SOLUTIONS & ANSWERS FOR AIEEE- 2010 VERSION – A [PHYSICS, CHEMISTRY & MATHEMATICS] PART – A – PHYSICS 1. The initial shape of the wavefront of the ----Ans: Planar Sol: Initially parallel, cylindrical beam will have planar wavefront. 2. The speed of light in the medium is Ans: Minimum on the axis of the beam Sol: Intensity is maximum along the axis refractive index maximum along the axis speed minimum along the axis. 3. As the beam enters the medium ------Ans: Converge Sol: Intensity is maximum at the centre and decreases with increasing radius light energy getting concentrated near centre beam is converging.. 4. The speed of daughter nuclei -----Ans: Mm2c Sol: Mass lost = (m) Energy released = (m)c2 By conservation of momentum and energy each has energy 2v2M21 Mm2cvcm21v2M2122 5. The binding energy per nucleon for the ----------Ans: E2 > E1 Sol: In radioactive decay, the parent nucleus decays to a more stable daughter nuclei. E2 > E1 6. Statement – 1 When ultraviolet light is incident on a photocell, its stopping potential -----Ans: Statement – 1 is true, Statement-2 is false. Sol: h = KEmax + if hincreases, KEm,ax increases Stopping potential increases Photoelectrons have various speeds. 7. Statement – 1: Two particles moving in the same direction do not ------Ans: Statement -1 is true, Statement – 2 is true; Statement-2 is the correct explanation of Statement-1 Sol: Linear momentum is conserved and initial momentum is not zero final momentum is not zero particles have speed after collision Particles have some K.E after collision. 8. The figure shows the position – time (x t) graph of ---Ans: 0.8 Ns Sol: Impulse = change in momentum = mv2 mv1 = 0.4 (1) 0.4 1 = 0.8 Ns 9. Two long parallel wires are at a distance 2d apart. They carry -------Ans: Sol: Hence graph (2) is correct. 10. A ball is made of a material of density where ------ Bmax Bmax Bmax Bmax B A B A X X‟ B d d Ans: Sol: Since water is greater than oil, oil shoule be above water. > oil it should sink in oil and float in water. Hence Answer 3. 11. A thin semicircular ring of radius r has positive charge q ----Ans: jˆr2q202 Sol: Taking symmetrical elements of charge as shown the sin components cancel out. The coscomponents add upto 2/02cosrdqK2 = 2/02rcosrdrq.K2 = 2/020dcorq41.2 = j ˆ r 2 q 2 0 2 12. A diatomic ideal gas is used in a Carnot engine as the working substance ----Ans: 0.75 Sol: In the adiabatic part of the cycle T1V1 1 = T2 V21 = T2 (32 V1)1 21TT= (32)1 = (32)7/5 1 = (32)2/5 = 4 T1 = 4 T2 = 43TTT121= 0.75 13. The respective number of significant figures for the numbers ------Ans: 5, 1, 2 14. The combination of gates shown below ------Ans: OR gate Sol: B.A= A + B OR gate 15. If a source of power 4 kW produces 1020 photons /second, the ----Ans: X -rays Sol: Energy of a photon = J10400020 1920106.11104000eV = 250 eV 2501242nm 5 nm (X-rays) 16. A radioactive nucleus (initial mass number A and atomic number Z) emits ------Ans: 8Z4ZA Sol: 23YX12A8ZAZ No. of neutrons = A 12 (Z 8) Ratio = 8 Z 4 Z A 17. Let there be a spherically symmetric charge distribution with charge -----Ans: Rr354r00 Sol: x = 0 Rx45 4x2dx Total charge upto r is r = Rx45r004x2 dx = 40 r0Rx45x2 dx E E oil water x r dx = 40 r04r03R4x3x45 = 40 R4r3r4543 Gauss‟s law is E.4r2 = R4r3r4544300 E = Rr 35 4 r0 0 18. In a series LCR circuit R = 200 and the voltage and the frequency of