inclusive). 36. The total number of carboxylic acid groups in the product P is . H 3O , 21 .O 3 3. H 2O2 Sol: (2) 37. A tetrapeptide has COOH groups on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), On complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) wih NH 2 group attached to a chiral center is Sol: (4) 38. Phe – Gly – Val – Ala Phe – Val – Gly– Ala Val – Phe – Gly– Ala Val – Gly – Phe – Ala EDTA 4 is ethylenediaminetetraacetate ion. The total number of N Co O bond angles in b Co EDTA g 1 complex ion is Sol: (8) 39. The total number of lone pairs of electrons in melamine is Sol: (6) 40. The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the Broglie wavelenth of He gas at 73 C is "M" times that of the de Broglie wavelength of Ne at 7270C M is Sol: (5) M MV He Ne 3 KT M M 3 K MT M Ne TNe M He THe 20 1000 5 4 200 [16] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS PART III MATHEMATICS Section – 1 (Only One option correct Type) This section contains 10 multiple choice questions Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 41. Let complex number and b 1 lie on circles x x 0 g by y g 2 b 2 r 2 and x x 0 0 g by y g 2 0 2 4r 2 , 2 respectively. If z 0 x 0 iy 0 satisfies the equation 2 z 0 r 2 2 , then (A) Sol. 1 2 1 2 1 7 (C) 1 3 (D) [C] z z0 2 r2 z0 2 z z0 2 2r 2 1 z0 2 also 1 z 0 2 z 0 cos 4 r 2 . . . (ii) (i) – (ii) 1 z 0 e F1 r 2 I e 1j r e1 4 j GH 2 JK 2 2 1 z0 42. (B) 2 je z 0 2 z 0 cos r 2 . . . (i) 4r 2 2 2 j e 1 r2 1 4 2 j 1 2 2 2 2 1 1 4 1 2 8 2 e 2 2 r2 j 1 2 2 z0 2 z 0 e 2 7 2 1 2 j 2 2 1 cos 4 r 2 e1 j r e1 4 j 2 2 2 2 1 7 Four persons independenly solve a certain problem correctly with probabilities 1 3 1 1 , , , . Then the 2 4 4 8 probability that the problem is solved correctly by at least one of them is (A) Sol. [A] 235 256 b (B) g 21 256 b gb gb gb g P A B C D 1 P A ' P B' P C' P D' 1 1 1 3 7 21 235 1 2 4 4 8 256 256 (C) 3 256 2 (D) 253 256 [17] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS 43. LM 1 , 1OP R (the set of all real numbers) be a positive, non-constant and differentiable function N2 Q F 1I such that f ' b xg 2 f b xg and f G J 1 . Then the value of z f b xgdx lies in the interval H 2K Let f : 1 1 2 b (A) 2e 1, 2e Sol. [D] also g b bg bg d ee f bxgj 0 dx f ' x 2f x e bg f x e 2 (C) FG e 1 , e 1IJ H2 K FG 1 IJ 1 H 2 K f FG IJ 2 x bg bg 1 e 0 f x e 2 x 1 bg f x e 2 x 1 z zbg z 1 bg f x 0 1 0 z 1 2 44. Le f b xg dx M MN 2 1 2 zbg 1 0 f x dx 1 2 1 2 b 1 2 e 1 2 g The number of points in , , for which x 2 x sin x cos x 0, is (A) 6 Sol. OP PQ 2 x 1 1 1 0 dx f x dx e 2 x 1dx 1 2 1 2 x 2 x 2 x e f x H 2K FG H (D) 0, bg bg e f ' b x g 2e f b x g 0 e f b xg is strictly decreasing f ' x 2f x 0 2 x 2 x g (B) e 1, 2e 1 [C] (B) 4 (C) 2 bg f ' b xg 2x b x cos x sin xg sin x xb2 cos xg Let f x x 2 x sin x cos x f’ f bg f 0 1 bg also f is continuous and lim I x x bg f x 0 for exactly two values of x. + – 0 (D) 0 IJ K e 1 2 [18] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS 45. d (A) 4 Sol. LM OP is N 2Q 2 d 2 1i The area enclosed by the curves y sin x cos x and y cos x sin x over the interval 0, [B] i 2 1 (B) 2 2 2 A z cb d g i d 2 1 (C) 2 i 2 1 (D) 2 h cos x sin x cos x sin x dx 0 4 z cb g 4 h cos x sin x cos x sin x dx 2 0 g b gh cos x sin x cos x sin x dx 0 4 z b 4 sin x dx 4 cos x g 4 0 FG H 4 1 0 46. z cb IJ 2 2 e 2K 1 j 2 1 FG IJ . Let the slope of the curve at each point (x, H 6K A curve passes through the point 1, FG IJ H K 1 F yI (A) sinG J log x H xK 2 F 2y I (C) secG J log x 2 H xK y y sec , x 0 . Then the equation of the curve is x x Sol. [A] dy y y sec dx x x x dt sec t dx cos t dt FG y IJ log x 2 H xK 1 F 2y I (D) cosG J log x H xK 2 (B) cosec by tx g dx x sin t ln x k x 1, y 6 FG y IJ ln y k H xK sin k 1 2 FG y IJ ln x 1 H xK 2 sin y) be [19] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS F cot FG1 2kIJ I is GH H K JK 23 47. The value of cot n 1 n 1 (A) 23 25 (B) [B] 25 23 (C) (D) 24 23 n 1 cot n 1 23 k 1 23 1 n 1 e 1 1 n 1 j cot tan 1 24 tan 1 1 cot tan 1 48. 23 24 F F II GH cot GH1 2kJK JK F I F I cot G cot c1 nb n 1ghJ cot G e tan b n 1g tan njJ H K H K 23 Sol. k 1 23 25 25 23 For a b c 0, the distance between (1, 1) and the point of intersection of the lines ax by c 0 and bx ay c 0 is less than 2 2. Then (A) a b c 0 Sol. [A] 49. Intersection of given lines FG c , c IJ H a b a bK c 1 ab (C) a b c 0 (B) a b c 0 FG H 2 1 (D) a b c 0 IJ K c 2 2 ab abc Perpendiculars are drawn from points on the line abc 0 x 2 y 1 z to the plane x y z 3. the feet of 2 1 3 perpendiculars lie on the line (A) Sol. [D] x y 1 z 2 5 8 13 (B) x y 1 z 2 2 3 5 (C) x y 1 z 2 4 3 7 i.e. FG1, 5 , 9 IJ H 2 2K Intersection of line and plane x 2 y 1 z 2 1 3 3 2 1 3 2 1 3 2 Foot of perpendicular of point (–2, –1, 0) x 2 y 1 2 2 1 3 1 1 1 11 1 Equation of projection line is given by x0 y 1 z2 1 0 5 1 9 2 2 2 x y 1 z 2 2 7 5 i.e. (0, 1, 2) (D) x y 1 z 2 2 7 5 [20] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS 50. Let PQ 3i j 2 k and SQ i 3j 4 k determine diagonals of a parallelogram PQRS and PT i 2 j 3k be another vector. Then