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IIT JEE Exam 2021 : : JEE Mains AITS 1 (SOLVED)

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AITS 9 (Main) Physics PART (A) : PHYSICS SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 1. 1. A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity 3 m/s towards right. The speed of end B of the rod when rod makes an angle of 600 with the ground is (A) 3 m/s (B) 2 m/s (C) 5 m/s (B) Let the length of rod is and co-ordinate of B is (x,y) v B = v i + v j = 3 i + v j x y (D) 2.5 m/s y x 2 + y 2 = 2 2xv x 2yv y 0 3 y vy 0 x 3 tan 600 v y 0 v y = -1m/s v B 3 i 1j |v B | 3 1 2m/s 2. A thin prism P1 with angle 4 0 and made from glass ( = 1.54) is combined with another prism P2 made of another glass of = 1.72 to produce dispersion without deviation. The angle of prism P2 is 2. (A) 53.30 (B) 4 0 (C) For no deviation ( 1 1) A1 ( 2 1) A2 (C) 30 (D) 2.6 0 40 (1.54 1) (1.72 1)A2 A2 3. 3. 4 0.54 30 0.72 A light emitting diode (LED) has a voltage drop of 2V across it and passes a current of 10mA. When it operates with a 6 V battery through a limiting resistor R, the value of R is (A) 40 (B) 4 (C) 200 (D) 400 (D) The potential drop across R is 4 V while current is 10 mA hence, 10 10 3 R 4 R 400 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 1 AITS 9 (Main) 4. 4. 5. 5. 6. 6. Physics A mirror ( = 3/2) is 10 cm thick. An object is placed 15 cm in front of it from the silvered surface. The position of image from the front surface is (A) 15 cm (B) 21.67 cm (C) 28.34 cm (D) 35 cm (C) The mass of a body is measured as 10.1 kg. The possible percentage error in the measurement is (A) 1% (B) 0.1% (C) 10% (D) 0.01% (A) m 0.1 m 0.1kg; m 10 m Error = 100 1% m Hydrogen (H), deuterium (D), singly ionised helium ( He) and double ionised (Li ) all have one electron round the nucleus. Consider n = 2 to n = 1 transition. The wavelength of emitted radiations are 1 , 2 , 3 and 4 respectively. Then approximately (A) 1 2 4 3 = 9 4 (B) 4 1 2 2 2 3 = 4 (C) 1 2 2 2 2 3 = 3 2 4 (D) 1 2 2 3 = 3 4 (A) hc 1 RhCZ2 1 4 1 1 1 1 : : : 1:1: 4 : 9 1 2 3 4 1 : 2 : 3 : 4 1 : 1 : 7. 7. 1 1 : 4 9 A tunnel is made across the earth passing through its centre. A ball is dropped from a height h in the tunnel. The motion will be periodic with time period. (A) 2 R 0 2h g (C) 2 R 2h 4 g g (B) 2 2R h 4 g g (D) 2 2R g h g (C) T = Tshm 4 2h g CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 2 AITS 9 (Main) 8. An open organ pipe of length l is sounded together with another organ pipe of length l + x in their fundamental tones (x << l ). The beat frequency heard will be (speed of sound is v) (A) 8. vx 4l 2 (C) vx 2l 2 (D) vx 2 2l RT M A particle is hanging from a fixed point O by means of a string of length. There is small nail O in the same horizontal line O at a distance (<L) from O. The minimum velocity with which particle should be projected from its lowest position in order that it may make a complete revolution round the nail. 3gL (B) 5gL (C) g (5L 3 ) (D) g (5 3L ) (C) From conservation of mechanical energy. 1 2 1 mv mgL m3( L ) g 2 2 v 10. 4l 2 vx n RT n ; f2 2 M 2( x) (A) 9. (B) (C) f1 9. Physics g (5L 3 ) In LCR circuit shown in figure, then choose incorrect option (A) current will lead the voltage (B) rms value of current is 20 A 1 2 (D) voltage drop across resistance is 200V (A) (C) Power factor of the circuit is 10. 