8- Solution Report (8)

Paper Code : 1001CT103516015 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE-III to VII (SCORE-I) ANSWER KEY : PAPER-1 TEST DATE : 05-03-2017 Test Type : PART TEST PART-1 : PHYSICS Test Pattern : JEE-Advanced Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. A,B,D A,B A,B,D B,D B,D A,D A,B B,D A C Q. 11 12 13 14 A. D A D C SECTION-II Q.1 A B C D P,Q,R,S P,Q,R,S P,Q,R,S,T R,S SECTION-IV Q. 1 2 3 4 5 A. 3 8 2 8 2 PART-2 : CHEMISTRY Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. A,C,D B,C A,C,D A,C A,B A,B,C,D B B,C,D A C Q. 11 12 13 14 A. D D C D SECTION-II Q.1 ABCD Q,R P,S P,R,T Q,T SECTION-IV Q. 1 2 3 4 5 A. 5 2 6 4 7 PART-3 : MATHEMATICS Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. B,C,D A,B,C A,B,C,D A,C B A,B A,C A,B,D C B Q. 11 12 13 14 A. C C A D SECTION-II ABCD Q.1 P,Q,R,S,T Q,S P,Q,R,S,T P,R,S,T SECTION-IV Q. 1 2 3 4 5 A. 6 3 9 7 7 Paper Code : 1001CT103516016 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE-III to VII (SCORE-I) ANSWER KEY : PAPER-2 TEST DATE : 05-03-2017 Test Type : PART TEST Test Pattern : JEE-Main Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans . 3 2 2 3 2 3 1 2 3 3 1 4 1 4 2 2 4 1 1 4 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans . 1 3 2 1 3 4 3 2 2 3 4 4 2 4 4 2 2 4 4 3 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans . 2 1 3 4 1 3 4 2 3 1 3 1 4 3 1 1 2 3 4 2 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans . 4 1 1 3 3 1 4 2 1 1 1 1 1 3 4 3 2 2 3 2 Que. 81 82 83 84 85 86 87 88 89 90 Ans . 3 2 4 4 2 3 2 4 4 3

Paper Code : 1001CT103516015 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE- III to VII (SCORE-I) Test Type : PART TEST Test Pattern : JEE-Advanced TEST DATE : 05 - 03 - 2017 PAPER-1 PART-1 : PHYSICS SOLUTION SECTION-I  V0  2L2  R2 = 0, 1. Ans. (A,B,D)  L  1 2  R2 Sol. Perimeter is decreasing at a rate of 2v  C    d (2r ) = 2v  dr = v VRL  V0 × C dt dt  r = (r0 – v 1  t) VC = i0 × C   = B · r2   = d dr =  dt = B · 2 · r dt VRL = V0 vv v   = 2B(r0 – t) · = 2Bv(r0 – t)     V0  2Bvr Bv  2LC  1 2  2R2C2 I = R =  · 2r =  2. Ans. (A,B) = 0 Sol. Just after switch is closed, L is open circuit VC  V0 LC and C is short circuit  Just after =  E 5. Ans. (B,D) +– R3 Sol. Applying ampere's law R0 i=0 R1  R1 B 2r = µ0 ienclosed R2 Option (B) ienclosed = 0 B=0 Long time E Option (D) ienclosed = 0 6. Ans. (A,D) R3 R1 i = 0 7. Ans. (A,B) R2 3. Ans. (A,B,D) Sol. p0 + h11g = p0 + h21g + h33g 4. Ans. (B,D) Sol. ad Sol. VRL  i0  2L2  R2 Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/12 +91-744-5156100 [email protected] www.allen.ac.in