the volume of the parallepiped determined by the vectors PT, PQ and PS is (A) 5 Sol. (B) 20 (C) 10 [C] PR PQ PS 3i j 2 k , SQ PQ PS i 3j 4 k 3i j 2 k j e i 3j 4 k j e PQ 2 i j 3k , 2 volume of parallelopiped 1 L O M PT PQ PSP 2 N Q 1 2 (D) 30 PS i 2 j k 3 1 3 10 2 1 SECTION – 2 : (One or more options correct Type) The section contains 5 multiple choice questions. Each questions has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct 4n 51. b g Let S n 1 k 1 b g k k 1 2 k 2 . Then S n can take value(s) (A) 1056 (B) 1088 4n Sol. [A, D] S n b1g b g k k 1 2 (C) 1120 k 2 12 2 2 32 4 2 52 62 ... k 1 n 1 r 0 n 1 n 1 n 1 n 1 b4r 1g b4r 2g b4r 3g b4r 4g 2 r 0 b32r 20g 32 r 0 S8 1056 2 r 0 2 r 0 bn 1gn 20bng 4nb4n 1g 2 S 9 1332 2 (D) 1332 [21] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS 52. For 3 × 3 matrices M and N, which of the following statement(s) is(are) NOT correct? N T M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric (A) (B) M N – N M is skew symmetric for all symmetric matrices M and N (C) M N is symmetric for all symmetric matrices M and N (D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N Sol. e [C, D](A) N T M N b g (C) M N T j T b g N M M (D) badjMgbadjN g adj b N Mg N T MT N (B) M N N M NT MT N M bg T T T T NT N M M N bg 53. Let f x x sin x, x 0 . Then for all natural numbers n, f ' x vanishes at Sol. FG 1 IJ H 2K (C) a unique point in the interval b n, n 1g [B, C] f ' bA g sin x x cos x 0 (A) a unique point in the interval n, n FG H IJ K 1 (B) a unique point in the interval n , n 1 2 b g (D) two points in the interval n, n 1 tan x x 0 FG H x n, n FG H IJ K 1 2 , tan x x 0 IJ K 1 x n , n 1 , tan x x 0 (once) 2 54. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are (A) 24 (B) 32 (C) 45 (D) 60 Sol. [A, C] 55. A line l passing through the origin is perpendicular to the lines b g b g b g : b3 2sgi b3 2sgj b2 sgk , s 1 : 3 t i 1 2 t j 4 2 t k , t 2 Then, the coordinate(s) of the point(s) on l 2 at a distance of 17 from the point of intersection of l and l1 is(are) (A) FG 7 , 7 , 5IJ H 3 3 3K b g (B) 1, 1, 0 (C) (1, 1, 1) (D) FG 7 , 7 , 8 IJ H 9 9 9K [22] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS Sol. [B, D] d. r of line i j k 1 2 2 2 i 3j 2 k 2 2 1 Equation of line x y z 2 3 2 The point of intersection of and 1 is (2, –3, 2). b g Any point of 2 is 3 2s, 3 2s, 2 s b3 2s 2g b3 2s 3g b2 3 2g 17 b9s 10gbs 2g 0 9s 28s 20 0 2 2 2 s 2 10 , 2 9 SECTION – 3 : (Integer value correct Tyep) Integer Answer Type This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive) b g 56. The coefficients of three consecutive terms of 1 x Sol. [6] n 5 Cr : br 1g bn r 5g n 5 n 5 n 5 C r 1 C r 2 b r 2g C r 1 : n 5 C r 2 5 : 10 : 14 1 2r 2 n r 5 2 5 are in the ratio 5 : 10 : 14. Then n = bn 5g! br 1g!bn 4 r g! 1 r !b n 5 r g! 2 bn 5g! 3r n 3 bn 5g! br 2g!bn 3 r g! 5 7 b gb g bn 5g! 10 14 b n r 4g 7 5n 4b n 3g 6 0 n 5 r 1 ! n r 4 ! 7r 14 5n 5r 20 5n 12 r 6 0 5n 4 n 12 6 0 n6 [23] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS 57. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k 20 Sol. [5] b g 1224 k bk 1g n n 1 2 By observation n 50 58. 50 51 1224 k k 1 2 k 25 k 20 5 Of the three independent events E1 , E 2 and E 3 , the probability that only E 1 occurs is , only occurs in and only E 3 occurs is . Let the probability p that none of events E 1 , E 2 or E 3 occurs satisfy the b g b g equations 2 p and 3 p 2 . All the given probabilities are assumed to lie in the interval (0, 1). Pr obability of occurrence of E1 Pr obability of occurrence of E 3 Then Sol. [6] Let probability of E1 , E 2 and E 3 is X1 , X 2 and X 3 respectively then X b1 X gb1 X g b gb g X b1 X gb1 X g and P b1 X gb1 X gb1 X g Now b 2gP cX b1 X gb1 X g 2X b1 X gb1 X ghb1 X gb1 X gb1 X g X b1 X gb1 X gX b1 X gb1 X g X1 1 X 2 1 X 3 3 1 1 l.ly 2 3 1 3 1 3 2 2 1 3 1 1 2 2 3 3 X1 2 X 2 . . . (i) b 3 gP 2r cX b1 X gb1 X g 3bX gb1 X gb1 X ghb1 X gb1 X gb1 X g 2 X b1 X gb1 X gX b1 X gb1 X g 2 1 2 2 2 1 2 3 1 3 3 X 2 3X 3 . . . (ii) using (i) and (ii) X1 6 X 3 X1 6 X3 3 1 1 2 2 1 2 3 3 [24] IIT JEE ADVANCED – 2013 / Paper-I MATHEMATICS 59. A vertical line passing through the point (h, 0) intersects the ellipse x2 y2 1 at the points P and Q. Let 4 3 bg bg h the tangents to the ellipse at P and Q meet at the point R. If h area of the triangle PQR, 1 max 1 2 bg 8 1 8 2 5 and 2 1min h , then 2 Sol. h 1 h 1 [9] (h, k) (4/h, 0) O 2 ½ 1 (h, –k) bg Max IJ K 1 4 2k h 2 h 1 FG 1 IJ , H 2K 8 5 60. b gFGH h 2 1 8 2 8 3 5 3 4 h2 2 F 4 h I 3 e4 h j GH h JK 2h 3 2 2 2 bg min 1 8 5 F 1I 3G 4 J H 4K 3 2 8 bg 3 3 2 3 2 15 15 3 8 3 3 45 36 9 8 2 o l qt Consider the set of eight vectors V ai bj ck : a , b, c 1, 1 . Three noncoplanar vectors can be chosen from V in 2p ways. Then p is Sol. Consider a cube whose centre is at origin and of side length two unit and faces are parallel to coordinate axes then all of the vertices of cube represent the given eight vectors. Now three vectors of them will be coplaner if we select two vertices along body diagonal and third vertex from remaining. [5] Number of ways to select 3 noncoplanar vectors 8 C3 4. 