
	Trending ▼ ICSE CBSE 10th ISC CBSE 12th CTET GATE UGC NET Vestibulares ResFinder

IIT-JEE Chemistry-2012

22 pages, 0 questions, 0 questions with responses, 0 total responses,

	mohan28	+Fave Message
	Profile Timeline Uploads

Home > mohan28 >

Formatting page ...

32 IIT/ELITE 2012 PET I/PET IV/CPM/P(II)/SOLNS BRILLIANT S PROGRESSIVE EVALUATION TEST FOR STUDENTS OF OUR ONE/TWO-YEAR POSTAL COURSES TOWARDS IIT-JOINT ENTRANCE EXAMINATION, 2012 PAPER II SOLUTIONS CHEMISTRY PHYSICS MATHEMATICS PART A: CHEMISTRY SECTION I 1. (A) 1 1 1 = =R 2 2 n2 n1 For first Lyman transition, 1 1 1 3 = L = R = R 2 1 22 4 For first Paschen transition, 1 1 7 1 = p = R = R 2 2 3 4 144 L : P = 3 7 : 4 144 = 3: 7 36 = 108 : 7 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 33 2. (D) Species Hybridization + 5 + 4 1 8 3 = = 4 sp 2 2 5+0+1 6 2 = = 3 sp 2 2 + 5 + 0 1 4 = = 2 sp 2 2 NH4 NO3 NO2 3. (A) In the reaction, H + Li LiH Li undergoes oxidation and H undergoes reduction ( H + e H ) . Hence hydrogen acts as an oxidising agent. 1 mol. wt 4. (A) Density of the solution, d = M + 1000 m 98 1 d = 11.07 + 21.91 1000 = 11.07 [0.0456 + 0.098] = 11.07 0.1436 = 1.589 g/mL 5. (B) 1 3 N 2 + H2 2 2 NH3 Partial pressure of NH3 = 24.91 100 = 24.91 atm 100 Pressure of (N2 + H2) = 100 24.91 = 75.09 atm Partial pressure of H2 = 3 75.09 = 56.32 atm 4 Partial pressure of N 2 = 1 75.09 = 18.77 atm 4 Kp = = PNH3 P1/2 N2 3/2 PH 2 = 24.91 1/ 2 (18.77 ) ( 56.32 )3 / 2 24.91 1 = 0.0136 atm 4.3 421 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 34 d RT 6. (C) For gas A, PA = A A MA d RT For gas B, PB = B B MB PA d A TA MB = PB dB TB M A [since MA = MB] PA 1 3 3 = 1 = 2 PB 2 1 or PA : PB = 3 : 2 7. (A) NO2 group exhibits I and M effect. These ve effects are greater for NO2 group than for CN group. Also I effect is more when the group is CH2 NO2 present in the ortho position. Thus is the most stable carbanion among them. 8. (C) I and M effects increase the strength of acid. Since NO2 group present in NH3 para position exhibits M and I effect, is most acidic. NO2 SECTION II 9. (A), (C) 2Na 2S2O3 + I2 2NaI + Na 2S4 O6 (tetrathionate) Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 35 10. (A), (B), (D) NaI < NaCl < BaO < CaO (Lattice energy) BaCO3 > SrCO3 > CaCO3 > MgCO3 (Thermal stability) Dipole moment of NH3 is greater than that of NF3. Nuclear spins are in opposite direction in para hydrogen. 11. (A), (B), (D) G = H T S When H is ve and S is +ve, G becomes ve. So exothermic reactions are spontaneous at all temperatures. When H is +ve, and S is ve, G is + ve. Hence such endothermic reactions are not spontaneous at all temperatures. When H is +ve and S is +ve, then G is ve only when H < T S i.e., G is ve and spontaneous when temperature is high. 12. (A), (C) Reductive ozonolysis CH3 C C CH2 CH3 O O3 CH3 C C CH2 CH3 O O H O / Zn 2 CH3 C C CH2 CH3 + ZnO O O dil H SO 1% HgSO4 2 4 CH3 C C CH2 CH3 CH3 C = C CH2 CH3 H OH Rearrange CH3 CH2 C CH2 CH3 O OH CH3 C C CH2 cold alkaline CH3 KMnO4 OH CH3 C C CH2 CH3 OH OH 2H O 2 CH3 C C CH2 CH3 O O (i) O3 CH3 C C CH2 CH3 CH3 COOH + CH3CH2 COOH (ii) H2O Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 36 SECTION III OH CH3 CH3 OH2 + H heat 13. (5) CH3 CH3 CH3 CH3 Ring + C H expansion CH3 H O 2 CH3 H CH3 CH3 14. (6) 2 All the six carbon atoms are in the sp hybrid state. 