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AIEEE 2009 Exam Papers

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Tags : AIEEE,2003,Paper,Exam,Chemistry,Maths,Physics,IIT,JEE,IIT JEE,Joint Entrance Examination, iit jee solved question paper with answers and detailed solutions, online question paper free,important questions,aieee exam paper, jee mains, iit jee papers. FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 1 AIEEE–2009, BOOKLET CODE(A) Note: (i) The test is of 3 hours duration. (ii) The test consists of 90 questions. The maximum marks are 432. (iii) There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. Part A − Physics (144 marks) − Question No. 1 to 2 and 9 to 30 consists FOUR (4) marks each and Question No. 3 to 8 consists EIGHT (8) marks each for each correct response. Part B − Chemistry (144 marks) − Question No. 31 to 39 and 46 to 60 consists FOUR (4) marks each and Question No. 40 to 45 consists EIGHT (8) marks each for each correct response. Part C − Mathematics(144 marks) − Question No. 61 to 82 and 89 to 90 consists FOUR (4) marks each and Question No. 83 to 88 consists EIGHT (8) marks each for each correct response. (iv) Candidates will be awarded marks as stated above for correct response of each question. 1/4th marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. (v) * marked questions are from syllabus of class XI CBSE. Physics PART − A 1. This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement – 1: For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. Statement-2: The net work done by a conservative force on an object moving along a closed loop is zero (1) Statement-1 is true, Statement-2 is false (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (3) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true Sol: (2) Work done by conservative force does not depend on the path. Electrostatic force is a conservative force. 2. The above is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions: (i) A + B → C + ε (ii) C → A + B + ε (iii) D + E → F + ε and (iv) F → D + E + ε where ε is the energy released? In which reactions is ε positive? (1) (i) and (iv) (2) (i) and (iii) (3) (ii) and (iv) (4) (ii) and (iii) Sol: (1) 1st reaction is fusion and 4th reaction is fission. FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 2 3. A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit. (1) (2) (3) (4) Sol: (3) Given figure is half wave rectifier 4. The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. Pick out the correct output waveform. (1) (2) (3) (4) Sol: (1) Truth Table A B Y 1 1 1 1 0 0 0 1 0 0 0 0 *5. If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time? (1) a2T2 + 4π2v2 (2) aT x (3) aT + 2πv (4) aT v Sol: (2) 2 2 2 2 aT xT 4 T 4 x x T T ω π π = = × = = constant. FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 3 6. In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45o with the x-axis meets the experimental curve at P. The coordinates of P will be (1) (2f, 2f) (2) f , f 2 2       (3) (f, f) (4) (4f, 4f) Sol: (1) It is possible when object kept at centre of curvature. u = v u = 2f, v = 2f. *7. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of (1) 1 2 2 3 g l ω (2) 1 6 g lω (3) 1 2 2 2 g l ω (4) 1 2 2 6 g l ω Sol: (4) T.Ei = T.Ef 2 1I mgh 2 ω = 2 2 1 1m mgh 2 3 × l ω = ⇒ 1 2 2 h 6 g ω = l h • 8. Let 4 P(r) Q r R = π be the charge density distribution for a solid sphere of radius R and total charge Q. for a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is (1) 0 (2) 2 o 1 Q 4πε r (3) 2 1 4 o Qr 4πε R (4) 2 1 4 o Qr 3πε R Sol: (3) r1 2 4 2 2 0 1 1 4 0 0 Q r4 r dr R Qr E4 r E 4 R π π π = ⇒ = ε πε ∫ . r P R FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 4 9. The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from (1) 2 → 1 (2) 3 → 2 (3) 4 → 2 (4) 5→ 4 Sol: (4) IR corresponds to least value of 2 2 1 2 1 1 n n    −    i.e. from Paschen, Bracket and Pfund series. Thus the transition corresponds to 5 → 3. *10. One kg of a diatomic gas is at a pressure of 8 × 104 N/m2. The density of the gas is 4 kg/m-3. What is the energy of the gas due to its thermal motion? (1) 3 × 104 J (2) 5 × 104 J (3) 6 × 104 J (4) 7 × 104 J Sol: (2) Thermal energy corresponds to internal energy Mass = 1 kg density = 8 kg/m3 ⇒ Volume = 3 mass 1m density 8 = Pressure = 8 × 104 N/m2 ∴ Internal Energy = 5 2 P × V = 5 × 104 J 11. This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1: The temperature dependence of resistance is usually given as R = Ro(1 + αΔt). The resistance of a wire changes from 100 Ω to 150 Ω when its temperature is increased from 27oC to 227oC. This implies that α = 2.5 ×10−3 /o C. Statement 2: R = Ri (1 + αΔT) is valid only when the change in the temperature ΔT is small and ΔR = (R – Ro) << Ro. (1) Statement-1 is true, Statement-2 is false (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (3) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true Sol: (1) Directions: Question numbers 12 and 13 are based on the following paragraph. A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30o. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin. FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 5 12. The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is (1) zero (2) ( ) o b a 24ab μ − (3) oI b a 4 ab μ  −  π   (4) o ( ) ( ) I 2 b a a b 4 3 μ  π   − + +  π   Sol: (2) Net magnetic field due to loop ABCD at O is B = BAB + BBC + BCD + BDA = o o I I 0 0 4 a 6 4 b 6 μ π μ π + × + − × π π = o o I I 24a 24b μ μ − = oI ( ) b a 24ab μ − 13. Due to the presence of the current I1 at the origin (1) The forces on AB and DC are zero (2) The forces on AD and BC are zero (3) The magnitude of the net force on the loop is given by o 1 ( ) ( ) II 2 b a a b 4 3 μ  π   − + +  π   (4) The magnitude of the net force on the loop is given by o 1 ( ) II b a 24ab μ − Sol: (2) The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC is parallel to elementary length of the loop. 14. A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’s double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is (1) 393.4 nm (2) 885.0 nm (3) 442.5 nm (4) 776.8 nm Sol: (3) 3λ1 = 4λ2 ⇒ 2 1 3 3 590 4 4 λ = λ = × = 1770 4 = 442.5 nm 15. Two points P and Q are maintained at the potentials of 10V and -4V respectively. The work done in moving 100 electrons from P to Q is (1) −19 ×10−17 J (2) 9.60 ×10−17 J (3) −2.24 ×10−16 J (4) 2.24 ×10−16 J Sol: (4) W = QdV = Q(Vq - VP) = -100 × (1.6 × 10-19) × (– 4 – 10) = + 100 × 1.6 × 10-19 × 14 = +2.24 10-16 J. 16. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 eV nm) (1) 3.09 eV (2) 1.41 eV (3) 151 eV (4) 1.68 eV Sol: (2) 2 o 1mv eV 1.68eV 2 = = ⇒ h hc 1240evnm 400nm ν = = λ = 3.1 eV ⇒ 3.1 eV = Wo + 1.6 eV ∴ Wo = 1.42 eV FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 6 *17. A particle has an initial velocity 3ˆi + 4ˆj and an acceleration of 0.4ˆi + 0.3ˆj . Its speed after 10 s is (1) 10 units (2) 7 2 units (3) 7 units (4) 8.5 units Sol: (2) u = 3ˆi + 4ˆj r ; a = 0.4ˆi + 0.3ˆj r u = u + at r r r = 3ˆi + 4ˆj + (0.4ˆi + 0.3ˆj)10 = 3ˆi + ˆj + 4ˆi + 3ˆj = 7ˆi + 7ˆj Speed is 72 + 72 = 7 2 units *18. A motor cycle starts from rest and accelerates along a straight path at 2 m/s2. At the starting point of the motor cycle there is a stationary electric sire. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (speed of sound = 330 ms-1). (1) 49 m (2) 98 m (3) 147 m (4) 196 m Sol: (2) Motor cycle, u = 0, a = 2 m/s2 Observer is in motion and source is at rest. ⇒ O S v v n n v v − ′ = + ⇒ O 94 330 v n n 100 330 − = ⇒ 330 – vO = 330 94 100 × ⇒ O v 330 94 33 10 × = − = 33 6 m/s 10 × s = v2 u2 9 33 33 9 1089 2a 100 100 − × × × = = ≈ 98 m. *19. Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time the height as function of time will be (1) (2) (3) (4) Sol: (3) 2 h 1 gt 2 = , v = - gt and after the collision, v = gt. (parabolic) (straight line) Collision is perfectly elastic then ball reaches to same height again and again with same velocity. t1 2t2 3t1 t -v1 +v1 v t h y FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 7 20. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then the Q/q equals (1) −2 2 (2) -1 (3) 1 (4) 1 2 − Sol: (1) Three forces F41, F42 and f43 acting on Q are shown Resultant of F41 + F43 = 2 Feach = 2 o 2 1 Qq 4πε d Resultant on Q becomes zero only when ‘q’ charges are of negative nature. ( ) 4,2 2 o F 1 Q Q 4 2d × = πε ⇒ 2 2 2 dQ Q Q d 2d × = ⇒ 2 q Q Q 2 × × = ∴ q Q 2 2 = − or Q 2 2 q = − Q q Q F42 F43 F41 q *21. A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature θ along the length x of the bar from its hot end is best described by which of the following figure. (1) (2) (3) (4) Sol: (2) We know that dQ KA d dt dx θ = In steady state flow of heat d dQ. 1 .dx dt kA θ = ⇒ θH - θ = k x ′ ⇒ θ = θH - k x ′ Equation θ = θH - k′ x represents a straight line. FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 8 22. A transparent solid cylindrical rod has a refractive index of 2 3 . It is surrounded by air. A light ray is incident at the mid point of one end of the rod as shown in the figure. θ The incident angle θ for which the light ray grazes along the wall of the rod is (1) sin 1 1 2 −       (2) sin 1 3 2 −       (3) sin 1 2 3 −       (4) sin 1 1 3 −       Sol: (4) SinC = 3 2 ….. (1) Sin r = sin (90 – C) = cosC = 1 2 2 1 sin sinr θ μ = μ sin 2 1 3 2 θ = × sin 1 1 3 −   θ =     θ C *23. Three sound waves of equal amplitudes have frequencies (v – 1), v, (v + 1). They superpose to give beats. The number of beats produced per second will be (1) 4 (2) 3 (3) 2 (4) 1 Sol: (3) Maximum number of beats = ν + 1− (ν −1) = 2 *24. The height at which the acceleration due to gravity becomes g 9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is (1) 2R (2) R 2 (3) R 2 (4) 2 R Sol: (1) ( )2 g GM R h ′ = + , acceleration due to gravity at height h ⇒ ( ) 2 2 2 g GM. R 9 R R h = + = 2 g R R h    +    ⇒ 2 1 R 9 R h =    +    ⇒ R 1 R h 3 = + ⇒ 3R = R + h ⇒ 2R = h FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 9 *25. Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area A and wire-2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount? (1) F (2) 4F (3) 6F (4) 9F Sol: (4) A1l1 = A2l2 ⇒ 1 1 1 1 2 2 A A A 3A 3 × = = = l l l l ⇒ 1 2 = 3 l l Δx1 = 1 1 F A × γ l ….. (i) 2 2 2 F x 3A Δ = γ l …..