11. A external agent moves the block m slowly from A to B, along a smooth hill such that every time he applies the force tangentially. Find the work done by agent in this interval. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 3 AITS 9 (Main) (A) 11. m2g2H 2 L Physics (B) mgH 2 L (C) mg(H + L) (D) mgH (D) Wagent - mgH = 0 Wagent = mgH 12. The total flux of electric lines of forces of a charge q placed at the mid point of the edge of a side of rectangular box, as shown in figure through the box is (A) 12. q 2 0 (B) q 4 0 (C) q 8 0 (D) q 16 0 (B) The charge q is completely closed by four identical given box. Therefore, the required flux = 13. 13. 1 q 4 0 1 gm of a radioactive substance takes 50 sec to loose 0.01 gm then the half life of the sample will be 50In(2) 50In(2) 50In(2) 50In(2) (A) (B) (C) (D) ln100 ln 99 ln 0.99 100 ln 99 (A) N N 0e t 100 0.99 1e t t ln 99 100 50sec ln 99 t1/ 2 14. 50ln (2) ln (100 / 99) Two smooth identical stationary spheres are kept touching each other on a smooth horizontal floor as shown. A third identical sphere moving horizontally with a constant speed hits both station sphere CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 4 AITS 9 (Main) Physics symmetrically. If after collision the third sphere moves in same direction with one fourth of its initial speed, the coefficient of restitution will be 14. (A) 2/3 (C) (B) 1/3 v 2mv1 cos 300 m mv 4 v v1 cos 300 4 e v cos 300 15. (C) (D) 1/6 .(i) .(ii) Two immiscible liquids are poured in a U-tube having densities 1 and 2 . The ratio of height of the liquid above their interference ( (h1 / h2 ) is (A) (B) (C) (D) 15. directly proportional to their densities inversely proportional to their densities directly proportional to square of their densities equal (B) 1 gh1 2 gh2 h1 2 h2 1 16. A very long uniformly charged rod falls with a constant velocity v through the center of a circular loop. Then the magnitude of induced emf in loop is (charge per unit length of rod = ) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 5 AITS 9 (Main) (A) 16. Physics 0 V2 2 (B) 0 2 V2 (C) 0 V 2 (D) zero (D) B 0 B S 17. d B 0 dt There is a uniform time varying magnetic field in a circular region as shown in the figure. Find out the potential difference across 2 point along an elliptical path as shown in figure. (A) 17. R2 2 2 B0 (C) R2 4 B0 (D) R2 5 B0 (C) V = d dB R R (Area) (B0 ) dt dt 2 2 VAB 18. (B) B0 R2 R2 4 B0 The figure shows the variation of photo current with anode potential for a photo sensitive surface for three different radiations. la , lb , and lc be the intensities and f a , fb , and fc be the frequencies for the curves a, b and c respectively (A) f a fb and la lb (B) f a f c and la lc CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 6 AITS 9 (Main) Physics (C) f a fb and la lb (D) fb f c and lb lc 18. (A) Saturation for A and B are different hence intensities are different. Stopping potential for A and B are equal hence frequency are equal. 