41- Solution Report (41)

Paper Code : 1001CT103516011 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III, IV & V) ANSWER KEY : PAPER-1 TEST DATE : 18-12-2016 Test Type : MINOR PART-1 : PHYSICS Test Pattern : JEE-Advanced Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. A,B,C A,B,C A,B A,B,D A,C B,C,D A,D B,C,D A or D A Q. 11 12 13 14 A. A A A B ABCD SECTION-II Q.1 Q,R P,T P,Q,R,S T or P,T SECTION-IV Q. 1 2 3 4 5 A. 3 5 6 5 4 PART-2 : CHEMISTRY Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. B,D B,D B,C,D A,B,D B,D B,C,D A,B A,B B B Q. 11 12 13 14 A. C C A C SECTION-II Q.1 A B C D S,T P,T T P,T SECTION-IV Q. 1 2 3 4 5 A. 6 5 3 4 2 PART-3 : MATHEMATICS Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. A,C Bonus B,C,D B,D A,B,C C,D B,C A,C C A Q. 11 12 13 14 A. B C D B SECTION-II Q.1 A B C D S R,T Q,T P,S SECTION-IV Q. 1 2 3 4 5 A. 2 2 1 1 5 Paper Code : 1001CT103516012 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III, IV & V) ANSWER KEY : PAPER-2 TEST DATE : 18-12-2016 Test Type : MINOR 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 Bonus 3 3 3 2 1 1 3 2 4 4 1 2 2 2 3 2 1 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans. 2 3 1 4 4214222432344414 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. 2 4 1 or 4 3 2 3 3 4 4 1 4 2 3 3 4 4 4 3 4 4 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans. 2 4 3 1 4334222313122143 Que. 81 82 83 84 85 86 87 88 89 90 Ans. 3 3 3 Bonus 2 2 1 4 1 4

Paper Code : 1001C T103516011 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE Test Type : MINOR PHASE : III, IV & V Test Pattern : JEE-Advanced TEST DATE : 18 - 12 - 2016 PAPER-1 PART-1 : PHYSICS SOLUTION SECTION-I t 2 1. Ans. (A,B,C)  = 10 sin ; Sol. Since the elevator may have an acceleration e 1 d 1 t   R R dt R 22 even though the block may remain Now i =   × 10 cos stationary w.r.t. the elevator. 2. Ans. (A,B,C) = 5 cos t 22 3. Ans. (A,B) Sol. (A) mg – T = ma d  dt is zero (from fig.) at t = 1, 3, 5 sec T.R = MR2   i = zero at t = 1, 3, 5 sec 2 a = R b 6 5 cos t 0R 2 (C) angular momentum about centre of mass  Total charge Q = idt = 0 not conserve because tension force provide 0 torque 6. Ans. (B,C,D)  dL  Sol. Taking torque about R.H. side of loop (D) dt  torque = r F   2 T1 about centre of mass mg   ibB  0 r  iˆ 7. Ans. (A,D)  F  ˆj (tension force) L 1  ML2  Sol. Mg 2  2  3  02    kˆ    iˆ  ˆj 3g  4. Ans. (A,B, D) 2L Sol. For equilibrium m2 ( sin ) sin  = mg cos  v  L  r e  L  r 0  1 1 g cos  COAM, ML2 0  ML2   mv L  r  = sin   3 3 Due to stable equilibrium, it will execute 8. Ans. (B,C,D) SHM. 5. Ans. (A,C) 9. Ans. (A or D) Sol. From figure  = 10 sin t,   ˆj Bˆi   MB  Sol. = ni r 2 where  = 2   =  n i r2Bkˆ 4 2 Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/13 +91-744-5156100 [email protected] www.allen.ac.in

Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-1 10. Ans. (A) 1. Ans. 3 SECTION-IV Sol. M.P. Sol.  d  5  niB cos   2  5 niB sin  x d 7 m 2 7m 2kx 11. Ans. (A) +Q F  B a Sol. F A FC F –Q +Q +Q f F D –Q 2kx + f = Ma ....(1) Q  Q0  (2kx – f)R = MR 2 ....(2)  A 0   ....(3)  2 acm = F  QQ0 a = R   5m 5m 5 0 mA 2kx  QQ0  f= 3  A 0   F.R  F.R 2FR 2   R  8kx I I I 3      Fnet = (2kx  f)  4mR2 T  2x 3M   3M 8k 2k    QQ0 2mAR 0 2. Ans. 5 12. Ans. (A) I mg Sol. Acceleration of point Sol. T  2 , I = m2 + m(2)2 = 5m2 QQ0 i  QQ0R aR 5 0 mA 2mAR 0 A= j 5m2 5 2mg 3 3g  2  2  Leq  5 3 = QQ0 i  QQ0 j acm 5 0 mA 2mA 0 2 13. Ans. (A) 3. Ans. 6  B y Sol.  E. d   A. dB dt Sol. t a sin t As B = 17 + (0.2) sin (t + ) 2 E (2r) = –r2 (0.2) cos (t + ) From the diagram flux through the given E = – r (0.2). cos (t + ) 2 loop is same as the flux through a r rectangular loop having width a sin t in Magnitude of the amplitude = 2 (0.2). xy frame. w = 240 mN/C     a sin f B0 y a dy  B0 a2 sin2 t 4. Ans. 5 0a 2 Sol. For vmin to reach the origin it must cross the point x = 2L. E   B0a2 sin 2t  By work energytheorem 2 KE = W (or area) 14. Ans. (B) 0– 1 mv2 =– 1LE Sol. Input power = Out put power ( is constant) 2 2 2 v  LE0 v = 5 m/s  = m R 5. Ans. 4  = B02a4 sin2 2t Sol. For solid sphere under pure rolling motion 4R SECTION-II KE =translational 5  1 KXm2 ax 7 KEtotal 2 1.Ans. (A)-(Q,R); (B)-(P,T); (C)-(P,Q,R,S); (D)-(T or P,T)  Xmax = 4 cm HS-2/13 1001CT103516011

Leader Course/Phase-III, IV & V/18-12-2016/Paper-1 PART–2 : CHEMISTRY SOLUTION SECTION-I 12. Ans.(C) 1. Ans. (B,D) 13. Ans. (A) 2. Ans.(B, D) 14. Ans. (C) 3. Ans.(B, C, D) SECTION-II 4. Ans.(A, B, D) 5. Ans.(B, D) 1. Ans. (A)-(S,T); (B)-(P,T); (C)-(T); (D)-(P,T) (A) RNH2 (0.01M) kb = 10–10 M 6. Ans. (B,C,D) [OH–] = kb  c  10–10 102  106 7. Ans. (A,B) 8. Ans. (A,B) pOH = 6 9. Ans.(B) pH = 8 AgCl   Ag+ + Cl– (B) H2A (0.1 M)  ka1  ka2 (s+0.1) [H+] from 1st dissociation of H2A is considered. 10–10 = s(s + 0.1) s = 10–9 mol/L  s = 10–11 mol/10 mL [H+] = ka1  c = 143.5 10 –11 10 3 = 1.43 × 10–6 mg [H+] = 107  0.1  104 10. Ans.(B) AgCl(s)   Ag+ + Cl– pH = 4 10–2 = 10 4 (D) 0.1 107  102 106 x2 22 x2 = 10–2 SECTION-IV 1. Ans. 2.4 [OMR Ans. 6] x = 0.1 2. Ans. 5 3. Ans. 3 k × k [0.01] 2 4. Ans. 4 sp f = [NH3 ]2 [NH ]2 = 10 4 3 10 2 [NH ] = 0.1 5. Ans. 2 3 To dissolve 0.1 M Ag+ 0.02 M NH is required, 3 [NH ] = 0.1 + 0.02 = 0.12 M 3 11. Ans.(C) 1001CT103516011 HS-3/13

Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-1 PART-3 : MATHEMATICS SOLUTION SECTION-I   2  2 sin2   2  2 1. Ans. (A,C) 2   0  sin2   1 Let centre be (,)  p.d = radius 2 4  3  5  2    15,5 B C2   n   ;n  Z 5 C1 4 C1(15,15), C2(–5,–5) AB  800 16 A ƒ   32  4  2 AB = 28 2 C1C2 = 20 2  (2 + 2) (2sin2 – 3) = 0   = –2 4. Ans. (B,D) 1 P(x) = (x – 1)2Q1(x) + 1 Area of ABC1 = .28.2  28 and P(x) = (x + 1)2Q2(x) + 3 2 P(1) = 1, P(–1) = 3, P'(1) = 0 = P'(–1)  P'(x) = (x2 – 1) Area of AC2BC1 = 56 2. Ans. (Bonus)  sec2 x  3 dx  sec2 xdx 3 cos x dx  Px   x3     4  3 sin2 x   3 x  sec2 x  3 4  tan2 x   dt 3 d Px  x3  3 x  2   4  t2   4  2 22  n t  4  t2  3 sin1     C  2  y=ƒ(x)  n tan x  4  tan2 x  3 sin1  3 sin x   C –4 1x  2  –1 ƒ(x) = tanx, g(x) = sinx, k  3 5. Ans. (A,B,C) 3. Ans. (B,C,D)  ƒ x lim ƒ x  2x3 ƒ() = 2sin2 – cos2 + 1 lim  1  ex0 x4  e3 x0 2x3  (2 + 2)sin2 + 1 –  ƒ x  2x3 at sin2 = 0  ƒ() = 1 –   lim 2x4  3  ƒ x  6x4  2x3 x0 at sin2 = 1  ƒ() =    ƒ x  6x4  2x3  ƒ 1  8,ƒ ' 1  30 for  > 0 ƒx range of ƒ() = [1 – , 2 +  + 1] lim x4  6 x  [0,1] ƒ()|max = 3 6. Ans. (C,D) ƒ   2  2 2 3 n 2 3  .....  n      Let   2  2 sin2  1    2  2 P  e1/ n  2 e1/ n e1/ n e1/ n 2    e1/nP = 2 3 e1 / n 2 e1/ n (n – 1)(e1/n)n + ....+ n(e1/n)n HS-4/13 1001CT103516011

Leader Course/Phase-III, IV & V/18-12-2016/Paper-1 on subtraction 9. Ans. (C)  n  1 t2  3  a  4t  b      1  e1/n e1/ n  e1 / n 3a b n1 P   n e1/ n e1/ n 1  t  3b 2 a  put in eq. (1) 3   e1/n e  1 n e1/ n n1  4a2  9b2  36    P  e1/ n 1 2  e1/n 1  x2  y2  1  94 P(3,4)  R now lim  P   lim  e 1  e  e1 / n  n2   e1/ n   n n   e1/ n  12  1   1/n    1/ n   Q     (3,0)  e 1  e 1 1 12 equation of tangent from P 7. Ans. (B,C) y  mx  9m2  4 Given x  1 y y  xy ' x 4  3m   9m2  4 2  dy  2 y   2 P(x,y)  m  ,m  1 dx x x x 2 (0,y–xy') 0 x = 3, x – 2y + 5 = 0  1 point of contact Q(3,0) x2 i.f.  9 8   8 5  R  , y. 1  2 dx x 2  C x2 x3    P(3,4)  y  1  x2 8. Ans. (A,C) R(–9/5,8/5) Q(3,0) 1 6 perpendicular bisector of PQ : y = 2 For concurrency 2 1 3  0 perpendicular bisector of RQ : y4  3  x  3  2 5 5  5   32 + 16 – 35 = 0  ( + 7) (3 – 5) = 0 circumcentre (1,2)   + 3 = 7    7, 5  m  2 10. Ans. (A) 3 P for not forming triangles either lines are RA parallel or concurrant 1      2, 2  1  4 n  6 RQ = 2ae 2  1   2  2  PR + RQ = 2a 1   2  2   e  RQ  2 10 2  PR  RQ 5  3 5 1001CT103516011 HS-5/13

Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-1 Solutions for Question 11 & 12 put m  t 1 2y cosec2x. dx + ncotydx t + ntanx.dy – 2xcosec2ydy = 0  (2y cosec2xdx + ntanxdy) (t – 1)3 + (2 – h) (t – 1)t2 + kt3 = 0 + (–2xcosec2ydy + ncoty dx) = 0  d(y ntanx) + d(xncoty) = 0 has three roots 1 ,1 ,1  y ntanx + xncoty = c 1  m1 1  m2 1  m3  (tanx)y.(coty)x = k  summation of roots  89 22 ƒ   ,    0  k  0  5  h  89  67h  89k  157  4 2  k  h  3 22 ƒ(x,y) = (tanx)y(coty)x SECTION – II 1. Ans. (A)(S); (B)(R,T); (C)(Q,T); 11. Ans. (B) (D)(P,S) 12. Ans. (C) (A) a2 = 2 + 2, b2 = 2 + 1  cos x dx   2 e2 1 0x 2 + 1 = (2 + 2) (1 – e2)   2  2    cos3 x dx   3 cos x dx   cos3x dx e  1    1  LR  2.2  4 0x 04 x 0 4x 3 33   = 3  2 = 6 0  3 .   cos t dt (B) ||x| n|x||  1 4 2 4. t 3 k 1 3 ƒ(x) = xnx e ||x|n|x||  3. 1.   – 1 42 42 2 01 1 e ke Paragraph for Question 13 & 14 13. Ans. (D) 1 e m1,m2,m3 = –k m3 k    2 k3  k k>e 8  3    2  h   k  0  y2  4x  C and C1 are same  least prime k is 3.  infinite common tangents 14. Ans. (B) (C) centroid   cos   sin   2 , sin   cos   1  y = mx – 2m – m3  3 3  m3 + m(2 – h) + k = 0 3h 2  cos   sin   eliminate '  ' 3k 1   sin   cos      (m1 ) 0    89 x2  y2  4 x  2 y  1  0 22 333  (m2 )  (m3 ) +=1  0 0 (D) yxnx = nx – ny 1 1 1 89  1  m1 1  m2 1  m3 22 m1  exny. 1  ex.ny  ny  x .y '  .nx  1  1 y'  y  x y m3  2  h m  k  0 m2 x   m3  1  e.y '  1  y ' x  e,y  1 HS-6/13 ee  y' = 0 1001CT103516011

Leader Course/Phase-III, IV & V/18-12-2016/Paper-1 SECTION – IV x 1. Ans. 2 xƒ x   ƒ tdt  1  x ƒ x  ex 0 (cos5x) + (sinx) (cos4x) – (sin4x) + (sin3x) + (cosx) (sin2x) – (cos2x + sin4x) = 0 d.w.r. to x ƒ(x) + ƒ'(x) = e–x (cos4x + sin2x) (cosx + sinx) – (cos4x + sin2x) I.f. = ex  ƒ(x) = (x – 1)e–x =0  |ƒ(0) = 1|  (cos4x + sin2x) (cosx + sinx – 1) = 0  sin  x     1 2 solutions 4. Ans. 1  4  2 2k – 1 = sin–1x + tan–1x – sin–1(sinx) x 0,  2k 1   3 1, 3  1 4 4 2. Ans. 2 A k   3  1, 3  P 8 8  x2 + y2 = 1 B  [/2,] k  0.2,1.2  k  0,1  1 1 < r < 2 5. Ans. 5 Let x = rcos, y = rsin S1 : x2 + y2 = 1 r2  r2 sin 2  1 S2: x2 + y2 = 2 2 AB : x2 + y2 = 2 OP  1  2  1  r2  2  2  r2  2 ,2 2 h2  k2 sin 2 3 13 2 2 x3y + xy3 + 7  xy(x2 + y2) + 7  h2  k2  4  R  2  r2sincosr2 + 7  (1 – r2)r2 + 7 3. Ans. 1  –(r2)2 + r2 + 7 xx  minimum value occur on r2 = 2  ƒ(t)dt   (x  t)ƒ t  ex  1 5 00 d.w.r. to x 1001CT103516011 HS-7/13