8. Ans. (B,D) Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 12. Ans. (A) Sol. The geometrical construction shown in 13. Ans. (D) figure is important for developing the Sol. d = RN + Rp mathematical description of interference. It is subject to misinterpretation, however, R as it might suggest that the interference can only occur at the position of the screen. R = R0A1/3 = R0 141/3 + R0 41/3 A better diagram for this situation is figure. 14. Ans. (C) which shows paths of destructive and constructive interference all the way from Sol. 1 µvr2el = kq1q2 the slits to the screen. These paths would 2 d be made visible by the smoke. Since fringes are present everywhere in space they are 1 × 4m 14m v 2 = k  2  e  7e non-localised and their shapes are 2 4m  14m rel R0 (141/3  22/ 3 ) hyperboloid. 1 × 4m v 2rel = 18ke2 9. Ans. (A) 2 R0 (141 / 3  22 / 3 ) Sol. µdAE = dBC + µdCD SECTION-II µ(dAE – dCD) = dBC air A1 B 1. Ans. (A )  (P,Q,R,S); (B)  (P,Q,R,S); medium E 1 C (C)  (P,Q,R,S,T); (D)  (R,S) SECTION-IV D r 1. Ans. 3 2. Ans. 8 Sol. Process A  B Pdv = 3 T1/ 2dv  WAB = 2  C  rr = 3 T1/ 2  1 RT1/ 2dT v 23 (dAE – dCD) =  d BC On solving, WAB = 50 R = 50 × 8.3 = 415 J rr Process B  C 10. Ans. (C) U = 1 V1/ 2 2 3 RT = 1 V1 / 2  3PV1/2 = 1 2 2 i r 1  P= 3 V Sol.  Now WBC = 1600 1 dv 2 Pdv = 100 3 V = 3 V sin i   = 2 [40 10] = 2  30 = 20 J sin r 3 3 µe < 0, r < 0 so refractive index is negative, Total W = 415 + 20 = 435 so final wave return to it's 3. Ans. 2 same side where it is incident. Sol. Pab = heat energy 11. Ans. (D) time Sol. Fring width;   D msT d (Intensity) (cross section area) = If V decreases, then  increases & hence  t increases. (4 103 )(4200)(5) HS-2/12 t = (1000)(1.4 104 ) = 600 s = 10 min. 1001CT103516015

Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 4. Ans. 8 1 1 1  K  12 22  1 = R(z – 1)2  Sol. Energy available = 2 µ vrel2. = Q value. 13  th  = 1 × 7 1 × vrel2 = Q value. 10 (K – th) =   K  2  2 7 1 18 3 K =  13  1 th  2 × vrel2 = Q × 7 10  10 2  8 3  4 107   8  12.4 107 Ki = 1645 × 7 = 1880 keV   = 10  15.5 103 5. Ans. 2 10  3(z7 )2  hc 5000 Sol. th = eVa  8 = (z – 1)2 625 = (z – 1)2  z = 26 PART–2 : CHEMISTRY SOLUTION SECTION - I 13. Ans. (C) 1. Ans. (A,C,D) 14. Ans. (D) 2. Ans. (B,C) SECTION - II dN  M du 1. Ans. (A)(Q,R); (B)(P,S); (C) (P,R,T); NT (D)(Q,T) SECTION - IV 3. Ans. (A, C, D) 1. Ans. 5 4. Ans. (A, C) 5. Ans. (A, B) [NaCl] = 0.025 M = C1 V1 = 2V 6. Ans. (A,B,C,D) T1 = 273 K 7. Ans. (B) 8. Ans. (B,C,D) 9. Ans. (A) Gº = –nFEº Eº298K = 0.07 volt (1)(96500)(0.07) [Na2SO4] = 0.1 M 1000 =  V2 = 3V = –96.5 × 0.07 = –6.755 T2 = 273 K 10. Ans. (C) c1v1i1   c2v2i2   dEº  = v1  v2  22.4 Sº = –nF dT    dEº   5104  7 106  T  298 0.025 2v 2  0.1 3v 3   dT 298K  K =  22.4  5v  = 5 × 10–4 – 7 × 10–6 (0)  0.1v  0.9v  22.4 = 5 × 10–4 =  5v  22.4 = 5 Sº = +(1) (96500) (5 × 10–4)  = 4.48 = + 482500 × 10–4 2. Ans. 2 = +48.25 JK–1Mol–1 3. Ans. 6 11. Ans. (D) 4. Ans. 4 12. Ans. (D) 5. Ans. 7 1001CT103516015 HS-3/12

Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 PART-3 : MATHEMATICS SOLUTION SECTION-I 1. Ans. (B,C,D) y = m1x – 2m1 – m13 (–1,0) (1,0) 6  m1  m13  m1  2  m  1 (–1/2,0) 2 P(4,4)  r  5 e2  1  3  e  2 2. Ans. (A,B,C) 1 4. Ans. (A,C) 2k1  2k2  k1  2 k2 16 72 9 (A,0) k2 = 0,9,.......63  8 values so n(A  B) = 8. e 2k1   2k2  i  /4  9 .  16 72    / 36  Hyperbola and ellipse will be confocal with    focus 2 2,0 as 1st,10th, 19th roots are common. 1  e12  1  e1  22  e2  3 5.  k2 = 0,1,.......8 and k1 = 0,1,........16. 9 3 22 6. will give all possible z1z2  144 Ans. (B) A = (d1,d2,d3,d4)  x2 y2 1  A  8  A4  d14 , d 4 , d43 , d 4 I A2  B2 3 2 4 64 9  1  8  d14  d 4  d34  d44 I 9  8 9 2 B2   B2   d1 ,d2 ,d3 ,d4 are forth roots of unity as d1 + d2 + d3 + d4 = 0 9x2  9y2  1 64 8   2. 4!   4! = 36 ways are there to 3. Ans. (A,B,C,D)  2!2!  2 1 1 assign values to d1,d2,d3,d4.  1 2 0 Also d1d2d3d4 is product of 4th roots of unity 2 1 3 which is –1 or 1 when 1, –1, 1, –1 or i, –i, i, –i are used. Ans. (A,B)  2 3  2   3  1  2  0 a b c  32  2  3  0 Let A  d e f     3 2    2  0 g h i  2  AB = BA     1  2  1 a ab a  b  c a  d  g beh c  f  i  2   4  d de eh gh d  e  f    d g h f i  3 g     1 2 y2 1 g  h  i   g i   2 3  so conic is x    g = 0, d = h = 0, a = e = i, b = f  4 whose center is   1 ,0  a b c  2   A  0 a b 0 0 a HS-4/12 1001CT103516015

(A) a = b = c = 1  |A|= 1 Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 (B) trace(A) = 6  a = z |A|= 8 Paragraph for Question 9 & 10 (C) |A| = a3 which may or may not be zero (D) 3a + 2b + c = 6 x1  5  V 5,0 (5,0) 3 abc 4 matrices y2 = –4a(x – 5) 0 0,1,2,3  2 matrices 1 0,1  x2  ax  5 1 200 1 matrix 9  x2 – 9ax + 45a – 9 = 0 D = 81a2 – 4(45a – 9) = 0 = 9a2 – 4(5a – 1) = 0 7. Ans. (A,C) 2     9a2 – 20a + 4 = 0  a = 2, 9 C  A  B  A.C  0  2a + 2b – c = 0 at a = 2  x = 9 rejected for (A) and (B) a + b + c = 0  c = 0, a + b = 0 a  2  x 1  A  4 2  42  9 1, 3 ; B 1, 3   1 , 1 ,0  Normal at A  2 2    as a2 + b2 + c2 = 1   9x  4y  5 3 y5 1, 1 1 4 2/3  9x  2   2 2 ,0     9x 2  3y  5 2 a2  b2  c2  1 tangent at A 3x  9y 2  27  c 9,0 (C) and (D) 2a  2b  c  are not satisfied   ABC  1 .8. 8 2 32 2 for any triplet of integers.  23 3 8. Ans. (A,B,D) 9. Ans. (C) If n coplanar lines are there such that no 3 10. Ans. (B) are concurrent and no two are parallel then the number of parts in which they will Paragraph for Question 11 & 12 divide their plane is 2 + 2 + 3....n a + b = 13 or 17 n2  n  2 L4 passes through Q(12,12)  (9,8) quadrilateral 2 Q AB (n + 1)th plane will be intersected by (2,7) (6,4) (10,1) previous n planes in n2  n  2 lines (3,10) No quad. 2 A B  ƒ n 1  ƒ n  n2  n  2 11. Ans. (C) 12. Ans. (C) 2 Paragraph for Question 13 to 14  ƒ n  ƒ n 1  n 12  n 1  2 2  2  12 1  2  22  2  2  .....  n 12  n 1  2 nP  n  n  1P   n   n  2P ........   1n 1  n n 1 1P ...(i)  1   2    F n,P     nP  22 2  2  1  n n  1 2n  1   n n  1  2  n   46   4  36   4  26   4 16   1   2   3  1 F 4,6      2  6 2   46 n3  5n  6  4096  2916  384  4 1560 195    4096 4096 512 6 1001CT103516015 HS-5/12

Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 Also if P < n then probability is zero and if SECTION – IV P = n by (1) 1. Ans. 6 nn   n  n  1n  .....   1 n1  n n 1 1n n! x+1=y  1     nn (y + 2)2 (y + 3)3(x + 4)4     = y9 + a1y8 + a2y6......+a8y + a9 nn n1  n C11n nC2 2n nC3 3n n1 nCn nn      1   1    ....  1  n!  D is right. a2= [(2 + 2 + 3 + 3 + 3 + 4 + 4 + 4 + 4)2 2 13. Ans. (A) – (2.22 + 3.32 + 4.42)] 14. Ans. (D)  1 292  99  371 SECTION – II 2 1. Ans. (A)(P,Q,R,S,T); (B)(Q,S); 2. Ans. 3 (C)(P,Q,R,S,T); (D)(P,R,S,T)  .nˆ is minimum area. (s  a)(s  c) (A) ac    A(3,3,3)    AB  AC (s  b)(s  c) (s  a)(s  b) bc ab     3kˆ  3ˆi  3kˆ B(3,3,0) n^ b 2b 2  9ˆj C(0,3,0)  sb acb as a, b, c are in A.P.  nˆ  1 2ˆi  ˆj  2kˆ D(0,3,3) 3 (B) bcsinA = 2bc – (b2 + c2 – a2)  sinA = 1 [2bc  2bc cos A] .nˆ  3 bc 3. Ans. 9 = 2(1 – cosA) (1 + x5)20 + 20C1x7(1 + x5)19 + 20C2x14(1 + x5)18 +.....  4 sin2 A  cot A  2 20.19C2 = 20.19.9 22 4. Ans. 7 sinA = 2. 1  4 , cosA = 3 2 5 5 x3 + y3 = 8 1 1 (I) x,y both even  all favourable 4 so x = 5sinA,5cosA or x = 3,4 = 50 49 2 AD  1 2  131  50C2 2 2  (C)  9  49 A (II) x,y both odd  1 100  5 category Number 2 B C of numbers AB, DE = AB.DC D (i) 8 + 1  1, 9,......97  13 7.7 E (ii) 8 + 3  3, 11,......99  13 22  DE  49 (iii) 8 + 5  3, 13,......93  12  (iv) 8 + 7  7, 15,......91  12 5 20 [ AE] 5 49  7 x  (i) y  (iv)  (13 × 12) 20     x  (ii) y  (iii)  (13 × 12)  tan A tan B 1 Ans. 25 × 49 + 312 = 1537 2 2 (D)  4 5. Ans. 7 r r r r 4 3 3 3 3 3 3 7 4 7 4     .  .1  .1  1 p  1  1  3  1   3  1  1  6  1  1  3  1  1  9  1  1  6  6  6   6  6  6  6  6  6  6  6 7 r2 + 11r = 28 r 7 25  64  = r(11 + r) = 28 r HS-6/12 1001CT103516015