(ii) Here Δx1 = Δx2 2 1 2 1 F F 3A A = γ γ l l F2 = 1 1 1 2 3F × = 3F × 3 l l = 9F *26. In an experiment the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree(=0.5o), then the least count of the instrument is (1) one minute (2) half minute (3) one degree (4) half degree Sol: (1) Least count = value of main scale division No of divisions on vernier scale = 1 MSD 30 = 1 1o 1 o 30 2 60 × = = 1 minute 27. An inductor of inductance L = 400 mH and resistors of resistances R1 = 2Ω and R2 = 2Ω are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is (1) 6e−5tV (2) 3t 12e V t − (3) 6(1− e−t / 0.2 )V (4) 12e−5tV Sol: (4) 1 1 F I R = = 12 6A 2 = 2 2 2 dI E L R I dt = + × ( t / tc ) I2 Io 1 e= − − ⇒ o 2 I E 12 6A R 2 = = = 3 c t L 400 10 0.2 R 2 × − = = = ( t / 0.2 ) I2 6 1 e= − − Potential drop across L = E – R2I2 = 12 – 2 × 6 (1 – e-bt) = 12 e-5t FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 10 Directions: Question numbers 28, 29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle ABCDA, as shown in the P – T diagram. *28. Assuming the gas to be ideal the work done on the gas in taking it from A to B is (1) 200 R (2) 300 R (3) 400 R (4) 500 R Sol: (3) WAB = ΔQ - ΔU = nCpdT – nCvdT (at constant pressure) = n(Cp – Cv)dt = nRdT = 2 × R × (500 – 300) = 400 R A B D C P n = 2, γ = 1.67 300 K 500 K T 2 × 105 Pa 1 × 105 Pa *29. The work done on the gas in taking it from D to A is (1) – 414 R (2) + 414 R (3) – 690 R (4) + 690 R Sol: (1) At constant temperature (isothermal process) WDA = 1 2 P nRTln P       = 5 5 2.303 2R 300log 10 2 10   × ×    ×  = 2.303 600Rlog 1 2 ×       = 0.693 × 600 R = - 414 R. *30. The net work done on the gas in the cycle ABCDA is (1) Zero (2) 276 R (3) 1076 R (4) 1904 R Sol: (2) Net work done in a cycle = WAB + WBC + WCB + WBA = 400 R + 2 × 2.303 × 500 R ln 2 – 400R – 414 R = 1000R x ln 2 – 600R x ln 2 = 400R x ln 2 = 276R FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 11 CHEMISTRY PART − B 31. Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements in incorrect ? (1) Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character. (2) The ionic sizes of Ln (III) decrease in general with increasing atomic number. (3) Ln (III) compounds are generally colourless. (4) Ln (III) hydroxides are mainly basic in character. Sol: (3) Ln+3 compounds are mostly coloured. 32. A liquid was mixed with ethanol and a drop of concentrated 2 4 H SO was added. A compound with a fruity smell was formed. The liquid was : (1) 3 CH OH (2) HCHO (3) 3 3 CH COCH (4) 3 CH COOH Sol: (4) Esterification reaction is involved 3 ( ) CH COOH l + 2 5 ( ) C H OH H + → l 3 2 5( ) 2 ( ) CH COOC H +H O l l *33. Arrange the carbanions, ( )3 3 3 CH C, CCl , ( )3 2 6 5 2 CH CH, C H CH , in order of their decreasing stability : (1) ( ) ( ) 6 5 2 3 3 3 3 2 C H CH > CCl > CH C > CH CH (2) ( ) ( ) 3 2 3 6 5 2 3 3 CH CH > CCl > C H CH > CH C (3) ( ) ( ) 3 6 5 2 3 2 3 3 CCl > C H CH > CH CH > CH C (4) ( ) ( ) 3 3 3 2 6 5 2 3 CH C > CH CH > C H CH > CCl Sol: (3) 2o carbanion is more stable than 3o and Cl is –I effect group. *34. The alkene that exhibits geometrical isomerism is : (1) propene (2) 2-methyl propene (3) 2-butene (4) 2- methyl -2- butene Sol: (3) C=C H CH3 CH3 H C=C H H3C H CH3 cis Trans *35. In which of the following arrangements, the sequence is not strictly according to the property written against it ? (1) 2 2 2 2 CO < SiO < SnO < PbO : increasing oxidising power (2) HF< HCl < HBr < HI : increasing acid strength (3) 3 3 3 3 NH < PH < AsH < SbH : increasing basic strength (4) B < C < O < N : increasing first ionization enthalpy. Sol: (3) Correct basic strength is 3 3 3 3 NH > PH > AsH > BiH FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 12 36. The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is : (1) benzoic acid (2) salicylaldehyde (3) salicylic acid (4) phthalic acid Sol: (3) Kolbe – Schmidt reaction is OH ONa OH OH NaOH→ 2 o CO 6atm, 140 C → COONa 3 H O+ → COOH Salicylic Acid 37. Which of the following statements is incorrect regarding physissorptions ? (1) It occurs because of vander Waal’s forces. (2) More easily liquefiable gases are adsorbed readily. (3) Under high pressure it results into multi molecular layer on adsorbent surface. (4) Enthalpy of adsorption ( ) adsorption ΔH is low and positive. Sol: (4) Enthalpy of adsorption regarding physissorption is not positive and it is negative. 38. Which of the following on heating with aqueous KOH, produces acetaldehyde ? (1) 3 CH COCl (2) 3 2 CH CH Cl (3) 2 2 CH Cl CH Cl (4) 3 2 CH CHCl Sol: (4) 3 2 CH CHCl aq.KOH→ 3 /OH CH CH \OH 3 2 CH CHO H O → − *39. In an atom, an electron is moving with a speed of 600m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 ×10−34 kg m2s−1 , mass of electron, 31 me = 9.1×10− kg ) (1) 1.52 ×10−4m (2) 5.10 ×10−3m (3) 1.92 ×10−3m (4) 3.84 ×10−3m Sol: (3) x.m v h 4 Δ Δ = π Δx = h 4π mΔv Δv = 600 0.005 0.03 100 × = ⇒ 34 3 31 x 6.625 10 1.92 10 m 4 3.14 9.1 10 0.03 − − − × Δ = = × × × × × FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 13 40. In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is 3 2 2 2 CH OH( ) 3 O (g) CO (g) 2H O( ) 2 l + → + l At 298K standard Gibb’s energies of formation for 3 2 CH OH(l), H O(l) and 2 CO (g) are -166.2, -237.2 and -394.4 kJ mol−1 respectively. If standard enthalpy of combustion of methanol is -726kJ mol−1 , efficiency of the fuel cell will be (1) 80 % (2) 87% (3) 90% (4) 97% Sol: (4) 3 CH OH(l) + 2 2 2 3O (g) CO (g) 2H O( ) 2 → + l ΔH = −726kJ mol−1 Also o f 3 ΔG CH OH(l) = -166.2 kJ mol-1 o f 2 ΔG H O(l) = -237.2 kJ mol-1 o f 2 ΔG CO (l) = -394.4 kJ mol-1 Q G Δ = of G ΣΔ products of −ΣΔG reactants. = -394.4 -2 (237.2) + 166.2 = −702.6 kJ mol-1 now Efficiency of fuel cell = G 100 H Δ × Δ = 702.6 100 726 × = 97% 41. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively : (1) 200 and 300 (2) 300 and 400 (3) 400 and 600 (4) 500 and 600 Sol: (3) o o T X X Y Y P = P x + P x X x = mol fraction of X Y x = mol fraction of Y ∴ 550 = o o x Y P 1 P 3 1 3 1 3   +    +   +      = o o X Y P 3P 4 4 + ∴ 550 (4) = o o X Y P + 3P ………….. (1) Further 1 mol of Y is added and total pressure increases by 10 mm Hg. ∴ 550 + 10 = o o X Y P 1 P 4 1 4 1 4   +    +   +      ∴ 560 (5) = o o X Y P + 4P ………….(2) By solving (1) and (2) We get, o X P = 400 mm Hg o Y P = 600 mm Hg FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 14 42. The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2=0.301) : (1) 230.3 minutes (2) 23.03 minutes (3) 46.06 minutes (4) 460.6 minutes Sol: (3) 1 1/ 2 0.6932 0.6932 min t 6.93 Qλ = = − Also t = [ ] [ ] o 2.303 A log λ A [ ] o A = initial concentration (amount) [A] = final concentration (amount) ∴ t = 2.303 6.93 log100 0.6932 1 × = 46.06 minutes 43. Given : 3 o Fe /Fe E 0.036V, + = − 2 o Fe / Fe E + = -0.439V. The value of standard electrode potential for the change, 3 2 (aq) Fe + + e− → Fe + (aq) will be : (1) -0.072 V (2) 0.385 V (3) 0.770 V (4) -0.270 Sol: (3) Q Fe3+ + 3e− →Fe; Eo = −0.036V ∴ O o 1 ΔG = −nFE = −3F(−0.036) = +0.108 F Also Fe2+ + 2e− →Fe; Eo = -0.439 V ∴ O2 ΔG = -nFEo = -2 F( -0.439) = 0.878 F To find Eo for 3 2 (aq) Fe + + e− →Fe + (aq) ΔGO = -nFE o = -1FE o Q o o o 1 2 G = G −G ∴ Go = 0.108F – 0.878F ∴ -FEo = +0.108F – 0.878F ∴ EO = 0.878 – 0.108 = 0.77v *44. On the basis of the following thermochemical data : ( o (aq) ΔfG H+ = 0) 2 H O(l) →H+ (aq) + OH− (aq); ΔH = 57.32kJ 2 2 2 H (g) 1O (g) H O( ) 2 + → l ; ΔH = −286.20kJ The value of enthalpy of formation of OH− ion at 25oC is : (1) -22.88 kJ (2) -228.88 kJ (3) +228.88 kJ (4) -343.52 kJ Sol: (2) By adding the two given equations, we have 2(g) H + 2(g) (aq) (aq) 1O H OH 2 → + + − ; ΔH =-228.88 Kj Here of ΔH of (aq) H+ = 0 ∴ of ΔH of OH− = -228.88 kJ FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 15 45. Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom ? (1) 108 pm (2) 127 pm (3) 157 pm (4) 181 pm Sol: (2) For FCC, 2a = 4r (the atoms touches each other along the face- diagonal) r = 2a 2 361 4 4 × = = 127 pm 46. Which of the following has an optical isomer ? (1) ( ) 3 3 CO NH Cl +   (2) ( )( ) 2 3 2 CO en NH +   (3) ( ) ( ) 3 2 4 CO H O en +   (4) ( ) ( ) 3 2 3 2 CO en NH +   Sol: (4) It is an octahedral complex of the type ( ) 2 2 M AA X  Where AA is bidentate ligand. *47. Solid Ba ( ) 3 2 NO is gradually dissolved in a 1.0 ×10−4M 2 3 Na CO solution. At what concentration of Ba2+ will a precipitate begin to form ?(Ksp for Ba CO3 = 5.1 × 10 9 − ). (1) 4.1 ×10−5M (2) 5.1×10−5M (3) 8.1×10−8M (4) 8.1×10−7M Sol: (2) ( )3 2 3 3 3 Ba NO + CaCO → BaCO + 2NaNO Here 2 [ ] 4 3 2 3 CO−  = Na CO = 10− M 2 2 sp 3 K = Ba+  CO−  ⇒ 5.1×10−9 = Ba2+  (10−4 )⇒ Ba+2  = 5.1×10−5 At this value, just precipitation starts. 48. Which one of the following reactions of Xenon compounds is not feasible ? (1) 3 XeO + 6HF → Xe 6 2 F + 3H O (2) 4 2 3 3Xe F + 6H O → 2 Xe + XeO + 12 HF + 1.5 2 O (3) 2 2 2 2XeF + 2H O → 2Xe + 4HF + O (4) 6 7 XeF + RbF → Rb(XeF ] Sol: (1) Remaining are feasible *49. Using MO theory predict which of the following species has the shortest bond length ? (1) 22 O + (2) 2 O+ (3) 2 O− (4) 22 O − Sol: (1) Bond length 1 bond order α Bond order = no..of bonding e no.of antibonding e 2 − Bond orders of 2 2 2 2 O+ , O− , O− and 2 2 O+ are respectively 2.5, 1.5, 1 and 3. 50. In context with the transition elements, which of the following statements is incorrect ? (1) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 16 in complexes. (2) In the highest oxidation states, the transition metal show basic character and form cationic complexes. (3) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding. (4) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases. Sol: (2) In higher Oxidation states transition elements show acidic nature *51. Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 ×103ms−1 (Mass of proton = 1.67 ×10−27 kg and h = 6.63 ×10−34 Js ) : (1) 0.032 nm (2) 0.40 nm (3) 2.5 nm (4) 14.0 nm Sol: (2) h mv λ = 34 27 3 6.63 10 0.40 1.67 10 10 − − × = ≡ × × nm 52. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ? (1) The solution formed is an ideal solution (2) The solution is non-ideal, showing +ve deviation from Raoult’s law. (3) The solution is non-ideal, showing –ve deviation from Raoult’s law. (4) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law. Sol: (2) The interactions between n –heptane and ethanol are weaker than that in pure components. *53. The number of stereoisomers possible for a compound of the molecular formula ( ) 3 CH − CH = CH − CH OH −Me is : (1) 3 (2) 2 (3) 4 (4) 6 Sol: (3) About the double bond, two geometrical isomers are possible and the compound is having one chiral carbon. *54. The IUPAC name of neopentane is (1) 2-methylbutane (2) 2, 2-dimethylpropane (3) 2-methylpropane (4) 2,2-dimethylbutane Sol: (2) Neopentane is 3 3 3 3 CH | HC C CH | CH − − *55. The set representing the correct order of ionic radius is : (1) Li+ > Be2+ > Na+ > Mg2+ (2) Na+ > Li+ > Mg2+ > Be2+ (3) Li+ > Na+ > Mg2+ > Be2+ (4) Mg2+ > Be2+ > Li+ > Na+ Sol: (2) Follow the periodic trends 56. The two functional groups present in a typical carbohydrate are : (1) -OH and -COOH (2) -CHO and -COOH (3) > C = O and – OH (4) – OH and -CHO FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 17 Sol: (3) Carbohydrates are polyhydroxy carbonyl compounds. *57. The bond dissociation energy of B – F in BF3 is 646 kJ mol−1 whereas that of C-F in 4 CF is 515kJ mol−1 . The correct reason for higher B-F bond dissociation energy as compared to that of C- F is : (1) smaller size of B-atom as compared to that of C- atom (2) stronger σ bond between B and F in 3 BF as compared to that between C and F in 4 CF (3) significant pπ – pπ interaction between B and F in 3 BF whereas there is no possibility of such interaction between C and F in 4 CF . (4) lower degree of pπ – pπ interaction between B and F in 3 BF than that between C and F in 4 CF . Sol: (3) option itself is the reason 58. In Cannizzaro reaction given below 2 Ph CHO ( ) 2 2 ( ) : OH Ph CH OH PhCO − − → + && the slowest step is : (1) the attack of : ( ) OH − at the carboxyl group (2) the transfer of hydride to the carbonyl group (3) the abstraction of proton from the carboxylic group (4) the deprotonation of Ph 2 CH OH Sol: (2) Hydride transfer is the slowest step. 59. Which of the following pairs represents linkage isomers ? (1) ( ) [ ] 3 4 4 Cu NH  Pt Cl and ( ) [ ] 3 4 4 Pt NH  CuCl (2) ( ) ( ) 3 2 2 Pd P Ph NCS  and ( ) ( ) 3 2 2 Pd P Ph SCN  (3) ( ) 3 5 3 4 CO NH NO SO and ( ) 3 5 4 3 CO NH SO NO (4) ( ) 2 3 4 2 PtCl NH Br and ( ) 2 3 4 2 PtBr NH Cl Sol: (2) NCS- is ambidentate ligand and it can be linked through N (or) S 60. Buna-N synthetic rubber is a copolymer of : (1) 2 2 Cl | H C = CH − C = CH and 2 2 H C = CH − CH = CH (2) 2 2 H C = CH − CH = CH and 5 6 2 H C − CH = CH (3) 2 H C = CH − CN and 2 2 H C = CH − CH = CH (4) 2 H C = CH − CN and 2 2 3 H C CH C CH | CH = − = Sol: (3) FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 18 Mathematics PART − C 61. Let a, b, c be such that b(a + c) ≠ 0 . If ( )n 2 ( )n 1 ( )n a a 1 a 1 a 1 b 1 c 1 b b 1 b 1 a 1 b 1 c 1 0, c c 1 c 1 1 a 1 b 1 c + + + − + + − − + − + − − + = − + − − − then the value of ‘n’ is (1) zero (2) any even integer (3) any odd integer (4) any integer Sol: (3) ( )n a a 1 a 1 a 1 b 1 c 1 b b 1 b 1 1 a 1 b 1 c 1 c c 1 c 1 a b c + − + + − − + − + − − − + − + − ( )n a a 1 a 1 a 1 a 1 a b b 1 b 1 1 b 1 b 1 b c c 1 c 1 c 1 c 1 c + − + − = − + − + − + − − − + − + ( )n 1 a a 1 a 1 a 1 a a 1 b b 1 b 1 1 b 1 b b 1 c c 1 c 1 c 1 c c 1 + + − + − = − + − + − + − − − + − + ( )n 2 a a 1 a 1 a a 1 a 1 b b 1 b 1 1 b b 1 b 1 c c 1 c 1 c c 1 c 1 + + − + − = − + − + − − + − − + − + This is equal to zero only if n + 2 is odd i.e. n is odd integer. 62. If the mean deviation of number 1, 1 + d, 1 + 2d, ….. , 1 + 100d from their mean is 255, then the d is equal to (1) 10.0 (2) 20.0 (3) 10.1 (4) 20.2 Sol: (3) ( ) n Mean x sum of quantities n = = (a l) 2 n + 1[1 1 100d] 1 50d 2 = + + = + [ ] i M.D. 1 x x 255 1 50d 49d 48d …. d 0 d …… 50d 2 n 101 = Σ − ⇒ = + + + + + + + + = d 50 x 51 101 2       d 255 x 101 10.1 50 x 51 ⇒ = = *63. If the roots of the equation bx2 + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is (1) greater than 4ab (2) less than 4ab (3) greater than – 4ab (4) less than – 4ab Sol: (3) bx2 + cx + a = 0 Roots are imaginary ⇒ c2 − 4ab < 0⇒ c2 < 4ab⇒ −c2 > −4ab 3b2x2 + 6bcx + 2c2 since 3b2 > 0 Given expression has minimum value Minimum value = ( )( ) ( ) 2 2 22 2 2 2 2 2 4 3b 2c 36b c 12b c c 4ab 4 3b 12b − = − = − > − . FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 19 *64. Let A and B denote the statements A: cosα + cosβ + cos γ = 0 B: sinα + sinβ + sinγ = 0 If cos( ) cos( ) cos( ) 3 2 β − γ + γ − α + α − β = − , then (1) A is true and B is false (2) A is false and B is true (3) both A and B are true (4) both A and B are false Sol: (3) cos( ) cos( ) cos( ) 3 2 β − γ + γ − α + α − β = − ⇒ 2cos(β − γ) + cos(γ − α) + cos(α − β) + 3 = 0 ⇒ 2cos(β − γ) + cos(γ − α) + cos(α − β) + sin2 α + cos2 α + sin2 β + cos2 β + sin2 γ + cos2 γ = 0 ⇒ ( )2 ( )2 sinα + sinβ + sinγ + cosα + cosβ + cos γ = 0 *65. The lines p(p2 +1)x − y + q = 0 and ( ) ( ) p2 + 1 2 x + p2 +1 y + 2q = 0 are perpendicular to a common line for (1) no value of p (2) exactly one value of p (3) exactly two values of p (4) more than two values of p Sol: (2) Lines must be parallel, therefore slopes are equal ⇒ p(p2 + 1) = −(p2 +1) ⇒ p = – 1 66. If A, B and C are three sets such that A ∩B = A ∩C and A ∪B = A ∪C , then (1) A = B (2) A = C (3) B = C (4) A ∩B = φ Sol: (3) 67. If u, v, w r r uur are non-coplanar vectors and p, q are real numbers, then the equality 3u pv pw − pv w qu − 2w qv qu = 0 r r uur r uur r uur r r holds for (1) exactly one value of (p, q) (2) exactly two values of (p, q) (3) more than two but not all values of (p , q) (4) all values of (p, q) Sol: (1) (3p2 − pq + 2q2 )u v w = 0 r r uur But u v w ≠ 0 r r uur 3p2 − pq + 2q2 = 0 2 2 2 2p2 p2 pq q 7q 0 2p2 p q 7 q2 0 2 4 2 4 + − +   + = ⇒ +  −  + =         ⇒ p = 0, q = 0, p q 2 = This possible only when p = 0, q = 0 exactly one value of (p, q) 68. Let the line x 2 y 1 z 2 3 5 2 − − + = = − lies in the plane x + 3y − αz + β = 0 . Then (α, β) equals (1) (6, – 17) (2) ( – 6, 7) (3) (5, – 15) (4) ( – 5, 15) Sol: (2) Dr’s of line = (3, − 5, 2) FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 20 Dr’s of normal to the plane = (1, 3, − α) Line is perpendicular to normal ⇒ 3(1) − 5(3) + 2(−α) = 0 ⇒ 3 −15 − 2α = 0 ⇒ 2α = −12⇒ α = −6 Also (2, 1, − 2) lies on the plane 2 + 3 + 6(−2) + β = 0⇒β = 7 ∴ (α, β) = (−6, 7) *69. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such arrangements is (1) less than 500 (2) at least 500 but less than 750 (3) at least 750 but less than 1000 (4) at least 1000 Sol: (4) 4 novels can be selected from 6 novels in 6 4 C ways. 1 dictionary can be selected from 3 dictionaries in 3 1 C ways. As the dictionary selected is fixed in the middle, the remaining 4 novels can be arranged in 4! ways. ∴ The required number of ways of arrangement = 6 3 4 1 C x C x 4! = 1080 70. [ ] 0 cot x dx π∫ , [•] denotes the greatest integer function, is equal to (1) 2 π (2) 1 (3) – 1 (4) 2 π − Sol: (4) Let [ ] 0 I cotx dx π = ∫ …(1) ( ) 0 cot x dx π = ∫  π −  [ ] 0 cot x dx π = ∫ − …(2) Adding (1) and (2) 2I = [ ] 0 cot x dx π∫ + [ ] 0 cot x dx π ∫ − = ( ) 0 1 dx π = ∫ − [x] [ x] 1if x Z 0 if x Z  + − = − ∉     = ∈  Q [ ]0 x π = − = −π ∴ I 2 π = − 71. For real x, let f (x) = x3 + 5x + 1, then (1) f is one-one but not onto R (2) f is onto R but not one-one (3) f is one-one and onto R (4) f is neither one-one nor onto R Sol: (3) Given f (x) = x3 + 5x + 1 Now f ‘(x) = 3×2 + 5 > 0, ∀x ∈R ∴ f(x) is strictly increasing function ∴ It is one-one Clearly, f(x) is a continuous function and also increasing on R, Lt x → −∞ f (x) = −∞ and Lt x → ∞ f (x) = ∞ ∴ f(x) takes every value between −∞ and ∞ . Thus, f(x) is onto function. FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 21 72. In a binomial distribution B n, p 1 4  =      , if the probability of at least one success is greater than or equal to 9 10 , then n is greater than (1) 4 3 10 10 1 log − log (2) 4 3 10 10 1 log + log (3) 4 3 10 10 9 log − log (4) 4 3 10 10 4 log − log Sol: (1) n n 3 4 1 q 9 3 1 n log 10 10 4 10 − ≥ ⇒   ≤ ⇒ ≥ −     4 3 10 10 n 1 log log ⇒ ≥ − *73. If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y + 2p − 5 = 0 and x2 + y2 + 2x + 2y − p2 = 0 , then there is a circle passing through P, Q and (1, 1) for (1) all values of p (2) all except one value of p (3) all except two values of p (4) exactly one value of p Sol: (1) Given circles S = x2 + y2 + 3x + 7y + 2p − 5 = 0 S’ = x2 + y2 + 2x + 2y − p2 = 0 Equation of required circle is S + λS’ = 0 As it passes through (1, 1) the value of ( ) ( 2 ) 7 2p 6 p − + λ = − If 7 + 2p = 0, it becomes the second circle ∴it is true for all values of p 74. The projections of a vector on the three coordinate axis are 6, – 3, 2 respectively. The direction cosines of the vector are (1) 6, − 3, 2 (2) 6, 3, 2 5 5 5 − (3) 6, 3, 2 7 7 7 − (4) 6, 3, 2 7 7 7 − − Sol: (3) Projection of a vector on coordinate axis are 2 1 2 1 2 1 x − x , y − y , z − z 2 1 2 1 2 1 x − x = 6, y − y = −3, z − z = 2 ( )2 ( )2 ( )2 2 1 2 1 2 1 x − x + y − y + z − z = 36 + 9 + 4 = 7 The D.C’s of the vector are 6, 3, 2 7 7 7 − *75. If Z 4 2 z − = , then the maximum value of Z is equal to (1) 3 +1 (2) 5 +1 (3) 2 (4) 2 + 2 Sol: (2) Z Z 4 4 Z Z =  −  +     Z Z 4 4 Z Z ⇒ = − + Z Z 4 4 Z Z ⇒ ≤ − + Z 2 4 Z ⇒ ≤ + FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 22 ⇒ Z 2 − 2 Z − 4 ≤ 0 ( Z − ( 5 + 1))( Z − (1− 5 )) ≤ 0⇒1− 5 ≤ Z ≤ 5 + 1 *76. Three distinct points A, B and C are given in the 2 – dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point ( – 1, 0) is equal to 1 3 . Then the circumcentre of the triangle ABC is at the point (1) (0, 0) (2) 5 , 0 4       (3) 5, 0 2       (4) 5, 0 3       Sol: (2) P = (1, 0) ; Q(−1, 0) Let A = (x, y) AP BP CP 1 AQ BQ CQ 3 = = = ..(1) ⇒ 3AP = AQ⇒ 9AP2 = AQ2 ⇒ 9(x −1)2 + 9y2 = (x +1)2 + y2 ⇒ 9×2 −18x + 9 + 9y2 = x2 + 2x + 1+ y2 ⇒ 8×2 − 20x + 8y2 + 8 = 0 x2 y2 5 x 1 0 2 ⇒ + − + = …(2) ∴ A lies on the circle Similarly B, C are also lies on the same circle ∴ Circumcentre of ABC = Centre of Circle (1) = 5, 0 4       *77. The remainder left out when ( )2n 1 2n 8 62+ − is divided by 9 is (1) 0 (2) 2 (3) 7 (4) 8 Sol: (2) ( )2n 1 2n 8 62+ − = ( )n ( )2n 1 1 63 63 1 + + − − ( )n ( )2n 1 ( n n ( )2 ( )n ) ( (2n 1) (2n 1) ( )2 ( )( )(2n 1) ) 1 2 1 2 1 63 1 63 1 c 63 c 63 …. 63 1 c 63 c 63 …. 1 63 = + + − + = + + + + + − + + + + + − + (n n ( ) ( )n 1 (2n 1) (2n 1) ( ) ( )(2n) ) 1 2 1 2 2 63 c c 63 …. 