6 C1 32 25 [1] JEE Advanced – 2013/ Paper - 2 PHYSICS Part I Section–I : One or More Options Correct Type The Section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 1. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is are (A) In the region 0 < r < R, the magnetic field is non-zero. (B) In the region R < r < 2R, the magnetic field is along the common axis. (C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis. (D) In the region r > 2R, the magnetic field is non-zero. Ans. [A, C, D] For cylinder, BC 0, r R, 0I to axis r R 2r For solenoid, BS 0 ni along the axis, 0 r 2 R; 0, r 2 R Accordingly, options A, C and D are correct. 2. Two vehicles, each moving with speed u on the same hoizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f1. An observer in the other vehicle hears the frequency of the whistle to be f2. The speed of sound in still air is V. The correct statement(s) is are (A) If the wind blows from the observer to the source, f2 > f1. (B) If the wind blows from the source to the observer, f2 > f1. (C) If the wind blows from observer to the source, f2 > f1. (D) If the wind blows from the source to the observer, f2 > f1. Ans. [A, B] When wind blows from observer to source, f2 FG V v HV v IJ K u f1 f1 w u w When wind blows from source to observer, f2 FG V v HV v w w IJ K u f1 f1 . u [2] JEE Advanced – 2013/ Paper - 2 PHYSICS 3. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are) (A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4 GM . L GM . L (C) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is (B) The miminum velocity of the mass m to escape the gravitational field of the two bodies is 2 2GM . L (D) The energy of the mass m remains constant. Ans. [B, D] For m to escape, its total energy must become zero or positive v M × o m M 2L LM N OP Q i.e. 1 2 2GM mv m 0 2 L or v2 GM L Also during the motion, its total energy remains constant. 4. A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity u0. When the speed of the particle is 0.5 u0, it collides elasticallywith a rigid wall. After this collision (A) the speed of the particle when it returns to its equilibrium position is u0. (B) the time at which the particle passes through the equilibrium position for the first time is t (C) the time at which the maximum compression of the spring occurs is t 4 3 m . k m . k (D) the time at which the particle passes through the equilibrium position for the second time is t 5 3 m . k [3] JEE Advanced – 2013/ Paper - 2 PHYSICS Ans. [A, D] Since the particle collides elastically so its speed first after collision is 0.5 u0. So option (A) is correct. bg u0 T t . 2 6 So the particle passes through the equilibrium position first time at v t u0 cos t t2 FG T IJ T 2 H 6K 3 3 m k so option (B) is wrong. Maximum compression of the spring occurs at T T 7T 7 m 4 3 12 6 k so option (C) is wrong. Particle passes through the equilibrium position for the second time at t t T T 5T 5 2 3 6 3 m k Hence, option (D) is correct. 5. The figure below shows the variation of specific heat capacity (C) of a solid as a function of temprature (T). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the following statement(s) is (are) correct to a reasonable approximation. (A) the rate at which heat is absorbed in the range 0-100 K varies linearly with temperature T. (B) heat absorbed in increasing the temperature from 0-100 K is less than the heat required for increasing the temperature from 400-500 K. (C) there is no change in the rate of heat absorption in the range 400-500 K. (D) the rate of heat absorption increases in the range 200-300 K. Ans. [A, B, C, D] From observing the figure, options B, C and D are correct. Also C is varying approximately linearly with temperature in range 0-100 K. dQ dT dT mC and since = constant dt dt dt dQ C and is varying linearly with temperature T. dt [4] JEE Advanced – 2013/ Paper - 2 PHYSICS 6. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the Bohr radius. Its 3h . It is given that h is Planck constant and R is Rydberg constant. The 2 possible wavelength(s), when the atom de-excites, is(are) orbital angular momentum is (A) 9 32R Ans. [A, C] L (B) 9 16R (C) 9 5R (D) 4 3R nh suggests electron is in third orbit. 2 n2 a0 z = 2. 2 Hence possible transitions are Also using rn (i) (ii) (iii) LM OP N Q 1 L1 1O 9 4RM P 3 2, 5R N4 9Q 1 L 1O 3 4 R M1 P 2 1, R N 4Q 3 1, 1 1 9 4R 1 9 32R Hence [A, C] are correct. 7. Using the expression 2d sin = , one calculates the values of d by measuring the corresponding angles in the range 0 to 90°. The wavelength is exactly known and the error in is constant for all values of . As increases from 0°, (A) the absolute error in d remains constant. (B) the absolute error in d increases. (C) the fractional error in d remains constant. (D) the fractional error in d decreases. bg d d cosec cot Ans. [D] Given d 2 sin so d 2 Absolute error in d, Fractional error in d, b g 2 cosec cot d d bd g cot d d d d Both decrease with increase in . [5] JEE Advanced – 2013/ Paper - 2 PHYSICS 8. Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities and , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region, (A) the electrostatic field is zero. (B) the electrostatic potential is constant. (C) the electrostatic field is constant in magnitude. (D) the electrostatic field has same direction. Ans. [C, D] The field at any point say P is given by r1 P r2 d r1 r2 E 3 0 3 0 b g r1 r2 d 3 0 3 0 i.e. field is same at all points. Hence, its magnitude and direction both remains same. As electric field is non-zero so potential is not constant. [6] JEE Advanced – 2013/ Paper - 2 PHYSICS Section–II : Paragraph Type This section contains 4 paragraphs each describing theory, excperiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices. (A), (B), (C) and (D). Paragraph for Questions 9 and 10 A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energydissipation. In the method usingtransformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values. 9. If the direct transmission method with a cable of resistance 0.4 km 1 is used, the power dissipation (in %) during transmission is (A) 20 (B) 30 (C) 40 (D) 50 Ans. [B] 10. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V , the ratio of the number of turns in the primaryto that in the secondary in the step-down transformer is (A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1 Ans. [A] For step up transformer at plant, VP N P VS NS VS 40,000V At consumer side 40,000 N P 200 NS N P 200 NS 1 Also current flowing during transmission iT P VT (assuming first transformer is 100% efficient) 600 10 3 15A 40,000 b g b 2 g Hence PLoss iT 2 R 15 0.4 20 1800 W % loss 1800 600 103 100% 30% . [7] JEE Advanced – 2013/ Paper - 2 PHYSICS Paragraph for Questions 11 and 12 A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2). 11. The speed of the block when it reaches the point Q is (A) 5 ms1 (B) 10 ms1 (C) 10 3 ms1 (D) 20 ms1 Ans. [B] 12. The magnitude of the normal reaction that acts on the block at the point Q is (A) 7.5 N (B) 8.6 N (C) 11.5 N (D) 22.5 N Ans. [A] For the motion till point Q, applying work energy theorem b g 1 mVQ 2 Wmg W f mg R sin 30 150 J 2 VQ 10 m / s. For the circular motion, N mg sin 30 mv 2 R N mg sin 30 mv 2 7.5N . R [8] JEE Advanced – 2013/ Paper - 2 PHYSICS Paragraph for Questions 13 and 14 The mass of a nucleus AZ X is less than the sum of the masses of (A – Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if bm m g M . Also two light nuclei of masses m and m can undergo complete fusion and form a heavy nucleus of mass M only if bm m g M . The masses of some neutral atoms are given in the table below. 1 2 3 3 2 1 6 3 152 64 H Li 4 2 1 7 3 206 82 1007825 . u 6.015123 u H Li Gd 151919803 . u 4 2.014102 u 7.016004 u Pb 205.974455 u 3 1 70 30 209 83 H Zn 3.016050 u 69.925325 u Bi 208.980388 u 4 2 82 34 210 84 He Se 4.002603 u 81916709 . u . Po 209.982876 u c1 u 932 MeV / c h 2 13. The correct statement is (A) The nucleus 63 Li can emit an alpha particle 210 (B) The nucleus 84 Po can emit a proton (C) Deuteron and alpha particle can undergo complete fusion. (D) The nuclei 70 30 can undergo complete fusion Zn and 82 34 Se Ans. [C] 14. 210 The kinetic energy (in keV ) of the alpha particle, when the nucleus 84 Po at rest undergoes alpha decay, is (A) 5319 (B)5422 (C) 5707 (D) 5818 Ans. [A] For decay of e Q m 214 84 Po 214 84 Po j me Now energy taken away by a particle j me Hej c = 5422 KeV. F A 4 IJ Q 5319 KeV G H AK 206 82 Pb K , energy released 4 2 2 For second part, only deutron and alpha particle undergo fission is feasible. [9] JEE Advanced – 2013/ Paper - 2 PHYSICS Paragraph for Questions 15 and 16 A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity . This can be Q . A uniform magnetic field along the positive 2 z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant . considered as equivalent to a loop carrying a steady current. 15. The magnitude of the induced electric field in the oribit at any instant of time during the time interval of the magnetic field change is (A) BR 4 (B) BR 2 (C) BR (D) 2BR Ans. [B] 16. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is (A) BQR BQR 2 (B) 2 2 BQR 2 (C) 2 (D) BQR 2 Ans. [B] The time varying increasing magnetic field in + z direction produces induced electric field as shown. The magnitude of induced electric field is E R dB RB . 2 dt 2 Z E The angular impulse imported by induced electric field is equal to change in the angular momentum of moving charge particle. z z L QER dt Q FG R dB IJ Rdt QR B H 2 dt K 2 As L so L 2 QR 2 B 2 [10] JEE Advanced – 2013/ Paper - 2 PHYSICS Section – III Matching List Type This section contains 4 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct mathcing with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS. 17. Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists: List I List II O 15 7 N . . . (P) Alpha decay (1) 15 8 (Q) decay (q) 238 92 U (R) Fission (r) 185 83 Bi 184 82 Pb . . . (S) (s) 239 94 Pu 140 57 La . . . Proton emission P Q R S (A) 4 2 1 3 (B) 1 3 2 4 (C) 2 1 4 3 (D) 4 3 2 1 Ans. [C] 234 90 Th . . . [11] JEE Advanced – 2013/ Paper - 2 PHYSICS 18. One mole of a monatomic ideal gas is taken along two cyclic processes E F G E and E F H E as shown in the pV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic Match the paths in List I with the magnitudes of the work done in list II and select the correct answer using the codes given below the lists: Column I Column II GE (1) 160 p0V0 ln2 (Q) G H (q) 36 p0V0 (R) FH (r) 24 p0V0 (S) FG (s) 31 p0V0 (P) Codes P Q R S (A) 4 3 2 1 (B) 4 3 1 2 (C) 3 1 2 4 (D) 1 3 2 4 Ans. [A] [12] JEE Advanced – 2013/ Paper - 2 PHYSICS 19. Math List I with List II and select the correct answer using the codes given below the lists : List I List II P. Boltzmann constant 1. ML2 T 1 Q. Coefficient of viscosity 2. ML1T 1 R. Planck constant 3. MLT 3 K 1 S. Thermal conductivity 4. ML2 T 2 K 1 Codes : P (A) 3 (B) 3 (C) 4 (D) 4 Q 1 2 2 1 R 2 1 1 2 S 4 4 3 3 Ans. [C] 20. A right angled prism of refractive index 1 is placed in a rectangular block of refractive indux 2, which is surrounded bya medium of refractive index 3, as shown in the figure. A rayof light e enters the rectangular block at normal incidence. Depending upon the relationships between 1, 2 and 3, it takes one of the four possible paths ef , eg , eh or ei . Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists : List I List II P. 1. e f 1 2 2 Q. e g 2. 2 1 and 2 3 R. eh 3. 1 2 S. ei Codes : P (A) 2 (B) 1 (C) 4 4. 2 1 2 2 and 2 3 Q 3 2 1 R 1 4 2 S 4 3 3 (D) 3 4 1 Ans. [D] 2 [13] JEE Advanced – 2013/ Paper -2 CHEMISTRY Part II SECTION – I : One or more options correct Type This section contains 8 multiple choice questions. Each question has fcour choices (A), (B), (C) and (D), out of which ONE or MORE are correct. 21. The major product(s) of the following (are) OH Br2 b 3.0 equivalents g aqueous ? SO3H (A) P Ans: (B) 22. (B) Q (C) R (D) S After completion of the reaction (I and II), the organic compound(s) in the reaction mixtures is (are) Reaction I: Reaction II : Br2 b1.0 mol g aqueous / NaOH 2 b1.0 mol g Br CH 3COOH CHBr3 P Q R S (A) Reaction I : P and Reaction II : P (B) Reaction I : U, acetone and Reaction II : Q acetone (C) Reaction I : T, U, acetone and Reaction II : P (D) Reaction I : R, acetone and Reaction II : P Ans: (A, C) T U [14] JEE Advanced – 2013/ Paper -2 CHEMISTRY 23. The Ksp of Ag 2 CrO 4 is 11 . 1012 at 298 K. The solubility(in mol/L) of Ag 2 CrO 4 in a 0.1 M AgNO 3 is solution is (A) 11 . 1011 Ans: (B) CrO 24 (B) 11 . 1010 K eq Ag 24. 2 11 . 10 12 . g b010 2 11 . 10 10 (C) 11 . 1012 (D) 11 . 109 M In the following reaction, the product(s) formed is(are) CHCl 3 ? OH P (A) P (major) Ans: (B, D) 25. Q (B) Q (major) R (C) R (minor) S (D) S (major) The carbon-based reduction method is NOT used for the extraction of (A) tin from SnO 2 (B) iron from Fe 2 O 3 (C) aluminium from Al 2 O 3 (D) magnesium from MgCO 3 CaCO 3 Ans: (C, D) 26. The thermal dissociation equilibrium of CaCO 3 ( s) is studied under different conditions. CaCO 3 (s) CaO ( s) CO 2 (g) For this equilibrium, the correct statement(s) is (are) (A) H is dependent on T (B) K is independent of the initial amount of CaCO 3 (C) K is dependent on the pressure of CO 2 at a given T (D) H is independent of the catalyst, if any Ans: (A, B, D) [15] JEE Advanced – 2013/ Paper -2 CHEMISTRY 27. The correct statement(s) about O3 is (are) (A) O O bond lengths are equal. (B) Thermal decomposition of O3 is endothermic (C) O3 is diamagnetic in nature. (D) O3 is diamagnetic in nature Ans: (A, C, D) 28. In the nuclear transmutation 9 4 Be X 84 Be Y bX, Yg is (are) (A) b , ng b g (B) p, D b g (C) n, D b g (D) , p Ans: (A, B) Section–II : Paragrah Type This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D). Paragraph for Questions 29 and 30 P and Q are isomers of dicarboxylic acid C4 H 4 O 4 . Both decolorize Br2 / H 2 O . On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO 4 , P as wall as Q could produce one or more than one from S, T and U. S 29. T U Compounds formed from P and Q are, respectively (A) Optically active S and optically active pair (T, U) (B) Optically inactive S and optically inactive pair (T, U) (C) Optically active pair (T, U) and optically active S (D) Optically inactive pair (T, U) and optically inactive S Ans: (B) [16] JEE Advanced – 2013/ Paper -2 CHEMISTRY 30. In the following reaction sequences V and W are, respectively 2 / Ni Q H V 3 b anhydrous g + V AlCl . Zn Hg / HCl 1 W 2 . H 3PO4 (A) and (B) (C) and (D) and and Ans: (A) Paragraph for Questions 31 and 32 An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H 2S in a dilute mineral acid medium. However, it gave a precipitate (R) with H 2S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H 2 O 2 in an aqueous NaOH medium. 