Paper Code : 1001C T103516012 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE Test Type : MINOR PHASE : III, IV & V Test Pattern : JEE-Main TEST DATE : 18 - 12 - 2016 PAPER-2 SOLUTION 1. Ans. (3) 6. Ans. (3) Sol. Component of velocity of A along common Sol. Dielectric will don't exert any effect on the normal is v cos 60° and this velocity outside electric field so potential at B KQ B= vcos60° b A 60° 7. Ans. (2) v vsin60° Sol.     B0et  a2 ; B.A 2 of A after collision with B is interchanged. Hence A moves along v sin 60° which    d  B0eta2 & I   is normal to common normal. dt 2 3R 2. Ans. (2) Sol. 4m x = m(3 – x) 8. Ans. (1) 9. Ans. (1) 3  x = 5 = 0.6m × C× 3. Ans. (Bonus) ×× ×× Sol. × ×× × iB Sol. × × mg × a × × HGF IJK FGH JKIdeflection y = 1 2 2 at2 = 1 eE v0 × 2 m ×× ×× y = 1 HFG e JIK FHG  KJI2 ×× ×× 2 md v0 = 400 16. 10 19 100 10 4 = 1.76 mm mg – ilB = ma ...(i) 2 2 10 2 91. 10 31 10 16 4. Ans. (3) q  Vv  dq  i  Bc dv ...(ii) c dt dt V Sol. Electric field at P and P' = D 5. Ans. (3) mg From (i) & (ii), a = m  B22c HS-8/13 Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in

Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 10. Ans. (3) 15. Ans. (2) Sol. I  BV Sol. By applying condition for circular motion net 15 force toward centre at point C should be I equal to mv2/R By applying work energy theoram at point I1 = I2 = 2 A and C 11. Ans. (2) P 1 mv2 – 1 mu2 = work done by all the forces Sol. B 2 2 M Q v R 45°  1 mv2 – 1 mRg = Welectric field + Wgravity 2 2 2  1 mv2 – 1 mRg = 0 – 2mgh 2 2 Projection of PQ perpendicular to velocity  v  5Rg  5 1 10  50 m/s 16. Ans. (2) R is PM = R – 2 KQ 2 KQ2  KQ .Q KQ 2 KQ2    R Sol. Ei = 2a 4a 2a = a 4a  emf across PQ = Bv R  2  = 5KQ2   4a 12. Ans. (4) Ef = K (2Q)2 = KQ2 KQ2 4a Ei – Ef = H = 4a a Sol. A 17. Ans. (2) Sol. Energy = 1 0 E2  volume  2 BC  8.85 × 10–6 = 1  8.851012 E2 106 AB    BC    EA   R2  2 E.d E.d Ed 4 E = 2 106 V / m R2 flux () = EA 0 +  + 0 = = 2 106 104  100 2 V  m  4 18. Ans. (3) 19. Ans. (2) 13. Ans. (4) Sol. Magnetic field can do no work  KEf = KEi Sol. I ML2 sin2   0.6  ML2 sin   ML 2 7.2 v = v0 12 12 2  14. Ans. (1)   Sol. X1 X2 I  ML2 sin2  , vB – vA = 1 B  X 2  X12  3 2 2 ,I ML2  3 ML2  7.2 at   3  3 4   4 4  1.8 kg-m2 = C  X 2  X12  [where C  B ] 2 2 v = C [12 – 02], C[22 – 1], C[32 – 22], C C(1), C(3) , C(5), C(7)  POtential difference is in A.P of comoon HS-9/13 difference 2 1001CT103516012

Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 20. Ans. (1) 3kx1 + kx2 = keqx 21. Ans. (2) T  2 m k eq 22. Ans. (3) 31. Ans. (2) 23. Ans. (1) 32. Ans. (4) 33. Ans. (3) 24. Ans. (4) 34. Ans. (2) Sol. Volume of domain H2C2O4 NaHC2O4    1 2  10 = 10  100 100 X  1 = Y.5 1 × 2 = X. 1  1 × 2 = Y. 5 mass = 7.87 × 10 = 78.7  gm N  78.7 2  6.0231023 7 55.87 X5 Y1 Magnetic moment = N × 1.8 × 10–23 35. Ans.(3) 25. Ans. (4) Sol. 0I1  0I2 1  kw 1  C2 2R k b .C1 2 C1 22R Sol. 26. Ans. (2) q v  B   e B  v   kw 0.01   0.4   2 Sol. Force on electron = k b .C2 2  0.1  27. Ans. (1) Sol. Torque due to magnetic force 0.01 = 2 = 0.005  L = iL2B 36. Ans. (4) | d | (idrB)r 02 37. Ans. (4) 38. Ans. (4) dr 39. Ans. (1) ir 40. Ans. (4) 41. Ans. (2) In equilibrium iL2B  (kx)  L sin 30 42. Ans. (4) 2 43. Ans. (1 or 4) 5iLB or x = 44. Ans. (3) 8k 45. Ans. (2) 28. Ans. (4) 46. Ans. (3) Sol. Mq 47. Ans. (3)  48. Ans. (4) L 2m    M  q  mr 2     M B  q r 2B sin   clockwise 49. Ans. (4) 2m 2 4 50. Ans. (1) 29. Ans. (2) 51. Ans. (4) v02   3 107 m  qE0 2  4v02 52. Ans. (2)  qE0 m  Sol.   53. Ans. (3) v0 = 107 m/s 54. Ans. (3) 30. Ans. (2) 55. Ans. (4) Sol. Let the elongation in spring are x1 & x2 56. Ans. (4) x1 + x2 = 2x 57. Ans. (4) 3kx1 = kx2 58. Ans. (3) HS-10/13 1001CT103516012

Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 59. Ans. (4) let 3x8 + 8x3 + 24x = t 60. Ans. (4) 61. Ans. (2)  1 dt  1 . 3 t2/3 C 24 24 2 adding all we have 6(a + b + c + d + e) = 6.31 t1/ 3 a = 31  c = –7 62. Ans. (4)   1 3x8  8x3  24x 2/3  C n(m sinx + 4) > 0  m sinx + 4 > 1 16 msinx > –3  m  [–3,3] 7 possible integral values. 69. Ans. (2) 63. Ans. (3) (x3 + 1)dy + 3x2ydx = xdx P(x,y) P  y(x3 + 1)  x2  C OP = 3PM O y=2x 2 x2  y2  3 2x  y   y  x2 5 2 x3  1  5(x2 + y2) = 9 (4x2 + y2 – 4xy)  lim 2 x3  1  2 31x2 + 4y2 – 36xy = 0 pair of st. line Aliter focus is (0,0) and y = 2x directrix, x 0 e=3 64. Ans. (1) focus lies on directrix therefore locus is pair of st. line. (2,3) 70. Ans. (2) 24   3  3 74 tan A  24  3   117 , 7 51 65. Ans. (4) 1  24 4 .3 74  lim 1 r2 r 1 n2 sin n  0 x2 sin xdx 3 3 n n 4 1 9 9 x2  cos x  2x  sin x  2 cos x10 tan B   13 = (–cos1 + 2sin1 + 2cos1) – 2 4 = cos1 + 2sin1 – 2 3  24 7 66. Ans. (3) tan C   45 9 1  3. 24 65   x x2  4  ƒ 'x  1  x2 3/2 , – 13 + 7 0 71. Ans. (2) 67. Ans. (3)  3 Using L' Hopitals rule  ex2 x  42  x  42 dx 0 H'x  x3 '   3 sin t3dt  1 x 1   ex216x dx  8ex2 lim  H ' 1  x2  3   x 1   8 e3 1 00 x3 72. Ans. (3) H'x   sin t3dt  x  1 3x2 sin x9  2x sin x6  bc ac ab  (b3 )(c3 )  (a3 )(c3 )  (a3 )(b3 ) x2 at x = 1 H'(1) = 2sin1 68. Ans. (4) a2  b2  c2 (a  b  c)2  2ab  a2b2c2   x7  x2 1 (abc)2   3x8  8x3  24x 1/3 dx ,  0  16  16 (1)2 1001CT103516012 HS-11/13

Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 73. Ans. (1) 79. Ans. (4) (x)  Aex  x 5 r,r 12 1  A  et Aet  t dt 0  A e2t  tet  et 1 2   13 r  r  20  r  240 0 12 80. Ans. (3)  A (e2  1)  (e  0)  (e  1) 2 A  A (e2  1)  1  3x  4y  2 1  4x  3y  2  32  42  5  42  32   A   e2  1  1  A  2 LR  1 1   e2 5  2  3 ( x )  2 ex  x 81. Ans. (3)  e2 3 Qt2  (n(e2  3))  2  n e2  3 t1 P 74. Ans. (3) S  11 2   12 2   13 2  .... R  7  7 t4     7    T t3  1 (11)2  122  ......212  t1t3 = t2t4 = –1 72   PQ : y(t1 + t2) = 2x + 4t1t2  1  21.22.43  10.11.21  72  6 6  3(t1 + t2) = – 4 + 4t1t2 ......(1) 1 . 21 .1186 10  11 .76  11 .38 TR : y(t3 + t4) = 2x + 4t3t4 ......(2) 49 6 14 7 from (1) 3  1  1  4  4  t3 t4  t3t4 75. Ans. (1)   (xy  1)2  2x(xy  1)  2x2  3(t3 + t4) = 4t3t4 – 4 ......(3) using (2) and (3) we can say that TR x(xy  1) passes through (–2, 3)  xy  1  2x  2  2 2  2 x (xy  1) 82. Ans. (3) 76. Ans. (2) X2 = 4Y, X2  Y2  1 2 x2  5x  17  x2  4x  2 8 y = mx – m2  2x2  x  1  0  x  1 y  mx  m2  2  84 m2  m2  2  P1 and P2 touches each other  m4 – m2 – 2 = 0 77. Ans. (2) m2 = 2, –1  m 2x + 3y = 26 m12  m22  4 78. Ans. (1) 83. Ans. (3) a – 2, a + 2 y = x2, z = y3  z = x6 a–2<1&a+2>6 logxz = 6 a < 3 & a > 4 not possible HS-12/13 1001CT103516012

Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 84. Ans. (Bonus) 87. Ans. (1) 3y2y' + 2y + 2xy' + 3x2 = 3(x – 1)2 2 cos ec2x  2 cos2 x at (1, –1) 1  cos4 x 1  1 + cos2x = cos2x  No solution 3y' – 2 + 2y' + 3 = 0  y' =  5 Normal (y + 1) = 5(x – 1) m = 0. 5x – y = 6. 85. Ans. (2) 88. Ans. (4) y=cos–1x  ƒ' = 2x(x2 – 1)(x2 – 4)(x2 – 9) /4 /4 ++ + +  (cos y  sin y)dy –3 –2 –1 0 1 2 3 ƒ(x) have 7 critical points and ƒ'' = 0 have 3 0 siny  cos y / 4  2 1 1/2 0 y=sin–1x positive and 3 negative solutions 89. Ans. (1) ƒ' is odd function there for ƒ\"' is also and b2  1  e2 and a2e2 = b2(1 – e12) odd function so ƒ\"'(0) = 0 a2 as there are 5 roots of ƒ\"'(x) = 0 there for it  1  e12  1 e2  e12 1  1 e2 will have 2 positive and 2 negative roots.  e2  e2 86. Ans. (2) 1  2e2 1  e2 a sin b 1  e1  a cos b  1 a2  2  a   2 90. Ans. (4) I-Case : a  2  sin b  1  b  3 axsec – bycosec = a2 – b2 cos b  2  4 axsec – bycosec = –(a2 – b2) 1  L  2(a2  b2 )  2  a2 sec2   b2 cosec 2  a2sec2 + b2cosec2 > (a + b)2 II-Case : a   2  sin b  1  b  7 cos b    4 L  2(a  b)  2  1  2  2, 3    2, 7   4   4  1001CT103516012 HS-13/13