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PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS
SOLUTIONS & ANSWERS FOR AIEEE-2010 VERSION – A [PHYSICS, CHEMISTRY & MATHEMATICS] PART – A – PHYSICS Ans: Statement – 1 is true, Statement-2 is false. 1. The initial shape of the wavefront of the ---- Sol: hυ = KEmax + φ ∴ if hυ increases, KEm,ax increases Ans: Planar ∴ Stopping potential increases Sol: Initially parallel, cylindrical beam will have Photoelectrons have various speeds. planar wavefront. 7. Statement – 1: 2. The speed of light in the medium is Two particles moving in the same direction do Ans: Minimum on the axis of the beam not ------ Sol: Ans: Statement -1 is true, Statement – 2 is true; Statement-2 is not the correct explanation Ιr < ΙC (µr < µC) of Statement-1 ΙC Sol: Statement – 1 is true because the energy (µr < µC) considered is not kinetic energy alone. Statement – 1 is correct. Statement -2 is correct but not the explanation for wavefront Statement – 1. 3. As the beam enters the medium ------ Ans: Diverge 8. The figure shows the position – time (x − t) graph of --- Sol: From the above diagram the beam on coming out into the medium, diverges. Ans: 0.8 Ns 4. The speed of daughter nuclei ----- Sol: Impulse = change in momentum = mv2 − mv1 2 ∆m = 0.4 × (−1) − 0.4 × 1 = −0.8 Ns Ans: c M 9. Two long parallel wires are at a distance 2d Sol: Mass lost = (∆m) apart. They carry ------- 2 Energy released = (∆m)c Ans: By conservation of momentum and energy 1M 2 B each has energy v 2 2 • • 2(∆m ) X X’ v = (∆m ) c 2 ⇒ v = c 1M 2 1 ∴ 2 2 2 M d d Sol: 5. The binding energy per nucleon for the ---------- Ans: E2 > E1 Bmax Sol: In radioactive decay, the parent nucleus B Bmax decays to a more stable daughter nuclei. • • ∴ E2 > E1 A A B Bmax 6. Statement – 1 Bmax When ultraviolet light is incident on a photocell, its stopping potential ----- Hence graph (2) is correct. 10. A ball is made of a material of density ρ where --- 13. The respective number of significant figures for --- the numbers ------ Ans: Ans: 5, 1, 2 oil 14. The combination of gates shown below ------ Ans: OR gate water Sol: A.B = A + B Sol: Since ρwater is greater than ρoil, ρoil shoule OR gate be above water. ρ > ρoil it should sink in oil and float in 15. If a source of power 4 kW produces 10 20 photons water. / second, the ---- Hence Answer 3. Ans: X - rays 11. A thin semicircular ring of radius r has positive charge q ---- 4000 Sol: Energy of a photon = J 1020 q 4000 1 Ans: − ˆ j = × eV 2 2π ε0r 2 10 1.6 × 10 −19 20 = 250 eV Sol: 1242 ⇒ nm ≅ 5 nm (X- rays) 250 • • • • 16. A radioactive nucleus (initial mass number A and θ atomic number Z) emits ------ A−Z−4 θ E Ans: E Z−8 Taking symmetrical elements of charge as Sol: ZX A →Z −8 Y A −12 + 3α + 2β+ shown the sinθ components cancel out. The cosθ components add upto No. of neutrons = A − 12 − (Z − 8) π / 2 K dq A−Z−4 ∴ Ratio = 2 0 ∫ r2 cos θ Z−8 π/2 q cos θ = 2 ∫ 0 K. rdθ 2 πr r 17. Let there be a spherically symmetric charge distribution with charge ----- 1 q π/2 = 2. 