63 c c 63 …. 63 = + + + + − − + + + + − ∴ Reminder is 2 *78. The ellipse x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is (1) x2 +16y2 = 16 (2) x2 +12y2 = 16 (3) 4×2 + 48y2 = 48 (4) 4×2 + 64y2 = 48 Sol: (2) FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 23 ( ) 2 2 x2 4y2 4 x y 1 a 2, b 1 P 2, 1 4 1 + = ⇒ + = ⇒ = = ⇒ = Required Ellipse is 2 2 2 2 2 2 2 2 x y 1 x y 1 a b 4 b + = ⇒ + = (2, 1) lies on it 2 2 2 4 1 1 1 1 1 3 b 4 16 b b 4 4 3 ⇒ + = ⇒ = − = ⇒ = ∴ 2 2 2 2 x y 1 x 3y 1 x2 12y2 16 16 4 16 4 3 + = ⇒ + = ⇒ + =       1 A’ 2 A 2 V’ V P (2, 1) (4, 0) *79. The sum to the infinity of the series 2 3 4 1 2 6 10 14 …… 3 3 3 3 + + + + + is (1) 2 (2) 3 (3) 4 (4) 6 Sol: (2) Let 2 3 4 S 1 2 6 10 14 …. 3 3 3 3 = + + + + + …(1) 2 3 4 1S 1 2 6 10 …. 3 3 3 3 3 = + + + + …(2) Dividing (1) & (2) 2 3 4 S 1 1 1 1 4 4 4 …. 3 3 3 3 3  −  = + + + + +     2 2 2S 4 4 1 1 1 …… 3 3 3 3 3 = +  + + +      2 2 2S 4 4 1 4 4 3 4 2 6 3 3 3 1 1 3 3 2 3 3 2 3     ⇒ = +   = + = + =  −    2S 6 S 3 3 3 ⇒ = ⇒ = 80. The differential equation which represents the family of curves c2x 1 y = c e , where 1 c and 2 c are arbitrary constants is (1) y ‘ = y2 (2) y ” = y ‘ y (3) yy” = y’ (4) ( )2 yy” = y’ Sol: (4) c2x 1 y = c e …(1) c2x 2 1 y ‘ = c c e 2 y’ = c y …(2) 2 y ” = c y ‘ From (2) 2 c y’ y = So, ( ) ( ) 2 y ‘ 2 y” yy” y’ y = ⇒ = 81. One ticket is selected at random from 50 tickets numbered 00, 01, 02, …., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals (1) 1 14 (2) 1 7 (3) 5 14 (4) 1 50 Sol: (1) FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 24 S = { 00, 01, 02, …., 49 } Let A be the even that sum of the digits on the selected ticket is 8 then A = { 08, 17, 26, 35, 44 } Let B be the event that the product of the digits is zero B = { 00, 01, 02, 03, …., 09, 10, 20, 30, 40 } A ∩B = {8} Required probability = ( ) ( ) ( ) 1 P A B 50 1 P A/B P B 14 14 50 ∩ = = = 82. Let y be an implicit function of x defined by x2x − 2xx cot y −1= 0 . Then y ‘(1) equals (1) – 1 (2) 1 (3) log 2 (4) – log 2 Sol: (1) x2x − 2xx cot y −1= 0 …(1) Now x = 1, 1 – 2 coty – 1 = 0 ⇒ coty = 0 ⇒ y = 2 π Now differentiating eq. (1) w.r.t. ‘x’ 2x2x (1 logx) 2 xx ( c osec2y) dy cot y xx (1 logx) 0 dx + −  − + +  =   Now at 1, 2  π      ( ) ( ) 1, 2 2 1 log1 2 1 1 dy 0 0 dx  π        + −  −   +  =           1, 1, 2 2 2 2 dy 0 dy 1 dx  π  dx  π          ⇒ +   = ⇒   = −         83. The area of the region bounded by the parabola ( )2 y − 2 = x −1, the tangent to the parabola at the point (2, 3) and the x-axis is (1) 3 (2) 6 (3) 9 (4) 12 Sol: (3) Equation of tangent at (2, 3) to ( )2 y − 2 = x −1 is 1 S = 0 ⇒ x − 2y + 4 = 0 Required Area = Area of ΔOCB + Area of OAPD – Area of ΔPCD ( ) ( ) ( ) 3 2 0 1 4 x 2 y 4y 5 dy 1 1 x 2 2 2 = + ∫ − + − 3 3 2 0 4 y 2y 5y 1 4 9 18 15 1 3   = +  − +  − = − − + −   = 28 −19 = 9sq. units 0 D (0, 3) P (2, 3) 2y = x + 4 A (1, 2) B (-4, 0) C (0, 2) A (1, 2) (or) FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 25 Area = ( ) ( ) 3 3 2 2 0 0 ∫ 2y − 4 − y + 4y − 5 dy = ∫ −y + 6y − 5 dy ( ) ( ) 3 3 3 2 0 0 y 3 27 3 y dy 9 sq.units 3 3  −  = − − =   = =   ∫ 84. Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P'(x) = 0 . If P(−1) < P(1) , then in the interval [−1, 1] (1) P(−1) is the minimum and P(1) is the maximum of P (2) P(−1) is not minimum but P(1) is the maximum of P (3) P(−1) is the minimum and P(1) is not the maximum of P (4) neither P(−1) is the minimum nor P(1) is the maximum of P Sol: (2) P(x) = x4 + ax3 + bx2 + cx + d P'(x) = 4x3 + 3ax2 + 2bx + c Q x = 0 is a solution for P'(x) = 0 , ⇒ c = 0 ∴ P(x) = x4 + ax3 + bx2 + d …(1) Also, we have P(−1) < P(1) ⇒1− a + b + d < 1+ a + b + d⇒ a > 0 Q P'(x) = 0 , only when x = 0 and P(x) is differentiable in ( – 1, 1), we should have the maximum and minimum at the points x = – 1, 0 and 1 only Also, we have P(−1) < P(1) ∴ Max. of P(x) = Max. { P(0), P(1) } & Min. of P(x) = Min. { P(-1), P(0) } In the interval [ 0 , 1 ], P'(x) = 4x3 + 3ax2 + 2bx = x(4x2 + 3ax + 2b) QP'(x) has only one root x = 0, 4x2 + 3ax + 2b = 0 has no real roots. ∴ ( ) 2 3a 2 32b 0 3a b 32 − < ⇒ < ∴ b > 0 Thus, we have a > 0 and b > 0 ∴ P'(x) = 4×3 + 3ax2 + 2bx > 0, ∀x ∈(0, 1) Hence P(x) is increasing in [ 0, 1 ] ∴ Max. of P(x) = P(1) Similarly, P(x) is decreasing in [-1 , 0] Therefore Min. P(x) does not occur at x = – 1 85. The shortest distance between the line y − x = 1 and the curve x = y2 is (1) 3 2 8 (2) 2 3 8 (3) 3 2 5 (4) 3 4 Sol: (1) x − y + 1= 0 …(1) x = y2 1 2y dy dx = dy 1 dx 2y ⇒ = = Slope of given line (1) 1 1 y 1 2y 2 = ⇒ = ⇒ 2 y 1 x 1 1 2 2 4 = ⇒ =   =     ⇒ (x, y) 1, 1 4 2 =       FIITJEE Solutions to AIEEE – 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 26 ∴ The shortest distance is 1 1 1 4 2 3 3 2 1 1 4 2 8 − + = = + Directions: Question number 86 to 90 are Assertion – Reason type questions. Each of these questions contains two statements Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions also have four alternative choices, only one of which is the correct answer. You have to select the correct choice 86. Let ( ) ( )2 f x = x +1 −1, x ≥ −1 Statement-1 : The set {x : f (x) = f −1 (x)} = {0, −1} Statement-2 : f is a bijection. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Sol: (3) There is no information about co-domain therefore f(x) is not necessarily onto. 87. Let f (x) = x x and g(x) = sinx . Statement-1 : gof is differentiable at x = 0 and its derivative is continuous at that point. Statement-2 : gof is twice differentiable at x = 0. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Sol: (3) f (x) = x x and g(x) = sinx gof (x) = sin(x x ) 2 2 sin x ,x 0 sin x ,x 0 − < =   ≥ ∴ ( ) ( ) 2 2 gof ' x 2x cos x ,x 0 2xcos x ,x 0 − < =   ≥ Clearly, L(gof )'(0) = 0 = R(gof )'(0) ∴ gof is differentiable at x = 0 and also its derivative is continuous at x = 0 Now, ( ) ( ) 2 2 2 2 2 2 gof " x 2cos x 4x sinx ,x 0 2cosx 4x sinx ,x 0 − + < =   − ≥ ∴L(gof )"(0) = −2 and R(gof )"(0) = 2 ∴L(gof )"(0) ≠ R(gof )"(0) ∴ gof(x) is not twice differentiable at x = 0. *88. Statement-1 : The variance of first n even natural numbers is n2 1 4 − Statement-2 : The sum of first n natural numbers is n(n 1) 2 + and the sum of squares of first n natural numbers is n(n 1)(2n 1) 6 + + (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 27 (4) Statement-1 is false, Statement-2 is true Sol: (4) Statement-2 is true Statement-1: Sum of n even natural numbers = n (n + 1) ( ) n(n 1) Mean x n 1 n + = = + ( ) ( ) ( ) ( ) 2 2 2 2 2 2 i Variance 1 x x 1 2 4 ..... 2n n 1 n n =   − =  + + +  − +     Σ 2 2 2 2 ( )2 ( )( ) ( )2 1 4 n n 1 2n 1 2 1 2 ..... n n 1 n 1 n n 6 + + =  + + +  − + = − + (n 1) 2(2n 1) 3(n 1) (n 1)[4n 2 3n 3] (n 1)(n 1) n2 1 3 3 3 3 +  + − +  + + − − + − − = = = = ∴ Statement 1 is false. 89. Statement-1 : ~ (p ↔~ q) is equivalent to p ↔ q. Statement-2 : ~ (p ↔~ q) is a tautology. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Sol: (3) p q p ↔ q ~q p ↔~ q ~ (p ↔~ q) T T T F F T T F F T T F F T F F T F F F T T F T 90. Let A be a 2 x 2 matrix Statement-1 : adj(adj A) = A Statement-2 : adj A = A (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Sol: (2) n 1 2 1 adj A A A A − − = = = ( ) n 2 0 adj adj A A A A A A − = = = FIITJEE Solutions to AIEEE - 2009 FIITJEE Ltd., FIITJEE House, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26515949, 26569493, Fax: 011-26513942. 28 AIEEE–2009, ANSWER KEY Test Booklet Code-A Test Booklet Code-B Test Booklet Code-C Test Booklet Code-D PHY CHE MAT CHE MAT PHY MAT PHY CHE CHE PHY MAT 1. (2) 2. (1) 3. (3) 4. (1) 5. (2) 6. (1) 7. (4) 8. (3) 9. (4) 10. (2) 11. (1) 12. (2) 13. (2) 14. (3) 15. (4) 16. (2) 17. (2) 18. (2) 19. (3) 20. (1) 21. (2) 22. (4) 23. (3) 24. (1) 25. (4) 26. (1) 27. (4) 28. (3) 29. (1) 30. (2) 31. (3) 32. (4) 33. (3) 34. (3) 35. (3) 36. (3) 37. (4) 38. (4) 39. (3) 40. (4) 41. (3) 42. (3) 43. (3) 44. (2) 45. (2) 46. (4) 47. (2) 48. (1) 49. (1) 50. (2) 51. (2) 52. (2) 53. (3) 54. (2) 55. (2) 56. (3) 57. (3) 58. (2) 59. (2) 60. (3) 61. (3) 62. (3) 63. (3) 64. (3) 65. (2) 66. (3) 67. (1) 68. (2) 69. (4) 70. (4) 71. (3) 72. (1) 73. (1) 74. (3) 75. (2) 76. (2) 77. (2) 78. (2) 79. (2) 80. (4) 81. (1) 82. (1) 83. (3) 84. (2) 85. (1) 86. (3) 87. (3) 88. (4) 89. (3) 90. (2) 1. (1) 2. (4) 3. (1) 4. (3) 5. (3) 6. (1) 7. (1) 8. (1) 9. (2) 10. (1) 11. (2) 12. (4) 13. (3) 14. (3) 15. (2) 16. (2) 17. (2) 18. (1) 19. (1) 20. (2) 21. (2) 22. (2) 23. (2) 24. (1) 25. (1) 26. (1) 27. (3) 28. (2) 29. (2) 30. (2) 31. (2) 32. (1) 33. (2) 34. (3) 35. (2) 36. (2) 37. (1) 38. (4) 39. (1) 40. (3) 41. (4) 42. (1) 43. (2) 44. (2) 45. (3) 46. (2) 47. (1) 48. (4) 49. (2) 50. (1) 51. (2) 52. (2) 53. (4) 54. (1) 55. (2) 56. (1) 57. (1) 58. (4) 59. (4) 60. (3) 61. (2) 62. (2) 63. (1) 64. (3) 65. (1) 66. (1) 67. (2) 68. (4) 69. (1) 70. (4) 71. (4) 72. (1) 73. (3) 74. (4) 75. (3) 76. (2) 77. (1) 78. (1) 79. (2) 80. (1) 81. (4) 82. (3) 83. (3) 84. (1) 85. (4) 86. (1) 87. (3) 88. (4) 89. (2) 90. (2) 1. (4) 2. (4) 3. (4) 4. (3) 5. (2) 6. (1) 7. (4) 8. (4) 9. (4) 10. (3) 11. (1) 12. (3) 13. (4) 14. (3) 15. (2) 16. (4) 17. (3) 18. (3) 19. (4) 20. (1) 21. (2) 22. (3) 23. (3) 24. (3) 25. (4) 26. (1) 27. (2) 28. (3) 29. (2) 30. (2) 31. (4) 32. (2) 33. (3) 34. (4) 35. (2) 36. (3) 37. (1) 38. (4) 39. (2) 40. (2) 41. (1) 42. (4) 43. (2) 44. (1) 45. (4) 46. (3) 47. (1) 48. (2) 49. (3) 50. (3) 51. (3) 52. (4) 53. (1) 54. (3) 55. (1) 56. (2) 57. (1) 58. (3) 59. (3) 60. (2) 61. (4) 62. (3) 63. (4) 64. (1) 65. (3) 66. (4) 67. (3) 68. (4) 69. (4) 70. (4) 71. (3) 72. (4) 73. (1) 74. (4) 75. (4) 76. (3) 77. (4) 78. (4) 79. (3) 80. (2) 81. (2) 82. (3) 83. (3) 84. (4) 85. (1) 86. (1) 87. (2) 88. (3) 89. (4) 90. (1) 1. (1) 2. (1) 3. (1) 4. (1) 5. (4) 6. (2) 7. (1) 8. (1) 9. (1) 10. (4) 11. (4) 12. (4) 13. (4) 14. (1) 15. (3) 16. (3) 17. (1) 18. (1) 19. (1) 20. (4) 21. (4) 22. (1) 23. (2) 24. (4) 25. (2) 26. (4) 27. (2) 28. (1) 29. (2) 30. (4) 31. (4) 32. (3) 33. (4) 34. (4) 35. (1) 36. (3) 37. (3) 38. (4) 39. (3) 40. (1) 41. (4) 42. (4) 43. (2) 44. (4) 45. (2) 46. (2) 47. (2) 48. (2) 49. (1) 50. (1) 51. (3) 52. (3) 53. (1) 54. (3) 55. (4) 56. (4) 57. (1) 58. (2) 59. (3) 60. (2) 61. (4) 62. (4) 63. (1) 64. (1) 65. (2) 66. (3) 67. (2) 68. (4) 69. (4) 70. (2) 71. (3) 72. (3) 73. (1) 74. (4) 75. (4) 76. (1) 77. (2) 78. (1) 79. (1) 80. (3) 81. (1) 82. (4) 83. (4) 84. (1) 85. (1) 86. (3) 87. (4) 88. (3) 89. (1) 90. (1)