15. (2) CH CH + 2 RMgX 2R H + XMg C C MgX Acetylene 16. (4) + A (g) Initial conc. 2B(g) 2C(g) + D(g) 1.5 1 At (1 x) equilibrium conc. (1.5 2x) 2x x At equilibrium since [A] = [D], (1 x) = x or x = 0.5 Kc = = [C]2 [D] [A] [B]2 = (2 0.5)2 (0.5) (1 0.5) [1.5 1]2 1 0.5 1 = =4 0.5 0.25 0.25 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 37 2 C O2 + 2H O 2CO3 + 4H+ + 2e 2 2 4 3 17. (4) MnO + 4H + + 3e MnO2 + 2H2O 4 2 Add 2 2 3C2O4 + 2MnO4 + 2H2O 6CO3 + 2MnO2 + 4H+ Since the reaction takes place in the basic medium, 2 2 3C2O4 + 2MnO4 + 2H2O + 4OH 6CO3 + 2MnO2 + 4H2O 2 2 or 3C2O4 + 2MnO4 + 4[OH ] 6CO3 + 2MnO2 + 2H2O T M 18. (1) v rms v rms (H v rms (O 2 2 at 50 K) = at 800 K) 50 32 =1 2 800 SECTION IV 19. (A) (p), (q); (B) (p), (r), (t); (C) (p), (s), (t); (D) (p), (q), (t) 3 (A) 50 ml of 0.1 M HNO3 + 50 ml of 0.1 M KOH gives 50 10 0.1 mole of KNO3. Complete neutralisation takes place. KNO3 does not undergo hydrolysis. It simply ionises. Hence the pH of solution is 7. (B) CH3 COOH + NaOH (moles) 50 0.2 10 3 50 0.2 10 CH3COONa + H2O 3 gives 50 0.2 10 3 moles of CH3 COONa which is present in 100 ml of solution. 50 0.2 10 3 [CH3COONa] = 1000 moles / lit 100 = 10 10 3 10 = 0.1 M CH3COONa undergoes hydrolysis in solution pH = 1 [ pK w + pK a + log [salt]] 2 = 1 [14 + 4.8 + log 0.1] 2 = 1 1 [14 + 4.8 1] = (17.8) 2 2 = 8.9 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 38 (C) + HCl (moles) 50 0.1 10 50 0.1 10 [NH4 Cl] = 100 50 0.1 10 3 NH4 OH 3 3 + NH4 Cl 50 0.1 10 3 H2O in 100 ml 1000 = 5 10 2 M NH4Cl undergoes salt hydrolysis. 1 [ pK w pK b log (salt)] 2 pH = = 1 14 4.8 log 5 10 2 2 = 1 [14 4.8 + 2 0.6990] 2 = 1 [11.2 0.6990] 2 ( ) = 5.6 0.3495 = 5.25 3 (D) 5 10 moles of NH4OOC CH3 is formed in 100 ml. Hence [CH3COONH4 ] = 5 0.1 10 3 1000 = 5 10 2 M 10 This salt undergoes hydrolysis. + CH3COO + NH4 CH3COONH4 + CH3COO + NH4 pH = = + H2O CH3COOH + NH4 OH 1 [pK w + pK a pK b ] 2 1 [14 + 4.8 4.8] = 7 2 20. (A) (p), (r), (t); (B) (p), (t); (C) (q); (D) (s) ii (A) S F4 sp3 d ii (B) XeF6 sp3d3 (C) CCl4 sp 3 3 2 (D) SF6 sp d Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 39 PART B: PHYSICS SECTION I 21. (D) Let the initial velocity be v i along the +x direction and the final velocity v f along the direction making an angle with v i . y Now v i = vi v f = v cos i + v sin j vf Change in velocity = v f v i i.e., v = v cos i + v sin j vi x vi = v (cos 1) i + v sin j v = v2 (cos 1)2 + (v sin )2 = v cos2 2cos + 1 + sin 2 = v 2 2 cos = v 4 sin 2 = 2v sin 2 2 