Aakash JEE Advance Solution Paper1

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Aakash JEE Advance Solution Paper1

  1. 1. (1)Answers & SolutionsforforforforforJEE (Advanced)-2013Time : 3 hrs. Max. Marks: 180INSTRUCTIONSQuestion Paper FormatThe question paper consists of three parts (Physics, Chemistry and Mathematics). Each partconsists of three sections.Section 1 contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.Section 2 contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONE or MORE are correct.Section 3 contains 5 questions. The answer to each question is a single-digit integer, ranging from0 to 9 (both inclusive).Marking SchemeFor each question in Section 1, you will be awarded 2 marks if you darken the bubble correspondingto the correct answer and zero mark if no bubbles are darkened. No negative marks will be awardedfor incorrect answers in this section.For each question in Section 2, you will be awarded 4 marks if you darken all the bubble(s)corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. In all othercases, minus one (–1) mark will be awarded.For each question in Section 3, you will be awarded 4 marks if you darken the bubble correspondingto only the correct answer and zero mark if no bubbles are darkened. In all other cases, minusone (–1) mark will be awarded.DATE : 02/06/2013PAPER - 1 (Code - 1)CODE1Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-47623456 Fax : 011-47623472
  2. 2. (2)PART–I : PHYSICSSECTION - 1 : (Only One Option Correct Type)This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of whichONLY ONE is correct.1. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zeroof the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisionsequivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scaledivisions. The diameter of the cylinder is(A) 5.112 cm (B) 5.124 cm(C) 5.136 cm (D) 5.148 cmAnswer (B)Hint : 1 MSD = 5.15 – 5.10 = 0.05 cm1 VSD =2.4550= 0.049 cmLC = 1 MSD – 1 VSD= 0.001 cmReading = 5.10 + L.C. × 24= 5.10 + 0.024= 5.124 cm2. A ray of light travelling in the direction 1 ˆ ˆ( 3 )2i j is incident on a plane mirror. After reflection, it travelsalong the direction1 ˆ ˆ( – 3 )2i j . The angle of incidence is(A) 30° (B) 45°(C) 60° (D) 75°Answer (A)Hint :               ˆ ˆ3 – 3.2 2cos(180 2 )ˆ ˆ3 – 32 2i j i ji j i j i 32j –i 32180° – 2^j^  (1 3)4cos21  1cos22 1cos22  2 60 = 30°
  3. 3. (3)3. In the Youngs double slit experiment using a monochromatic light of wavelength , the path difference (interms of an integer n) corresponding to any point having half the peak intensity is(A)(2 1)2n (B)(2 1)4n(C)(2 1)8n (D)(2 1)16nAnswer (B)Hint :    2max cos2I I    21cos2 2 cos  = 0    3 5 7, , ,2 2 2 2   3 5, ,4 4 4x  (2 1)4x n4. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partialpressures, 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 : 9Answer (D)Hint :RTPM 1 1 22 1 2P MP M124 33 21289
  4. 4. (4)5. Two rectangular blocks, having identical dimensions, can be arranged either in configuration-I or inconfiguration-II as shown in the figure. One of the blocks has thermal conductivity  and the other 2. Thetemperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 sto transport a certain amount of heat from the hot end to the cold end in the configuration-I. The time totransport the same amount of heat in the configuration-II isConfiguration-I Configuration-II 2x2(A) 2.0 s (B) 3.0 s(C) 4.5 s (D) 6.0 sAnswer (A)Hint : 1 22T TQl ltA A[in case (i)]      1 22( )Q A AT Tt l l  29tt t = 2 s6. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulseis 30 mW and the speed of light is 3 × 108 ms–1. The final momentum of the object is(A) 0.3 × 10–17 kg ms–1 (B) 1.0 × 10–17 kg ms–1(C) 3.0 × 10–17 kg ms–1 (D) 9.0 × 10–17 kg ms–1Answer (B)Hint :      –3 –9–17 –1830 10 100 1010 kg ms3 10E P tpC C7. 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 completely inelastic collision with another identical particle,which was thrown vertically upward from the ground with the same initial speed u0. The angle that thecomposite system makes with the horizontal immediately after the collision is(A)4(B) 4(C)–2(D)2Answer (A)Hint : Speed of first particle at highest point = u0cosu0cosu0Hu0Speed of second particle at highest point = 20 2u gHNow,2 20 sin2uHg Speed of 2nd particle = u0cos Final momentum =   0 0ˆ ˆcos cosmu i mu j Angle =4
  5. 5. (5)8. The work done on a particle of mass m by a force,    2 2 3/2 2 2 3/2ˆ ˆ( ) ( )x yK i jx y x y(K being a constant ofappropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circularpath of radius a about the origin in the x – y plane is(A)2Ka(B)Ka(C)2Ka(D) 0Answer (D)Hint :      2 2 3/2 2 2 3/2ˆ ˆ( ) ( )xi yjF kx y x y .dW F dx    2 2 3/2( )xdx ydydW kx ylet x2 + y2 = r2xdx + ydy = r2  3 2krdr kdW drr r    21rrkWrNow, r1 = a, r2 = a  W = 09. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontalthin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends,the ratio of the elongation in the thin wire to that in the thick wire is(A) 0.25 (B) 0.50(C) 2.00 (D) 4.00Answer (C)Hint : Force will be sameNow, FLAY  2122(2 )22L RLR10. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real is one-third the size of the object. The wavelength of light inside the lens is23times the wavelength in free space.The radius of the curved surface of the lens is(A) 1 m (B) 2 m(C) 3 m (D) 6 mAnswer (C)Hint :    32airmedfcf
  6. 6. (6)Now,  = +8 m,     124 m3m uu 1 1 1f u  1 1 1 48 24 24ff = 6 m 1Rf  6 3 m0.5RRSECTION - 2 : (One or More Options Correct Type)This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONE OR MORE are correct.11. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming  = 3.14, the correct statement(s) is (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 HzAnswer (B, C)Hint : No. of nodes = 6, 2k=2 3.140.1 m62.8Length =50.25 m2The mid-point is antinode. Its maximum displacement = 0.01 mf = 20 Hz2 2vl k l12. A solid sphere of radius R and density  is attached to one end of a mass-less spring of force constant k. Theother end of the spring is connected to another solid sphere of radius R and density 3. The completearrangement is placed in a liquid of density 2 and is allowed to reach equilibrium. The correct statement(s)is (are)(A) The net elongation of the spring is 343R gk(B) The net elongation of the spring is 383R gk(C) The light sphere is partially submerged(D) The light sphere is completely submerged
  7. 7. (7)Answer (A, D)Hint : At equilibrium, for upper sphere R R 32     3 34 4(2 )3 3kx R g R g 343R gxkThe system is completely submerged as total weight = Total Buoyant force.13. A particle of mass M and positive charge Q, moving with a constant velocity  11ˆ4 msu i , enters a regionof uniform static magnetic field, normal to the x-y plane. The region of the magnetic field extends from x = 0to x = L for all values of y. After passing through this region, the particle emerges on the other side after10 milliseconds with a velocity   12ˆ ˆ2( 3 )msu i j . The correct statement(s) is (are)(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503MQunits(D) The magnitude of the magnetic field is1003MQunitsAnswer (A, C)Hint :  MtqBClearly  = 30° =6CF x L=ˆ2jvˆ2 3iˆ4ix = 0      3100 506 36 10 10M M MBq QqB must be in –z direction.14. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 and 2respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere,along the line joining the centres of the spheres, is zero. The ratio12can be(A) –4 (B) 3225(C)3225(D) 4Answer (B, D)