19. Pressure versus density of an ideal gas is shown in the graph. Then the correct temperature versus density graph of gas is (A) (B) (C) (D) 19. (A) Process A B is isothermal Process B C isochoric Process C A is isobaric 20. A resistance of frustum shape is shown in figure. If a current I passes through the resistance, the electric field at A and B are related as (A) EA > EB 20. (B) EB > EA (C) EA = EB (D) there is no relation (A) From ohm s law Electric field current density CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 7 AITS 9 (Main) Physics NUMERICAL VALUE TYPE This section contains 10 questions. Attempt any 5 questions out of 10.Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. 21. 21. 10v Two coherent point sources of frequency f where v is speed of sound in air are placed at a d distance d apart as shown in figure. The receiver is free to move along the dotted line shown in the figure. Find total number of maxima heard by receiver. (6) f v 10 v d d 10 Maximum path diff. can be 22. 22. d . Hence, number of maximum will be 6. 3 A boat which has a speed of 6km/h in still water crosses a river of width 1 km along the shortest possible path in 20 min. The velocity of the river water in km/h is (5.19) 1 T = 20 min = hr 3 1 1km = (6)2 V 2 hr 3 V = 3 3 km/hr 23. A uniform solid brass sphere is rotating with angular speed 0 about a diameter. If its temperature is now increased by 1000 C . What will be its new angular speed. (Given abrass 2.0 10 5 per 0 C) 23. (0.99) Variation of moment of inertia with temperature I = I0 (1 + 2a T ) C.O.A.M. Ii i = If f 24. 24. A cube with a mass m = 20 g wettable by water floats on the surface of water. Each side of the cube has length l = 3cm.The angle of contact between water and glass is zero degree and the surface tension of the water is 7.5 10 2 N/m. The distance between the lower face of the cube and the surface of the water is (2.32) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 8 AITS 9 (Main) 25. 25. Physics The dimensional formula of a physical quantity x is [M-1L3T-2 ]. The error in measuring the quantities M, L and T are 2%, 3% and 4%. The maximum percentage error that occurs in measuring the quantity x is (19.00) 26. A 6 volt battery o internal resistance 1 is connected across a uniform wire AB of length 100 cm. The positive terminal of another battery of emf 4V and internal resistance 1 is joined to the point A as shown. Find the distance of point D from A if there is no current through galvanometer. (resistance of AB = 5 ) 26. (80.00) 4 100 80cm 5 A brass scale is graduated at 100 C . What is the true length of a zinc rod which measures 60.00 cm on this scale at 30 0 C . Coefficient of linear expansion of brass = 18 10 6 K 1. 60.02 Since the scale is graduated at 100 C . 1 cm of the scale at 100 C = exactly 1 cm cm of the scale at 30 0 C . Distance AD = 27. 27. = exactly (1 + 8 10 6 20)cm 60 cm of the scale at 30 0 C . = exactly 60.00 (1 + 36 10 5 ) = 60.02 cm CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 9 AITS 9 (Main) 28. 28. Physics A very small sphere of mass 80 g having a change q is held at a height 9 m vertically above the center of a fixed non conducting sphere a radius 1 m, carrying an equal charge q. When released it falls until it is repelled just before it comes in contact with the sphere. The charge q is 7K. The value of K is (in C ) [ g 9.8m / s 2 ] (4) Keeping in mind that here both electric and gravitational potential energy are changing and for external point a charged sphere behaves as whole of its charge were concentrated at its centre, applying conservation of energy between initial and final we have qq 1 q2 mg 9 mg 1 4 0 9 4 0 1 1 80 10 3 9.8 109 Or q = 28 C or q 2 = 29. 