2v sin v average acceleration a = t = 2 t 22. (C) Since the rod is in rotational equilibrium, torque on it is zero. N mg = 0 F.B.D. for the rod N = 39.2 N N Taking moments about O, 60 0.5 0.5 m sin = N cos m 2 2 O sin N = cos 60 = mg 39.2 60 P = 0.65 = tan 1 (0.65) Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 40 1 23. (B) Elastic potential energy U = stress strain per unit volume 2 = = But strain 1 B (strain)2 , where B is bulk modulus 2 y = (a cos kx cos t) x x = ak sin kx cos t U = = 1 B ( ak sin kx cos t)2 2 1 B a 2 k 2 sin 2 kx cos2 t 2 The velocity of longitudinal wave is given by c= B B = c 2 Also c = U = k 1 a 2 2 cos2 kx cos2 t 2 U (x, 0) = In units of 1 a 2 2 cos2 kx 2 a 2 2 2 2 U(x, 0) = cos kx 24. (A) For x-motion, x = vt (i) But v = 2gh from Torricelli s theorem For y-motion, (H h) = t = 1 2 gt at the initial moment. 2 2(H h) g Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 41 From (1), 2(H h) g x = 2gh = 2 h(H h) 1 4 =2 25. (A) P = k F 2 a b 3 [ML T 3 3 4 = 2 m c ] = [MLT 2 a b ] [L] [ML 3 c ] a+c=1 a + b 3c = 2 2a = 3 Solving, we get, a= 3 1 , b = 1 and c = 2 2 3 1 P = kF 2 1 2 Let the power be P1 when weight, length and density are all altered. Then, 3 /2 P1 = k ( 2F ) 2 1 1 ( 2 ) 2 P 1 3/2 1 = ( 2) (2)1 2 = 2 2 2 1 2 =4 26. (A) Centre of mass lies along the x-axis. x cm = x dm dm L = x 0 x e L dx 0 L 0 x e L dx 0 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 42 Let u = x and dv = v = L x cm = x e L dx x e L x x L eL L x L L e L dx 0 0 L x e L dx 0 x L e L eL 2 = 2 L 0 L L x e L 0 = L2e L2 e + L2 L (e 1) = L L = (e 1) (2.718 1) = 0.582 L 27. (C) Frequency of source = 1000 Hz At t = 3 sec, velocity of detector v = a t = 10 3 = 30 m/s The frequency detected = 1100 Hz at t = 3 sec c + 30 1100 = 1000 , where c is the velocity of sound. c i.e., 11c = 10 c + 300 c = 300 m/s 28. (C) From conservation of momentum we have m1 v1i + m2 v2i = m1 v1f + m2 v2f (1.60) (4.00) + (2.10) ( 2.50) = (1.60) (v1f) + (2.10) ( 1.75) Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 43 v1f = 3.00 m/s Again by law of conservation of energy 1 1 1 1 1 2 2 2 2 m1 v1i + m 2 v2i = m1 v1f + m 2 v2f + kx 2 2 2 2 2 2 On substituting the numerical values and simplifying, we get compression in the spring as x = 0.17 m SECTION II 29. (A), (B), (D) dQ dQ in the steady state, where K is coefficient of thermal conductivity. = AK dt dx Since dQ is constant, dx dQ A . Hence rate of flow of heat decreases from A to C but increases from dt C to B. The section at P at a distance x from A and the section at Q are at the same distance from C. Since temperature gradient is uniform, temperature at C is 50 . Since the rod is symmetrical about C, rate of flow of heat through P and Q is the same. 30. (A), (C), (D) Since blocks move with uniform velocity, acceleration a = 0 for the system. The F.B.D. for B is T = N = mBg a (1) T = mA g T N The spring is stretched by tension