  8. 8. (8)Hint : EP = 0 3122004( )334 (2 )RRR P PR 2R2R124 0PE    3 31 22 20 04 4(2 )3 34 (2 ) 4 (5 )R RR R 12322515. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switchS1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to chargethe capacitor C2. After some time, S2 is released and then S3 is pressed. After some time,(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 0S12V0C1V0S2 S3C2(D) The charge on the upper plate of C2 is –CV0Answer (B, D)Hint : When S1 is pressed and released2V0C1 V0S2 S3C2+–+2CV0When S2 is pressed and released2V0 V0S2 S3C2+–CV0S1+CV0When S3 is pressed2V0C1 V0C2+–CV0 CV0CV0–+
  9. 9. (9)SECTION - 3 : (Integer Value Correct Type)This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive).16. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stoppingpotential versus frequency plot for Silver to that of Sodium isAnswer (1)Hint : hfVe eSlope = .heIt is same for both.17. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Giventhat ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) thatwill decay in the first 80 s after preparation of the sample isAnswer (4)Hint :  0tN N e0tNeN ln28013860NeN0.693 8013860NeN 0.040NeN    0.0401NN eFraction of nuclei decayed =      0.04011 1NN e= 0.04 = 4%18. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W tothe particle. If the initial speed (in ms–1) of the particle is zero, the speed (in ms–1) after 5 s isAnswer (5)Hint : As power is constant, P × t = kE    210.5 5 (0.2) 02v 2.5 = 0.1 v2 v = 5 m/s19. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s–1about 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 theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s–1) ofthe system is
  10. 10. (10)Answer (8)Hint : By conservation of angular momentum,             2 2 21 150 (0.4) 10 50 (0.4) 2 2 6.25 (0.2)2 2   408 rad/s4 120. A bob of mass m, suspended by a string of length l1, is given a minimum velocity required to complete a fullcircle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspendedby a string of length l2, which is initially at rest. Both the strings are mass-less and inextensible. If the secondbob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, theratio12llisAnswer (5)Hint : Speed of first bob at highest point = 1glFor elastic collision between objects of same mass, velocities are exchanged  speed of 2nd1bob = glbut 1 25gl gl 125llPART–II : CHEMISTRYSECTION - 1 : (Only One Option Correct Type)This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of whichONLY ONE is correct.21. The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is(A) Benzoic acid (B) Benzenesulphonic acid(C) Salicylic acid (D) Carbolic acid (Phenol)Answer (D)Hint : Carbolic acid (Phenol) is weaker acid than carbonic acid and hence does not liberate CO2 on treatmentwith aq. NaHCO3 solution. Benzoic acid, benzenesulphonic acid and salicylic are more acidic than carbonicacid and hence will liberate CO2 with aq. NaHCO3 solution.22. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of(A) NO (B) NO2(C) N2O (D) N2O4Answer (B)Hint : Conc. HNO3 slowly decomposes as4HNO3  4NO2 + 2H2O + O2It acquires yellow-brown colour due to the formation of NO2.