29. A hollow sphere and a solid sphere have equal mass and equal moment of inertia about the respectively diameter. Find the ratio of their square of radii is given by (0.60) 2 2 MR12 MR 22 3 5 30. For the circuit shown in the figure, the current through the inductor is 0.6A, while the current through the capacitor is 0.4A. Find the current (in A) drawn from the generator. 30. (0.20) IL and IC will be in opposite phases Inet = IL - IC = 0.6 0.4 = 0.2A CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 10 AITS 9 (Main) Chemistry PART (B) :CHEMISTRY SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 31. In terms of mole fraction, equilibrium constant is written as K x . The relationship between K c and K x is RT (A) K c = K x P (C) K c RT = Kx P a (B) K c = K x n (D) Kx = K c P RT n P RT n 31. (B) 32. For HCl (ag) + NaOH(aq) NaCl(aq) + H 2O(l), H = -13.6 Kcal for the reaction 32. 1 1 H 2SO4 (aq) + NaOH Na 2SO4 (aq) + H 2O(L), H is 2 2 (A) 6.8 Kcal (B) 13.6 Kcal (C) +13.6 Kcal (B) 33. (D) +6.8 Kcal If the energy of H-atom in the ground state is E, the velocity of photo-electron emitted when a photon having energy E p strikes a stationary Li 2 ion in ground state, is given by: (A) v 2( E p E ) m (B) v 2( E p 9 E ) m (C) v 2( E p 9 E ) m (D) v 2( E p 3E ) m 33. (C) 34. Among the following metal carbonyls, the C-O bond order is lowest in (A) [Mn(CO)6 ]+ (B) [Fe(CO)5 ] (C) [Cr(CO) 6 ] (D) [V (CO ) 6 ] 34. (B) 35. The Nernst equation E E RT / nF ln Q indicates that the Q will be equal to equilibrium constant Kc when: 35. 36. (A) E E (C) (B) RT / nF 1 (C) E zero (D) E 1 The standard oxidation potential of the electrodes Ag / Ag , Sn / Sn 2 , Ca / Ca 2 , Pb / Pb 2 are 0.8, 1.36, 2.86 and 10.12V respectively. The most powerful oxidizing agent among these metals is (A) Pb (B) Ca (C) Sn (D) Ag CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 1 AITS 9 (Main) Chemistry 36. (D) 37. 37. How many 9.65 minutes will it take to plate out 5.2 g of Cr from a Cr2 ( SO4 )3 solution using a current of 9.65 A? (Atomic mass: Cr = 52.0) (A) 200 (B) 50 (C) 100 (D) 103 (B) 38. For the reaction between CO 2 graphite CO 2 (g) + C(g) 2CO(g) H = 170kJ S = 170J/K 38. Reaction is spontaneous at (A) 1200K (B) 900K (A) (C) 500K (D) 298K 39. The melting point of a solid is 27 0 C and its latent heat of fusion is 600 calories/mole the entropy change for the fusion of one mole of the solid in cal/K is (A) 2.22 (B) 2 (C) 22.2 (D) 0.5 (B) 40. The number of lone pairs is same in PCl3 and 39. (A) BCl3 (B) NCl3 40. (B) 41. 41. The bond angle and hybridization in ether is (A) 1060 51 ,sp3 (B) 1040 31 ,sp3 (D) 42. Imidazole has basic site/s 42. (A) Two (C) 43. Predict the product of the reaction below (B) Three (C) CCl 4 (D) PCl5 (C) 1090 28 ,sp 3 (D) None of these (C) One (D) None CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 2 AITS 9 (Main) (A) (B) (C) (D) Both (A) and (B) 43. (D) 44. Acidic nature is more for (A) o-amino phenol (C) p-amino phenol (B) 44. 45. 45. 46. 