the main supply ------Ans: 242 W Sol: Since the lag by removing the capacitance is equal to the lead by removing the inductor XC = XL. The circuit is in resonance condition. Power dissipated is 200220RV22 = 242 W 19. In the circuit shown below, the key K is closed at t = 0. --------Ans: 2RVat t = 0 and 2121RRRRVat t = Sol: At the instant of switching on there is no current through L. Therefore current at t = 0 is 2RV At t = , VL = 0 Reff = 21212121RRRRVRRRR 20. A particle is moving with velocity -------Ans: y2 = x2 + constant Sol: jˆxiˆyKv jˆKxiˆKydtrd CjˆKxtiˆKytr r2 = K2y2t2 + K2x2t2 + constant (x2 + y2) = K2y2t2 + K2x2t2 + constant y2 [1 K2t2] = x2[K2t2 1] + constant y2 = 22222tK11tKx+ constant = x2 + current = x2 + constant y2 = x2 + constant 21. Let C be the capacitance of a capacitor discharging through ------Ans: 41 Sol: Q = R0 et/RC 21.C2QC2Q2021 Q1 = 2Q0 /t001e.Q2Q /t1e21 1/tt2lne21 12etlog21 ---(1) /t002eQ4Q /t2e41 22etlog2 ---(2) 41tt)ii()i(21 22. A rectangular loop has a sliding connector PQ of length l and -----Ans: 1 = 2 = R3vB2,R3vB Sol: Motional emf, E= Blv Reffective (external) = R || R = 2R Internal resistance = R Total resistance = R + 2R32R = R3vB2R3223REE 1 = 2 = R3vB2 Aliter R + 1R = Blv ----(i) R + 2R = Blv ----(ii) (i) (ii) 1R 2R = 0 1 = 2 = 1 + 2 = 22 (ii) 3 2 R = Blv 2 = R3vB 1 = 2 = R 3 v B ; = R3vB2 23. The equation of a wave on a string of linear mass density 0.04 Ans: 6.25 N Sol: y = 0.02 sin 50.0x204.0t2 Compare with y = A sin ( t kx) = 50.02kand04.02 v = 04.05.0k= 12.5 m s1 But v = TT = v2 T = (12.5)2 0.04 = 6.25 N 24. Two fixed frictionless inclined planes making an angle -------Ans: 4.9 m s2 in the vertical direction. Sol: Acceleration of A along the plane, . a = g sin = g sin60 Component of acceleration of A in the vertical direction, av = a sin= g sin260(av)A = g sin2 60Similarly (av)B = g sin2 30(av)AB = (av)A (av)B = g [sin2 60sin2 30] = 9.8 4143 = 4.9 m s2 in the vertical direction. 25. For a particle in uniform circular motion, the acceleration ------Ans: jˆsinRViˆcosRV22 Sol: ac = RV2 (ac)X = iˆcosRV2 (ac)Y = jˆsinRV2 jˆsinRViˆcosRVa22c 26. A small particle of mass m is projected at an angle -----Ans: kˆcos2gtmv20 Sol: LprL is in the kˆ direction iˆcostvx0 jˆgt21sintvy20 jˆ2gtsinvtiˆcostvr00 jˆviˆvvyx = v0 cosiˆ+ (v0 singt)jˆ jˆgtsinvmiˆcosmvmvp00 prL = jˆ2gtsinvtiˆcostv00 jˆgtsinvmiˆcosmv00 = kˆ2gtsinvcostmvkˆgtsinvcostmv0000 = kˆcosgtmvcossintmv2020 kˆ2gtmvcossintmv2020 = k ˆcos 2gt mv 2 0 27. Two identical charged spheres are suspended by strings of Ans: 2 Sol: l sin 15= 202d4q mg d l 15152 02d 4 q R V P(R, ) X Y l cos 15 = mg tan 15= 0224dmgq ----(i) In liquid g‟ = g2g6.18.01g1 = 0 K Again tan 15= K4mgdq24d'mgq02222 ------(ii) From (i) and (ii) 0222024mgdqKmgd4q2 1K2 K = 2 28. A point P moves in counter-clockwise direction on a circular path as shown ----Ans: Sol: S = t3 + s Speed v = dtds= 3t2 At t = 2 s, v = 12 m s 1 ac = 2012rv22= 7.2 m s2 Tangential acceleration, at = dtdv= 6t At t = 2 s, at = 6 2 = 12 m s2 a = 222t2c122.7aa = 14484.51 = 84.195 14 m s2 29. The potential energy function for the force between two atoms -----Ans: 0 a4ba4b22 Sol: U(x) = 612xbxa F = dxdUx= [12 ax13 + 6bx7] = 12ax13 6 bx7 At equilibrium, F = 0 0 = 12ax13 6bx7 12ax13 = 6bx7 1 = 137xx.ab126 = 6x.a2b x = 6/1ba2at equilibrium U(x) = = 0 Uat equilibrium = 6/66/12ba2bba2a = a4ba2ba4ab2222 D = 0 a 4b a 4b 2 2 30. Two conductors have the same resistance at 0 C but their temperature Ans: Sol: In series R0 = R1 + R2 Rt = R1‟ + R2‟ = R1 + R1 1t + R2 + R22t = (R1 + R2) + t[R1 1 + R2 2] ------(i) But Rt = R0 + R0 t = (R1 + R2) + (R1 + R2)t -----(ii) From (i) & (ii) = 212211RRRR = 2121RR2 In parallel R0 = 2R Rt = t1t1Rt1Rt1R2121 = tt2t1t1R2121 ----(ii) But Rt = t12R ---(ii) From (i) & (ii) (1 + t) = tt2t1t122121 t = 1tt2ttt122122121 = tt2ttt2122121 = t2tt212121 = t2t212121; At t = 0, = 221 PART B CHEMISTRY 31. In aqueous solution the ionization constants ......... Ans : The concentration of H+ and 3HCO are approximately equal. Sol : H2CO3 H+ + 3 HCO 3HCO H+ + 23CO Since the k2 value is very low compared to that of k1, the H+ obtainable from 3 HCO is negligibly small. 32. Solubility product of silver bromide is 5.0 × 10–13....... Ans : 1.2 × 10–9 g Sol : ksp(AgBr) = [Ag+] [Br–] [Br–] = 05.010513= 1 × 10–11 moles/L No. of moles of KBr = 10–11 Wt of KBr = 120 × 10–11 = 1.2 × 10–9 g 33. The correct sequence which shows decreasing order of ............... Ans : O2– > F– > Na+ > Mg2+ > Al3+ Sol : For isoelectronic species the radii decreases with increase in atomic number. 34. In the chemical reactions, ................ Ans : benzene diazonium chloride and fluorobenzene Sol : NH2NaNO2HCl,278KN2Cl(A) HBF4heatF 35. If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, .................. Ans : 1.27 × 10–3 mol Sol : PV = nRT n = )K(300)molKJ(314.8)m(101)Pa(31701133 = 1.27 × 10–3 mol 36. From amongst the following alcohols the one that would react fastest with.............. Ans : 2–Methylpropan–2–ol Sol : CH3COHCH3CH32-Methylpropan-2-ol(3°alcohol) Order of reactivity of alcohols with con.HCl/ZnCl2 (Lucas reagent) is 3 > 2> 137. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change ................ Ans : 0.0558 K Sol : Tf = i × kf × m = 3 × 1.86 × 0.01 = 0.0558 K 38. Three reactions involving 42POH are given below:.................. Ans : (ii) only Sol : 42POH act as H+ donor in reaction (ii). 39. The main product of the following reaction is ..... Ans : H5C6CHCHCH(CH3)2 Sol : C6H5CH2CHCHCH3OHCH3H+H2O C6H5CH2CHCHCH3CH3(2°carbocation) ionhydrideofmigration2,1 C6H5CH2CHCH3CH3CH(morestable2°benzyliccarbocation)H H5C6 C H C HCH(CH3)2 40. The energy required to break one mole of Cl – Cl bonds in Cl2 is 242 kJ mol–1........... Ans : 494 nm Sol : E = 2331002.610242J molecule–1 E = ch = )moleculeJ(1002.610242)ms(103)Js(10626.612331834 = 0.494 × 10–6 m = 494 nm 41. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl‟s method ................ Ans : 23.7 Sol : % of N = 1000w100N)VV(14121 = 10000295.01001.0)1520(14= 23.7 42. Ionisation energy of He+ is 19.6 × 10–18 J atom–1. The energy ............... Ans : –4.41 × 10–17 J atom–1 Sol : E 22nz HeLiE49E2 = 118atomJ106.1949 = –4.41 × 10–17 J atom–1 43. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures ........... Ans : 72.0 kPa Sol : nA = 10025= 0.25 nB = 11435= 0.31 xA = 56.025.0= 0.45 p = B0BA0Ax.px.p = 105 × 0.45 + 45 × 0.55 = 72 kPa 44. Which one of the following has an optical isomer? ......... Ans : [Co(en)3]3+ Sol : [Co(en)3]3+ is chiral. 45. Consider the following bromides:............ Ans : B > C > A Sol : Order of SN1 reactivity is related to the relative stability of carbocation formed by ionisation (B) gives allylic secondary carbocation, (C) gives secondary carbocation and (A) gives primary carbocation on ionisation. 46. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde ................ Ans : 2–butene Sol : butene233CHCHCHCHozonolysis Ethanal3CHOCH2 Molecular mass : 44 u 47. Consider the reaction: Cl2(aq) + H2S(aq) S(s) + 2H+(aq) + 2Cl–(aq) ................... Ans : A only Sol : Slow step is the rate determining step. According to A; rate = K[Cl2][H2S] According to B; rate = ]H[]SH][Cl[K22 48. The Gibbs energy for the decomposition of Al2O3 at 500 C is as follows:.................. Ans : 2.5 V Sol : G = –nFE E965006)J(10966233 E = 2.5 V 49. The correct order of increasing basicity of the given conjugate bases .................. Ans : RHNCHCORCO2 Sol : Acidic strength of the corresponding conjugate acid is CH3 – COOH > CH CH > NH3 > CH4 Hence the basicity of the conjugate base must be the reverse. 50. The edge length of a face centered cubic cell of an anionic substance is 508 pm............ Ans : 144 pm Sol : 2(r(+) + r(–)) = a r(+) + r(–) = 2508= 254 r(–) = 254 – 110 = 144 pm 51. Out of the following, the alkene that exhibits optical isomerism is ............. Ans : 3–methyl–1–pentene Sol : CH3CH2CCHCH2CH3H3-Methyl-1-pentene It contains a chiral carbon atom. 52. For a particular reversible reaction at temperature T, H and S were found to be ........... Ans : T > Te Sol : At equilibrium, H = Te S G = H – TS = S (Te – T) G will be negative when T > Te. 53. Percentages of free space in cubic close packed structure and in body centered .............. Ans : 26% and 32% Sol : For ccp and bcc percentages of free space are 26% and 32% respectively. 54. The polymer containing strong intermolecular forces e.g. hydrogen bonding ............... Ans : nylon–6, 6 Sol : Nylon–6,6 is a fibre having strong intermolecular forces due to hydrogen bonding. 55. At 25C, the solubility product of Mg(OH)2 is 1.0 × 10–11. At which pH, will Mg2+ ions start precipitating ................ Ans : 10 Sol : ])OH(Mg[sp2k= [Mg2+] [OH–]2 [OH–]2 = 3111010= 10–8 [OH–] = 10–4 pOH = 4 pH = 10 56. The correct order of 0M/M2Evalues with negative sign for the four successive elements ............... Ans : Mn > Cr > Fe > Co Sol : Mn > Cr > Fe > Co Standard reduction potential values of Mn2+/Mn = –1.18 V Cr2+/Cr = –0.91 V Fe2+/Fe = –0.44 V Co2+/Co = –0.28 V 57. Biuret test is not given by ................ Ans : carbohydrates Sol : Biuret test is not answered by carbohydrates. 58. The time for half life period of a certain reaction A Products is 1 hour. When the initial concentration of the reactant „A‟,.................. Ans : 0.25 h Sol : For a zero order reaction, 21t a 2.0 mol L–1 1.0 mol L–1; 21t = 1 hour 0.5 mol L–1 0.25 mol L–1; 21t = 0.25 hour 59. A solution containing 2.675 g of CoCl3.6NH3 (molar mass = 267.5 g mol–1) is passed through a cation exchanger. .................. Ans : [Co(NH3)6]Cl3 Sol : No. of moles of AgCl = 5.14378.4 0.03 i.e., 0.01 moles of the compound gives 0.03 moles of AgCl No. of moles of Cl– per unit = 3 Formula of the complex is [Co(NH3)6]Cl3 60. The standard enthalpy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 from its atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1. .......... Ans : +352 kJ mol–1 Sol : N2 + 3H2 2NH3 2 × –46 = +712 + 3 × +436 – (6 × N – H) N – H = +352 kJ mol–1 PART – C -MATHEMATICS 61. Consider the following relations : R = {(x, y)|x, y are real numbers and ….. Ans: S is an equivalence relation but R is not an equivalence relation. Sol: x Ry = x = wy x Rx R is reflexive xRy x = wy and y Rx y = w‟x where w ‟ = w1, this is possible only if w 0 ie x R0 0 Rx ie; R is not symmetric R is not an equivalence relation. pnmqSqpnm nmnmS exists by the definition so S is reflexive . pnmqSqpnm pn = mq nmqpS S is symmetric. Again, srqpqpnmS,S mq = pn and ps = qr ie; mq.ps = pn.qr ms = nr srnmS S is transitive S is an equivalence relation but is not an equivalence relation. 62. The number of complex numbers z such that |z 1| = |z + 1| = ….. Ans: 1 Sol: z is a point equidistant from 3 given points. z is the centre of the circle passing through 1, 1, i. 63. If and B are the roots of the equation x2 x + 1 = 0, …… Ans: 1 Sol: , 2 2009 = ()2009 = 2007 . 2 = 2 2009 = (2)2009 = 4018 = 4017 = 2 = (2 + ) = 1. 64. Consider the system of linear equations : x1 + 2x2 + x3 = 3 2x1 + 3x2 + x3 = 3 ……………………… Ans: No solution. Sol: A = 0A253132121 Ax1 = .0Ax2511331231 The given system has no solutions. 65. There are two urns. Urn A has 3 distinct red balls ……. Ans: 108 Sol: A 3 distinct red balls B 9 distinct blue balls .108363CC2923 66. Let f : ( 1, 1) R be a differentiable function with ……… Ans: 4 Sol: g(x) = [f(2f(x) + 2)]2 g‟ (x) = 2f(2f(x) + 2) 2f‟ (x) g „(0) = 2f(2f(0) + 2) 2f‟(0) = 4 1 f(2 2) = 4 f(0) = 4. 67. Let f : R R be a positive increasing function with 1xfx3flimx….. Ans: 1 Sol: Given 1xfx3fLimx since f(x) is an increasing function, xfx2fLimx is also equal to 1. 68. Let p(x) be a function defined on R such that p‟(x) = p‟(1 x), ……. Ans: 21 Sol: f(x) = p(x) + p(1 x) f‟(x) = p‟(x) p‟(1 x) = 0(given) f‟(x) = 0 f(x) = k constant when x = 0, p(0) + p(1) k = 42 p(x) + p(1 x) = 42 101042x1pdxxp 2 1042dxxp 1021dxxp. 69. A person is to count 4500 currency notes. ….. Ans: 34 minutes Sol: In the first 9 minutes the person counts 9 150 = 1350 notes Total left notes = 4500 1350 = 3150 He counts in A.P with d = (2) and a = 150 3150 = 21n3002n = n[150 n + 1] 3150 = 151n n2 n2 151n + 3150 = 0 n = 250or2252 n = 25 Total time = 25 + 9 = 34 = 34 mts. 70. The equation of the tangent to the curve y = x + 2x4, …… Ans: y = 3 Sol: y = x + 2x4 0x810dxdy3 x = 2 y = 3 Equation of tangent y = 3. 71. The area bounded by the curves y = cos x and y = sin x …… Ans: 224 Sol: Required area = 40454dxxcosxsindxxsinxcos + 2545dxxsinxcos = 122212 = 2 2 4 . 72. Solution of the differential equation cos x dy = y (sinx y) dx, …… Ans: secx = (tanx +c)y Sol: Consider dy = y(sinx y)dx consider 2yxsinydxdy xsecyxtanydxdy2 xsecyxtanydxdy2 xsecxtany1dxdyy12 z = y1 dxdyy1dxdz2 xsecxtanzdxdz I. F elogsecx = secx zsecx = Cxtanxsec2 Cxtanyxsec secx = y(tanx +C). 73. Let kˆjˆa and kˆjˆiˆc. …… Ans: k2ji Sol: (a b) + c = 0 a (a b) + a c = 0 (a . b)a (a . a)b + a c = 0 0kji2b2k3j3 k4j2i2b2 k2jib 74. If the vectors kˆ2jˆiˆa, ……. Ans: (3, 2) Sol: 0c.a 1 + 2 = 0 + 2= 1 -------(1) 0c.b 2+ 4 + = 0 2+ = 4 ----(2) Solving = 3 and = 2 75. If two tangents drawn from a point P to the parabola y2 = 4x are at right angles, …. Ans: x = 1 Sol: Locus of p is directrix of y2 = 4x x = 1 76. The line L given by 1by5x passes …….. Ans: 1 Sol: 1by5x passes through (13, 32) = 20 Equation is 4x y = 20. It is parallel to 13ycx c = 43. ie; equation of line k becomes 4x 3y = 3. The distance between them = 116320 = 1723. 