  11. 11. (11)23. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25°C. For this process, thecorrect statement is(A) The adsorption requires activation at 25°C(B) The adsorption is accompanied by a decrease in enthalpy(C) The adsorption increases with increase of temperature(D) The adsorption is irreversibleAnswer (B)Hint : The adsorption of methylene blue on activated charcoal is physiosorption which is exothermic, multilayerand does not have energy barrier.24. Sulfide ores are common for the metals(A) Ag, Cu and Pb (B) Ag, Cu and Sn(C) Ag, Mg and Pb (D) Al, Cu and PbAnswer (A)Hint : Silver, copper and lead are commonly found in earths crust as Ag2S (silver glance), CuFeS2 (Copperpyrites) and PbS (Galena)25. The arrangement of X– ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radiusof X– is 250 pm, the radius of A+ isX–A+(A) 104 pm (B) 125 pm(C) 183 pm (D) 57 pmAnswer (A)Hint : Cation A+ occupies octahedral void formed by anions X–. The maximum radius ratio for a cation toaccommodate a octahedral void without distortion is 0.414. Radius of anion X– is 250 pm.AXR0.414R    AR 0.414 250 103.50 104 pm26. Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is(A) Fe(III) (B) Al(III)(C) Mg(II) (D) Zn(II)Answer (D)Hint : Upon treatment with ammoniacal H2S, Zn2+ ion gets precipitated as ZnS. Fe3+ ion and Al3+ ions also getprecipitated as hydroxides but not as sulphide.27. The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 25°C are –400 kJ/mol, –300 kJ/moland –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25°C is(A) +2900 kJ (B) –2900 kJ(C) –16.11 kJ (D) +16.11 kJAnswer (C)
  12. 12. (12)Hint : C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)H° = 6 (–400) + 6(–300) – (–1300)H° = –2900 kJ/mol     2900H 16.11 kJ / gm18028. Consider the following complex ions, P, Q and R.P = [FeF6]3–, Q = [V(H2O)6]2+ and R = [Fe(H2O)6]2+The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is(A) R < Q < P (B) Q < R < P(C) R < P < Q (D) Q < P < RAnswer (B)Hint : The electronic configuration of central metal ion in complex ions P, Q and R are  3 36P [FeF ] ; Fe :3dQ = [V(H2O)6]2+; V2+:3dR = [Fe(H2O)6]2+; Fe2+3dThe correct order of spin only magnetic moment is Q < R < P29. In the reaction,P + Q R + SThe time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Qvaries with reaction time as shown in the figure. The overall order of the reaction is[Q]0[Q]Time(A) 2 (B) 3(C) 0 (D) 1Answer (D)Hint : The order of reaction with respect to P is one since t3/4 is twice of t1/2. From the given graph the orderof reaction with respect to Q is zero. Therefore, overall order of reaction is one.30. KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary asH C – Cl3PClOClClQ R S(A) P > Q > R > S (B) S > P > R > Q(C) P > R > Q > S (D) R > P > S > Q
  13. 13. (13)Answer (B)Hint :PClOClQClRSCompounds: CH –Cl3 : :Relative reactivitiestowards S reactionN21,00,000 200 79 0.02: : :SECTION - 2 : (One or More Options Correct Type)This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONE OR MORE are correct.31. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)(A) [Cr(NH3)5Cl]Cl2 and [Cr(NH3)4Cl2]Cl (B) [Co(NH3)4Cl2]+ and [Pt(NH3)2(H2O)Cl]+(C) [CoBr2Cl2]2– and [PtBr2Cl2]2– (D) [Pt(NH3)3](NO3)Cl and [Pt(NH3)3Cl]BrAnswer (B, D)Hint : The pair of complex ions [Co(NH3)4Cl2]+ and [Pt(NH3)2(H2O)Cl]+ show geometrical isomerism. The pair ofcomplexes [Pt(NH3)3(NO3)]Cl and [Pt(NH3)3Cl]Br show ionisation isomerism. The other pairs given do nothave same type of isomerism.32. Among P, Q, R and S, the aromatic compound(s) is/areClPQAlCl3NaHR(NH ) CO4 2 3100-115 °CO OSHClO(A) P (B) Q(C) R (D) SAnswer (A, B, C, D)Hint : (i) ClAlCl3AlCl4 (P)(AROMATIC)(ii)NaH(Q) + H2(AROMATIC)Na(iii)O O(NH ) CO4 2 3100–115 CºR(NH ) CO4 2 32NH + CO + H O3 2 2
