32- Solution Report (32)

Paper Code : 1001CT103516003 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III, IV & V) ANSWER KEY : PAPER-1 TEST DATE : 28-08-2016 Test Type : MINOR PART-1 : PHYSICS Test Pattern : JEE-Advanced Q. 1 2 3 4 5 6 7 8 9 10 A. A,B,C SECTION-I Q. B A,B,C,D A,B,C A,C A,B,C,D A,B,D C A B A. 11 SECTION-II C 12 SECTION-IV Q.1 A P,R,T D Q. 1 A. 5 BC D Q,S P P,Q,R,S,T 23 4 567 Q. 1 05 4 884 A. A,C Q. 11 PART-2 : CHEMISTRY A. A 2 3 4 5 6 7 8 9 10 Q.1 A SECTION-I R,T A,B,C A,B or A,B,C A,B,C,D A,B,C A,B,C,D B,D B,C A B Q. 1 SECTION-II A. 9 12 SECTION-IV D BCD Q S P,T 2 3 4567 Q. 1 5 5 4288 A. A,C Q. 11 PART-3 : MATHEMATICS A. C 2 3 4 5 6 7 8 9 10 Q.1 A SECTION-I P A,B,C,D A,D A,B,D B,C B,D B,C,D A,B,C,D D B Q. 1 SECTION-II A. 5 12 SECTION-IV D BCD P,Q,R,S,T P,Q,T P,Q,R,S,T 234567 452622 Paper Code : 1001CT103516004 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III, IV & V) ANSWER KEY : PAPER-2 TEST DATE : 28-08-2016 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 4 2 3 1 1 2 1 or 3 3 4 4 4 2 4 1 4 3 3 Bonus 2 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans. 4 3 2 2 2 3 1 3 4 4 2 2 2 2 2 3 2 2 3 3 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. 1 3 3 4 4 4 4 3 1 1 3 2 4 3 4 1 3 2 3 3 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans. 3 3 4 2 2 3 4 2 2 2 4 2 1 4 2 3 1 4 4 2 Que. 81 82 83 84 85 86 87 88 89 90 Ans. 2 3 4 3 2 4 4 4 3 3

Paper Code : 1001C T1035160 03 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 : 28 - 08 - 2016 PAPER-1 PART-1 : PHYSICS SOLUTION SECTION-I 3. Ans. (A,B,C,D) 1. Ans. (A,B,C) 4. Ans. (A,B,C) Sol. For (A) : Particle will return to its Sol. Hmax  u2 sin2  , R = 2u2 sin  cos  starting point if its displacement is zero. 2g g  If R = Hmax then tan  = 4, x = 0  t2 – t3 = 0  t =  if Hmax > R then tan  > 4 For (B) : Particle will come to rest if and if Hmax < R then tan  < 4 . 5. Ans. (A,C) v = 0; v = dx  2t  3t2 0t  2 dt 3 Sol. Relative to cyclist For (C & D) : vcar = 25 m/s; scar = 150 m 6. Ans. (A,B,C,D) v  2t  3t2  Initial velocity=0 ; 7. Ans. (A, B, D) Sol. Impulse exist in action reaction pair a  dv  2  6t during collision so magnitude will dt remain the same. (a) Elastic collision principle  Initial acceleration = 2 0 2. Ans. (B) (b) Internal force Sol. (d) If m1 >>> m2 direction of velocity of motion does not change ax v1' = (1 + e)        vC  ev1 vC v1 v1'   vC x2 = 4a(A) [Since y = a 8. Ans. (C) x2 = 4a2 9. Ans. (A) x  2a 10. Ans. (B) the slope of curve at point (2a, a) Sol. (9 & 10) Frictional force between two blocks will m = tan= 2x 2  2a 1 oppose the relative motion. For 1 kg block 4a  friction support the motion & for 2 kg friction oppose the motion. Let common velocity be 4a v then for 1 kg v = 2 + a1t g so tangential acceleration = g sin  2 Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/12 +91-744-5156100 [email protected] www.allen.ac.in