31. The precipitate P contains (A) Pb2 (B) Hg2 2 (C) Ag (D) Hg2 (C) ZnSO4 (D) Na 2 CrO 4 Ans: (A) 32. The coloured solution S contains b g (A) Fe SO4 3 (B) CuSO4 Ans: (D) Paragraph for Questions 33 and 34 The reactions of Cl 2 gas with cold-dilute and hot concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl 2 gas reacts with SO 2 gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compounds S. On hydrolysis, S gives an oxoacid of phosphorus, T. [17] JEE Advanced – 2013/ Paper -2 CHEMISTRY 33. P and Q, respectively, are the sodium salts of (A) hypochlorus and chloric acids (C) chloric and perchloric acids Ans: (A) (B) hypochlorus and chlorus acids (D) chloric and hypochlorus acids SO 2 Cl 2 SO 2 Cl 2 10 SO 2 Cl 2 () P4 4 P Cl5 10 SO 2 PCl5 4 H 2 O H 3PO 4 HCl 34. R, S and T, respectively, are (A) SO 2 Cl 2 , PCl5 and H 3PO4 (B) SO 2 Cl 2 , PCl 3 and H 3PO3 (C) SOCl 2 , PCl 3 and H 3PO2 (D) SOCl 2 , PCl5 and H 3PO4 Ans: (A) Paragraph for Questions 35 and 36 A fixed mass 'm' of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure L N M Pressure K Volume 35. The succeeding operations that enable this transformation of states are (A) Heating, cooling, heating, cooling (B) Cooling, heating, cooling, heating (C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling Ans: (C) 36. The pair of isochoric processes among the transformation of states is (A) K to L and L to M (B) L to M and N to K (C) L to M and M to N (D) M to N and N to K Ans: (B) [18] JEE Advanced – 2013/ Paper -2 CHEMISTRY Section–III : Matching List Type This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 37. The standard reduction potential data at 250C is given below. c h E c Fe , Feh 0.44 V E cCu , Cuh 0.34 V; E cCu , Cuh 0.52 V E O bgg 4 H 4e 2 H O 1.23 V; E O bgg 2 H O 4e 4OH 0.40 V E cCr , Cr h 0.74 V; E cCr , Cr h 0.91 V E 0 Fe 3 , Fe 2 0.77 V; 0 2 0 2 0 0 2 2 0 2 2 0 3 0 3 Match E0 of the redox pair in list I with the values given in List II and select the correct answer using the code given below the lists. List I List II c (P) E 0 Fe 3 , Fe h 0 4H 4OH (Q) E 4H 2 O c E cCr (R) E 0 Cu 2 Cu 2Cu (S) 0 3 , Cr 2 h h Codes: (A) (B) (C) (D) Ans: (D) P 4 2 1 3 Q 1 3 2 4 R 2 4 3 1 S 3 1 4 2 1. 018 . V 2. 0.4 V 3. 0.04 V 4. 0.83 V [19] JEE Advanced – 2013/ Paper -2 CHEMISTRY 38. An aqueous solution of X is added slowlyto an aqeuous solution of Y as shown in the List I. The variationin conductivity of these reactions is given in List II. Match List I with List II and select the correct answer using the code given below the lists. List I List II b g (P) C2 H 5 3 N CH 3COOH X Y (Q) CH 3COOH KOH X Y (R) CH 3COOH KOH X Y (S) NaOH HI X Y Codes: P Q R (A) 3 4 2 (B) 4 3 2 (C) 2 3 4 (D) 1 4 3 1. Conductivity decreases and then increases 2. Conductivity increases and then does not change much 3. Conductivity increases and then does not change much 4. Conductivity does not change much and then incresaes S 1 1 1 2 Ans: (A) 39. Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists. List I List II bg b gb g (P) 1. i Hg OAc , ii NaBH 4 (Q) 2. NaOEt (R) 3. Et Br (S) 4. i BH 3 , ii H 2 O 2 / NaOH bg Codes: (A) (B) (C) (D) Ans: (A) P 2 3 2 3 Q 3 2 3 2 R 1 1 4 4 S 4 4 1 1 bg [20] JEE Advanced – 2013/ Paper -2 CHEMISTRY 40. The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists. List I List II (P) PbO2 H 2SO4 ? PbSO4 O2 other product 1. NO (Q) Na 2S2O3 H 2 O ? NaHSO4 other product 2. I2 (R) N2H4 ? N 2 other product 3. Warm (S) XeF2 ? Xe other product 4. Cl 2 Codes: (A) (B) (C) (D) Ans: (D) P 4 3 1 3 Q 2 2 4 4 R 3 1 2 2 S 1 4 3 1 [21] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS Part III Section – 1 : (One or more options correct Type) This section contains 8 multiple choice questions. Each question has fcour choices (A), (B), (C) and (D), out of which ONE or MORE are correct. 41. Two lines L1 : x 5, y z y z and L 2 : x , are coplanar. Then can take value(s) 3 2 1 2 (A) 1 Sol. [A,D] (B) 2 (C) 3 (D) 4 x5 y z 0 3 2 x y z 0 1 2 Given lines are coplanar. 5 0 0 0 3 2 0 1 2 0 b 5g b3 gb2 g 2 0 b 5ge 5 4j 0 2 b 5g 6 5 2 0 b 5gb 4gb 1g 0 2 1, 4, 5 42. Sol. 1 In a triangle PQR, P is the largest angle and cos P . Further the incircle of the triangle touches the sides 3 PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is(are) (A) 16 (B) 18 (C) 24 (D) 22 [B, D ] cosP 1 3 NP s p 2n 0 QL s q 2 n 2 0 RM s r 2 n 4 0 s 6n 6 3s 2s 6n 6 p 4n 6, q 4n 4, r 4n 2 cos P 1 q 2 r 2 p2 3 2qr [22] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS b g b g b b gb g bn 1g 2 1 3 4b n 1gb2 n 1g 2 g b2n 2g b2n 1g b2n 3g 4b n 1gb2 n 1g 2 4 n 4 4n 2 4n 6 1 3 2 4n 4 4n 2 2 2 2 2 b g b gb g 4b n 1gb2n 1g 2 4 n 1 4 n 4 2 2 n 1 3n 9 2 n 8 n 4 43. For a R (the set of all real numbers), a 1, c1 2 ... n h 1 lim bn 1g bna 1g bna 2g ...bna ng 60 a a a a 1 n then a (A) 5 Sol. [B, D] (B) 7 lim n ra n a 1 IJ z K 1 1 FG a 1 H 2 x a dx 0 2 1 2a 2 3a 119 0 a 2a 17 7 2a 17 0 g b (D) 17 2 1 60 1 b 15 2 1 60 a 7, a Sol. FG1 1 IJ LMa 1 FG1 1 IJ OP H nK N 2 H nKQ b2a 1g ba 1g 60 120 b2a 1gba 1g 44. 1 a 1 (C) g 17 2 Circle(s) touching x–axis at a distance 3 from the origin and having an intercept of length 2 7 on y–axis is(are) (A) x 2 y 2 6x 8y 9 0 2 2 (B) x y 6x 7 y 9 0 (C) x 2 y 2 6x 8y 9 0 (D) x 2 y 2 6x 7 y 9 0 [A, C] r 9 7 4 b C 3, 4 g [23] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS equation of circle bx 3g by 4g 2 45. 