  14. 14. (14)O O+ NH3O O NH2 O NH2OH IMPENH–2H O2NHHO OHIMPENHO OHH(R) AROMATICH(iv) OHClOH Cl (S)(AROMATIC)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 delocalisationsAnswer (A)Hint : In hyperconjugation   p (empty) electron delocalization for tert-butyl carbocation and   * electrondelocalization for 2-butene will take place.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) Ssystem is positive(C) Ssurroundings = 0 (D) H = 0Answer (B, C, D)Hint : Benzene and naphthalene form an ideal solution. For an ideal solution, H = 0, Ssystem > 0 andSsurroundings = 0 because there is no exchange of heat energy between system and surroundings.35. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strongacid (HX, 1M), at 25°C. The Ka of HA is(A) 1 × 10–4 (B) 1 × 10–5(C) 1 × 10–6 (D) 1 × 10–3Answer (A)Hint : Rate with respect to weak acidR1 = K[H+]WA[ester]and rate with respect to strong acidR2 = K[H+]SA[ester] WA12 SA[H ]R 1R 100[H ]       SAWA[H ] 1H 0.01 M C100 100 = 0.01Ka for weak acid = C2= 1(0.01)2= 1 × 10–4
  15. 15. (15)SECTION - 3 : (Integer Value Correct Type)This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive).36. The total number of carboxylic acid groups is the product P isOO OOO1. H O ,+32. O33. H O2 2PAnswer (2)Hint : OOOOOH O3+OOOOOHOO(i) O3(ii) H O2 2OOHOOCHOOC(P)OHOH (–H O, –CO )2 2So, final product (P) has two carboxylic acid Heating causes evaporation of water and due to which concentration of acid increases and inconcentrated acidic medium decarboxylation as well as dehydration will take place.37. A tetrapeptide has —COOH group 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 (primarystructures) with —NH2 group attached to a chiral center isAnswer (4)Hint : According to question C – Terminal must be alanine and N – Terminal do have chiral carbon means itsshould not be glycine so possible sequence is :Val Phe Gly AlaVal Gly Phe AlaPhe Val Gly AlaPhe Gly Val AlaSo, answer is (4).38. EDTA4– is ethylenediaminetetraacetate ion. The total number of N—Co—O bond angles in [Co(EDTA)]1–complex ion isAnswer (8)Hint :CoO – C = OO – C = OOC CH2 CH2OCH2CH2NO NC CH2OCH2IIIIIIVIIII
  16. 16. (16)So, bond angles are(1) NI CoOI(2) NI CoOII(3) NI CoOIII(4) NI CoOIV(5) NII CoOI(6) NII CoOII(7) NII CoOIII(8) NII CoOIVSo, total asked bond angles are 8.39. The total number of lone-pairs of electrons in melamine isAnswer (6)Hint : Structure of melamine isN NH2NNH2H2NNSo, melamine has six lone pair of electrons.40. The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the de Broglie wavelengthof He gas at –73ºC is "M" times that of the de Broglie wavelength of Ne at 727ºC M isAnswer (5)Hint :He Ne NeNe He HeM VM VNeNeNeHeHeHe3RTMM3RTMMNeNeNeHeHeHeTMMTMM Ne NeHe HeM TM T205 10004 200  He Ne5