4πε0 πr 2 ∫ coθ dθ 0 ρ0r 5 r q Ans: 4ε 0 3 − R = − 2 2ˆ j 2π ε0r Sol: 12. A diatomic ideal gas is used in a Carnot engine as the working substance ---- dx Ans: 0.75 x r Sol: In the adiabatic part of the cycle T1V1γ − = T2 V2γ − 1 1 = T2 (32 V1)γ − 1 T ∴ 1 = (32)γ − = (32) − = (32) 5 x 1 7/5 1 2/5 ρx = ρ0 − 4πx dx 2 T2 4 R = 4 ⇒ T1 = 4 T2 Total charge upto r is T −T 3 η = 1 2 = = 0.75 T1 4 y [1 − K t ] = x [K t − 1] + constant 2 2 2 2 2 2 5 x r ρr = ∫ρ0 − 4πx dx [ ] 2 0 4 R x 2 K 2t 2 − 1 [1− K t ] + constant 2 y = 2 2 x 2 r5 = 4πρ0 ∫ − x dx 04 R = −x + current = x + constant 2 2 ∴ y = x + constant 2 2 x3 r x4 r 5 = 4πρ0 − 4 3 0 4 R 0 21. Let C be the capacitance of a capacitor discharging through ------ 5 r3 r4 = 4πρ0 − 4 3 4 R 1 Ans: Gauss’s law is 2 4πρ0 5 r 3 r4 − Q = R0 e− 2 t/RC E.4πr = Sol: ε0 4 3 4 R Q12 Q02 1 = . ρ0r 5 r 2C 2C 2 ⇒E= − 4ε 0 3 R Q0 ⇒ Q1 = 2 18. In a series LCR circuit R = 200 Ω and the voltage Q0 ∴ = Q0 .e − t1 / τ and the frequency of the main supply ------ 2 1 Ans: 242 W = e −t1 / τ 2 2 = e t1 / τ ln 2 = 1 t Sol: Since the lag by removing the capacitance ---(1) τ is equal to the lead by removing the Q2 2 Q0 1 2 inductor XC = XL. = . 2C 2C 4 The circuit is in resonance condition. V 2 (220 )2 Q0 Q2 = Power dissipated is = 2 R 200 Q0 = 242 W ∴ = Q0 e − t 2 / τ 2 ∴ 2 = et 2 / τ 19. In the circuit shown below, the key K is closed at t2 t = 0. -------- ∴ ln2 = τ V V (R1 + R2 ) ∴ t1 ln 2 1 = = Ans: at t = 0 and at t = ∞ t2 ln 2 2 R2 R1R2 Sol: At the instant of switching on there is no 22. A rectangular loop has a sliding connector PQ of current through L. Therefore current at length l and ----- V t = 0 is R2 Bλv 2Bλv Ans: Ι1 = Ι2 = ,Ι= At t = ∞, VL = 0 3R 3R R1R2 V (R1 + R2 ) Reff = ⇒Ι= Sol: Motional emf, E = Blv R1 + R2 R1R2 R Reffective (external) = R || R = 2 20. A particle is moving with velocity ------- Internal resistance = R 2 2 R 3R Ans: y = x + constant Total resistance = R + = 2 2 Sol: i( v = K y ˆ + xˆj ) ∴Ι= E = 2E = 2 Bλv 3R 3 R 3R dr = Ky ˆ + Kx ˆ i j dt 2 Ι Bλv ⇒ r = Kyt ˆ + Kxtˆ + C i j Ι1 = Ι2 = = 2 2 2 2 2 2 2 2 3R ⇒ r = K y t + K x t + constant 2 2 2 2 2 2 2 2 (x + y ) = K y t + K x t + constant Aliter V2 ac = R ΙR + Ι1R = Blv ---- (i) V2 ΙR + Ι2R = Blv ---- (ii) (ac)X = − cos θ ˆ i R (i) − (ii) ⇒ Ι1R −Ι2R = 0 ⇒ Ι1 = Ι2 V2 Ι = Ι1 + Ι2 = 2Ι2 (ac)Y = − sin θˆ j ∴ (ii) ⇒ 3 Ι2 R = Blv R Bλv V2 V2 ⇒ Ι2 = ∴ ac = − cos θ ˆ − i sin θ ˆ j 3R R R Bλv 2 Bλv Ι1 = Ι2 = ;Ι= 3R 3R 26. A small particle of mass m is projected at an angle θ ----- 23. The equation of a wave on a string of linear mass mv 0gt 2 ˆ density 0.04 Ans: − cos θ k 2 Ans: 6.25 N Sol: ˆ L = r × p ⇒ L is in the − k direction 2πt 2πx Sol: y = 0.02 sin − x = tv cos θ ˆ 0 i 0.04 0.50 1 Compare with y = A sin (ωt − kx) y = tv 0 sin θ − gt 2 ˆ j 2 2π 2π ⇒ω= and k = gt 0.04 0.50 r = v 0 t cos θ ˆ + t v 0 sin θ − ˆ i j