46. Chemistry (B) m-amino phenol (D) all have equal Ka s Anti Markovnikoffs addition of HBr is not observed in (A) Propene (B) Butene (C) Butene-2 (C) (D) Pentene-2 A carbohydrate which cannot be hydrolyzed to simpler compounds is called: (A) Monosaccharide (B) Polysaccharide (C) Disaccharide (D) Trisaccharide (A) 47. A is: (A) (B) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 3 AITS 9 (Main) (C) 47. Chemistry (D) (B) 48. Z is: (A) (B) (C) (D) 48. (C) 49. A nitrogeneous substance X is treated with HNO 2 and the product so formed is further treated with NaOH solution, which produces blue colouration X can be (A) CH 3CH 2 NH 2 (B) CH3CH 2 NO 2 (C) CH 3CH 2 ONO 2 (D) (CH 3 ) 2 CHNO 2 49. (D) 50. What is A? 50. (A) (B) (C) (D) None (A) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 4 AITS 9 (Main) Chemistry NUMERICAL VALUE TYPE This section contains 10 questions. Attempt any 5 questions out of 10.Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. 51. 51. An unknown volume and unknown concentration of weak acid HX is titrated with NaOH of unknown concentration. After addition of 10.0 cm3 of NaOH solution pH of solution is 5.7 and after the addition of 20.0 cm3 of NaOH solution, the pH is 6.3. Calculate the pK 8 for the weak acid HX. (given antilog of 0.6 4) (6) 52. One mole ideal monoatomic gas is heated according to path AB and AC. If temperature of state B q and state C are equal. Calculate AC 10 . q AB 52. (8) 53. How many faradays of electricity are required for the reduction of 1 mole nitrobenzene into Nphenyl hydroxylamine? 53. (4) 54. 54. A large irregularly shaped closed tank is first evacuated and then connected to a 5.5 litre cylinder containing compressed nitrogen gas. The gas pressure in the cylinder, originally at 2.13 atm, falls to 1.562 atm after it is connected to the evaculated tank. Calculate the volume of the tank in litres. (2) 55. What is the maximum value of van t Hoff factor for AlCl3 ? 55. (4) 56. 56. 22.21 ml solution of HIO3 is reacted with excess of an aqueous solution of SO2 . The excess of SO 2 and I 2 formed are removed by heating the solution. The remaining solution in neutralized by 35.5 ml of 0.16 (N)NaOH solution. The strength of HI O3 in gm/lt. (Round off the answer to the nearest whole number) is ..? (9) 57. How many lone pairs are present on the central atom in [ICl 4 ] 57. (2) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 5 AITS 9 (Main) Chemistry 58. Fructose is subjected to bond cleavages by HI O4 . The number of HCHO unit(s) per unit fructose is 58. (2) 59. Nitration of given compound will take place which one of the carbons indicated by number. 59. (3) 60. When 1-bromomethylcyclohexene under-goes solvolysis in ethanol, what will be the number of major product? (3) 60. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 6 AITS 9 (Main) Mathematics PART (C) : MATHEMATICS SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 61. 61. If a, b, c, d R+ then the equation(x2 + ax - 3b)(x2 + cx + b)(x2 + dx + 2b) = 0 has (A) 6 real roots (B) At least 2 real roots (C) 4 real roots (D) 3 real roots (B) The discriminants of the given equation are D1 a 2 12b ; D2 c 2 4b and D3 d 2 8b D1 D2 D3 a 2 c 2 d 2 0 At least one of D1 , D2 , D3 is non-negative. Hence the equation has at least two real roots. 100 62. The value of [cot 1 x] [tan 1 x] dx is [where [.] denotes greatest integer function) 0 62. (A) 100 + 1cot1 (B) 100 [cot 1 (B) 100 + 2cot2 (C) 2cot2 (D) 100 2cot2 x] [tan 1 x] dx cot1 (100 tan1) 0 1 tan 2 1 100 100 2 cot 2 tan1 63. sin 2 x sin 5 x sin 3x dx (A) n sin 3 x n sin 5 x c (C) 63. 