  17. 17. (17)PART–III : MATHEMATICSSECTION - 1 : (Only One Option Correct Type)This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of whichONLY ONE is correct.41. Let complex numbers and1lie on circles    2 2 20 0( ) ( )x x y y r and    2 2 20 0( ) ( ) 4x x y y r ,respectively. If  0 0 0z x iy satisfies the equation  2 202 2,z r then  (A)12(B)12(C)17(D)13Answer (C)Hint : , lies on    2 2 20 0( ) ( )x x y y r ... (1)( , )x y0 0r21  lies on    2 2 20 0( ) ( ) 4x x y y r ... (2)Let  be 1 + i1 lies on 2 2 2 2 20 0 0 0( ) ( ) 2( )x y x y xx yy r     2 2 20 1 0 1 02( )z x y r       ... (3)From eq. (2), 2 21 0 1 002 2( )12 4x yz r     2 2 220 1 0 1 01 2( ) 4z x y r        ... (4)From (3) – (4),  2 2 2 2201 (1 ) 1 4z r        2 2 220( 1)(1 ) 1 4z r      ...(5)Now,  2202 1r z 22021rzSubstituting in (5),2 21 2(1 4 )     2 21 2 8     27 1 17 
  18. 18. (18)42. Four persons independently solve a certain problem correctly with probabilities1 3 1 1, , ,2 4 4 8. Then theprobability that the problem is solved correctly by at least one of them is(A)235256(B)21256(C)3256(D)253256Answer (A)Hint :    1 3 1 1( ) , ( ) , ( ) , ( )2 4 4 8P A P B P C P DP(A  B  C  D) =    1 ( )P A B C D=    1 ( )P A B C D= 1 ( ) ( ) ( ) ( )P A P B P C P D=    1 1 3 712 4 4 8= 211256=23525643. Let   1: ,12f (the set of all real numbers) be a positive, non-constant and differentiable function suchthat f(x) < 2f(x) and   112f . Then the value of 11/2( )f x dx lies in the interval(A) (2e – 1, 2e) (B) (e – 1, 2e – 1)(C)   1, 12ee (D)   10,2eAnswer (D)Hint :  2dyydx 2 22x xdye yedx2( ) 0xdyedx ye–2x is decreasing functionAs,   112xe–1 > ye–2x > y(1)e–2e2x–1 > y > y(1)e2x–2
  19. 19. (19)     1 1 12 1 2 21/2 1/2 1/2(1) 0x xe dx ydx y e 11/2102eydx44. The number of points in  (– , ) , for which   2sin cos 0x x x x , is(A) 6 (B) 4(C) 2 (D) 0Answer (C)Hint :   2( ) sin cosf x x x x x( ) 2 cos sin sinf x x x x x x    = x(2 – cosx)f(x) is increasing x > 0f(x) is decreasing x < 0f(0) = –1( )f   (– )f   45. The area enclosed by the curves  sin cosy x x and  cos siny x x over the interval   0,2is(A) 4( 2 1) (B) 2 2( 2 1)(C) 2( 2 1) (D) 2 2( 2 1)Answer (B)Hint :/2 /4 /20 0 /4(sin cos ) (cos sin ) (sin cos )x x dx x x dx x x dx            =/2 /2 /4 /4 /2 /20 0 0 0 /4 /4cos sin sin cos cos sinx x x x x x              =1 1 1 1(0 1) (1 0) 1 0 12 2 2 2                        O /4 /2=1 12 2 1 12 2       = 2 2 2 2   = 4 2 2= 2 2( 2 1)
  20. 20. (20)46. A curve passes through the point   1,6. Let the slope of the curve at each point (x, y) be    sec , 0y yxx x.Then the equation of the curve is(A)    1sin log2yxx(B)    cosec log 2yxx(C)    2sec log 2yxx(D)    2 1cos log2yxxAnswer (A)Hint :   secdy y ydx x xPut y = vx    secdvv x v vdx cosdxv dvx sin v = ln x + c It passes through   1,6 12c  1sin ln2yxx47. The value of2311 1cot cot 1 2nn kk           is(A)2325(B)2523(C)2324(D)2423Answer (B)Hint :  2311cot 1 ( 1)nk k =231 11tan ( 1) tan ( )nk k  
  21. 21. (21)= tan–1(24) – tan–1(1)Now,  1 1cot (tan (24) tan (1) =1 24 1cot tan1 24 1       =      1 23cot tan25=252348. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines    0ax by c and   0bx ay c is less than 2 2 . Then(A) a + b – c > 0(B) a – b + c < 0(C) a – b + c > 0(D) a + b – c < 0Answer (A)Hint :    0ax by c ... (1)   0bx ay c ... (2)Solving,cxa bAlso from (1) & (2)y = x Point of intersection lies on y = xcya bGiven,2 21 1 2 2c ca b a b              2 1 2 2ca b     2a b ca b  2 2a b c a b    0a b c  