2 42 x 2 y 2 6x 8y 9 0 Let be a complex cube root of unity with 1 and P p ij be a n × n matrix with p ij i j . Then P 2 0, when n = (A) 57 (B) 55 (C) 58 (D) 56 Ans. [B,C,D] LM MM PM MM . MM . N 2 3 4 5 3 ..... n 1 . 4. n 1 . . . . 2n . OP PP PP PP PQ P2 0 (2)2 + (3)2 + (4)2 + (5)2 + (6)2 + . . . + (n+1)2 0 + 1 + 2 + + 1 + 2 ... 0 n must not be multiple of 3. 46. bg The function f x 2 x x 2 x 2 2 x has a local minimum or a local maximum at x = (A) –2 (B) 2 3 (C) 2 R| 2x x 2 cbx 2g 2xh, x 2 ||2x x 2 bx 2 2xg, 2 x 23 2 f b xg S 2 x x 2 b x 2 2 x g, x0 || 3 || 2x 2xxx21bx bx222x2g,xg, 0xx2 2 T R| 2x 4, x 2 ||2x 4, 2 x 23 S 4 x, 2 x 0 || 3 4x 0 x 2 |T 2x 4, x 2 (D) Ans. [A, B] Points of local maxima or minima are 2, 2 ,0 3 2 3 Y –2 – 2/3 O X 2 [24] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS 47. 3 i 2 w Let and RS T H 2 z C : Re z o t P w n : n 1, 2 , 3,... . Further UV W RS T H 1 z C : Re z 1 2 2 (B) 6 (C) 2 3 (D) 5 6 Ans. [C, D] i 3i e 6 2 Y H2 H1 o X Total 12 points in P Required angles = 00, 1200, 1500, 1800 48. If 3x 4 x 1 , then x = 2 log 3 2 (A) 2 log 2 1 3 Ans. [A, B, C] 3x 4 x 1 b g x log 2 3 x 1 2 x log 2 3 2x 2 2 x 2 log 2 3 x 2 2 log2 3 2 log 2 3 2 log 2 3 1 and 1 , where C is the set of all complex numbers. If z1 P H 1 , z 2 P H 2 and 2 O represents the origin, then z1Oz 2 (A) UV W 2 (B) 2 log 3 2 1 (C) 1 log 3 4 2 log 2 3 (D) 2 log 3 1 2 [25] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS Section – 2 : (Paragrah Type) This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questioons on each paragraph. Each question of a paragraph has only one correct answer among the four cohices (A), (B), (C) and (D), out of which ONE ONE is correct. Paragraph for Questions 49 and 50 Let PQ be a focal chord of the parabola y 2 4ax. The tangents to the parabola at P and Q meet a point lying on the line y 2 x a , a 0 . 49. 50. Length of chord PQ is (A) 7a (C) 2a (D) 3a 2 If chord PQ subtends an angle at the vertex of y 4ax, then tan (A) Sol. (B) 5a 2 7 3 (B) 2 7 3 (C) 2 5 3 (D) e j Qeat , 2at j [B, D] P at 12 , 2at 1 2 2 2 For focal chord t 1t 2 1 b c gh c b Intersectionof tangents at P and Q is at 1t 2 , a t 1 t 2 i.e. a , a t 1 t 2 y a t 1 t 2 1 bt g bt t g 4t t b1g 4b1g 5 PQ a e t t j c2b t t gh a 5b t t g 1 t2 2 2 1 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 t 2 t1 t1 t 2 2 5 tan 2 2 t 1t 2 4 3 1 t1 t 2 b b is obtuseg 2 1 g 2 5a gh 2 5 3 [26] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS Paragraph for Questions 51 and 52 Let f : 0, 1 R (the set of all real numbers) be a function. Suppose the function f is twice differentiable, bg b g bg bg bg f 0 f 1 0 and satisifes f " x 2 f ' x f x e x , x 0, 1 . 51. Which of the following is true for 0 x 1 ? bg (B) (A) 0 f x (C) bg bg 1 1 f x 2 2 bg 1 f x 1 4 (D) f x 0 bg bg bg cf bxg f bxghe e cf bxg f bxgh 1 d e c f b xg f b xgh 1 dx Ans. [D] f x 2f x f x e x or x x x d2 dx 2 ee . f bxgj 1 x bg bg Let g x e x x then but 52. bg gb0g gb1g 0 gb xg 0 x b0, 1g e f b xg 0 x b0, 1g f b xg 0 x b0, 1g d2 dx 2 cgbxgh 1, g x is concave up x bg If the function e x f x assumes its minimum in the interval [0, 1] at x (A) f ' x f x , bg bg 1 3 x 4 4 bg bg 1 4 (C) f ' x f x , 0 x bg FG 1 IJ H K Ans. [C] g x 0, x 0, 4 1 , which of the following is true? 4 bg bg 1 4 bg bg 3 x 1 4 (B) f ' x f x , 0 x (D) f ' x f x , [27] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS F 1I e f b xg e f b xg 0, x G 0, J H 4K F 1I f b xg f b xg , x G 0, J H 4K x x Paragraph for Questions 53 and 54 A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls. 53. If 1 ball is drawn from each of the boxes B1 , B2 and B3 , the probability that all 3 drawn balls are of the same colour is (A) 82 648 90 648 (B) (C) 558 648 (D) 566 648 Ans. [A] E: Three balls drawn are of same colour bg PE 54. 1 Favourable case Sample space C1 .2 C1 .3 C1 3C1 .. 3C1. 4 C1 2 C1.4 C1 .5 C1 6 C1 . 9 C1 12 C1 82 648 If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2 is (A) Ans. [D] 116 181 (B) 126 181 (C) E1: balls are drawn from B1 E 2 : balls are drawn from B2 E 3 : balls are drawn from B3 E : One white and one red ball is drawn FG E IJ PdE i HE K FE I PG J H E K PbE g. PF E I PbE g . PF E I PbE g . PF E I GH E JK GH E JK GH E JK P 2 2 2 1 2 3 1 1 3 F GH 1 . 3 1 C13C1 6 C2 I 1F JK 3 GH F GH 2 2 2 C1 . 3 C 1 9 C2 3 3 I JK I F JK GH C1 . C 1 1 9 3 C2 3 C1 C1 12 C2 4 I JK 55 181 65 181 (D) 55 181 [28] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS Paragraph for Questions 55 and 56 Let S S1 S2 S3 , where m r S1 z C : z 4 R| S| T LM z 1 3i OP 0U|V MN 1 3i PQ |W l q S 2 z C : Im and S 3 z C : Re z 0 55. Area of S = (A) 56. 