ω 0 .5 2 ∴v= = = 12.5 m s− 1 k 0.04 v = v ˆ+v ˆ ix jy T = v0 cosθ ˆ + (v0 sinθ − gt) ˆ ⇒T=v µ 2 i j But v = µ p = mv = mv 0 cos θ ˆ + m(v 0 sin θ − gt )ˆ i j ∴ T = (12.5) × 0.04 2 L = r×p = 6.25 N gt = v 0 t cos θ ˆ + t v 0 sin θ − ˆ × i j 24. Two fixed frictionless inclined planes making an 2 angle ------- [ mv cos θ ˆ + m(v sin θ − gt )ˆ 0 i 0 j ] Ans: Zero = gt ˆ mv 0 t cos θ(v 0 sin θ − gt )k − tmv 0 cos θ v 0 sin θ − k ˆ Sol: Vertical acceleration of A and B are g. 2 Hence relative vertical acceleration of A w.r.t B is zero. [ = mv t sin θ cos θ − mv gt cos θ k 2 0 2 0 ˆ ] 2 mv 0gt 2 ˆ − mv 0 t sin θ cos θ − k 25. For a particle in uniform circular motion, the 2 acceleration ------ mv 0gt 2 ˆ = − cos θ k 2 V2 V2 Ans: − cos θ ˆ − i sin θ ˆ j R R 27. Two identical charged spheres are suspended by strings of Sol: Y Ans: 2 V Sol: P(R, θ) R θ X 15° 15° l q2 • • 4πε0d2 d mg q2 6 b x −7 l sin 15° = 1= . 4πε0d 2 12 a x −13 l cos 15° = mg b 6 = .x 2a q2 tan 15° = ---- (i) 1/ 6 mg d2 4 πε0 2a ∴x= at equilibrium σ b 0 .8 g In liquid g’ = g 1 − = g1 − = U(x) = ∞ = 0 ρ 1 .6 2 a b ε = ε0 K Uat equilibrium = 12 / 6 − 6/6 2a 2a Again tan 15° q2 2q2 b b = = ------(ii) 2 mg' d 4 πε mgd2 4πε0K ab2 b2 b2 = − =− 4a2 2a 4a 2q2 q2 From (i) and (ii) = b2 b2 4πε0Kmgd2 mgd2 4πε0 ∴ D = 0 − − = 4a 4a 2 ⇒ =1 ⇒ K = 2 K 30. Two conductors have the same resistance at 28. A point P moves in counter-clockwise direction 0 °C but their temperature on a circular path as shown ---- Ans: Ans: Sol: In series 3 R0 = R1 + R2 Sol: S=t +s ds 2 Rt = R1’ + R2’ Speed v = = 3t dt = R1 + R1α1t + R2 + R2α2t At t = 2 s, v = 12 m s− 1 = (R1 + R2) + t[R1 α1 + R2 α2] ------ (i) v 2 (12)2 But Rt = R0 + R0 αt = = 7.2 m s− 2 ac = = (R1 + R2) + (R1 + R2)αt ----- (ii) r 20 From (i) & (ii) Tangential acceleration, dv α= (R1α1 + R2α2 ) at = dt = 6t (R1 + R2 ) α1 + α 2 At t = 2 s, at = 6 × 2 = 12 m s− 2 = (Θ R1 = R2 ) 2 ∴a= ac 2 + a t 2 = (7.2)2 + (12)2 = 51.84 + 144 In parallel R = 195.84 R0 = 2 ≅ 14 m s− 2 R[1 + α1t ]R[1 + α 2 t ] Rt = R{[1 + α1t ] + [1 + α 2t ]} 29. The potential energy function for the force R(1 + α1t )(1 + α 2t ) between two atoms ----- = ----(ii) [2 + α1t + α2t] Ans: b2 b2 0 − − = But Rt = R [1 + αt ] --- (ii) 4a 4a 2 From (i) & (ii) (1 + αt) 2[1 + α1t ](1 + α 2t ) = Sol: U(x) = a − b [2 + α1t + α2t] [ ] 12 x x6 dU(x ) 2 1 + α1t + α 2 t + α1α 2t 2 αt = −1 F= − dx = −[−12 ax− + 6bx− ] 13 7 [2 + α1t + α2t] = 12ax− − 6 bx− 13 7 α1t + α 2t + α1α 2 t 2 = At equilibrium, F = 0 ⇒ 0 (2 + α1t + α2t ) = 12ax− − 6bx− t(α1 + α 2 + α1α 2t ) 13 7 = −13 ∴ 12ax = 6bx −7 2 + (α1 + α 2 )t α1 + α 2 + α1α 2t F ⇒α= ; At t = 0, 2 + (α1 + α 2 )t α1 + α 2 HBF4 α= 2 heat PART B − CHEMISTRY –4 3 3 35. If 10 dm of water is introduced into a 1.0 dm 31. In aqueous solution the ionization constants flask at 300 K, .................. ......... –3 Ans : 1.27 × 10 mol − Ans : The concentration of H and HCO3 are + Sol : PV = nRT 3170 (Pa ) × 1 × 10 −3 (m3 ) approximately