  22. 22. (22)49. Perpendiculars are drawn from points on the line  2 12 1 3x y zto the plane x + y + z = 3. The feet ofperpendiculars lie on the line(A)  1 25 8 13x y z(B)  1 22 3 5x y z(C)  1 24 3 7x y z(D)  1 22 7 5x y zAnswer (D)Hint :       2 1, 32 1 3x y zx y zAny point on this line is (2 – 2, – – 1, 3)( , , )  BAC(–2, –1, 0)This point will satisfy the plane(2 – 2) + (– – 1) + (3) = 3 4 – 3 = 3  32So, point of intersection of plane is   5 91, ,2 2C .Now, point on line is (–2, –1, 0) and direction ratio of AB (from figure) is      2 11 1 1k .Any general point on line AB is (k – 2, k – 1, k).This will satisfy equation so, (k – 2) + (k – 1) + k = 3. k = 2Therefore, (, , )  (0, 1, 2).So, equation of line passing through BC is  1 27 512 2x y z  1 22 7 5x y z50. Let   ˆ ˆ ˆ3 2PR i j k and   ˆ ˆ ˆ3 4SQ i j k determine diagonals of a parallelogram PQRS and   ˆ ˆ ˆ2 3PT i j kbe another vector. Then the volume of the parallelepiped determined by the vectors  , andPT PQ PS is(A) 5 (B) 20(C) 10 (D) 30Answer (C)
  23. 23. (23)Hint :         1ˆ ˆ ˆ3 2PR a b i j k d S b( )PRQ a( )        2ˆ ˆ ˆ3 4SQ a b i j k d   ˆ ˆ ˆ2 3a i j k   ˆ ˆ ˆ2b i j k Volume of the required parallelopiped = 2 1 31 2 11 2 3= |2(6 – 2) + 1(3 – 1) – 3(2 – 2)|= 10 cubic unitsORArea of parallelogram =  1 21| |2d d Volume of parallelepiped = 3 1 211 3 421 2 3= 10 cubic units.SECTION - 2 : (One or More Options Correct Type)This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONE OR MORE are correct.51. Let ( 1)4 2 21( 1)k knnkS k . Then Sn can take value(s)(A) 1056 (B) 1088(C) 1120 (D) 1332Answer (A, D)Hint : ( 1)4 2 21( 1)k knnkS k2 2 2 2 2 2 2 2 2 2–1 – 2 3 4 5 6 ........ (4 3) (4 2) (4 1) (4 )nS n n n n            Sn =2 2 2 2 2 2 2 2 2 2 22 2 2 2(3 1 ) (4 2 ) (7 5 ) (8 6 ) (11 9 ) (12 10 )...... (4 1) (4 3) (4 ) (4 2)n n n n                 Sn = 2(1 3) 2(4 2) 2(7 5) 2(8 6) ...... 2(4 1 4 3) 2(4 4 2)n n n n              Sn =2 4 (4 1)2[1 2 3 ..... 4 ]2n nn     From A, 4n(4n + 1) = 10564n2 + n = 2644n2 + n – 264 = 0n = 8
  24. 24. (24)From B, 4n(4n + 1) = 1088 (Not possible)From C, 4n(4n + 1) = 1120 (Not possible)From D, 4n(4n + 1) = 1332n = 952. For 3 × 3 matrices M and N, which of the following statement(s) is (are) NOT correct?(A) NTMN is symmetric or skew symmetric, according as M is symmetric or skew symmetric(B) MN – NM is skew symmetric for all symmetric matrices M and N(C) MN is symmetric for all symmetric matrices M and N(D) (adj M) (adj N) = adj(MN) for all invertible matrices M and NAnswer (C, D)Hint :(NTMN)T = – NTMT(NT )TNTMTN(A) If M is skew symmetric, (NTMN)T = –NTMN,Hence skew symmetric.If M is symmetric, (MTMN)T = NTMN,Hence symmetric. Option (A) is correct(B) (MN – NM)T = (MN)T – (NM)T= NTMT – MTNT= –(MTMT – NTMT)= –(MN – NM) Skew symmetric option (B) is correct.(C) (MN)T = NTMTSymmetricity and skew symmetricity depend on nature of M and N, option (C) is incorrect.(D) adj(MM) = adj(N) adjM,Option (D) is incorrect.53. Let f(x) = x sin x, x > 0. Then for all natural numbers n, f (x) vanishes at(A) A unique point in the interval   1,2n n(B) A unique point in the interval    1, 12n n(C) A unique point in the interval (n, n + 1)(D) Two points in the interval (n, n + 1)Answer (B, C)
  25. 25. (25)Hint : f(x) = x sinxf (x) = sin x + x cos x = 0 – tan x = x0 1 322 5212Clear f (x) has one root in1, 12n n    , also f (x) has one root in (n, n + 1).54. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted intoan open rectangular box by folding after removing squares of equal area from all four corners. If the totalarea of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of therectangular sheet are(A) 24 (B) 32(C) 45 (D) 60Answer (A, C)Hint :(8 2 )(15 2 )V x x x    x 158=3 2 24 46 120x x x   2 212 92 120 0dVx xdx      , at x = 5 260 230 150 0    26 23 15 0    (6 5)( 3) 0    For  = 3 lengths of sides are 45, 2455. A line l passing through the origin is perpendicular to the lines     1ˆ ˆ ˆ: (3 ) ( 1 2 ) (4 2 ) ,l t i t j t k    t    2ˆ ˆ ˆ: (3 2 ) (3 2 ) (2 ) ,l s i s j s k    sThen, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and l1 is(are)(A)   7 7 5, ,3 3 3(B) (–1, –1, 0)(C) (1, 1, 1) (D)   7 7 8, ,9 9 9