10 3 (B) 20 3 (C) 16 3 (D) 32 3 (B) 2 3 2 (C) 3 3 2 (D) 3 3 2 min 1 3i z z S (A) 2 3 2 F Z 1e1 i 3 j I Ans. [B, C] ImG GH 1 i 3 JJK 0 F Z IJ 1 0 ImG H 1 i 3 K If Z x iy then Im S bx iyg 0 e1 i 3j F F x iy IJ ei i 3jIJ 0 ImG G HH 4 K K o P (1, –3) yx 3 0 Required area S 60 e j FGH 6 IJK 1 2 4 2 FG 5 IJ 20 H 6K 3 8 b g Now Min of |1 3i Z| Min of | Z 1 3i | = minimum distance between region S and point P (1, –3) 4, 0 y x 3 [29] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS = Perpendicular distance from (1, –3) to y x 3 0 3 3 3 3 2 2 Section – 3 : (Matching list Type) This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 57. x4 y3 z3 x 1 y z 3 L2 : , and the planes 1 1 2 2 1 1 P1 : 7 x y 2 z 3 , P2 : 3x 5y 6z 4. Let ax by cz d be the equation of the plane passing Consider the lines L1 : through the point of intersection of lines L1 and L 2 , and perpendicular to planes P1 and P2 . Match List – I with List – II and select the correct answer using the code given below the lists: List I List II P. a= 1. 13 Q. b= 2. –3 R. c= 3. 1 S. d= 4. –2 Codes : Sol. P Q R S (A) 3 2 4 1 (B) 1 3 4 2 (C) 3 2 1 4 (D) 2 4 1 3 [A] Point of intersection of L1 and L 2 is b2 1, , 3g b 4, 3, 2 3g 2 1 4 , 3 and 3 2 3 2 3 , 3 Solving 3 6 2 3 1 [30] JEE ADVANCED – 2013 / Paper -2 b MATHEMATICS g Point of intersection of lines L1 and L 2 5, 2, 1 normal to req. plane i j 7 1 k 2 16i 48j 32 k 3 5 6 d.r. of normal to plane a, b, c > is 1, 3, 2 Hence equation of required plane is x 3y 2 z d b g which passes through P 5, 2, 1 58. b g b g 5 3 2 2 1 d d 13 a 1, b 3, c 2 d 13 Match List – I with List – II and select the correct answer using the code given below the lists: List I P. Volume of parallelepiped determined by vectors a , b and c is 2. Then the volume of the parallelepiped determined by vectors 2 a b ,3 b c and c a is d id i Q. 1. 100 2. 30 3. 24 4. 60 b g Volume of parallelepiped determined by vectors a , b and c is 5. Then the volume of the parallelepiped determined by vectors 3 a b , b c and 2 c a is d id i R. List II b g Area of a triangle with adjacent sides determined by vectors a and b is 20. Then the volume of the parallelepiped determined by vectors 2a 3b and a b is d i d i S. Area of parallelogram with adjacent sides determined by vectors a and b is 30. Then the area of the parallelogram determined by vectors a b and a is d i [31] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS Codes : Sol. P Q R S (A) 4 2 3 1 (B) 2 3 1 4 (C) 3 4 1 2 (D) 1 4 3 2 [C] 2 ab 3 bc Q a b c 5 3 ab e j e j bc ag 6 a b b c c a e 6a bc 2 6 2 2 24 b c c a 6 2 a b c 12 5 60 j e b c j 2bc a g 6 a b S 1 a b 20 a b 40 2 1 1 1 2a b 3b a 5 a b 5 40 100 Req. area 2 2 2 a b 30 Area of required parallelogram a b a a b 3 0 R 59. a bc 2 P e j b g A line L : y mx 3 meets y–axis at E 0, 3 and the arc of the parabola y 2 16x, 0 y 6 at the b g b g b g point F x 0 , y 0 . The tangent to the parabola at F x 0 , y 0 intersects the y–axis at G 0, y1 . The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum. Match List – I with List – II and select the correct answer using the code given below the lists: List I List II P. m= 1. 1 2 Q. Maximum area of EFG is 2. 4 R. y0 3. 2 S. y1 4. 1 [32] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS Codes : P Q R S (A) 4 1 2 3 (B) 3 4 1 2 (C) 1 3 2 4 (D) 1 3 4 2 yt x 4 t 2 Ans. [A] tangent at F is Y G : x 0 y 4t b g hence G 0, 4 t 0 Area of triangle EFG 1 0 2 2 4t 3 1 4t 1 8t 1 e b gj 2 t b3 4 t g dA 12 tb1 2 t g dt 1 2 4t 3 4t 2 (0, 3) E F (4t2, 8t) G (0, 4t) X O 2 area is maximum when t = 1/ 2 so y1 4 t 2 bx , y g e4t , 8t j b1, 4g 2 0 0 y0 4 Area of triangle EFG 2 FG 3 1 IJ 2FG 3 2 IJ 1 H 4 2K H 4 K 2 and y mx 3 passes through (1, 4). m=1 dA dt + – ½ [33] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS 60. Match List – I with List – II and select the correct answer using the code given below the lists: List I List II P. F 1 F cosctan yh y sinctan yhI GG y G cot sin y tan sin y J H H c h c hK Q. If cos x cos y cos z 0 sin x sin y sin z then possible 1 1 2 value fo cos R. 1 2 I y J JK 4 1 1/ 2 takes value 1. 1 5 2 3 2. 2 xy is 2 FG xIJ cos 2x sin x sin 2x sec x cos x sin 2x sec x cosFG xIJ cos 2x H4 K H4 K If cos 3. 1 2 4. 1 then possible value fo sec x is S. e e d ij j 1 2 1 If cot sin 1 x sin tan x 6 , x 0 , then possible value fo x is Codes : P Q R S (A) 4 3 1 2 (B) 4 3 2 1 (C) 3 4 2 1 (D) 3 4 1 2 Ans. [B] LM e j e j OP MN cot esin yj tanesin yj PQ 1 1 1 cos tan y y sin tan y [P] y2 e 1 j 1 e j e j for cos tan 1 y y sin tan 1 y let y tan then e j cos tan 1 y y sin tan 1 y = 2 y4 [34] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS bg 1 sec 1 y 2 cos cos tan sin and e j e 1 1 for cot sin y tan sin y j let sin 1 y than cot tan 1 1 sin cos y 1 y 2 F G 1 G 1 y 1 y G GH y 1 y 2 so [Q] 2 2 I JJ JJ K 2 e1 y j y y4 4 4 1 cos x cos y cos z . . .(1) sin x sin y sin z . . .(2) (1) 2 ( 2) 2 gives b g 2 2 cos x cos y sin x sin y 1 [R] b g 2 c1 cosb x y gh 1 F x y IJ 1 4 cos G H 2 K F x y IJ 1 cosG H 2 K 2 F I F I cosG xJ cos 2x cosG xJ cos 2x sin 2xb1 tan xg H4 K H4 K bcos x sin xg cos 2 xe 2 sin xj sin 2 x 2 2 cos x y 1 2 cos x e 2 sin x cos xj sin 2xbcos x sin xg bsin 2 xgFGH cos22x bcos x sin xgIJK 0 cos 2 x bsin 2xgbcos x sin xgLMFGH cos x 2sin x IJK 1OP 0 N Q x 0, 4 Posible value of sec x is 1, 2 [35] JEE ADVANCED – 2013 / Paper -2 MATHEMATICS [S] F F 1 x I I sinetan ex 6 jj KK H H F F x6 F x I cot G cot sinG sin G J H 1 x K H H 1 6x cot sin 1 1 2 1 1 2 x 1 x 2 x 6 1 6x 2 e1 6x j 6e1 x j b x 0g 12 x 2 5 x 2 2 5 1 5 12 2 3 2 II JK JK