34- Solution Report (34)

Paper Code : 1001CT103516005 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III, IV & V) ANSWER KEY : PAPER-1 TEST DATE : 25-09-2016 Test Type : MINOR PART-1 : PHYSICS Test Pattern : JEE-Advanced SECTION-I Q. 1 2 3 4 5 6 7 8 9 10 A. A,B,D A,B,D A,D C,D A,C,D A,B,C,D A,B,C A,B,C A,C A,B,C,D SECTION-II AB CD A B CD Q.1 Q.2 P,R,T P,R Q,R R,S,T P,Q,R,S,T P,T T P,Q,T or P SECTION-IV Q. 1 2 3 4 5 6 7 8 A. 6 5 2 1 4 6 6 2 PART-2 : CHEMISTRY Q. 1 2 3 4 5 6 7 8 9 10 S ECTIO N-I A,B B,C ,D A ,C,D in E C A,C ,D o r B A. C or or A,B ,C A ,B ,C,D C A,B A,C in H A,B,D B ,D S ECT IO N- II Q .1 A B C D Q .2 A B C D S ECTIO N-IV P T P ,S ,T Q ,R ,T P ,T Q ,T R ,T S ,T Q. 1 2 A. 6 7 3456 7 8 9 9 5 A,B,D SECTION-I Q. 1 2 8203 10 SECTION-II A. A,B,D A,C,D 7 8 D A,B,C SECTION-IV PART-3 : MATHEMATICS A,C,D B,D P,Q,R Q.1 A B P,S Q 3456 B C Q. 2 P,Q R,T A. 1 7 B,C A,C,D A,B,C,D A,B,C 8 0 7 2 CD Q.2 A 5 R,T R,T P,S 3456 2543 Paper Code : 1001CT103516006 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III, IV & V) ANSWER KEY : PAPER-2 TEST DATE : 25-09-2016 Test Type : MINOR PART-1 : PHYSICS Test Pattern : JEE-Advanced Q. 1 2 3 4 5 6 7 8 9 10 A. B,D SECTION-I Q. 11 A,B C,D A,C A,B,D B,C,D A,B,C,D A,B,C,D A,C,D A,C SECTION-IV A. B,C Q. 12 A. 1 8 A,D 2345678 Q. 1 9229425 A. C Q. 11 PART-2 : CHEMISTRY A. A,D Q. 1 2 3 4 5 6 7 8 9 10 A. 3 SECTION-I A,B,C or BC A,B,C,D Bonus C,D A,C,D A,C B,C,D C D SECTION-IV 12 A,D 2 345678 Q. 1 2 303554 A. B,C,D Q. PART-3 : MATHEMATICS A. 11 Q. B,C,D 2 3 4 5 6 7 8 9 10 A. SECTION-I 1 B,C,D A,B,C B,C A,B,C,D B,C A,C,D A,D A,B,D A,B SECTION-IV 6 12 A,B,D 2345678 4629212