  26. 26. (26)Answer (B, D)Hint :    x y za b c(Equation of line l)Equation of a line l1, 3 1 41 2 2x y zt    Equation of a line l2,3 3 22 2 1x y zs    Direction ratio of a line l is given by,ˆ ˆ ˆ1 2 22 2 1i j k= ˆ ˆ ˆ2 3 2i j k  Equation of a line l is2 3 2x y z    Point of intersection of l and l1,2 3 t    ...(1)3 2 1t   ...(2)Put the value of t, 3 2( 2 3) 1     3 4 6 1     7 7  1  Point of intersection is (2, –3, 2)So, 2 2 2(3 2 2) (3 2 3) (2 2) 17s s s         2 2 24 4 1 36 24 4 17s s s s s       29 28 20 0s s  102,9s   i.e., intersection points are (–1, –1, 0) and   7 7 8, ,9 9 9SECTION - 3 : (Integer Value Correct Type)This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive).56. The coefficients of three consecutive terms of (1 + x)n+5 are in the ratio 5 : 10 : 14. Then n =Answer (6)Hint : Let consecutive terms be tr+2, tr + 1, trSo,1 105rrtt
  27. 27. (27)( 5) ( 1) 121n rr    n – 3r + 3 = 0 …(i)also,211410rrtt 5n – 12r + 6 = 0 …(ii)Solving, 6n 57. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the packand the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removedcards is k, then k – 20 =Answer (5)Hint : The smallest value of n for which( 1)12242n n n(n + 1) > 2448 n > 49For n = 50( 1)2n n  1275So, k + (k + 1) = 1275 – 1224 = 51k = 25k – 20 = 558. Of the three independent events E1, E2 and E3, the probability that only E1 occurs is , only E2 occurs is and only E3 occurs is . Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations( – 2)p =  and ( – 3)p = 2. All the given probabilities are assumed to lie in the interval (0, 1).Then 13Probability of occurrence ofProbability of occurrence ofEEAnswer (6)Hint :  1 2 3( ) ( ) ( )P E P E P E ...(i) 1 2 3( ) ( ) ( )P E P E P E ...(ii) 1 2 3( ) ( ) ( )P E P E P E ...(iii) 1 2 3( ) ( ) ( )P E P E P E ...(iv)Divide (i) by (iv),         11( ) 2( )2P EP E11( )( ) 2P EP E  
  28. 28. (28)111 ( )( ) 2P EP E   111( ) 2P E   11( ) 2P E     12( )P E    …(v)Now,22 3      22 3        – 3 = 2 – 4  = 5– 4 =5 4  …(vi)Divide (iii) by (iv)          33( ) ( 2 )( )2P EP E33( ) 25 4( )P EP E    33( ) 5 4( ) 2P EP E    331 ( ) 5 4( ) 2P EP E     31 6 6 6( )( ) 2 2P E           P(E3) =26( )    132( )2( )6( )P EP E        = 6
  29. 29. (29)59. A vertical line passing through the point (h, 0) intersects the ellipse  2 214 3x yat the points P and Q. Letthe tangents to the ellipse at P and Q meet at the point R. If (h) = area of the triangle PQR,1 = 1/2 1max ( )hh and 2 = 1/2 1min ( )hh , then   1 2885=Answer (9)Hint :2 2+ = 14 3x ySLet P and Q be (h, ) and (h, –)So, R is   4, 0hNow,  =    1 4× 2 × –2hhRQ(h, – )h,( 0)P h,( )=   243 1 – × –4hhh= 23/24 –32hhSo,ddh< 0i.e.  is decreasingi.e. 1 =   12=15 458and 2 = (1) =92Now 1 –5  288 9
  30. 30. (30)60. Consider the set of eight vectors V =     ˆ ˆ ˆ : , , { 1,1}ai bj ck a b c . Three non-coplanar vectors can be chosenfrom V in 2p ways. Then p isAnswer (5)Hint : Total 8 vectors are shown in the figure.Total number of vectors = 8C3 = 56Number of coplaners = 2 × (6 × 2) = 2456 – 24 = 32  25

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