equal. n= + − 8.314 (J K −1 mol−1) × 300(K ) Sol : H2CO3 H + HCO3 –3 = 1.27 × 10 mol − HCO3 H + CO3 − + 2 Since the k2 value is very low compared to 36. From amongst the following alcohols the one that + − that of k1, the H obtainable from HCO3 is would react fastest with.............. negligibly small. Ans : 2–Methylpropan–2–ol 32. Solubility product of silver bromide is –13 CH3 5.0 × 10 ....... Sol : CH3 C OH 2-Methylpropan-2-ol Ans : 1.2 × 10 –9 g alcohol) (3° CH3 + – Order of reactivity of alcohols with Sol : ksp(AgBr) = [Ag ] [Br ] con.HCl/ZnCl2 (Lucas reagent) is 5 × 10 −13 – [Br ] = –11 = 1 × 10 moles/L 3° > 2° > 1° 0.05 ∴ No. of moles of KBr = 10 –11 –11 –9 37. If sodium sulphate is considered to be completely Wt of KBr = 120 × 10 = 1.2 × 10 g dissociated into cations and anions in aqueous solution, the change ................ 33. The correct sequence which shows decreasing order of ............... Ans : 0.0558 K 2– – + 2+ 3+ Ans : O > F > Na > Mg > Al Sol : ∆Tf = i × kf × m = 3 × 1.86 × 0.01 Sol : For isoelectronic species the radii = 0.0558 K decreases with increase in atomic number. 38. Three reactions involving H2PO− are given 4 34. In the chemical reactions, ................ below:.................. Ans : benzene diazonium chloride and Ans : (ii) only fluorobenzene H2PO− act as H donor in reaction (ii). + Sol : 4 NH2 N2Cl 39. The main product of the following reaction is ..... NaNO2 Sol : Ans : H5C6 H HCl, 278K C C (A) H CH(CH3)2 Sol : z2 CH3 Sol : E α n2 + H 9 CH 2 CH CH CH3 ELi2 + = × EHe+ C6H5 H2 O 4 9 = × − 19.6 × 10 −18 J atom−1 OH CH3 4 –17 –1 CH 3 = –4.41 × 10 J atom C6H 5 CH2 CH CH (2°carbocation) 43. On mixing, heptane and octane form an ideal 1, 2 −migration solution. At 373 K, the vapour pressures ........... → of hydride ion CH 3 Ans : 72.0 kPa −H+ C6H5 CH CH2 CH CH3 → 25 Sol : nA = = 0.25 (more stable 100 2°benzylic carbocation) 35 H5C6 nB = = 0.31 H 114 C C 0.25 CH(CH3)2 xA = = 0.45 H 0.56 0 p = p 0 . x A + p B .x B A 40. The energy required to break one mole of Cl – Cl –1 bonds in Cl2 is 242 kJ mol ........... = 105 × 0.45 + 45 × 0.55 = 72 kPa Ans : 494 nm 44. Which one of the following has an optical isomer? ......... 242 × 103 –1 Sol : E = J molecule 23 6.02 × 10 Ans : [Co(en)3] 3+ h×c E= λ 3+ Sol : [Co(en)3] is chiral. ∴λ 6.626 × 10 −34 (Js) × 3 × 108 (ms −1) 45. Consider the following bromides:............ = 242 × 103 ( J molecule −1) Ans : B > C > A 6.02 × 1023 –6 = 0.494 × 10 m Sol : Order of SN1 reactivity is related to the = 494 nm relative stability of carbocation formed by ionisation (B) gives allylic secondary 41. 29.5 mg of an organic compound containing carbocation, (C) gives secondary nitrogen was digested according to Kjeldahl’s carbocation and (A) gives primary method ................ carbocation on ionisation. Ans : 23.7 46. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde ................ 