1 1 n sin 3 x n sin 5 x c 3 5 (B) 1 1 n sin 3 x n sin 5 x c 3 5 (D) 3 n sin 3 x 5 n sin 5 x c (C) sin 2 x sin(5 x 3 x ) sin 5 x sin 3x dx sin 5x sin 3x sin 5 x cos 3 x cos 5 x sin 3 x sin 5 x sin 3 x CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 1 AITS 9 (Main) Mathematics 1 1 n sin 3 x n sin 5 x c 3 5 64. 64. The general solution of differential equation ( x 6 y 4 x 2 ) dy (1 x 5 y 5 xy ) dx is (A) In |x| = xy x4 y 4 C 4 (B) In |y| = xy x4 y 4 C 4 (C) In |x| = xy x5 y5 C 5 (D) In |y| = xy x5 y5 C 5 (C) ( x 6 y 4 x 2 ) dy (1 x 5 y 5 xy ) dx x ( xdy ydx ) dx x 5 y 4 ( xdy ydx ) 0 dx (1 x 4 y 4 )d ( xy ) x ln | x | xy x5 y5 C 5 65. 65. (A) 66. lim h 0 x h x 1 4 4 sin tdt sin tdt = h a a (A) sin 4 x 66. h 0 (D) sin 5 x 5 sin 4 ( x h) 0 sin 4 x 1 If , be the roots of x 2 x 1 0 and A n = n + n then A,M. of A n - 1 and An is (A) A n +1 67. (C) 0 (A) Use L Hospital rule, we get = lim 67. (B) 4 sin 3 x cos x (B) (1/2)An 1 (C) 2An -2 (D) 1 ( An 2 ) 2 (B) 1 and 1 AM of A n - 1 and A n = = n 1 n 1 An - 1 + An n 2 n 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 2 AITS 9 (Main) = Mathematics n 1 (1 ) n 1 (1 ) 2 ( ) n 1 ( 2 ) 2 1 1 = ( n 1 n 1 ) An 1 2 2 = 68. n 1 2 2 1 and 2 1 Solution of the differential equation (2 x 10 y 3 ) dy y 0 is dx (A) ( x 2 y ). y c 2 y3 2 (B) x y c 5 (C) ( x 2 y3 ). y c 2 y3 2 (D) x y c 5 3 68. 2 (A) (2 x 10 y 3 ) dy y 0 dx dx 2 x 10 y 3 0 dy dx 2 x 10 y 2 0 dy y y 2 I.F. e y dy y2 xy 2 10 y 4 dy xy 2 2 y5 c y2 ( x 2 y3 ) c 69. If , are the roots of the equation ax 2 bx c 0 then (A) 69. 2 c 2 a (C) 2 a (D) 2 b (D) 10 n (10 1) 2n 9 (B) b a a b a a b a , 70. (B) a b a b c a 2 a b a b a The sum to n terms of the series 11 + 103 + 1005 + 10007 +. . is 1 1 10 n (A) (10n 1) n 2 (B) (10 n 1) 2n (C) (10 1) n 2 9 9 9 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 3 AITS 9 (Main) 70. Mathematics (C) Let S n denote the sum to n terms of the given series. Then Sn = 11 + 103 + 1005 + to n terms S n (10 1) (10 2 3) (103 5) ..... {10 n (2n 1)} S n (10 10 2 ..... 10 n ) {1 3 5 .... (2n 1)} Sn 71. 71. 10(10 n 1) n 10 (1 2n 1) (10 n 1) n 2 (10 1) 2 9 The set of values of p for which both the roots 4 x 2 20 px (25 p 2 15 p 66) 0 are less than 2 will be (B) (2, ) (A) (4/5, 2) (D) (C) (-1, 4/5) of the quadratic equation (D) ( , 1) 4 x 2 20 px (25 p 2 15 p 66) 0 x 2 5 px (25 p 2 15 p 66) 0 4 Case 1: (i) D 0 (25 p 2 15 p 66) 0 4 -15p + 66 0 15 p 66 0 25 p 2 4.1 66 15 22 p 5 p 22 p , 5 ii) f(2) > 0 16 40p + 25 p 2 15 p 66 0 (i) 25 p 2 25 p 50 0 p2 p 2 0 p2 2 p p 2 0 p ( p 2) 1( p 2) 0 p ( , 1) (2, ) .(ii) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 4 AITS 9 (Main) Mathematics b 2 2a 20 p 16 4 2 p p 2.4 20 5 4 p , 5 (1) (2) (3) (3) p , 1 /2 72. The value of x cot x dx is 0 72. (A) n 2 (B) - n2 (C) ( n 2) / 2 (C) Integrating by parts taking cot x as second function, given integral I x log sin x 0 /2 (D) 2 n2 /2 log sin x dx 0 2 0 lim ( x log sin x) x 0 1 log sin x dx 2 log 2 0 log sin x As lim x log sin x lim x 0 x 0 1/ x x2 cot x lim lim 2 x 0 1/ x x 0 tan x x lim x, 0.1 0 x 0 tan x 73. 73. The degree and order of differential equation of family of all parabolas whose axis is x-axis are respectively (A) 1,2 (B) 2,1 (C) 1,1 (D) 2,2 (A) By repeated differentiation y 2 4a ( x h) yy1 2 a yy2 y12 0 Degree is 1 and order 2 74. 