 1g 0.4 10 Target : JEE (Main + Advanced) 2017/28-08-2016/Paper-1 2. Ans. 0 where a1 = 1 = 1 = 4 ms-2 Sol. 2T = K1x1, As m2 > m1 for 2 kg v = 8 - a2 t T + k3x3 = m2g, T–K2x2=m1g  1g 0.4 10  K1x1 – K2x2=m1g 2 where a2 = 2 = 2 = 2 ms-2  2 + 4t = 8 - 2 t  6t = 6  t = 1s  x2 = K1x1  2m1g and v = 2 + 4 t = 2 + 4 × 1 = 6 ms-1 11. Ans. (C) 2K 2 12. Ans. (D) SECTION-II 200  20 102  2  2 10 = =0 1. Ans. (A)  (P,R,T); (B)  (Q,S); 2 1000 (C) (P); (D) (P,Q,R,S,T) 3. Ans. 5 Vf  Vi Sol. Avg acceleration = tt  ti aB  (A) Vf must be in the direction upward Sol. 2a = a vertical to downward vertical B AB aA aAB Average velocity = Displacment time aA = aB 2  aAB 2 ; aA = 5 m/s2 Direction of average velocity in the direction 4. Ans. 4 5. Ans. 8 of displacment. Sol. From work energy theorem W = KE (B) Particle should be in the 3rd and 4th quardent (C) Distance moved from starting point to quarter half (t = 1) sec. 1 1 2 2 t=1 W + W =gravity mv 2  mv12 air friction 2 R t=0  Wair  1 4 142  4 10 10  8J 2 (D) Magnitude R 2 will be when particle friction is at circumference point of circle on y-axis. Work done against air friction = 8J SECTION-IV 6. Ans. 8 1. Ans. 5 Sol. mucos30 = 2× mvx a k2 2 3u Sol. vk s  v2  k2s  4 vx  usin30° ucos3300° °  vA  3u 30°  W = Fs cos 0° = (ma)  1 at2  W  mk4t2 4  2  8 u  vB  3u2 u2 u 7 =8   7. Ans. 4 16 16 4 vB  7  Sol. i  f t  80  2 vA 3   5 1001CT103516003 2 HS-2/12

PART–2 : CHEMISTRY Leader Course/Phase-III, IV & V/28-08-2016/Paper-1 SECTION - I SOLUTION 1. Ans. (A,C) 11. Ans.(A) G = H – TS 12. Ans.(D) T = H  –10000 2500 = 625 K SECTION - II S –160 4 1. Ans. (A)(R,T); (B)(Q); (C)(S); 2. Ans. (A,B,C) (D)(P,T) 3. Ans. (A,B or A,B,C) SECTION - IV 4. Ans. (A, B, C, D) 1. Ans. 9 Sol. Based on Fajan's Rules. B(crystalline)  B(amorphous) ; H = +0.4 kcal mol–1 5. Ans. (A, B, C) for 247.5 gm = 247.5  0.4 9 Sol. (A) is unfavorable because it places +ve 11 charge on the electronegative oxygen atom 2. Ans.(5) and also has adjacent +ve charges. (B) and (C) should be bent and are energetically unfavorable when forced to be in resonance with (D). Also, (C) has widely separated charges.  V1  ln  V2  6. Ans.(A, B, C, D) G = nRT  = 200 × 0.7 = 140  7. Ans. (B,D) 8. Ans. (B,C) 3. Ans.(5) 9. Ans. (A) 5 millimole per lit per sec. A  d[I2 ] 2.5mM s 1 dt 2atm A P 1 d[HI]   d[I2 ] T=300K 1atm T=600K BC 2 dt dt V 2V 4V V d[HI]  2 × 2.5 = 5 mM –1 s–1 Process AB dt U = 0, qAB = –w = nRT ln V2 V1 4. Ans. (4) qAB = 1 × R × 300 ln 2 = 210 R Sol. 8 1 –1 –8 Process BC CayK[F(S i4 O10 )2 ]8H2O qBC = H = nCp T +2 × y = +8 ; y = 4 5. Ans. (2) qBC = 1 × 5 R(300) = 750 R 2 qTotal = 210 R + 750 R = 960 R Sol. H2SO5 H2O H2SO4 + H2O2 10. Ans. (B) SAB = q AB  nR ln  VB  = 1 × R ln 2 HSO H2O 2H SO + HO T  VA  22 8 24 22   SAB = 0.7 R 6. Ans. (8) SBC = nCpm ln T2 = 1.75 R 7. Ans. (8) T1 Ssystem = 0.7 R + 1.75 R = 2.45 R 1001CT103516003 HS-3/12