14 × ( V1 − V2 ) N1 × 100 Sol : % of N = w × 1000 Ans : 2–butene 14 × (20 − 15 ) × 0.1 × 100 = = 23.7 ozonolysis 0.0295 × 1000 Sol : CH3 − CH = CH − CH3 → 2 − butene 42. Ionisation energy of He+ is 19.6 × 10 –18 J atom . –1 2CH3 − CHO Ethanal The energy ............... Molecular mass : 44 u –17 –1 Ans : –4.41 × 10 J atom 47. Consider the reaction: 52. For a particular reversible reaction at Cl2(aq) + H2S(aq) → S(s) + 2H (aq) + 2Cl (aq) temperature T, ∆H and ∆S were found to be + – ................... ........... Ans : A only Ans : T > Te Sol : Slow step is the rate determining step. Sol : At equilibrium, ∆H = Te ∆S According to A; rate = K[Cl2][H2S] ∴ ∆G = ∆H – T∆S K[Cl 2 ][H 2 S] = ∆S (Te – T) According to B; rate = [H + ] ∆G will be negative when T > Te. 53. Percentages of free space in cubic close packed 48. The Gibbs energy for the decomposition of Al2O3 structure and in body centered .............. at 500°C is as follows:.................. Ans : 26% and 32% Ans : 2.5 V Sol : For ccp and bcc percentages of free Sol : ∆G = –nFE 3 space are 26% and 32% respectively. × 966 × 10 3 (J) = 6 × 96500 × E 2 54. The polymer containing strong intermolecular E = 2.5 V forces e.g. hydrogen bonding ............... 49. The correct order of increasing basicity of the Ans : nylon–6, 6 given conjugate bases .................. Sol : Nylon–6,6 is a fibre having strong − − − − Ans : RCO O < HC ≡ C < N H2 < R intermolecular forces due to hydrogen bonding. Sol : Acidic strength of the corresponding conjugate acid is 55. At 25° C, the solubility product of Mg(OH)2 is –11 2+ CH3 – COOH > CH ≡ CH > NH3 > CH4 1.0 × 10 . At which pH, will Mg ions start Hence the basicity of the conjugate base precipitating ................ must be the reverse. Ans : 10 50. The edge length of a face centered cubic cell of 2+ – 2 an anionic substance is 508 pm............ Sol : k sp[Mg(OH)2 ] = [Mg ] [OH ] 10 −11 ∴ [OH ] = – 2 –8 Ans : 144 pm = 10 10 − 3 – –4 [OH ] = 10 Sol : 2(r(+) + r(–)) = a pOH = 4 508 r(+) + r(–) = = 254 pH = 10 2 r(–) = 254 – 110 = 144 pm 56. The correct order of E0 2 + values with negative M /M 51. Out of the following, the alkene that exhibits sign for the four successive elements ............... optical isomerism is ............. Ans : Mn > Cr > Fe > Co Ans : 3–methyl–1–pentene Sol : Mn > Cr > Fe > Co H Standard reduction potential values of 2+ CH3 CH2 C CH CH2 Mn /Mn = –1.18 V 2+ Sol : Cr /Cr = –0.91 V CH3 2+ Fe /Fe = –0.44 V 2+ Co /Co = –0.28 V 3-Methyl-1-pentene It contains a chiral carbon atom. 57. Biuret test is not given by ................ if w ≠ 0 ie x R0 ⇒ 0 Rx ie; R is not symmetric Ans : carbohydrates ∴ R is not an equivalence relation. m S p ⇒ mq = pn n q Sol : Biuret test is not answered by carbohydrates. ∴ m Sm exists by the definition so S is n n reflexive . 