Paper Code : 100 1CT103516005 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 : 25 - 09 - 2016 PAPER-1 PART-1 : PHYSICS SOLUTION SECTION-I Case II 5:3 1. Ans. (A,B,D) 2. Ans. (A,B,D) a = F2 Sol. Kinetic energy will remain conserve hence 8 speed remains constant. f2max = 5 g Tmax = mv 2  = mv20 5g 0 Tmax a2max = 5 = g F2max = 8g   F1max : F2max Angular momentum = mv0 = mv0 mv 2  mv03 5. Ans. (A,C,D) 0 Tmax Tmax x0 kx0 3. Ans. (A,D) Sol. For (A) since loss in potential energy in both Sol. N.L. mg sin  case is same. fr=0 hence kinetic energy at 0 = 0 will be same equili. in both case. For (D) in case (2) slender bar will have At equilibrium kx0 = mg sin  more speed because it has only k(x + x0) – mg sin  – fr = ma translational kinetic energy. 4. Ans. (C,D) k(x0+x0) x Sol. Case I Equi. a a a = F1 R 8 mg sin  fr 3F1 a 8 fr × R = (MR2)  f1 = < f1max R f1max = 5g from here we get a= k x, 0  k max. acceleration with which the blocks 2m 2m If instead solid cylinder is used then can move together, a1max = f1max = 5g I  MR 2 then a  2k x , '  2k  0 33 2 3m 3m 40g amplitude does not depend on physical F1(max) = 3 system. 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) 2017/25-09-2016/Paper-1 6. Ans. (A,B,C,D) 9. Ans. (A, C)   Sol. At t = 0 x1 = 0 x2 = 2 3 Sol. by   I x = x1 + x2 = 2 3 m.   Fr kˆ mk 2  Resultant Amp. A    aA    r  acm  aC (2)2  (4)2  2(2)(4) cos   = 3 Fr 2 ˆi  F ˆj  2rˆj  m = mk2 A = 28 10. Ans. (A,B,C,D) Max. speed = A = 10 28 = 20 7 m/s SECTION-II Max. acc. = A2 = 100 28 = 200 7 m/sec2 1. Ans. (A)-(P,Q,R,S,T); (B)-(P,T); (C)-(T); (D)-(P,Q,T or P) 11 Energy = m2A2 = (20 × 10–3) × (10)2 2. Ans. (A)-(P,R,T); (B)-(P,R); (C)-(Q,R); (D)-(R,S,T) 22 Sol. (A) W = k = Positive  v   28 2 = 28 J P = Fv, v  P velocity in positive 7. Ans. (A,B,C) direction Sol. Mg = kx (B) W = positive  v 6000  0.1 M = = 60 kg P = Fv at x = 2 f = 0  P = 0 10 (C) F = 0  v = const. amax = A2  g (for not loosing contact) P = Fv = 0  P = const. 10 (D) F = negative  speed is decreasing A = 100 = 0.1 m P = Fv v  v 8. Ans. (A,B,C) v +ve  Sol. T cos  = mg  T > mg SECTION-IV kq2 1. Ans. 6 T sin  – r2 = m2r Sol.  m × g  9 × 109 × q1q2 r2 kq2 T sin  = r2 + m2r  String will not become slack hence kq2  m2r v = 4gR = 4 10 0.9 = 3.6 10 r2 T = vmin = 6 m/s sin  2. Ans. 5 or we can say that tension is MR 2 aa more than electrostatic repulsive force Sol. FR = R also T is constant if  is constant 2 mv2 kq2 mv2  T sin  = , prev T sin  = r2  r r  v should be increased HS-2/13 1001CT103516005

 = 50 rad/s2 Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 R = 10 m/s2 6. Ans. 6 s  1 ×R × t2 = 1 10 22 = 20 m Sol. u2 sin2   10   = 45° ; 22 2g 3. Ans. 2 T= 2u sin   2 2 g 1 2 2 2 2 Sol. T  2 1 x + y = 1 + 1 = 2  S = 20 cos 45 × 2 2 + 2 = 48 1g 7. Ans. 6 4. Ans. 1 Sol. N = Tcos Sol. A= QE = 104 100 = 10–5 m µN = Tsin k 103 N k  A = 1000 × 10–5 m/s µN m 0.1 Vmax = A= = 10–3 m/s = 1 mm/s  T  tan = µ = 3/4 5. Ans. 4 d Sol. E = 6iˆ  8ˆj u2 sin = 3 = 2 20 ^j 1.6 sect. 5 5   E a  q = 6ˆi  8ˆj  d = 6m = 600 cm m 8. Ans. 2      = 20ˆj  6tiˆ  8tˆj =(8t – 20) ˆj + 6tˆi Sol. When the maximum speed is achieved, the v u at propulsive force is equal to the resistant force. Let F be this propulsive force, then v2 = 36t2 + 64t2 + 400 – 320t = 100t2+ 400 – 320t F = aV and FV = 400 W Eliminating F, we obtain For minimum value dv  0 t = 1.6 sec. dt V2 = 400 = 100 m2/s2 Hence minimum value of v a v2 = 100 × (1.6)2 + 400 – 320 × 1.6 and the maximum speed on level ground min with no wind  vmin = 12 m/s v = 10 m/s 1001CT103516005 HS-3/13

Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 PART–2 : CHEMISTRY SOLUTION SECTION - I 2. Ans. 7 1. Ans. (C) 2. Ans. (A,B or A,B,D) k1  A1 e  Ea2  Ea1  k2 A2  1000  Ssys= nRln2 Ssys= (½)(2)(ln2) = 0.7 cal K–1 1100 = e Ea2 Ea1  Ssurr= –0.7 cal K–1  1000  q = (0.7) (300) = 210 cal   sys 10 qsurr= –210 cal 1000 = e Ea2 Ea1  Stotal= 0  1000  3. Ans. (B,C,D or B,D) ln10 =3  Ea 2 Ea1  4. Ans. (A, B, C)  1000  5. Ans. (A, B, C, D) 3ln10 1000  (Ea2  Ea1 ) 1000 6. Ans. (C) 3 × 7  (Ea2  Ea1 ) 3 7. Ans. (A, B) [Ea – Ea = 7] 8. Ans. (A,C) 21 3. Ans. 8 9. Ans. (C) A  2B 10. Ans. (A,C,D in E & A,C,D or B in H) t=mol 4 – SECTION - II t=mol 4–x 2x t = 4min 1. Ans. (A)-(P); (B)-(T); (C)-(P,S,T); rt = 20 (4 – x) + 2x (–10) = 0 (D)-(Q,R,T) 80 – 20x – 20x = 0 2. Ans.(A)(P,T); (B)(Q,T); (C)(R,T); 40x = 80 (D)(S,T) x=2 SECTION - IV t3/4 = 2t1/2 = 8 min 1. Ans. 6 1 4. Ans. 2 W = – (5 + 3) (1) = –4J 5. 6. Ans. 0 II 2 (U)I = (U)II Ans. 3 (U)II = q + W 7. Ans. 9 II II 8. Ans. 5 2 = qI + (–4) q = 6J II HS-4/13 1001CT103516005

PART-3 : MATHEMATICS Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 SECTION-I SOLUTION 1. Ans. (A,B,D) 10 200  x x  100 2 5 8 A is wrong  ƒ  x    x x 100,100 10 but B and C are true otherwise ƒ(x) = 10 x  200 x  100 will have one more root in (10,12] 4. Ans. (A,C,D) y y=1  x xn 1 –200 0 200 x ƒn x  e x1 y = –1 x for |x| < 1 g x  ex1 g'x  x x 1 e x 1  12 2. Ans. (A,C,D) 5. Ans. (A,B,C,D) 2(a – b) = ab + ab a – 3b = ab Let 3x  4  t  ƒ t  2t 10 b a 3x  4 3t  3 3a ...(1) 2x 10 8 2 3x  3 3 3  ƒ x dx   dx   n|x  1| x  c 2ab = a ab a  8,b  2 b 33  2ab2 = a + ab – b2 6. Ans. (A,B,C) b2(2a + 1) = a(1 + b)  a2 2a  1  a 1  a  by (1) y2  ex  y   3  a2 3 a y '  ex  y2  y 2y  1 2y 1 a(2a + 1) = (3 + 2a) (3 + a) 2a2 + a = 2a2 + 9a + 9 also y2 – y – ex = 0 8a = –9  a   9 y  1  1  4ex  2y  1  1  4ex 8 2 b   9/8  3 7. Ans. (A,C,D) 15 / 8 5 AP is 15 , 27 ,3  9 ,   9  3 for A,B x 1  1 1/5 1 dx 8 40 5 8  8 5  x4 x6   75 , 27 ,  21 ,  69 ,117 let 1 1  t  dt  4 dx 40 40 40 40 40 x4 x5 3. Ans. (B,C)  t1/ 5 dt  1 t6/ 5  5 t6/ 5 4 4 6 / 5 24  (C)  tan2 x 1 tan x 1dx so D is wrong tan x  12 25 8 10 C 2 1001CT103516005 HS-5/13

Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1  (D)  tan x sec2 x  sec2 x dx  89  3 89 sin2 2k 1 tan2 x  tan x  C 4 k1 2 = 89  3 [(sin22 + sin292º) + (sin24º + sin294º) or sec2 x  tan x   C  1  4 2 2  + ......... + (sin288º + sin2178º) + sin290º] 8. Ans. (B,D)  89  3 45  356 135  221  cos1 tan1 x  0  tan1 x  1  x  tan1 4 44 sin1 x2  cos1 x x  0  S  221  27 5 88 x2  1  x4  x4  x2 1  0 / 2   5.3.1 .   15  5  x2  1  5 cos6 2 0 6.4.2 2 96 32  2  5 1  2  2 sin18 SECTION – II 2 9. Ans. (A,B,D) 1. Ans. (A)(P,S); (B)(Q); (C)(R,T); (D)(R,T)   2x2  2x  1 ex2dx (A) y = –2n|x| as |x| < 1   ex2dx   x  1 2xex2dx   ex2dx  x  1 ex2  ex2 dx  y'  2  y '   1   4   2   x  1 ex2  C x (B) ƒ 1  1  g  1   1 2  2  g \"  y     ƒ\"x  ƒ x  x  1 ex2  ƒ(0) = 1 ƒ 'x3 (A)  ƒ x ex2 dx  x  12 C g \"  1    ƒ \"1  2  2  ƒ '13 (B) ƒ(1) = 2e ƒ '  3 x2 1 2  1  x2   (C,D) lim ƒ x 1/ x  lim x  1 ex2 1/ x also x 0 x0 2x  lim 1  x1/ x ex  e  ƒ \"  3x  1  x2 2 x 0 10. Ans. (A,B,C) 89 ƒ'(1) = 2, ƒ\"(1) = 3  1  5 2 2 Let  cos6 k  S 5 k 1 g \"  1    2  5 89  2  16 23  S   sin6 k k 1 (C) ny = xxnx  89 1 y '  xx 1  nx nx  xx  y '1  1 2S   cos6 k  sin6 k yx k1 yy \" y '2  89  y2   1  3sin2 k cos2 k 1001CT103516005 k 1 HS-6/13

Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 xx 1 nx 2 nx xx  1 2nx  SECTION – IV  x x      1. Ans. 0 nx < x xx 1  nx xx 0  x100nx  x101  x  x2 at x = 1  y\" – 1 = 0 + 1 + 1 – 1  y\" = 2 e4 x tan 1    sin2 x  e4x tan1    sin2 x   3   3  /2 & lim x101 0  limit is 0. ex 4x (D) tan3 x 16 sin4 x cos4 xdx 0 2. Ans. 7 x 8 / 2 x + 1 = 99 sin((x + 1)) 0 16 sin7 x cos x dx  2  sin  2  Also (S) e4 nx dx  1 n2x e4  4 1 23 98 99 1 2x 4 1 3 t  sin t (T) H.M of 3 and is 2 99 98 solution from (0,99] 2 98 solution from [–99,0) and t = 0 2. Ans. (A)(P,S); (B)(P,Q); (C)(R,T); so total 197 solution n – 190 = 7 (D)(P,Q,R) Ans. 2 (A) 4a + 5 = 4b + a2 ...(1) 3. (x + 1)3 + 2016(x + 1) = sin2   1 4a = 2b  b = 2a ...(2) 2 a2 + 4a – 5 = 0  a = –5,1 (y + 1)3 + 2016(y + 1) = cos2   1 2 (B) 1 xƒ ' 2x  1  1 ƒ 2x  1 2 0 4 0  1 ƒ'2  1 ƒ2  ƒ0  4 (x + 1)3 + (y + 1)3 + 2016 [x + y + 2] = 0 24 (C) 4.  (x+y+2)[(x+1)2 + (y+1)2–(x + 1)(y +1)+2016] = 0  x + y + 2 = 0  x + y = –2 ƒ(x) = 0 Ans. 5  will have at least 4 roots but may have any even numbers roots greater 1  2 cos 2  sin 3 than or equal to 4. 3 3sin  n sin 3. x k1 3k    x  lim x 3k n 3 sin 1  x2 x4 x6  x3 x5  sin x sin x sin x xn  2 3 ...  x  x 3! 5! 3sin x 3 sin 3n1 (D) lim      ...    lim 3 .....  3 x x0 n 32 x 3n 3 sin lim 1 x4 1  1  x6 1  1  ....  lim sin x sin x xn   6 2   3 5!   x 0 n 3n sin x 3n x exist for any n  {1,2,...4} y = |sinx| + ||x– 2| – 1| not diff. at ,2,1,2,3 1001CT103516005 HS-7/13

5. Ans. 4 Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 ƒ(4) < 0  4a – b < 4  ,1  1, 6. Ans. 3 and ,2  1, ...(2) 1  z 1  x 1  y by (1), (2) z 1  x  x 1  y  y 1  z [–3,–2]  [1,3] 3  5 integers in the domain.  1  z 1  x 1  y 1/3 8. Ans. 2   z 1  x  x 1  y y 1  z      x2 2x2  3x  3 dx 2  2x  2  x  1 req. exp > 3 x2  2x  2  dx x2  2x  2 7. Ans. 5 12 1 2x  2 dx 2 x2  2x  2 x2 < 9 ...(1)  2 x   1dx   and 1  log 2  x x 1   1  2 x 1 x2  2x  2  1 n x 1  x 12 1      2  2  x  1 and x 2  x2  2x  2  c  2 x 1 x 1 x  1 x 2 0  x x2  2x  2  n x  1  x  12  1  c x  1 x 1   0 and HS-8/13 1001CT103516005

Paper Code : 100 1CT103516006 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 : 25 - 09 - 2016 PAPER-2 PART-1 : PHYSICS SOLUTION SECTION-I 10. Ans. (A,C) 1. Ans. (B,D) 11. Ans. (B,C) 2. Ans. (A,B) Sol. U(x) = –ax2 + bx4 Sol. p = mv = (x)v =   at2  at F =  dU  0  2  dx p = 1 a2t3  a 1/2 2  x = x0 =  2b  F = dp  3 a2t2 Umin = U(x0) = a2 dt 2  4b t  d2U   W = Fdx  Fvdt Angular frequency :    dx2  0 x x 0 t  3 a2t2    m 0  2  W = at dt 12. Ans. (A,D) W = 3 a3t4 SECTION-IV 8 1. Ans. 8 3. Ans. (C,D) 2. Ans. 9 4. Ans.(A,C) 3. Ans. 2 Sol.   cm  0 y2  2 y1 net 3  4. Ans. 2 FR 2  FL 2  0 .......(i) 5. Ans. 9 FR + FL = mg .......(ii) 5. Ans. (A,B,D) vy M 6. Ans. (B,C,D) TT 7. Ans. (A,B,C,D) Sol. vy 8. Ans. (A,B,C,D) vy 9. Ans. (A,C,D) vx vx 1  mAmB  v02 vy = v/3 2  mA  mB  Sol.   2mAgL 1 mv2 1 3m  2  1 mv2x  2 2 y  2    v  2 v0 = 2 gL   mA  v 1 mB  vx = 3   Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-9/13 +91-744-5156100 [email protected] www.allen.ac.in

Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-2 2T 6. Ans. 4 Acceleration of M is a0 = m Sol. Kmax = U In the frame of M Kmax = Uf – Ui T  ma0 K = Kmax = mg 2 5 + mg  vx 4 22 2  mg x =2 2 T+ ma0 = mv 5 1  mv2  mg 3T = 3 K= 4 10  2 mv2 7. Ans. 2 T  9 8. Ans. 5 PART–2 : CHEMISTRY SOLUTION SECTION-I SECTION-IV 1. Ans.(C) 1. Ans.(3) 2. Ans.(A,B,C or B,C) 2. Ans.(2) 3. Ans.(A, B, C, D) 3. Ans.(3) 4. Ans.(Bonus) q = nCmT 4 = 2C (–2) 5. Ans.(C, D) m 6. Ans. (A,C,D) Cm = –1 7. Ans. (A,C) 2 8. Ans. (B,C,D) – 1 = 3 + 1x 2 9. Ans.(C) –4 = 1  x –4 + 4x = 2 10. Ans.(D) 4x = 6  dA  2 dB  3 dC x  3 dt 3 dt 4 dt 2  2K1  3K2  t  ln  A0  2x = 3  At    t1/2 = 1/8 4. Ans.(0) % yield of B = 3K1 × 100 5. Ans.(3) 3K1  4K2 6. Ans.(5) = 300  300 7. Ans. (5) 38 11 11. Ans. (A,D) (II) , (III) , (IV) , (V) , (VI) 12. Ans. (A,D) 8. Ans. (4) HS-10/13 1001CT103516006

Leader Course/Phase-III, IV & V/25-09-2016/Paper-2 PART–3 : MATHEMATICS SOLUTION SECTION-I apply king 1. Ans. (B,C,D)     2017 1 2017r cos2017r x sinr x dx Put x = 1 in the given quadratic equation, we get I  ar. (2 – 2a2 + b) + 2p(1 – 2a) = 0  p R 0 r0  if r  even I = 0  a  1 ,b   3 5. if r  odd I = depends upon a2r+1 2 2 Ans. (A,B,C,D) (A) (a – b) = 2 (A) Apply intermediate value theorem p2  2p  b 7 (B) ƒ(e) = 1; ƒ(e+) = 1; ƒ(e–) = 1 p2  2 4 (B)  1 1  1  p  (C)  = 13,   1    1  10 3 (C) sin + cos = –1    ,   , , 3 10 sin 2x 10 5 2 2   4  tan x  cot x2   (D) 4 2 (D) t + –t = 2  x = 0 (only solutoin) 2. Ans. (B,C,D)    5 ,    2  10  3k  2  3k  1    tan 1  1  3k  2 3k 1  6. Ans. (B,C)   sin1.sin 2......sin 44.sin 45  1 k1  cos 46.cos 47.........cos89  2 P   10 7. Ans. (A,C,D)    tan1 3k  2  tan1 3k 1 y k 1 0,127  6  ƒ(x) = x2 – 6x + 17  13  2   tan1  y=1 O x (0,0) 3. Ans. (A,B,C) V 3,–21 y  e x  e x  y '  e x  e x ...(A)  For solution to exist, 2x x2  6x  17  k , where  y'  y2  4 ...(B) 2 2x k  [–1,1] also y  e x  e x 8. Ans. (A,D)  2 2  8  ye x  e x  1  + 2d = 8  y'e x 2 e x  + 3d =   x  ye 8  2 3 2x 2x    y '  2e x  y  + 2 = 24 2x ...(C) on solving we get 4. Ans. (B,C) =8  = 12  = 8 or  = 6  2017 =8  = 10 0 r 0  I  ar . cos2017r x. sinr x  given    dx  = 12,  = 6,  = 10  > .......... A 1001CT103516006 HS-11/13

Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-2  x17 dx Let x18 = t, 18x17dx = dt 12. Ans. (A,B,D) dx x36  16  xx  2x  4x  6 1 tan1  x18  ..........(C) 72  4  ..........(D) dx Paragraph for Question 9 and 10    x2  6x x2  6x  8 9. Ans. (A,B,D) 1 dx 1 dx  8  x2  6x  8  x2  6x  8 1 dt 1 n x  6  1 n x  4  C p  0 t2  2t cos   1 48 x 16 x  2  A = 48, B = 16  1 dt  SECTION – IV t  cos 2  sin2  1. Ans. 6 0 Either c2 = 3 or c2 = 3c – 2 or (3c – 2) = 7  1  tan 1 t  cos  1  sin   sin  0    c   3,1,2,3 1 tan1 1  cos      sin    sin    tan 1 cot 2. Ans. 4  x 1 n 1  1  tan1  cot    tan1 cot    ey  x  2  n 2 n  3, 4, 5,......., 99  sin   2      99  n  n  1  n3 n  n  1    y   n  2   y   n  2  2sin   n 98  4,5  S q  3 t2 sin 2t dx  0  [S] = 4 3 t2016 1 3. Ans. 6 odd function 7(a + (a + d – 1)d) = 4(a + (a + 4d – 1)) lim      0  q  Putting a = 1, we get d = 1  2 sin  0 lim0    4  10   10  ar 2 sin   0  q  3  aar     1  r 1 r 2 lim  1   1, lim0 1   2  4 75  64  100  64  36  6  p  p  0 3 10. Ans. (A,B) 4. Ans. 2 x2  2  0  x  2 sin  C   ,sin A  4 ,cos A  3 2sin   255 Paragraph for Question 11 and 12  sin B  3 ,cos B  4 55 P(x) = x3 – 1 = x(x – 2) (x – 3) (x – 4)(x – 6)  expression = 2 11. Ans. (B,C,D) 5. Ans. 9 9 I  /2 8 sin2016 x  5 cos2016 x  x x  2x  3x  4 x  6 1dx 0 sin2016 x  cos2016 x 3 apply king and add apply king and add 9  / 2 13 2I   2dx  I  12 2I  13dx  I  3 04 P(5) = 96, P(–1) = –420 a = 13, b = 4  a – b = 9 HS-12/13 1001CT103516006

6. Ans. 2 Leader Course/Phase-III, IV & V/25-09-2016/Paper-2 in x  (2,3) 8. Ans. 2 ƒ(x) = 2x–2 + (x – 2)2 ƒ'(x) = 2x–2n2 + 2(x – 2) diff. both sides LHS : ƒ '  5   2n2  1    2 –cosec2x(1 + cotx + cot2x +.....cot98x)  2  + 1 + cotx + cot99x + cot100x 7. Ans. 1    1  cot2 x cot99 1    1  cot x 1  cot99 x 1  cot x ƒ x  x3  3x  ƒ–1(0) = 0  2 cot2 x cot97 1  d 11  1  cot x dx x0  ƒ '0  3 ƒ1 x 1 1 RHS :   cot2 x cot97 x  1 2 3 ƒ1 h  tan1 h 2 1  cot x 1 3 lim sin1 h  lim  h0 h0  ƒ1 h  tan1 h   lim   1  1 h0 sin1 h  1001CT103516006 HS-13/13