Target : JEE (Main + Advanced) 2017/28-08-2016/Paper-1 PART-3 : MATHEMATICS SOLUTION SECTION-I  a2 , 2a2c2 ,c2 are in A.P. 1. Ans. (A,C) b2 1 , 1 ....... 1 are in A.P.  a2b2, 2a2c2, b2c2 are in A.P. x1 x2 x4001  1  1  1  1  .....  ab , 2ac , bc are in A.P. x1 x4001 x2 x4000 c b a  'D' is right and 'C' is wrong  1  1  2  50 Also (4a2 – b2)(4c2 – b2) x x x2000 = 16a2c2 + b4 – 4b2(a2 + c2) 2002 2001   1  2000.50  25  100025 = b4  'A' is right 'B' is wrong. xr 4. Ans. (A,B,D) also g = ƒ–1 h(gog)–1 = ƒoƒ(x) 1  1  1  1  ..........  1  1  d h(x) = ƒ3(x) + 5ƒ(x) + 3 x2 x1 x3 x2 x x4001 4000 h(1) = 93 + 45 + 3 = 729 + 48 = 777 x1x2  1 x1  x2  5. Ans. (B,C) d x2x3  1 x3  x2  (A) d (B) x = 0 is only value where equation hold ---------------------- and it satisfy ---------------------- (C)  x x4000 4001 1 x=0  d x4000  x4001 1 , d 10  x1  x 4001  1  1  4000d , x1  x4001  4000d x4001 x1 x1x4001  x1x4001 4000 = 10 (D) No solution  1 . 1  400  1  10, 1  40 x1 x4001 x1 x4001 6. Ans. (B,D) 2. Ans. (A,B,C,D) y ƒog  x   1  gx  g x Q  g2 x g x  Q ƒ(0) < 0 x so always true a  R  3. Ans. (A,D)   1  x xQ xQ 1  x 2 2.2a2 .2c2 4a2c2 b2  2a2  2c2  b2  a2  c2 ...(1) ƒog(x)  n2, n > 1 so into  a2  c2  2a2c2    also ƒog 1  a  ƒog 1  a os many one. 2 b2 HS-4/12 1001CT103516003

7. Ans. (B,C,D) Leader Course/Phase-III, IV & V/28-08-2016/Paper-1 (x + 3)2=mx 9. Ans. (D) 10. Ans. (B) Paragraph for Question 11 & 12 at x = 0 y=12x  2a  1 sin1 3  b cos1 7  2a 1  b 44 m = 12 and all negative values of m at x  4  7  b tan–1A – 2tan–14 = 0 8. Ans. (A,B,C,D) b=2 Paragraph for Question 9 & 10 3 a= 2,b=2 a+b+c=9 ...(1) range of ƒ(x) is ,  2 tan1 4  lim ax2  bx  c  9  4 2sin–13/4 x1 x  1 ax2  bx  c  3 y = –2tan–14 –7/2  lim a x 1  b  4  2a  b  24 ...(2) 4–7 x1 ax2  bx  c  3  a > 0, b2 – 4ac < 0 ...(3) 11. Ans. (C) 12, 0, 3   12. Ans. (D) 10,4,5 satisfy (1) & (2) both but not (3) 1. SECTION – II Ans. (A)(P); (B)(P,Q,R,S,T); 15, 6,0  (C)(P,Q,T); (D)(P,Q,R,S,T)   A,B & C are incorrect and D is correct. SECTION – IV by (1) (2) (3) 1. Ans. 5 b = 24 – 2a  c = 9 – a – (24 – 2a) = a – 15  (24 – 2a)2 – 4a(a – 15) < 0 lim  3 8x3x2 22x  3x 9x2  3x 1   a  1 x  b    x      1 1  12 2 (a – 12)2 – a(a – 15) < 0  79  a = –1 –9a + 144 < 0 12 a > 16  c > 1  1  1  b  79  b  6  a  b  5 for a = 16  b = –8, c = 1 12 2 12  4x 1  3 2. Ans. 4 ƒ x    x 1 , x 1 Let t cot 1  2x  3    7    4 , x 1 2  4x  2 , x1    t   t2  k t  (0,)  x 1 4  2  3  0 /4 , x1  4 ƒ t  2t2  t  2 4 y 4  2  ƒ 0   4 y=4 1/4 x ƒ     2  4  8 y=–4 2  k  2 32  k  32  k  4,5,6,7  range (–4,4] 834 8 4 1001CT103516003 HS-5/12