58. The time for half life period of a certain reaction m S p ⇒ mq = pn ⇒ pn = mq ⇒ p Sm A → Products is 1 hour. When the initial n q q n concentration of the reactant ‘A’,.................. ∴ S is symmetric. Ans : 0.25 h ⇒ mq = pn and ps = qr Again, m Sp , pS r n q q s Sol : For a zero order reaction, t 1 α a 2 ie; mq.ps = pn.qr ⇒ ms = nr ⇒ mSr → 1.0 mol L ; t 1 = 1 hour –1 –1 n s 2.0 mol L 2 ∴ S is transitive 0.5 mol L –1 → 0.25 mol L ; –1 ∴ S is an equivalence relation but is not t 1 = 0.25 hour an equivalence relation. 2 62. The number of complex numbers z such that 59. A solution containing 2.675 g of CoCl3.6NH3 |z − 1| = |z + 1| = ….. –1 (molar mass = 267.5 g mol ) is passed through Ans: 1 a cation exchanger. .................. Sol: z is a point equidistant from 3 given Ans : [Co(NH3)6]Cl3 points. ∴ z is the centre of the circle passing 4.78 through 1, −1, i. Sol : No. of moles of AgCl = ≅ 0.03 143.5 63. If α and B are the roots of the equation x − x + 1 = 0, …… i.e., 0.01 moles of the compound gives 2 0.03 moles of AgCl ∴ No. of moles of Cl per unit = 3 – Ans: 1 ∴Formula of the complex is [Co(NH3)6]Cl3 −ω, −ω 2 Sol: α = (−ω) 2009 2009 60. The standard enthalpy of formation of NH3 is = −ω 2007 .ω 2 –1 = −ω –46.0 kJ mol . If the enthalpy of formation of H2 2 –1 β = (−ω ) 2009 2 2009 from its atoms is –436 kJ mol and that of N2 is = −ω –1 4018 –712 kJ mol . .......... = −ω ×ω 4017 Ans : +352 kJ mol –1 = −ω −ω − ω = −(ω + ω) = 1. 2 2 64. Consider the system of linear equations : Sol : N2 + 3H2 → 2NH3 x1 + 2x2 + x3 = 3 2 × –46 = +712 + 3 × +436 – (6 × N – H) 2x1 + 3x2 + x3 = 3 –1 N – H = +352 kJ mol ……………………… Ans: No solution. PART – C -MATHEMATICS 1 2 1 61. Consider the following relations : R = {(x, y)|x, y are real numbers and ….. Sol: A = 2 3 1 ⇒ A = 0 3 5 2 Ans: S is an equivalence relation but R is not an equivalence relation. 3 2 1 Ax1 = 3 3 1 ⇒ Ax1 ≠ 0. Sol: x Ry = x = wy ⇒ x Rx 1 5 2 ∴ R is reflexive xRy ⇒ x = wy and y Rx ⇒ y = w’x ∴ The given system has no solutions. 1 where w ’ = , this is possible only w [300 + (n − 1)(− 2)] 65. There are two urns. Urn A has 3 distinct red n balls ……. ∴ 3150 = 2 = n[150 − n + 1] Ans: 108 3150 = 151n − n 2 Sol: A ⇒ 3 distinct red balls ∴ n − 151n + 3150 = 0 2 B ⇒ 9 distinct blue balls 252 50 3 C2 × 9C2 = 3 × 36 = 108. ⇒n= or 2 2 n = 25 66. Let f : (−1, 1) → R be a differentiable function ∴ Total time = 25 + 9 = 34 with ……… = 34 mts. Ans: −4 70. The equation of the tangent to the curve 4 Sol: g(x) = [f(2f(x) + 2)] 2 y = x + 2 , …… x g’ (x) = 2f(2f(x) + 2) × 2f’ (x) g ‘(0) = 2f(2f(0) + 2) × 2f’(0) Ans: y=3 = 4 × 1 × f(2 − 2) = 4 f(0) 4 = −4. Sol: y=x+ x2 67. Let f : R → R be a positive increasing function dy 8 = 0 ⇒ 1− 3 = 0 f (3 x ) dx x with lim = 1 ….. x →∞ f (x ) ⇒x=2 ∴y=3 Ans: 1 ∴ Equation of tangent y = 3. f (3 x ) 71. The area bounded by the curves y = cos x and Sol: Given Lim =1 f (x ) y = sin x …… x →∞ since f(x) is an increasing function, f (2x ) Ans: 4 2 −2 Lim is also equal to 1. x →∞ f (x ) Sol: 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) Required area = f’(x) = p’(x) − p’(1 − x) = 0(given) π 5π ∴ f’(x) = 0 4 4 ⇒ f(x) = k constant