77. A line AB in three-dimensional space makes…….. Ans: 60 Sol: cos2 45 + cos2 120 + cos2 = 1 1cos41212 cos2= 1 4143 cos= 21 = 60. 78. Let S be a non-empty subsets of R. ….. Ans: Every rational number x S satisfies x 0 Sol: “There is a rational number x S such that x > 0” means that “There exist some positive rational number in S”. Then its negation becomes “All rational numbers in S are negative”. That is “Every rational number x S satisfies x 0”. 79. Let 54cosand . ….. Ans: 3356 Sol: tan2 = tan(+ + ) = tantan1tantan = 12543112543 = .3356 80. The circle x2 + y2 = 4x + 8y + 5 …… Ans: 35 < m < 15 Sol: Perpendicular distance from (2, 4) < Radius 525m166 = 55m10 = |10 m| < 25 25 < 10 + m < 25 35 < m < 15. 81. For two data sets, each of size 5…… Ans: 211 Sol: ,nndndnnn212222112222112 212211nnxnxnx since n1 = n2 we get 2dd222122212 2xxx21 d12 = (2 3)2 = 1 3242x d22 = (4 3)2 = 1 211211542. 82. An urn contains nine balls of which….. Ans: 72 Sol: Three balls without replacement can be done in = 39C243 = 72. 83. For a regular polygon, let r and R be the….. Ans: There is a regular polygon with 32Rr Sol: Let n sided regular polygon is inscribed in a circle. From the figure it is clear that Rrncos There an possible integer value corresponding to 23and21,21 But cos = 32cos4321 n is not an integer. 84. The number of 3 3 non-singular matrices…… Ans: at least 7 Sol: Consider 100010011. The 1 on the non diagonal position can be shifted to 5 more positions. Further we can consider 001100011. at least 7 matrices are there. 85. Let f : R R be defined by…… Ans: 1 Sol: Since function has local minimum it must be continuous at x = 1 xfLimxfLim1x1x 1 = k + 2 k = 1. 86. Four numbers are chosen at random…… Ans: Statement 1 is true, Statement 2 is false. Sol: If four chosen numbers form an AP, the common differences can be 1, 2, 3, 4, 5 or 6. (e.g. 1, 7, 13, 19 is an AP with common difference 6) Statement 2 is not true. 87. Let 101jj101C1jjS, 101jj102CjS….. Ans: Statement 1 is true, Statement 2 is false. Sol: S1 = 101jj10C1jj S2 = 101jj10Cj S3 = 101jj102Cj S1 S3 = 101jj1022Cjjj = 101jj10Cj = S2 S1+ S2 = S3. n2 R r 1jj!j10!j!10 !j10!2j!8109!)j10(!2j!10 101j101j2j8j10C90C1jj = 90 28. . 88. Statement 1 : The point A (3, 1, 6) is the mirror image……. Ans: Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for statement 1 Sol: A (3, 1, 6) B = (1, 3, 4) Midpoint of AB is (2, 2, 5) 2 2 + 5 = 5 Statement 2 is true D. R‟s of AB are [2, 2, 2] or [1, 1, 1] which represent the D.R‟s of normal to the plane x y + z = 5 Statement 1 is true We used statement 2 to prove statement 1. 89. Let f : R R be a continuous function…… Ans: Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for statement 1 Sol: f(x) = 0xfe2e1xx f‟(x) = xx2xxe2ee2e1 f‟(x) = 0 ex = xe2 e2x = 2 x = 21log2 Checking the sign of f‟(x) as x crosses ,2log21 we note that f(x) is maximum at x = 2log21. Maximum value of f(x) = 21221 = 221 Statement 2 is true 4414.142221 = 0.3535 Since f(x) is continuous in R, f(x) has to assume all values between 0 and 0.3535 Since 31 is a number lying between o and 0.3535, statement 1 is also true. 90. Let A be a 2 2 matrix with non-zero…….. Ans: Statement 1 is true, Statement 2 is false. Sol: Let A = dcba A2 = |A| = 1 Statement 2 is NOT true. A2 = 1001dcbadcba 22dbccdabdabca )1(1bcd1bca22 )2(0cda0bda Since b or c 0 a + d = 0 a = (d) Tr(A) = 0 Statement – 1 is true.