3. Ans. 5 Target : JEE (Main + Advanced) 2017/28-08-2016/Paper-1 6. Ans. 2  tan 90  10n tan  1   10n1  tanx = tan  lim  lim x n 1    n tan 90  10 n1 tan  10n  x  n    x2  nx  1  0 x tan  1  1 1 x  n  n2  4  10n  2 1 10n1 10n 10n n2 + 4 must be perfect square  lim 1  1   10 n=0  10n  x =1,–1 n tan 7. Ans. 2 10n1 4. Ans. 2 2 tan1 x 2 r ƒr  lim x0 x r 1 1  e  elim x x 2  x3 x1 x 1 4 1   ƒ 2r 2 1 x  n  2r  lim 2 2r 2  n 5. Ans. 6  1  1  x2     n  lim x 2 x 4  e2  129  129  129  129  131 x x2  4x 1 129 4       333 xx  4 131 HS-6/12 1001CT103516003

Paper Code : 1001C T103516004 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 : 28 - 08 - 2016 PAPER-2 SOLUTION 1. Ans. (3) 4. Ans. (3) Sol. AT the top V mv2  vy gsin  vx T + mg = Sol. uy = 20cm/s u L  g gcos T < 10 mg v < 11gL 3gL  u  13gL ux = 10cm/s 2. Ans. (4) at any instant Sol. vH = u cos  = 6 at = g sin  an = g cos  vv = v2  u2 cos2   8  an = at t1  u sin   8 10 t2  u sin   8 vy 10 vx  tan  = 1  1 t2 – t1 = 82  1.6s 10 vy uy  gt 3. Ans. (2)  vx 1  ux =±1  N dR Sol.  Solving we get r t = 1 and 3 mg 5. Ans. (1) Centripetal Force = mass × centripetal acceleration V2 ....(i) 30° T y N cos  = m ....(ii) Sol. 60° T x r r —mrv–2 N sin  = mg rg mg (ii)/(i), tan  = V2 Force equation from ring frame along y-axis  V  rg T cos 30° + T cos 60° = mg tan   3 1 V g(R2  d2) r  R2  d2  T   2  = mg ....(i)   d  tan   d / r  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) 2017/28-08-2016/Paper-2 along x-axis 10. Ans. (4) Sol. Range on the inclined plane mv2 h2   R 2  u2 21 T sin 30° + T sin 60° = max  2 g 8  r   3 1 mv2 = T  2  r  ....(ii) 11. Ans. (4) (ii)/(i) we get Sol. work done by internal forces is equal to change in kinetic energy of the system. v = rg 12. Ans. (4) 6. Ans. (1) Sol. v=  x  dx  x x = 2t2 Sol. a = u1cos30 t = u2 cos60 t dt 4 & y = u2sin60 t – (1/2) gt2 dv  2 & – (h – y) = u1sin 30 t – (1/2)gt2 at = v dx =  x. 2  = constant  y = a tan60 – (1/2)gt2 x 2  y – h = a tan30 – (1/2)gt2  h = a (tan60 – tan30) aC = v2  2x 2 2t2 r r   r 4 aC t2  h = 2a/ 3 ac a at 7. Ans. (2) Sol. Let velocity of 1kg after collision is v 1 1  v 2    1 1 1 1 t 2  2 4 13. Ans. (2) v  1m /s Sol. The momentum of the system in horizontal 2 direction will not change. Now, conserving momentum in collision 14. Ans. (4) Sol. From the F.B.D. of m, 1 1 (1) (1) = (1)  2  + mv mv = 2 T1 Also v  1 3 2  1  v  m / s 12 T2 mg so m = 1 kg T1 = m2  mg , T2 = m2  mg 3 2 2 8. Ans. (1 or 3) 15. Ans. (1) Sol. vcm  20  6 10  3 = 3 m/s towards right. 2m  25 20  10 At the time of either maximum elongation Sol. 3m  40 mvx or compression, both the blocks will be moving with a common velocity = 3 m/s mvy towards right. vx = 120 m/s 9. Ans. (3) vy = 50 m/s v = 1202  502  130m / s Sol. Friction force is constant and is equal to the weight of the block. HS-8/12 1001CT103516004