74. The area bounded by parabola y2 = x , straight line y = 4 and y -axis is 16 64 (A) (B) (C) 7 2 3 3 (B) (D) 32 3 y = 4 meets the parabola y 2 x at A is (16,4) Required area = Area of rectangle OMAX Area OMA CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 5 AITS 9 (Main) Mathematics 16 16 4 16 x dx 64 0 64 75. 75. x 3/2 3/ 2 0 2 3 128 64 sq. units (4) 64 3 3 3 The ratio in which the area bounded by the curves y 2 12 x and x 2 12 y is divided by the line x = 3 is (A) 15:49 (B) 13: 480 (C) 13:37 (D) 1 : 1 (A) 3 A1 3 12 x dx 0 2 x3/2 12 3 3 0 12 A2 3 x2 0 12 dx 1 3 3 45 x 0 36 4 12 12 x dx x2 147 3 12 dx 4 (on simplification) Required ratio = 45 : 147, i.e. e., 15 : 49 76. lim n 1 1 ( n 2 12 )( n 2 22 )(n 2 32 ).....(n 2 4n 2 ) n is equal to 4 n (A) 2 n5 2(tan 2) 4 1 (C) 76. 2 ( n 2) 2 25e 2 tan (B) e4 (D) 1 2 2e /2 e2 (B) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 6 AITS 9 (Main) n S 1 n 2n n r 0 Mathematics n2 r 2 1 n2 n n 1 r 0 2 2n 2 = n(1 x 2 ) dx x n(1 x 2 ) 0 0 r n 2 2 2 x2 0 1 x 2 dx 2 2 dx = 2 n 5 2 1dx 2 n 5 4 2 tan 1 2 2 1 x 0 0 1 1 25 S e 2 n 5 4 2 tan 2 4 e 2 tan 2 e Paragraph for Question Nos. 77 to 78 dy f ( x, y ) where f and g are homogeneous function of x and y dx g ( x, y ) and of the same degree, is called homogeneous differential equation and can be solved easily by putting y vx. A differential equation of the form 77. 77. The solution of the differential equation dy x 2 xy y 2 is dx x2 x (A) tan 1 ny c y y (C) tan 1 nx c x (C) Put y = vx Differential equation becomes y (B) tan 1 ny c x y (D) sin 1 nx c x dv x 2 x 2 v v 2 x 2 1 v v2 2 dx x dv dv 1 v2 x v x y tan 1 v nx c tan 1 nx c x 78. 78. The solution of the differential equation dy x( x 2 3 y 2 ) 0 is dx y ( y 2 3x 2 ) (A) x 4 y 4 x 2 y 2 c (B) x 4 y 4 3 x 2 y 2 c (C) x 4 y 4 6 x 2 y 2 c (C) Differential equation becomes (D) x 4 y 4 9 x 2 y 2 c 4(v3 3v) dx v 4 6v 2 1 dv 4 x n(v 4 6v 2 1) 4 nx nc CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 7 AITS 9 (Main) Mathematics x4 y 4 6 x2 y2 c Paragraph for questions Nos. 79 to 80 We are giving the concept of arithmetic mean of mth power. Let a, b > 0 and a b and let m be a m real number. Then a m bm a b , if m > 1 2 2 m am bm a b 2 2 If a = b then equality comes in all cases. On the basis of above information answer the following questions. However if m (0,1) then 79. If a, b, be positive and a + b = 1 (a b) and If A = (B) A (A) A 22/3 79. 2/3 2 3 3 a 3 b then the correct statement is (C) A 22/3 (D) A 22/3 (C) 1/3 a1/3 b1/3 a b 2 2 1/3 1 2 a b 1 a1/3 b1/3 1 1/3 2 2 2 A 1/3 or A < 2 2/3 2 80. 80. If x, y be positive real numbers such that x 2 y 2 8 , then the maximum value of x +y is (A) 2 (B) 4 (C) 6 (D) 8 (B) 1/2 ( x 2 )1/2 ( y 2 )1/ 2 x 2 y 2 2 2 |x| | y| 2 2 Or |x| + |y| 4 x y 4 1/2 8 2 2 NUMERICAL VALUE TYPE This section contains 10 questions. Attempt any 5 questions out of 10. Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 8 AITS 9 (Main) Mathematics 81. f (x The value of n x n ) log x 0 81. dx is x (0) Here limits and type of function suggest that there is something which is reciprocal to each other. 1 Let, t 1/ x x 1/ t dx 2 dt t Also when x 0, t ; x , t 0 I f (x n 0 0 f (t n x n ) n x 1 t n ) n t f (t n t n ) n(t ) 0 dx x dt t2 1 t dt I 2I 0 I 0 t Paragraph For Question Nos. 25 to 26 k r f n, (for k N), then S = f ( x) dx. then S x n r 0 0 1 transforms to x, tansforms to and transforms to dx. n If S = lim 82. 