when x = 0, p(0) + p(1) ⇒ k = 42 ∫ (cos x − sin x )dx + ∫ (sin x − cos x)dx 0 π p(x) + p(1 − x) = 42 4 1 1 5π p(x )dx + p(1 − x ) = 42 ∴ ∫ ∫ 2 0 1 0 + ∫ (cos x − sin x )dx 5π ∴2 ∫ p(x )dx = 42 0 = 4 2 − 1+ 2 2 + 2 − 1 1 = 4 2 −2. p(x )dx = 21 . ∴ ∫ 0 72. Solution of the differential equation cos x dy = y (sinx − y) dx, …… 69. A person is to count 4500 currency notes. ….. Ans: secx = (tanx +c)y Ans: 34 minutes Sol: Consider dy = y(sinx − y)dx Sol: In the first 9 minutes the person counts dy consider = y sin x − y 2 9 × 150 = 1350 notes dx Total left notes = 4500 − 1350 dy = 3150 = y tan x − y 2 sec x dx He counts in A.P with d = (−2) and a = 150 dy −3 − y tan x = − y 2 sec x ∴c= . ie; equation of line k becomes dx 4 −1 dy 1 4x − 3y = −3. + tan x = sec x y 2 dx y ∴ The distance between them 1 dz −1 dy 20 − (− 3 ) z= ⇒ = = y dx y 2 dx 16 + 1 dz 23 ∴ + z tan x = sec x = . dx 17 ∴ I. F e logsecx = secx 77. A line AB in three-dimensional space ∫ sec 2 ∴z x = tan x + C secx = makes…….. sec x = tan x + C Ans: 60° y ∴ secx = y(tanx +C). Sol: cos 45 + cos 120 + cos θ = 1 2 2 2 1 1 ρ ρ + + cos2 θ = 1 j ˆ i j ˆ 73. Let a = ˆ − k and c = ˆ − ˆ − k . …… 2 4 3 1 cos θ = 1 − = 2 Ans: −i + j − 2k 4 4 1 Sol: (a × b) + c = 0 ∴ cosθ = 2 a × (a × b) + a × c = 0 ∴ θ = 60°. (a . b)a − (a . a)b + a × c = 0 3 j − 3k − 2b − 2i − j − k = 0 78. Let S be a non-empty subsets of R. ….. ∴ 2b = −2i + 2 j − 4k ∴ b = −i + j − 2k Ans: There is a rational number x ∈ S such that x ≤ 0. ρ ˆ 74. If the vectors a = ˆ − ˆ + 2k , ……. i j Sol: The negation of the given statement is – ‘There is no rational number x ∈ S such that Ans: (−3, 2) x > 0.’ The equivalent statement is given above. 79. Let cos (α + β ) = Sol: a.c = 0 4 and . ….. ⇒ λ − 1 + 2µ = 0 5 ⇒ λ + 2µ = 1 -------(1) 56 b.c = 0 Ans: ⇒ 2λ + 4 + µ = 0 33 ∴ 2λ + µ = −4 ----(2) ∴ Solving λ = −3 and µ = 2 Sol: tan2α = tan(α + β + α − β) tan(α + β ) + tan(α − β ) = 75. If two tangents drawn from a point P to the 1 − tan(α + β) tan(α − β ) 2 parabola y = 4x are at right angles, …. 3 5 + Ans: x = −1 = 4 12 3 5 1− × Sol: Locus of p is directrix of y = 4x 2 4 12 ∴ x = −1 56 = . 33 x y 76. The line L given by + = 1 passes …….. 2 2 80. The circle x + y = 4x + 8y + 5 …… 5 b Ans: 1 Ans: −35 < m < 15 x y Sol: Perpendicular distance from Sol: + = 1 passes through (13, 32) (2, 4) < Radius 5 b 6 − 16 − m ⇒ = −20 <5 ∴ Equation is 4x − y = 20. It is parallel to 25 x y −10 − m + =1 = <5 c 3 5 = |10 − m| < 25 −25 < 10 + m < 25 84. The number of 3 × 3 non-singular matrices…… − 35 < m < 15. Ans: at least 7 81. For two data sets, each of size 5…… 1 1 0 Sol: Consider 0 1 0 . The 1 on the non 11 0 0 1 Ans: 2 diagonal position can be shifted to 5 more positions. Further we can consider n1σ12 + n2σ22 + n1d12 + n2d22 Sol: σ2 = , 1 1 0 n1 + n2 0 0 1 . ∴ at least 7 matrices are there. n1 x1 + n2 x 2 1 0 0 x= n1 + n2 since n1 = n2 we get 85. Let f : R → R be defined by…… σ 2 + σ22 + d12 + d22 x + x2 σ2 = 1 x= 1 2 2 Ans: −1 2+4 d1 = (2 − 3) = 1 2 2 x= =3 Sol: Since function has local minimum it must 2 be continuous at x = −1 d2 = (4 − 3) = 1 2 2 ∴ Lim f (x ) = Lim f (x ) 4 + 5 + 1 + 1 11 x → −1+ x → −1− ∴ σ2 = = . 