T. By considering the F.B.D. for A, we have mBg (2) From (1) and (2) we get mB = 10 kg For the spring we have T = k = mAg 2 9.8 = 1960 = Energy stored = Brilliant Tutorials Pvt. Ltd. 1 2 = 10 m = 1 cm 100 1 1 2 kx2 = 1960 10 4 = 9.8 10 J 2 2 IIT/ELITE/PET/CPM/P(II)/Solns 44 31. (A), (B) Potential energy for the system is 1 kx 2 mgx sin 37 + mg sin 37 2 Up = For Up to be a minimum Up =0, x 2 Up x 2 >0 kx = mg sin 37 x = = mg sin 37 k 2 10 0.6 = 0.6 m 20 Value of Vmin = = 1 k (0.6)2 + mg sin 37 mg (0.6) sin 37 2 1 20 (0.6)2 + 2 10 0.4 0.6 2 2 = 10 (0.6) + 8 0.6 = 8.4 J Maximum elastic potential energy occurs when mg sin 37 = 1 1 kx 2 = 20 x2 2 2 2 10 (0.6) 1 = 10 x x= 2 1.2 m = 1.1 m 32. (A), (B), (C), (D) Volume of the block = 125 cm 3 Let h1 be the height of the block above mercury. Then (5 h1) 25 13.6 g = 125 7.2 g h = 5 2.65 = 2.35 cm Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 45 Let h2 be the height of water required to just submerge the block. Then wt. of block = wt. of (mercury + water) displaced i.e., 900 = h2 2.5 1 + (5 h2 ) 25 13.6 h2 = 2.54 cm After pouring water, depth of mercury = 5 2.54 = 2.46 cm SECTION III 33. (8) The kinetic energy of the system about BD is given by = K.E. about BC 1 2 I = K.E. of A about BD 2 A 1 = mv 2 2 I 2 m v2 = 60 B = 3 4 C m( sin 60 )2 2 m = D K.E. = 2 2 3 m 8 2 2 Aliter About BD about BC 1 2 1 I = mv 2 (K.E. of A) 2 2 v2 = m m = 3 4 K.E. = 2 2 4 2 2 2 1 3 m 2 4 Brilliant Tutorials Pvt. Ltd. 2 2 IIT/ELITE/PET/CPM/P(II)/Solns 46 34. (1) We have 1 = q , where q is the charge on the metal sphere. The charge 4 0 r1 induced on the shell is q = 4 0 r1 1 q Hence the potential acquired by the shell is 2 = r2 = 2 = 4 0 r1 1 4 0 r2 r1 1 r2 On comparison n = 1 T 35. (4) Average energy u = u dt 0 T dt 0 = 1 m 2 A 2 x2 2 ( But u = T 1 T u dt 0 ) But x = A sin ( t + ) u = = T But 1 2T T mA 2 2 cos2 ( t + ) dt 0 1 mA 2 2 2T T cos2 ( t + ) dt 0 cos2 ( t + ) dt = 0 u = 1 T 2 1 mA 2 2 4 n=4 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 47 36. (2) By the law of conservation of energy, 1 mgh , where h is measured from the surface of the earth. mv2 = h 2 1+ R Given v = kv e = k 2gR 1 mg (r R) , where r is the distance measured from the centre mk 2 (2gR) = 2 r R 1+ R of the earth. r R k 2 R 1 + = (r R) R 2 i.e., k r = (r R) r = R 1 k2 On comparison r = 2 37. (6) Due to extension produced in the cord, energy stored in it is converted into kinetic energy K, when the stone flies away. Assuming that there is no loss of 1 energy in this process, the K.E. of the stone is given by K = mv 2 = 4 J . This 2 must be equal to the work done in stretching the cord. Therefore W = F = Stress 1 F = 4J 2 4 2 = 40 N 0.2 F 40 40 = = = 1.415 106 Nm 2 A r 2 (3 10 3 )2 Strain = 20 = 0.476 42 Young s modulus Y = stress 1.415 106 = = 2.97 106 Nm 2 strain 0.476 On comparison n = 6 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 48 38. (4) The angular