Paper Code : 1001CT103516016 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE- III to VII (SCORE-I) Test Type : PART TEST Test Pattern : JEE-Main TEST DATE : 05 - 03 - 2017 PAPER-2 SOLUTION 1. Ans. (3) y  mg  20 Sol. Out-put is XOR GATE AY A & B two are input Total length is = 40 cm = 20 mg cm AY XOR out put = A . B + B. A 4. Ans. (3) 2. Ans. (2) Sol. Sine magnification is –2 so virtual object Sol. qE = qvB have virtual image. mv2 virtual object to virtual image then mirror qvB = is convex. 3. Ans. (2) r mv mE r = qB  qB2 5. Ans. (2) M Sol. Some of the characteristics of an optical fibre Sol. dx are as follows x (i) This works on the principle of total internal reflection. Tension is rod at a distnace x from lower (ii) It consists of core made up of glass/silica/ mxg plastic with refractive index n1, which is surrounded by a glass or plastic cladding end is  with refractive index n2 (n2 > n1). The Y is young modulus of elasticity refractive index of cladding can be either then change in length in dx element is dy changing abruptly or gradually changing Y × strain = stress (graded index fibre). dy T (iii) There is a very little transmission loss Y×  through optical fibres. dx A dy mgx Y × dx    A Y ydy 70 mg  xdx (iv) There is no interference from stray 30 A electric and magnetic field to the signals 0 through optical fibres. mg 702  302  6. Ans. (3)   Yy = A  2  7. Ans. (1) y  mg  2000 8. Ans. (2) AY 100 Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-7/12 +91-744-5156100 [email protected] www.allen.ac.in

Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-2 9. Ans. (3) 14. Ans. (4) Sol. Replacing it with string block system Sol. V voltage of source drop across resistance is VR, inductor is VL and capacitor VC v=0 v=0 Applying Krichhof's law NLP I.P. F.P. K m 2F V = 0 AB VR + VL = VC – V = 0 15. Ans. (2) F/K X Sol. Temperature of surrounding is 20°C Let at initial position 2F force is applied then 50°C 35°C W.E.T. from A to B Heater 100 W Heater XW WS + WF = 0 100 = K (50 – 20) x = K(35 – 20) X  3F K 100 = K × 30 x  100 15 30 Net elongation in spring = 3F  3CE K x = 50 10. Ans. (3) 16. Ans. (2) Sol. Eye recive all component of light which is Sol. F = pressure at centroid × area along the line. AP  P0 h h ah  a g  1 ab  3  2 IA F =    I/2 =  P0   h  2a  g  1 ab   3   2 –2I sin2    17. Ans. (4) 18. Ans. (1) P Sol. Maximum current thorugh capacitor = V0 And perpendicular component is I sin2  z 2 i  V0 Net intensity recieved by light is z I  I sin2  Q  V0 22 z 11. Ans. (1) Q  V0 Sol. I  B02  C z 2 19. Ans. (1)  1002  109 2 Sol. On increaing temperature of semiconductor I  2  4 107  3 108 bond between molecule break so more electron become free to move hence. I  104 1018  3  108 Conducting infreases and mobility 8 107 decreases. I  30  1.19426 20. Ans. (4) 89 Sol. Charge is conserved. In order to fully convert 12. Ans. (4) an electron into energy, a positron (the 13. Ans. (1) electron's antiparticle must be involved). Sol. R2B0 = E × 2r That is, electron + positron – > energy, NOT electron – > positron + energy.] E= R2 B0  R2B0 21. Ans. (1) 2r 2r a = qE  q R2B0  qB0R2 1001CT103516016 m m 2r 2mr HS-8/12