82. 1 n kn k f ( x) dx. 0 In this process r n 1 2 n sin sin 2 ..... sin 2 is equal to n n 2n n 2n (0.5) 1 n lim sin 2 sin 2 ..... sin 2 n n 2n 2n 2n lim sin 2 2 0 1 x dx cos 2 2 0 1 x dx 2 I 1 I 1 2 83. The area bounded by the curves 83. (0.33) The required area x y 1 and x + y = 1 is CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 9 AITS 9 (Main) Mathematics 1 = area of AOB 1 0 x 2 dx 1 1 x2 x 3/2 1 1 1 x 2 2 2 3 / 2 0 2 6 3 x e (1 t )2 dt and g(x) = f(h(x)) where h(x) is defined for all x R. If g (2) = e4and h (2) 1 84. Let f(x) = 84. then absolute value of sum of all possible values of h(2) is (2) 2 h( x) g ( x) 2 e (1 t ) dt g ( x) h ( x ) . e (1 h ( x )) 2 2 g (2) h (2) . e (1 h (2)) 2 2 e 4 1. e(1 h (2)) 1 h(2) 2 h (2) 3,1 sum = -2 |sum| = 2 21 85. a j 693 where a1 , a2 ,....a21 are in A.P. then j 1 65. 10 a i 0 2i 1 is (363) 21 a j 1 j 693 21 (a1 a21 ) therefore a1 a21 66 2 Now a11 A.M = 693/21 = 33 Also a2 a20 a3 a10 ...... a9 a13 a10 a12 66 10 a2i 1 5 ( a1 a21 ) a11 5 66 33 363 i 0 86. 86. If x1 , x2 , x3 ,......., x2008 are in HP and 2007 x x i 1 i i i x1 x2008 , then is (2007) Given x1 , x2 , x3 ,.......x2008 are in HP CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 10 AITS 9 (Main) 1 1 1 1 1 are in AP. , , ,......., , x1 x2 x3 x2007 x2008 1 1 1 1 1 1 ...... , D x2 x1 x3 x2 x2008 x2007 Then x x2008 x1 x2 x2 x3 ..... 2007 D x1 x2 x2 x3 x2007 x2008 or Mathematics ( x1 x2 ) ( x2 x3 ) D x1 x2 .... x2 x3 .... x2007 x2008 (By law of proportion) or x1 x2 .... x2 x3 .... x2007 x2008 2007 or xx i i 1 i 1 Now, 1 x2008 x1 x2008 D x1 x2008 D (i) 1 (2008 1)D x1 x1 x2008 2007 x1 x2008 .(ii) D From eqs. (i) and (ii) we get Or 2007 x x i 1 87. the 87. i i 1 2007 2007 x1 x2008 If 2nd, 5th, and 9th terms of an A.P. are in G.P., then the sum of all possible ratios of the first term to common difference of A.P. is (8) a + d, a + 4d, a + 8d are in G.P. (a 4d )2 (a d )(a 8d ) 88. a 2 8ad 16d 2 a 2 9ad 8d 2 a 8d 2 ad 8 d Let S1 , S2 ,......Sn be squares such that for each n 1, the length of a side of S n equals the length of a diagonals of S n 1. If the length of a side of S1 is 10 cm, then find the minimum value of n for which the area of Sn is less than 1 sq. cm? 88. (8) We have length of side of S n length of a diagonal of S n 1 . Length of a side of S n = Length of a side of Sn + 1 Length of a side of Sn 2 (length of a side of S n 1 ) 1 for all n 1 2 Sides of S1 , S2 ,.....Sn from a G.P. with common ratio 1 and first term 10. 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 11 AITS 9 (Main) Mathematics 1 sides of Sn = 10 2 n 1 10 2 ( n 1)/2 Area of S n ( side) 2 2 10 100 = n 1 n 1 . Now, area of S n 1 n 1 7 2 2 2 /2 89. The value of the integral 89. (0) a sin log d , a > 1 is a sin /2 a sin a sin f ( ) log for which f ( ) log a sin a sin a sin log f ( ) a sin Hence the integrand is an odd function. So, the given integral is zero. 90. 90. Find the area covered by the curve y max .{2 x, 2,1 x} with x-axis from x = -1 to x = 1 is (4.5) The Graph of (x) =max {2 x, 2, 1 + x} is shown as in the figure. 1 Therefore max {2 x ,2, 1 x}dx 1 = Area of the shaded region 1 9 = (2 3) 1 1 2 2 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 12

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