Leader Course/Phase-III, IV & V/28-08-2016/Paper-2 16. Ans. (4) 3 P12  mkx2 Sol. Velocity of approach = v1 – v2 = velocity of 4 separation (elastic collision) 3 m2v2  mkx2 After collision, velocity of particle is 4 v2 – (v1– v2) = 2v2 – v1. k  3 1  64  16 N / m 17. Ans. (3) 4 93 21. Ans. (4) Sol. NWWW 5W 2 44 ut Sol.  B vt B 5 W.  W . 3 3   4 22 5 From the figure, 22. Ans. (3) ut  50  vt = cos   Sol.F   r1 18. Ans. (3)  3 iˆ  ˆj  kˆ , dx  r2  iˆ  ˆj  kˆ  W = F.dx  50 3 J A 4Mg/3 23. Ans. (2) Sol. MT TT 2M Sol. T  2M1M2g mg mg M1  M2 T  2Mg Now downward force on the right mass is 3 more. Balancing torque 24. Ans. (2) L 4Mg  L Sol. The initial extension of spring is x0  mg 2 3 k M'g × = just after collision of B with A the speed of M'  4M v 23 combined mass is . M '  8M 2 3 For spring to just attain natural length the combined mass must rise up by : 19. Ans. (Bonus) x0  mg (see figure) and comes to rest. k Sol. Initial momentum = 35 Ns Change in momentum = area of graph Now, applying conservation of energy = 14 10  70Ns  12m   v 2  1 k x 2  2m  gx0 2  2 2 0 2   Final momentum = –35 Ns mg k 20. Ans. (2) & x0  Sol. 1 mv12  1 mv 2  1 kx2 se get, v  g 6m 2 2 2 2 k P12 – P22 = mkx2 1001CT103516004 HS-9/12

Target : JEE (Main + Advanced) 2017/28-08-2016/Paper-2 Alternate solving by SHM : 32. Ans.(2) In SHM v   A2  x2 33. Ans. (2) we can write, v  k  2mg 2   mg 2 34. Ans. (2) 2 2m  k  k 35. Ans. (2)     Given that , 3A(g) + 4B(g)  2C(g) 5 mol 8 mol v  g 6m Limiting reagent is A so if 5 mol of A reacts k then energy liberated = 102 kJ then for 3 mole is 60 kJ. 25. Ans. (2) 2u sin  T2 2u cos  H  E  n gRT g g Sol. T1 = ,  E = –60 kJ and R  u2 sin 2  R  T1T2g ng = –5 g 2 H = – 60 + (–5) R × 300 × 10–3 = –60 – 1500 × 10–3 R = – (60 + 1.5 R) 26. Ans. (3) 36. Ans. (3) Sol. 0 + mg (L cos  – ) = mg(L – )cos 37. Ans. (2) 27. Ans. (1) 38. Ans. (2) 39. Ans. (3) Sol. mv1 = 2 mv2 40. Ans. (3) 1 kx2  1 mv12  1  2m  v 2 Heat absorbed = 160 × 1.3 + 200 × 4.2 × 1.3 2 2 2 2  v1  2x k , v2 x k Hsoln.= 1300 = 26 kJ mol–1 6m 6m 4 / 80 41. Ans.(1) v1  v2  x 3k Sol. dxy + dxz no bond formation. 2m 42. Ans.(3) 28. Ans. (3) Sol. SO2 has 1 p –d bond  Sol. 1 mv2  kt 2 SOCl2 has 1 p –d bond v t  H2SO3 has 1 p –d bond  1 HClO3 has 2 p –d bond  acc.  t  29. Ans. (4) 43. Ans.(3) Sol. The slope of the displacement time graph 44. Ans.(4) gives velocity. 45. Ans.(4) The change in velocity is positive in the Sol. T.H.F. , Me3N , C5H5N give symmetrical region AB. cleavage. 30. Ans. (4) 46. Ans.(4) Sol. F   dm   = (V) V = V2 = 4 Sol. BiI5 and SH4 do not exist.  dt  Vr 47. Ans.(4) P = F.V. = 4 × 2 = 8 Sol. (1) CH3+ is planar but is not an odd e– species 31. Ans.(2)  H O(s) H2O() (2) C lO3 is non-planar and sp3 hybridised 2  H (3) C F3 is non-planar and sp3 hybridised S = 60.01 – 38.2 = 273  H = 5954.13 J mole–1 HS-10/12 1001CT103516004

Leader Course/Phase-III, IV & V/28-08-2016/Paper-2 48. Ans. (3) 67. Ans. (4) 49. Ans.(1) an+1 – an = n – 2 Sol. (2) Total 4 '2' are present a51 – a50 = 48 a50 – a49 = 47 (3) 3 > 2 > 1 (4) Total 20 unshared lone pairs are present a3 – a2 = 0 50. Ans.(1) a2 – a1 = –1 Sol. Due to inert pair effect, Pb+2 is more stable a51 – a1 = –1 + 0 + 1 + 2 +....+ 48 a51 – 5 = 1175  a51 = 1180 than Pb+4. 51. Ans. (3) 68. Ans. (2) 52. Ans. (2) lim tan(x  2)(x2  (a  2)x  2a)  7 53. Ans. (4) (x  2)2 54. Ans. (3) x2 55. Ans. (4) 56. Ans. (1) a+2=7a=5 69. Ans. (2)       4p2  8q2  4q2  4p2 2p 2, 2    4 4a q 57. Ans. (3)  2p 58. Ans. (2) q  0, 2 59. Ans. (3) 70. Ans. (2) 60. Ans. (3) tan 1  4   tan 1  r  1   tan 1  r  1  61. Ans. (3)  4r2    2   2  3 x2 – 14x + 45 > 0 n tan1  4  cot1  1   4r2    2   x  (–,5)  (9,) ....(1) r 1 3  4 – x > 0 & 4 – x 1  x < 4 & x  3 ...(2) 71. Ans. (4) from (1)  (2) x2 + 3|x| – 18 = 0 x  (–,4) – {3} |x| = 3  cos–1cos9 = 9 – 2 62. Ans. (3) 72. Ans. (2) a > 0; b < 0; c > 0. Do yourself. 63. Ans. (4) (2x – 1) (2x – 7) = 7  x = 3 only. 73. Ans. (1) 64. Ans. (2) Do yourself. E  1  tan 9  tan 54 74. Ans. (4) 1  tan 9 1 1 4  65. Ans. (2) x2 1    4 tan1 ( x2 1 )  elim tan1 x 1    x 1 lim x x 1 a 7 ar 3  lim 4 tan1 xtan11 tan 1  x 1  1r 1  r2  1x  ...(1) ...(2) lim 4 1    x1  x2 1 x1  x2 1 e  e  e 7 3 from (1) & (2)  a 4 and r 4 75. Ans. (2) 66. Ans. (3)   x3  x5   3x  x3 3  x 3! 5! ....  2 Tn = Sn – Sn–1 lim Tn = (4n2 + 6n) – (4(n – 1)2 + 6(n–1)) x0 2xn Tn = 8n + 2  T15 = 122  greatest value of n is 5. 1001CT103516004 HS-11/12

76. Ans. (3) Target : JEE (Main + Advanced) 2017/28-08-2016/Paper-2 tan2x + tanx – 6 = 0  tanx = –3 82. Ans. (3) logx > 0  x > 1 cos 2x  1  tan2 x  1  9   4 for x > 1, sin1 2x    2 tan1 x 1  tan2 x 1  9 5 1  x2 77. Ans. (1) log( – 2 tan–1x + 2 tan–1x) = 1. 83. Ans. (4)  1 n1tan x   e x e  k  lim x   11   x0    2  3 cos1 x 2  5x   k  e  1 n1tan x1  x2  5x  11  1  x2  5x  6  0  ex 1 22 lim  x0  x   Hence,  = 6 or  = –1.  1 n1tan x1 1   n 1  tan x   x  84. Ans. (3) ...(1)  ex   x2  sin–1x > 0   1  k  e lim  1  x x0 n 1 tan x  log 1  0  sin1 x    sin1 x 4 4 k  e    1    e x  0, 1  2  2  from (1)  (2)    2  78. Ans. (4)  a = –3 and b = 15 85. Ans. (2) ƒ(1) = 4 and ƒ(3) = 12. x3 – 2x2 + 4x + 5074 = (x – r1) (x – r2) (x – r3) put r = –2 79. Ans. (4) x2  3  –5050 = (r1 + 2) (r2 + 2) (r3 + 2). x 1 2 ,1 86. Ans. (4)   87. Ans. (4)  ƒ x   ,   ƒ(x)  ex2  cos x is even function and into.  3 2  88. Ans. (4) 80. Ans. (2) ƒ(x) is discontinuous at x = 0,3,4. ƒ x    sin1  2x     cos1 1  x2  89. Ans. (3) 2  1  x2  2  1  x2  Do yourself.     sin1  2x   cos1  1  x2   90. Ans. (3)   1  x2   1  x2    a  ƒ(x)  sgn  3 cos x  3  is continuous for all x  ƒ 1    sin1 1  cos1 0  0 3 cos x  a cos x  a 3 9   4   3   hence 0    5   5   ƒ 2     sin1  cos1   0 81. Ans. (2)  a  1 or a 1 9 9 Do yourself.  a  (–,–9)  (9,) HS-12/12 1001CT103516004