2 2 1=k+2 ∴ k = −1. 82. An urn contains nine balls of which….. 86. Four numbers are chosen at random…… 2 Ans: Ans: Statement 1 is true, Statement 2 is false. 7 Sol: Three balls without replacement can Sol: If four chosen numbers form an AP, the 3× 4×2 common differences can be ± 1, ±2, ±3, ±4, ±5 or be done in = 9 ±6. (e.g. 1, 7, 13, 19 is an AP with common C3 difference 6) 2 ∴ Statement 2 is not true. = . 7 10 10 83. For a regular polygon, let r and R be the….. 87. Let S1 = ∑ j (j − 1)10 C j , S2 = ∑ j 10C j ….. j =1 j=1 r 2 Ans: There is a regular polygon with = Ans: Statement 1 is true, Statement 2 is false. R 3 10 Sol: Let n sided regular polygon is inscribed in a circle. From the figure it is clear that Sol: S1 = ∑ j(j − 1) j =1 10 Cj 10 S2 = ∑j j =1 10 Cj 10 R 2π S3 = ∑j j =1 2 10 Cj n r ∑ (j ) 10 S1 − S3 = 2 − j − j2 ×10 C j j =1 π r 10 ∴ cos = n R =− ∑j j=1 10 Cj There an possible integer value 1 1 3 = − S2 corresponding to , and S1+ S2 = S3. 2 2 2 j( j − 1) 10! But cosθ = 2 π 2 ⇒ = cos −1 j! (10 − j)! 3 4 3 10! 8! = 9 × 10 × ⇒ n is not an integer. (j − 2)! (10 − j) ! (j − 2)! (10 − j)! 10 10 90. Let A be a 2 × 2 matrix with non-zero…….. ∑j =1 j(j − 1) 10C j = 90 ∑ j=1 8 C j− 2 Ans: Statement 1 is false, Statement 2 is true. = 90 × 2 . 8 . a b Sol: Let A = 88. Statement 1 : The point A (3, 1, 6) is the mirror c d image……. Given |A| = 1 ad − bc = 1 ------(1) Ans: Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for a b a b 1 0 A = c d c d = 0 1 2 statement 1 a2 + bc (a + d)b Sol: A (3, 1, 6) B = (1, 3, 4) (a + d)c bc + d2 Midpoint of AB is (2, 2, 5) 2 −2 + 5 = 5 a2 + bc = 1 (1) Statement 2 is true d + bc = 1 2 D. R’s of AB are [2, −2, 2] or [1, −1, 1] ⇒ which represent the D.R’s of normal (a + d)b = 0 (a + d)c = 0 ( 2) to the plane x − y + z = 5 ⇒ Statement 1 is true Case 1 We used statement 2 to prove b = 0 and c = 0 statement 1. a 0 A= 0 d 89. Let f : R → R be a continuous function…… Using (1) Ans: Statement 1 is true, Statement 2 is true; ±1 0 Statement 2 is a correct explanation for A= 0 ± 1 statement 1 It is obvious that for a given A, Trace (a) can be different from zero. ⇒ f (x ) > 0 1 Sol: f(x) = Therefore, statement 1 is not true. ex + 2e− x −1 [e x − 2e − x ] OR (e ) f’(x) = x + 2e −x 2 1 0 Take the 2 × 2 unit matrix 0 1 as A. f’(x) = 0 2 x 2 |A| = 1 and A = I e = x However, Trace (A) ≠ 0 e Statement 1 is not true. 2x 1 ⇒e =2⇒x= log2 2 Checking the sign of f’(x) as x crosses 1 log 2, we note that f(x) is maximum at 2 1 x = log 2 . 2 1 Maximum value of f(x) = 1 2 + 2× 2 1 = 2 2 Statement 2 is true 1 2 1.414 == = 0.3535 2 2 4 4 Since f(x) is continuous in R, f(x) has to assume all values between 0 and 0.3535 1 Since is a number lying between o and 3 0.3535, statement 1 is also true.