22. Ans. (3) Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-2 35. Ans. (4) Sol. 1st 2nd 3rd 4th 1 V.S. 0.95 0.90 0.85 0.80 Collision frequenciy Z11 = 2 (2) GAP 0.05 0.10 0.15 0.20   2Avg. Nº 2 difference is = 0.5 mm = 0.05 cm  N º P  KT  Net reading is = 3.1 + 0.05 = 3.15 cm  23. Ans. (2) 1 8 RT  P 2 2  Z11 =  Sol. Index error in u = +1cm 2 2  M  KT  u = 8 cm 1 T3/2 index error in v = –1cm Z11  v = 17 + 1 = 18 cm 36. Ans. (2) 111 37. Ans. (2)  P0 = 76 f 18 8 Ps = 38 f = 5.53 cm P0  Ps  1  i(1) 24. Ans. (1) Ps (2) i = 2 = 1 + (2)  25. Ans. (3) 26. Ans. (4) Sol. Damping coefficient = 2 km 2 = 1 = 2m 1 m =2 38. Ans. (4) mA = m m 39. Ans. (4) mB = 2 40. Ans. (3) bA  1 41. Ans. (2) bB 2 42. Ans. (1) 27. Ans. (3) 43. Ans. (3) Sol. PV  m RT 44. Ans. (4) M 45. Ans. (1) 46. Ans. (3) PVM  T T M Þ THe  TN2 47. Ans. (4) Rm 28. Ans. (2) 48. Ans. (2) 29. Ans. (2) 49. Ans. (3) 50. Ans. (1) 30. Ans. (3) 51. Ans. (3) 52. Ans. (1) Sol. min (p  )min  1500Å ; 53. Ans. (4) E  E1  12420 eV  8.28eV 54. Ans. (3) 1500 55. Ans. (1) 56. Ans. (1) Hence ionization potential is 8.28V 57. Ans. (2) 58. Ans. (3) 31. Ans. (4) 32. Ans. (4) 33. Ans. (2) 59. Ans. (4) 60. Ans. (2) 34. Ans. (4) 1001CT103516016 HS-9/12

Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-2 61. Ans. (4) minimum distance is  distance P(), Q(), R(), S(), T(,)  distance  6  4  2 RQ 187  4  1 .2    5 P 2 69. Ans. (1) S 187 = ( + 3) P(A) = 0.3 T  = 11, 3 +  = 17 P(A  B) = P(A  B) = P(A) + P(B) – P(A)P(B)  = 11 and  = 2  0.8 = 0.3 + P(B) – 0.3 P(B) 62. Ans. (1) x2  2x  1  0  PB  5  1 i 1 ,  1 i 1 7 2 2 2 2 PA  B 1  PA  B  = 1 – (P(A) – P(A)P(B)) = ei/4 = e–i/4  2  32  7  35 50 + 50 = ei25/2 + e–i25/2 = i + (–i) = 0 1  0.3   63. Ans. (1) 70. Ans. (1) 1 .a2 1  a n Cr2  36, nCr1  84 , n Cr  126 r 2 2 2 a a 2a Cn  84  nr2  7 a r 1 36 r 1 3 2 Cn r2    1 a2  1 . 4  2  4 2 a  3n  13  10r ...(1) 22 n Cr  126 n r 1 3 Cn 84  r  2 32 2 64. Ans. (3) r 1  2n + 2 = 5r ....(2) 15º tan15  30  n = 9, r = 4 d nC2r = 9C8 = 9 30 75º 71. Ans. (1)  d  15º 30 30. 3 1  d  Let |U| 2 cos iˆ  2 sin ˆj  2  3 3 1 65. Ans. (3)     2 2cos  2sin  0 2 It is always true for n > 5 [U V W]  2 1 1 66. Ans. (1) 1 03 P1 and P2 are x + 2y – 2z = 0 and 2x – 3y + 6z = 0 = |6cos – 14sin|2 Maximum value = 36 + 196 = 232 72. Ans. (1) A cos   2  6 12  16 sin   1 D1   3.7 21 3 1 C1 67. Ans. (4)  AC1 = 3 D2 C2 |adj 3P| = |3P|3 = 312|P|3 = –312.23 AC2 = 6 2 AD = 8 68. Ans.(2) Given conic C is parabola  BD  2 2 B DC focus : (1,–1) E : 3x – 4y = 0 Area = 1 .4 2.8  16 2 1001CT103516016 2 HS-10/12

Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-2 73. Ans. (1) (2,1,–1) x  2  y 1  z 1 |z – 3 – 4i| = 4 L: 2 2 8 ||z| – 5| < 4 29 P |z| < 9 K:  x 2  y3  z4 74. Ans. (3) (–2,3,–4) 2 2 8 12 b c cos   8  4  24  2 a 24 c  0 72 29 29 a b 36  sin   27 29 (12) (24) (36) – 12bc – 36ba + 2abc – 24ac = 0 79. Ans. (3) ....(i) 4 1 2 3  (1009  2n)4 4Cn (1)n a 12 b  24 c  36 n0 b  24 c  36  2a 12 c  36  3a 12 b  24  a 12b  24c  36 (1009)4–4(1007)2+6.(1005)4– 4(1003)4+(1001)4 (1005 + 4)4 + (1005 – 4)4 –4[(1005 + 2)4 + (1005 – 2)4] + 6(1005)4 on solve = 512 – 4 × 32 = 384 1 80. Ans. (2) 6 75. Ans. (4) 2p – 3q + 12r = 5 b = p2 + q2 + r2 6 × 5! c = pq – qp + qr – qr + 3r2 = 3r2 76. Ans. (3) b + c = p2 + q2 + 4r2 y = mx + 1 is tangent to ellipse    use : 2iˆ  3ˆj  6kˆ . piˆ  qˆj  2rkˆ  22  32  62 p2  q2  4r2 x2 + 4y2 = 1 in Ist quadrant  m < 0 25  p2  q2  4r2 1 49  1 = m2 + 4 81. Ans. (3) Q m  3 or  3 PQ2 + PR2 + QR2 = 2QR2 R 22 = 2((C1C2)2 – (r1 – r2)2) = 2[36 – 4] = 64 C1 P C1 (reject) 77. Ans. (2) 82. Ans. (2) Use : contrapositive of p  q is (~q)(~p) a R b  a = 2k.a it is true for k = 0  reflexive (2,1)  R but (1,2)  R  it is not symmetric 78. Ans. (2) if a  2k1 b and b  2k2 c , then a  2k1 k2 c (1,2,3) lie on  it is transistive. 114 83. Ans. (4) L  2  b  c P  2 19  19  b = –2, c = –8 10  9 45 Line L and k are parallel favourable : {(3,4),(3,5),......(3,10) (6,7), (6,8),....(6,10) 2  b  c  2  1   8 (9,10), a2d a d (1,4),(2,4)  a = –2,d = 8 (1,8),(2,8),(4,8),(5,8),(7,8)} 1001CT103516016 HS-11/12

84. Ans. (4) Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-2 87. Ans. (2) Let ellipse x2 y2 1 If circle intersect at 4 points then a2  b2 sum of x coor = 0 (0,5) tangent x  ye 1  Points (17,289), (–2,4), (13,169),(–28,784) b b directrix is y   1 – ba2,be a2,be b 4 sum of perpendicular distances 130,0   289  1    4  1   169  1    784  1   4   4  4   4   b 5 = 1247 e 88. Ans. (4) and b 10  e  2 2  A 2  A 2  A 2  B x y z 33 500 2a2 100 = (Area of PQR)2  b   a2  L.R =  iˆ ˆj kˆ 2 81 27 3 1 1   85. Ans. (2)   1   1 1  16   2 2 2 1  4 Determinant value of matrix = 1 – wc – aw + w2ac = 0 9 =2  (1 – aw) (1 – wc) = 0 a  1  w4  b and c each have 4 and 4 89. Ans. (4) w np = 2 options. npq = 1  p = q = 1 ,n  4 2 if c  1  w4 and a  w4 w Px 1  4C0  1 4  4C1  1 4 1 1 4  11 16 16 16  a have 3 and b have 4 options.    2   2   Total matrices = 4 × 4 + 3 × 4 = 28 90. Ans. (3) 4x3 + 4y3 = xy(xy + 16) 86. Ans. (3) 4x3 + 4y3 – x2y2 – 16xy = 0 (4x – y2) (x2 – 4y) = 0 Mean of i = (mean of yi) + k  it is combined equation of two parabola 55 = .48 + k ...(i)  given tangent is common tangent of parabola y2 = 4x and x2 = 4y standard deviation of  given tangent is x + y + 1 = 0 ==1 i = (standard deviation of yi) 15 = .12 ...(ii)  = 1.25 and k = –5 HS-12/12 1001CT103516016