impulse about the centre of mass J = Lf Li = I 0 But J = p 2 2 m p = 12 2 6p = m 6p = 2t m n m t = 48 p On comparison n = 4 SECTION IV 39. (A) (p), (r), (t); (B) (p), (s), (t); (C) (p), (q), (t); (D) (p), (t) Points x1 and x3 are points of zero displacement. Just to the left of x1, the displacement is negative indicating that the gas molecules are displaced to the left, away from point x1 at some instant. Just to the right of x1, the displacement is positive, indicating that the molecules suffer displacement to the right which is again away from x1. So at point x1, the pressure is minimum. At point x2, the pressure does not change because the gas molecules on both sides of that point have equal displacements in the same direction. Hence pressure is normal. At x3, the pressure is maximum, because the molecules on both sides of that point are displaced towards point x3. Since pressure p is proportional to density , pressure and density are in phase. 40. (A) (p), (q), (s); (B) (q), (r), (t); (C) (p), (q), (s); (D) (q), (r), (t) Process AB is an isothermal process with T = constant. Therefore (P V) graph is a hyperbola. (V T) graph is a straight line since T is constant. By ideal gas PM equation = . Therefore P. As T is a constant, ( T) graph is a straight RT line. Process BC is an isobaric process with P = constant. As P is constant, (P V) graph is a straight line. In isobaric process, V T and hence (V T) graph is a straight line. 1 When P is constant . Hence ( T) graph is a hyperbola. T For isothermal process U = 0, but for isobaric process U 0. What applies to process AB, also applies to CD. Similarly what applies to BC, also applies to DA. Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 49 PART C: MATHEMATICS SECTION I p + 1 p + 2 2 41. (A) ax + bx + c a (x ) (x ) a x x p p +1 1 1 a Putting x = 1, a + b + c = a = p p + 1 p(p + 1) = a ( ) = a ( + )2 4 = a 42. (B) b2 4ac a2 = b 2 4ac a a3 108 = 3 and = 3 1 r 13 1 r 1 + r + r2 1 2r + r 2 ( = a 1 rn Now 1 r 1 1 3 n > 1 1 13 r = or 3 . Here r = and a = 2 3 3 4 1 1 n 3 > 728 243 243 1 3 ) > 728 728 1 1 1 < = n>5 n 729 729 3 35 Minimum value of n = 6 43. (D) 5 x( log x 2 ) ( log7 x ) ( 5 2log7 x )+2 + 2log7 log7 x x = 24 = 24 2log7 x 2 = 4 log7 x = 2 x = 7 = 49 44. (C) x + y + z = 16 and x, y, z 3 Let u = x 2, v = y 2, w = z 2 Then u + v + w = 10, 9 The number of positive integral solutions = C2 = 36 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 50 P 45. (D) xA = 1, xM = 2, AP = 2 MAP = 60 Required area = area QRPMQ A 1 = 2 2 3 3 = 4 3 3 46. (D) f ( ) = 5 cos + 3 cos = R M Q cos 3 sin sin + 3 3 3 13 3 3 13 cos sin + 3 = 7 cos ( + ) + 3 where cos = 2 2 14 Range of f( ) = [3 7, 3 + 7] = [ 4, 10] 2 47. (B) P(a, a) must be outside the circle 2a 4a 6 > 0 either a > 3 or a < 1 a) P (a, PT1 = S1 = 2a 2 4a 6 and CT1 = r = 2 2 2 2 > 2 1 3 2a 4a 6 T1 1 < < < < tan > 3 6 2 2 2 3 a 2 2a 3 < 2 3 2 C T2 2 a 2a 15 < 0 3 < a < 5 a (3, 5) 48. (A) (q, p) satisfies the equation of the line. Hence AB is a diameter of the circle, origin lies on the circle. SECTION II 49. (A), (B) y = mx c where c = 4a 2 m2 4b2 is a tangent to the first hyperbola. It is also a tangent to x2 ( 4b2 ) y2 ( 4a 2 ) =1 c = 4b2 m2 + 4a 2 2 2 2 2 2 2 4a m 4b = 4a 4b m 2 m = 1 and m = 1 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 51 50. (A), (B) n Cr ( r + 2) C = r n ( 2) r r=0 2 (n + 1) (n + 2) n Cr (r + 2) (n + 2) Cr + 2 on simplification, 2 (n + 1) (n + 2) = Cr n (n + 2) ( 2) r Cr + 2 r= 0 = 1 (1 2)n + 2 1 2(n + 1) (n + 2) = 1 (2n + 3) + ( 1)n 2(n + 1) (n + 2) = 1 1 when n is even and when n is odd. n +1 n+2 { (n + 2) } C1 2 51. (A), (C) If we select any four vertices, they form a quadrilateral which gives one point of intersection. Total number of points of intersection = n C4 = 1820 n (n 1) (n 2) (n 3) = 1820 n = 16 24 Number of diagonals = n C2 n = 16 C2 16 = 104 52. (B), (D) Consider 1 z1 z2 < 1 . This is true if 1 z1 z2 z1 z2 ( which is true if 1 z1 ( (i.e.,) 1 z1 2 2 z2 2 + z1 2 z2 2 ) (1 z2 z1 ) < ( z1 z2 ) ( z1 z 2 ) <0 ) (1 z ) < 0 which is true. 2 2 Solution for (B) is similar. SECTION III (n 1) ! (n 1) 53. (2) n Pn 1 = k 1 n 2Cn 3 + Pn 1 2! (n 2) (n 1)! n! = k + (n 1)! 2 k=2 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns 52 54. (3) r2 + 2r + 4 2r2 + 1 0 2 1 = 1 . This r 2r 3 0 2 (r 3) (r + 1) 0 r [0, 3] maximum value of r = 3 2 4x 3 4x 3 4x 3 55. (3) f (x) = 1 cos + 2 cos = 1 + cos2 2 2 3 3 2 3 = 3 1 8x 6 2 3 3 + cos whose period = = 2 2 4 8 3 2 2 3 k=3 3 1 56. (8) 1 cos2 1 cos2 = 8 8 n sin2 3 1 sin 2 = 8 8 n sin2 1 4 1 4 cos2 = 2 sin = = 8 8 n 4 n 2 n 2 n=8 n(n + 1) 57. (7) L.H.S = 5 1 R.H.S = 25 28 = 556 n (n + 1) = 56 n=7 a2 2 2 2 58. (4) b = a (e 1) b 2 = 4 a = 2b Required ratio = Brilliant Tutorials Pvt. Ltd. 2a b2 2 a = a2 b2 =4 IIT/ELITE/PET/CPM/P(II)/Solns 53 SECTION IV 59. (A) (p); (B) (q); (C) (r); (D) (s) (1 + x) 200 2 = C0 + C1 x + C2 x + . + C200 x 200 Put x = i (1 + i) 200 = C0 + C1 i C2 C3 i + C4 + . + C200 Comparing real parts, C0 C2 + C4 C6 + C8 .... + C200 = Real part of 2100 cos + i sin 4 4 100 =2 100 cos 50 = 2 200 (1) Comparing imaginary parts C1 C3 + C5 C7 + .... C199 = 0 (2) 199 But C0 + C2 + C4 + C6 + C8 + . + C200 = 2 (3) Adding (1) and (3) C0 + C4 + C8 + ....... + C200 = 1 199 2 + 2100 = 299 299 + 1 2 ( ) Subtracting (1) from (3) C2 + C6 + C10 + . = 1 199 2 2100 = 299 299 1 2 60. (A) (q), (t); (B) (s); (C) (r); (D) (p) (4, 3) satisfies x2 y 2 + =1 32 18 2 2 (4, 3) is the centre of the circle x + y 8x 6y = 0 2 2 4 + 3 16 > 0 2 2 (4, 3) lies outside the circle x + y 16 = 0 2 16 24 < 0 (4, 3) lies inside the parabola x 8y = 0 9 16 16 + 4 < 0 2 (4, 3) lies inside the parabola y 4y 4x + 4 = 0 Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns

Formatting page ...

mohan28 chat

Horizontal lines at:
Guest Horizontal lines at:
AutoRM Data:
Box geometries: