14- Solution Report (14)

Paper Code : 1001CT103516013 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III, IV & V) ANSWER KEY : PAPER-1 TEST DATE : 14-01-2017 Test Type : MINOR PART-1 : PHYSICS Test Pattern : JEE-Advanced Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. D B D A DDCBBB Q. 11 12 13 14 15 A. B,D A,C A,B A,C C SECTION-IV Q. 1 2 3 4 5 A. 5 3 2 5 1 PART-2 : CHEMISTRY Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. C C B D DCBDCD Q. 11 12 13 14 15 A. Bonus C,D A,B,D A,B,C A,C,D Q. 1 2 3 4 5 SECTION-IV A. 7 3 4 8 7 PART-3 : MATHEMATICS Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. B C B C Bonus C B A C C Q. 11 12 13 14 15 A. A,B,C A,C,D A,B,C A,B,D A,D SECTION-IV Q. 1 2 3 4 5 A. 6 2 2 2 5 Paper Code : 1001CT103516014 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III, IV & V) ANSWER KEY : PAPER-2 TEST DATE : 14-01-2017 Test Type : MINOR PART-1 : PHYSICS Test Pattern : JEE-Advanced Q. 1 2 3 4 5 6 7 8 9 10 A. D C B A C B B A C B SECTION-I Q. 11 12 13 14 15 16 17 18 19 20 A. C B D A A B B C C D PART-2 : CHEMISTRY Q. 1 2 3 4 5 6 7 8 9 10 A. B D C D C B C B C B SECTION-I Q. 11 12 13 14 15 16 17 18 19 20 A. A C C D C A B C D B PART-3 : MATHEMATICS Q. 1 2 3 4 5 6 7 8 9 10 A. A C C D A C C B C D SECTION-I Q. 11 12 13 14 15 16 17 18 19 20 A. C D B A Bonus A D A D C

Paper Code : 100 1CT103516013 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 : 14 - 01 - 2017 PAPER-1 PART-1 : PHYSICS SOLUTION SECTION-I 8. Ans. (B) 1. Ans. (D) Sol. Sol. Taking m & M in the system,  phase . Fnet horizontal = 0 compression in spring = 0 2. Ans. (B) P /2   9. Ans. (B) Sol. C Sol.   Biˆ  Bˆj  Bkˆ  Bnet  3B  30i Bnet 2R 10. Ans. (B) P  1  C  1  v  P1 P2 2 2  r  3. Ans. (D) Sol. Pt = 1 mv2 v is doubled  t is 4 times 2 2 19 2 Blv Sol. 4 cm Blv 4. Ans. (A) Sol. Let they will be in phase after time 't' then t  t  1  t  21 s Q1 Q2 B = 1 Te s la 37 2 8 5. Ans. (D) p   Blv  0.1mA Sol.  E p2 20 Torque    then torque on 0 due to 11. Ans. (B, D)   Sol. At this point particle will lie on the y-axis  = p2  E1 but  E1 so   and moving in horizontal direction. p1 p2 21 6. Ans. (D) Sol. Internal loss = i2r i  12. Ans. (A,C) rR Battery cap. v COM v 2m battery life time = Current Sol. m 7. Ans. (C) VCM  2m  v  m(v)  v 3m 3 E 1  ER1  2 R1  R2 , E1 = 2  R1  R2  Sol. I= C1   , The blocks will come to rest in COM frame at maximum extension. 1  ER2  2 E1 R12C1  They both will have speed v/3 i.e. same 2  R1  R2  E2 R22C2 E2 = C2    as that of COM. Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/12 +91-744-5156100 [email protected] www.allen.ac.in

During this motion they both will sometime Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-1 SECTION-IV v have velocity in COM frame and at that 1. Ans. 5 3 Sol. The force required to keep the double time their individual absolute velocity will pulley system in equilibrium can be determined from the torque balancing. be zero. 13. Ans. (A, B) P P 2 2 14. Ans. (A,C) r2  F  r1 Sol. For given situation q  L di  0  F  P r1  r2  = 10g 0.1  5N C dt 2 2 1 d2q q d2q  dt2  LC  0  dt2  2q  0 2. Ans. 3  q = q0 cos t & i = –q0sint  h 2 According to given conditions  Sol. K  P2     2m 2m q2  1 Li2 2C 2 K  h2  9 8L2m q0 cos2 t 1  2C  2 Lq022 sin2 t 3. Ans. 2  cot2t = 1 Sol. Initially let no. of molecules of A = NA and that of B = NB  3 5 7 t = 4 , 4 , 4 , 4 ......... N A  25  t   LC , 3 LC , 5 LC , 7 LC s ..... NA  2NB 100 44 4 4  100 NA = 25NA + 50NB NA = 2 N B 15. Ans. (C) 3 Sol. t =  N 'A  NAet  2 NBet Now, 3 N'B  N e2t B 2 N et 75 3 3   B 100 4 E  2 NBet 2N B e2t 3 E 2E  3R 3R i   2 13  1  3et  4 1 = 9 e t  e t  9 v YZ  R  2E  2E So, t = n 9 hr  2n 3 hr  2hr 3R 3 n 3  t=0 4. Ans. 5 RR Sol. By Kepler's law, T2R3 5. Ans. 1 i E n 2  1 n 2R  2 E Sol.     1 vXY = 2  1 n   2  HS-2/12 1001CT103516013

Leader Course/Phase-III, IV & V/14-01-2017/Paper-1 PART–2 : CHEMISTRY SOLUTION SECTION-I 7. Ans. (B) 1. Ans. (C) 8. Ans. (D) 9. Ans. (C) 2. Ans. (C) 10. Ans. (D) Ni2+ + 2e–  Ni(s) 11. Ans. (Bonus) 12. Ans. (C,D) nF = 2 nNi+2 13. Ans. (A, B, D) nF = (126.5)(60)(5.15) = 2  M  225 14. Ans. (A,B,C) 96500 1000 15. Ans. (A,C,D) (7590)(5.15)(1000) SECTION-IV M = (96500)(450) 1. Ans. (7) 2. Ans. (3) (759)(51.5) 3. Ans. (4) = (965)(45) 4. Ans. (8) 5. Ans. (7) = 0.9 M 3. Ans. (B) 4. Ans. (D) 5. Ans. (D) 6. Ans. (C) PART-3 : MATHEMATICS SOLUTION SECTION-I I =  + I1 1. Ans. (B) I1 x2  (1 cos x  dx x+y+1>0&y–x+2>0 0      s in x)2  Squaring x  1 I 2 II   I1 x2 s in x)   x dx  (1   2 0 0 (1  s in x) A B (2,0) (0,0)  1 ,0  2  dx  2  I1 0  s in   (1 x) C  /2 dx   /2 dx (1  s in x) (1  cos  1 3  2 2 2 2  2 2  0 0 ,  I1   x) 133 9  /2 x dx  x  /2  2  2 0 Area (ABC)    2   s in2  2 2 tan 0 222 8 I1  2. Ans. (C) I1 = –2 + 2  RHL = sgn cot1 h  cot1 h2  sgn(ve)  1 4. Ans. (C) h > h2  cot–1h < cot–1h2 iˆ ˆj kˆ LHL = sgn(cot–1(–h) – cot–1h2) = sgn(+ve) = 1 Line has direction 2 3 1  iˆ  ˆj  kˆ 3. Ans. (B) 3 5 2 LetI   x2 (1  cos x)dx A(2,–3,) 0 (1  s in x)2  x2dx  x 2 . cos x dx 0 (1  s in x)2 0  s in x)2    (1 B(3µ+2,1–5µ,2µ–2) 1001CT103516013 HS-3/12

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-1 2 – 3µ – 2 = 5 – 3 – 1 =  + 2 – 2µ 9. Ans. (C) Solving 31 & 19 xx1  yy1 1 & xx2  yy2 1   a2 b2 a2 b2 3 3 Point B is  21,  92 , 32  m1  b2 x1 & m2  b2x2  3 3  a2y1 a2y2 5. Ans. (Bonus) m1m2 = –1 a  b  c  x1x2  a4 b  c (c  a) (a  b) y1y2  b4 a = (b – c) .........(1) 10. Ans. (C) b = (c – a) .........(2) x, y & x  y are linearly independent c = (a – b) .........(3) a2 + b + c = 0 a2 + b + c = 0 (1) + (2) + (3)  a + b + c = 0 a2 + b + c = 0 Quadratic becomes ax2 + bx + c = 0 | | a2  b2  c2 || 2(a2  b2  c2 )  2(ab  bc  ca) which is an identity and a + b + c = 0 By squaring a + b + c = 0 11. Ans. (A,B,C)     (a2  b2  c2 )  2 ab  bc  ca 21a  41b  54c  a(x1  x2  x3 )   | |= 1 b(2x1  3x2  x3 )  c(3x1  4x2  x3 ) || 3 6. Ans. (C) By comparison x1 + x2 + x3 = 21 (z – 1)(x – i)(z + 2i) = 0 2x1 + 3x2 + x3 = 41 3x1 + 4x2 + x3 = 54 z1 = 1, z2 = i, z3 = –2i Re(z1) + Re(z2) + Re(z3) = 1 By solving x1 = 6, x2 = 7, x3 = 8 7. Ans. (B) 12. Ans. (A,C,D) a  k ; b  k , c  k  m n (1,1) 1 k4  1 1  1   k2 Area = 2  2m2  m2n2 n22  2mn One solution x=2  2  m2  n2 1  3 8. Ans. (A) 2m2n2 (A) Let integral root is , where,  > 0 3 432 mn  1 3 = positive or zero 33  &  = (3 – a2 – b – c) 1  3 3 mn So  >  cotradicts (B) Let  < 0 and  = –µ ......(1) k2 3 3k2 µ4 + µ3 – µ2 + µ –  = 0 2mn  2 µ4 + µ2(µ – b) + (µ – ) = 0 13. Ans. (A,B,C) (y – 2)2 = 4(x – 1) Since µ is positive y2 = 4x a=1 µ –  > 0 & µ –  > 0 equation (1) is not possible. HS-4/12 1001CT103516013

Leader Course/Phase-III, IV & V/14-01-2017/Paper-1 (B) Direction is x – 1 = –1 3. Ans. 2 x=0 Let lines for first equation is (C) Cir cle x2 + y2 = 4 cuts at two different points. (y – m1x – c1)(y – m2x – c2) = 0 y2 – x(m1 + m2)xy + m1m2x2 + x(m1c2 + c1m2) (D) (y – 2)2 = –4(x – 1) – y(c1 + c2) + c1c2 = 0 y2 = –4x common tangent is one. By comparison m1m2  a , m1  m2  2h b  14. Ans. (A,B,D) b and c1c2 c  (t2,2t) b (–5,2) tan   2( h2  ab ) ; sin   2( h2  ab ) Middle point is h  t2  5 & k  t  1 (a  b) (a  b)2  4h2 2 Another pair has lines 2h = (k – 1)2 – 5 2h + 4 = k2 – 2k y – m1x + c1 = 0 & y – m2x + c2 = 0 (B) Put k = 1  h5 Area = P1P2  4|C1C2 | (a  b)2  4h2 2 sin  1  m12 1  m22 (2 h2  ab ) (C) When x = 0, y2 – 2y – 4 = 0 4. Ans. 2 5. D>0  b  4  b  8a 2a (D) When y = 0, x = –2 4ac  b2  2  c  2  b2 15. Ans. (A,D) 4a 4a (A) ƒ(x) > 0, ƒ'(x) > 0 & ƒ''(x) > 0 c = 2 + 16a (B) ƒ(x) > 0, ƒ'(x) < 0 & ƒ''(x) > 0 (C) ƒ(x) < 0, ƒ'(x) < 0 & ƒ''(x) < 0 (D) ƒ(x) < 0, ƒ'(x) > 0 & ƒ''(x) < 0 ƒ(a) = abc = –8a2(2 + 16a) SECTION – IV ƒ(a) = –16a2 – 128a3 1. Ans. 6 d  ƒ '(a)  0 when a  (1,3] da x2 + y2 + z2 = x2 + y2 + 4 ƒ(1) = –144 xy x2  y2  1  1  1  1  6 ƒ(3) = –3600 xy xy xy xy Difference = 3456 2. Ans. 2 Ans. 5 Case-I : when y < 0, then 3x 1  0 x(3x  2) x.10 cos   y.10 sin   1 49 – +– + 0 1/3 2/3 y cos   y sin   1 (1 / 5) (9 /10) Case-II : when y > 0, then (3x  1)  0 x(3x  2) 4 2   9 2 (1  e2 )  10    10  x  ( 1, 0)   1 , 2  1  e2  16  3 3  81 1/3 –1 0 2/3 1 1 32 e2  65 3 81 1001CT103516013 HS-5/12

Paper Code : 100 1CT103516014 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 : 14 - 01 - 2017 PAPER-2 PART-1 : PHYSICS SOLUTION SECTION-I 4. Ans. (A) 1. Ans. (D) 5. Ans. (C) Sol. Choosing an element of width dx of the Sol. Increasing frequency increases the energy triangular conductor at a distance of x from of the incident radiation whereas increasing its vertex, area of strip intensity increases the no. of radiations incident on the surface. dA   bx  dx  h  6. Ans. (B) Sol. J = E, i = JA = EA Writing d = Bda = {µ0Ibx/2h (a + x)}dx Intergrating between zero and h, we get q Now, E  M    0b {h – a ln| (a + h)/a|} I 2h K0A  i  q = 1.2 × 10–8 H K0 2. Ans. (C) At, t = 0, i = 1 A, K = 4.26 7. Ans. (B) E W1 Sol. W2 Sol. W3 E W3 will be maximum because charge moves opposite to electric field Since Enet = 0 inside cavity  Eind has to be W1 will be minimum because charge moves downwards. along electric field 8. Ans. (A) So W3 > W2 > W1 Sol. Since TP = 0 3. Ans. (B)   = constant P Sol. I  10 4  4 104 103 but 4104  120 ; 4 × 10–4 x + 4 = 120; x  104 x  116 104 ; x = 290 k f3 N 4 HS-6/12 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/14-01-2017/Paper-2 9. Ans. (C) k e   a0  r  a 4 k e  a0 Sol. Due to rotation of disc in M.F.,    r v  0  a30 v E.M.F will be induced between centre and a0 2 a0  r 3 rim of discs. k  e  a0  r e v a0  v r  a0  r       a0 2  Magnetic field at centre B  N0I k  1 1 r  2a  v  a0   Bb2 b2N0I  e a0    r    2 v     2 2a  a0 r 2 E.M.F. between 1 & 4 = 2 = N  0 b2 I  1  2a     a0 v 10. Ans. (B) a0 v  R   Sol. C 4 G 13. Ans. (D) D 1 Sol. For nuclei having N  1 , nucleus regains D' Z C' its stability by I If no current flows through G p  n + e+ + v then IR = D + D' 14. Ans. (A) IR  N0b2I Sol. Equation for k-capture is given by 2a p+ + e–  n° + v 11. Ans. (C) 15. Ans. (A) 16. Ans. (B) 1 17. Ans. (B) Sol. F  r2 and 18. Ans. (C) 19. Ans. (C) F e –t Sol. P   = 2 × 10–6 × 2 × 106 = 4 sec. –. r Q   = 2 × 3 = 6 sec F e V –  .r  F  e v so F = k. 1 r r2 ev 12. Ans. (B) m02a0  k e  a0 a20 v Sol. ... (i) R  = 1  1  100 When radius is changed to a0 + r (r << a0) LC 3L  2L 23 ma020  m a0  r 2  (conserva tion of I0 S  I  1.33 angular momentum) ... (ii) k e  a0  r   m2 a0  r v      Fnet  2 a0  r k e   a0 r  a40 02 20. Ans. (D) a0  r v a0  r 4 Sol. (P) U  6Kq2       m a0 r 2 R 1001CT103516014 HS-7/12

q q Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 2R R (Q)  12  5 10 Kq2  27 Kq2 4R 20R 20 R 54 Kq2 U = Uself + Uinteraction  40 R U  Kq 2  Kq 2   Kq  q 2R  4R   6 Kq2 9 2  2R  40 R U  Kq2  Kq2  Kq2 2R 2R R 4R (S) R 6Kq2 Wext = Uf – Ui U  3 Kq2  3 Kq2 6R 5 R 5 4R (R) U  3 Kq 2  Kq 2  Kq2 5 R 2R 2  2R  PART–2 : CHEMISTRY 5. Ans. (C) SOLUTION 6. Ans. (B) SECTION - I 7. Ans. (C) 1001CT103516014 8. Ans. (B) 1. Ans.(B) 9. Ans.(C) 10. Ans.(B) 2. Ans.(D) 11. Ans.(A) 12. Ans.(C) 3. Ans.(C) 13. Ans. (C) 14. Ans. (D) n  2nNa2S2O3 NaOCl 15. Ans. (C) 16. Ans. (A) n NaOCl = 0.1 1.95 = 192 105 17. Ans.(B) 2 1000 2 18. Ans. (C) 19. Ans. (D) = 97.5 × 10–5 20. Ans. (B) WNaOCl = 97.5 × 10–5 × 74.5 mass % = 97.5 105  74.5 100 1.356 975  745 102 = 1356 4. Ans. (D) HS-8/12

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 PART–3 : MATHEMATICS SOLUTION SECTION-I 4. Ans. (D)  1. Ans. (A) BA    1 iˆ  ˆj  = ^i + ^j + 2^k x1  P x3 + x2 + 1 = 0 x2 BC  iˆ  ˆj  3kˆ A(,1,1) B(1,2,1) x3 x1x2x3 = –1.    C(2,3,4)  g(–1) = 0. BA BC P  0 Also y = x2 – 1 = g(x)  1 1 0  1 1 3 0 1 12  x  y 1  ( – 1) (2 – 3) + 1(2 – 3) = 0  Equation with roots as g(x1), g(x2), g(x3) –( – 1) – 1 = 0 will be =0    3 2 5. Ans. (A) y 1  y 1 1  0 g x1  0 12 3  Min. no.of roots of ƒ(x) = 0 is 3.  y3  2y2  y  3  0 g x2  g x3   g(x1).g(x2).g(x3) = 3. 03  3 + 17g(–1) = 3. 2. Ans. (C) Min. no.of roots of g(x) = 0 is 2. x2 + (1 – 2)x +  – 3 = 0 Given equation is (ƒ'(x).g(x))' = 0 ƒ'(x) = 0 has minimum 2 roots x  2  1  1  8 2  ƒ'(x).g(x) = 0 has minimum 4 roots. k k 1 ƒ'(x).g(x) = 0 has minimum 3 roots. put   6. Ans. (C) 2 aˆ  bˆ  aˆ  c ...(i) 2 1  2k 12 aˆ  bˆ  c  aˆ x 2 Dot with   v  aˆ bˆ c   c2  aˆ .cˆ ...(ii) c  Roots will necessarily be rational. 3. Ans. (C) for (i) : dot with bˆ  0  aˆ.bˆ  bˆ.c  0  aˆ.bˆ  0 at x = 1 : y = 2 also y2 – x2y – 2x = 0  aˆ  bˆ ...(iii)  2yy' – 2xy – x2y' – 2 = 0  y' = 2 y  x2 for (i) : squaring x2  y y2  x      ƒy  C  aˆ  bˆ 2  1  2 0  c2 (by (iii))  c2  1 1  2 ...(iv) y  x2  ƒ 'y dy for (i) : dot with aˆ :aˆ 2  c.aˆ x2  y y2  x    dx ...(v) at x = 1; y = 2; dy  2  ƒ 'y  1 . put (iv) & (v) in (ii) : v=2–1=1 dx 30 1001CT103516014 HS-9/12

7. Ans. (C) Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 11. Ans. (C) ƒx  1  x2  ƒ 3  3 2  x2 6 12. Ans. (D) range  1 ,1   2   A  3 tan1 x dx   n3 8. Ans. (B) 1/3 x 2 a,b,c are in A.P. Solution for Question 13 & 14 2 = ab For minimum area A',B' and C' must be mid 2 = bc  ab, b2, bc will be in AP. points of QR, PR & PQ respectively. Solution for Question 9 to 10  A'(0,3,4), Q(0,6,0),R(0,0,8), C'(2,3,0) 5 7 P(4,0,0), B(0,3,0) 7  mL1   mL2 13. Ans. (B) 5 Volume  1 4  6  8  32 Let L2 : 7x + 5y = c  C  c ,0  ;D  0, c   7   5  6 14. Ans. (A)  equation of AD : x 5y  1;  Shortest distance is OB i.e. 3 7c Solution for Question 15 & 16 equation of BC : 7x y 1 c  x(x2 + (y–2)2 – 1) < 0 5 eliminating c, we get locus : x2 + y2 – 7x + 5y = 0 (–1,2)  S x2 + y2 – 7x + 5y = 0 /4  7 5  A  2 2  This is a circle with center ,  and 37 0 radius 2  Area  .37 sq. units. 15. Ans. (Bonus) 2 zA has min amplitude Also circle passes through origin  farthest point : (7,–5) OA  4 1  3 9. Ans. (C) 16. Ans. (A) 10. Ans. (D) Only point of intersection is (0,1) 17. Ans. (D) Solution for Question 11 to 12 (Q) The line makes equal angle with planes tan y   sec2 y   dy  y   0 put y = vx and hence with normals. x  x   dx x  dv  2a  2  36  a  4  36  tanv =–sec2v.x dx 14 14 dx sec2 v  a = 74; a = –2 (R) DC at x = 4.   x   tan v dv  nx  n tan v  nc  v y and at x =1, y   c = 1. x x 4 (S) ƒ x  x2   et ƒ x  t dt  nx  n tan y 0 x apply king property,  tan y 1  y  x tan1 1 x x  x ƒ x  x2   etx ƒ tdt x 0 HS-10/12 1001CT103516014

x Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 (Q) y2 = 4ax + b  2yy' = 4a  ƒ x  x2  ex. et ƒ t dt 0 ...(i) y '  2a  1  y  2a y  x  (2a)2 = 4ax + b  x  4a2  b 4a ƒ ' x  2x  ex .ex.ƒ x  ex. et ƒ t dt ...(ii) 0 by (i) & (ii) ƒ'(x) = 2x + ƒ(x) – (ƒ(x) – x2)  4a2  b , 2a   4a   ƒ'(x) = x2 + 2x  point of contact :  ƒ x  x3  x2  C  2a  4a2  b  2  b  8a  4a2 4a 3  ƒ(0) = 0  C = 0  k = 4.  ƒ x  x3  x2 (–4,7) P 3 Q (P) Let   r;  = r2 (R) (  are in G.P) 2..r  12  r  6 1  r ....(i)   r     16  r  r2  32 Least distance occur along PQ & PQ 23 3 is normal to parabola at Q. y = mx – 2m – m3  r 1  r  32 ...(ii) 7 = –4m –2m – m3  m3 + 6m + 7 = 0  m = –1. 3  Q(1(–1)2,–2(1)(–1))  Q(1,2) 16  PQ  25  25  5 2    5 . by (i) and (ii) : (1 + r)2 = 9  r  1;r  7 (reject) 33   = 24 ) B(x 2y 2 reqd GM  3 .r.r2  r A(x1y1) =8 18. Ans. (A) (S) C B mT at A =  b2 . x1 a2 y1 (P) mT at B =  b2 . x2 a2 y2 A(0,–b) mA.mB = –1 A lies on director circle of hyperbola  A lies on x2 + y2 = a2 – b2  b4 . x1x2  1  x1x2 a4  10.  0 + b2 = a2 – b2 a4 y1y2 y1y2  b4  2b2 = a2 1001CT103516014 HS-11/12

19. Ans. (D) Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 (P) 2ƒ(x).ƒ(y) = ƒ(x + y) + ƒ(x – y) (R) cos7x = 1 –sin4x at x = y = 0 : 2(ƒ(0))2 = 2ƒ(0)  ƒ(0)=1 x = 0 : 2ƒ(y) = ƒ(y) + ƒ(–y)  cos7x = cos2x(1 + sin2x)  ƒ(y) = ƒ(–y)  cosx = 0; cos5x = 1 + sin2x  ƒ(x) = ƒ(–x)  ƒ'(x) = ƒ'(–x) = 0.  cos x  1 sin x  0  x    ,0,  22  dx (S) 3  i  ac  bd  i bc  ad x2 1 x5 1   (Q)I ...(i) 0  ac  bd  3; bc  ad  1 1 0 t5dt tan1  a  tan1  c  1  ad  bc   b   d   bd  ac     x I      tan t  t2 1 t5 1    x5    tan 1   1       5  3  6 6 I dx ...(ii)   0 x2  1 x5 1 20. Ans. (C) 2I   dx  tan1 x   ƒ\"(x) = 6(x – 2) (i) 0 x2  0 2 + (ii)  1 ƒ'(x) = 3(x – 2)2 – 27 I  ƒ(x) = (x– 2)3 – 27x +  4 = x3 – 6x2 – 15x +  – 8 HS-12/12 1001CT103516014