6- Solution Report (6)

Paper Code : 1001CT103516017 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE-III to VII (SCORE-I) ANSWER KEY : PAPER-1 TEST DATE : 11-03-2017 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,C A,B,C B,C,D A,B B,C,D B,D A,B,D A,D C B Q. 11 12 13 14 15 A. A D A B C SECTION-IV Q. 1 2 3 4 5 A. 8 4 0 1 4 PART-2 : CHEMISTRY Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. A,C A,C A,B,D B,C,D B,C,D A,B,C A,C,D D B B Q. 11 12 13 14 15 A. D C D A D SECTION-IV Q. 1 2 3 4 5 A. 8 2 4 1 5 PART-3 : MATHEMATICS Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. C,D B,C,D C A,C A,B,C A,B,C,D B,C B,C,D B B Q. 11 12 13 14 15 A. C C C D B Q. 1 2 3 4 5 SECTION-IV A. 2 5 5 3 2 Paper Code : 1001CT103516018 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE-III to VII (SCORE-I) ANSWER KEY : PAPER-2 TEST DATE : 11-03-2017 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 . 2 2 4 1 4 3 2 4 1 4 2 2 2 1 1 1 4 3 1 2 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans . 2 1 4 4 2 1 3 4 1 2 3 2 1 2 3 3 4 1 1 4 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans . 4 4 1 4 4 2 1 1 4 3 3 3 1 2 3 3 2 2 1 4 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans . 2 3 1 2 2 3 2 4 3 4 2 2 3 3 1 1 4 2 3 2 Que. 81 82 83 84 85 86 87 88 89 90 Ans . 1 3 4 3 1 2 1 2 2 3

Paper Code : 1001CT103516017 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 : 11 - 03 - 2017 PAPER-1 PART-1 : PHYSICS SOLUTION SECTION-I 4m2 1. Ans. (A, C) Sol. Maximum angle which velocity vector at  2 3  2 8 3mg 9g any instant can make with initial direction of projection is 80° < 90°. So distance of 2 projectile from thrower always increases. 8 2. Ans. (A,B,C) T2  9  2 NT T1 2 13 Sol. A C 3 B Conserving angular momentum about W–T hinge, before and after the collision. Taking torque about A 1 m2  4 m2'  '   T(2) = (W – T)(4) 33 4 T = 2W – 2T   0  3g 2 3T = 2W ....(i) Also N+T=W–T '  0 new 9g N = W – 2T 8  2W  '   0 new 31 4   0 rew N = W – 2  3  = –ive  0   3 0  3. Ans. (B, C, D) 44 Sol. In case 2 requirement of centripetal force  0  0 requirement of centripetal force will not 12 new be fulfilled. In case 3, if requirement of 5. Ans. (B, C, D) centripetal force is fullfilled then, change Sol. Shielding property of conductor. in angular momentum will not be 6. Ans. (B, D) consistent. Sol. By Newton's thrid law F12 = F21 4. Ans. (A, B) So 11 = 22 2 7. Ans. (A, B, D) 3g Sol. T1  2 I  2 –2Q Q 4Q –3Q mg 0 –2Q 2Q –Q Q 3Q –3Q 0 1 m2  m2 Sol. A 2E E 3E B 3 T2  2  3  2m  g   4  dd d Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/13 +91-744-5156100 [email protected] www.allen.ac.in

Final distribution Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-1 SECTION-IV VA – VB = 0 0 –2Q 2Q 3Q –3Q 1. Ans. 8 A 2E 3E B Sol. L = 1 1 32 0.5 + (1) (0.5)(3)(3) d' d' 3 VA – VB  0 + (1)(2)2(0.5) 8. Ans. (A, D) = 1.5 + 4.5 + 2 = 8kg – m2/s Sol. R  V2 2. Ans. 4 P Sol.  = (4 × 106)(10–6) = 4S 3. Ans. 0 In series, bulb with higher resistance will Sol. No redistribution of charge takes place. glow more. In parallel bulb with lower resistance will glow more. So H = 0 4. Ans. 1 9. Ans. (C) i1 40V 10 Sol. In head-on elastic collision, velocities of Sol. V i2 5 0 particles will reverse in CM frame. 10. Ans. (B) i3 5V 5 Sol. sin (lab)max = V1C  m2  15  3 V  40  V  0  V  5  0 VCM m1 25 5 10 5 5 (lab)max = 37° V  40  2V  2V 10  0 10 11. Ans. (A) 5V = 50 V = 10 volt Sol. L sin  d  mg sin  Current in 5 is dt 10  5  1A L = mgl 5 5. Ans. 4 mg  = IS 12. Ans. (D) 510 0.5 /4  Sol.  = 2  100 2 rad/s 60 25  60 rad / s Sol. 400 = 25  60 60 =  rev/min 400 2   6400  0.15 sin 4 42400 25  60  60 0.6 {in radian} = 8002 rev/min = 11.4 rev/min nmax = 6.28  10.46 13. Ans. (A) 0.6 14. Ans. (B) 15. Ans. (C) n {3, 5, 7, 9}  2N + 1 = 9 N = 4 HS-2/13 1001CT103516017

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-1 PART–2 : CHEMISTRY SOLUTION SECTION-I 12. Ans. (C) 1. Ans. (A, C) 2. Ans.( A, C ) 13. Ans. (D) H = 3 × (–390) – 2 (–820) 14. Ans. (A) = 470 kJ/mol 15. Ans. (D) G < 0 H – T S < 0 for spontaneous reaction  T S > H SECTION - IV (S > 0) 1. Ans. 8 3. Ans. (A, B, D) We know 4. Ans. (B, C, D) K t10 5. Ans. (B, C, D) Kt = Temperature cofficient 6. Ans. (A,B,C) K 35 2  ln 2 I. K25 = 2  K35 = 40 t1/2 = 20 min. 35º No. of half lives = 3 7. Ans. (A,C,D)  fraction of A left = (1/2)3 8. Ans. (D) II. K 35 ln 2 9. Ans (B) K25 = 3  K35 = 3 × 30 8 t1/2 = 10 min. [HX] = 80 × 1000 = 1 M 35º 100 No. of half lives = 6 1 (since ka  10–3)  fraction of B left = (1/2)6 pH = 2 (pKa + logC) C  13 pH = 2  2 10. Ans.(B) Hence, at   1 6 = 23 =8 bt  2 2. Ans. 2 At equivalence point, M1V1  M2V2 , V2 = 400 ml Acid Base Sol. In 1st vessel at equilibrium PCl5  PCl3 + Cl2 HX + NaOH  NaX + H2O 1 atm 1atm 1atm 0.1mol 0.1 mol 0 0 hence Kp = 1 atm PCl5  PCl3 + Cl2 In 2nd vessel initially 3 atm 3atm 3atm 1 at equilibrium 3+x 3–x 3–x n 1000  1 10 [NaX] = = VT 400 100 5 (3  x)2 Kp = (3  x) = 1 1  pH = 2 (pKw + pKa + log c) hence x = 1 = 8.65 PCl2  2 atm 11. Ans. (D) 3. Ans. 4 1001CT103516017 4. Ans. 1 5. Ans. 5 HS-3/13

Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-1 PART-3 : MATHEMATICS SOLUTION SECTION-I 6. Ans. (A,B,C,D) 1. Ans. (C,D)  ƒ x 1 x2  x  1 2 The given equation simplifies to  ƒ x 1  x4  x2  1 cos4x = 0  2cos2(2x) – 1 = 0  ƒ(x) = x  1 , x  R – {0}  cos2 2x  1  sin2 2x x 2 Now verify it. 7. Ans. (B,C) Now verify it. 2. Ans. (B,C,D) Let a and d be the first term and common Put x = sin = sin–1x    ,   difference of A.P. Also, let b and r be the  2 2  first term and common ratio of G.P.  On solving, we get  y = sin–1(sin2) a  1 ,b  1 ,d  5,r  3     2sin1 x ; 1  x  1 22 ; 2  ; 20 1  x  1  2 sin1 x 22 Also,  ak  960 = 1  x 1 k 1  2 8. Ans. (B,C,D)      2 sin1 x ƒ x  x2 2  1   1 x2 1  x2  1   (even function)  Clearly, ƒ(x) is continuous for all y x  [–1,1] but non-differentiable at (0,2) x  1, 1 , 1 ,1 . 2 2 3. Ans. (C) A.M of roots = H.M of roots y=1 x   =  =  =  = 2(each) (0,0) x=1 where  are roots of given equation.  p = –8, q = 24 4. Ans. (A,C) Also, area Tn  cot 1   n n  1   1 1  1 1  dx  1    4 0  x2  4 4  Solutions for Question 9 & 10 1 ƒ(x) = x4 + x3 + x2 + x – 1 n n  4  ƒ'(x) = 4x3 + 3x2 + 2x + 1   Tn tan1 1    n1 n1   Sn    n  1  n ƒ\"(x) = 12x2 + 6x + 2 > 0  x  R  4  4 .  ƒ'(x) is strictly increasing function so, ƒ'(x) = 0 has exactly one real root.  n  tan1  n  1   tan 1  n   From Rolle's theorem, ƒ(x) = 0 has n1   4   4 atmost two distinct real roots.     Now, ƒ(0) = –1 < 0  tan1 so, ƒ(x) = 0 has only two distinct real roots.   S  lim Sn 4 9. Ans. (B) n so, a = 4 and b = 1 Let one root be  and other root be ( < ) 5. Ans. (A,B,C)  3   2  It is obvious. As ƒ  0 and ƒ(–1) < 0 HS-4/13 1001CT103516017

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-1     3 , 1  5  k  0 ...(4)  2 13  (3) (4) gives Also, ƒ  1  0 and ƒ(1) > 0 k   5 , 0   2   13    +   (–1,0) 13. Ans. (C) 10. Ans. (B) (P) Range of ƒ(g(x)) Let k(x) = xƒ(x) = [ƒ(–1), ƒ(1)] = [0,4] (Q) Range of g(ƒ(x)) = [–1,1]  k'(x) = xƒ'(x) + ƒ(x)  k(x) = 0 has 3 distinct real roots. (R) g h x  1 , when x = 1 min  k'(x) = 0 has minimum 2 distinct real roots.  (S) h g ƒ x  5 , when g(ƒ(x)) = –1. 2min. so, xƒ'(x) + ƒ(x) = 0 has minimum two 14. Ans. (D) distinct real roots. Now, let r(x) = xk'(x) = x(xƒ'(x) + ƒ(x)) 4x2  4x  1 ; x0 ; x0 so, r'(x) = x2ƒ\"(x) + 2xƒ'(x) + xƒ'(x) + ƒ(x) ƒ  x    x2  since, r(x) = 0 has minimum 3 distinct real  4x 1 roots, so r'(x) = 0 will have minimum  a = 4, b = 1; c = –3 and m = 12 2 distinct real roots. 15. Ans. (B) 11. Ans. (C) 5 1 D = 4 – 4k2   sin x2  3dx   sin x  62  3 dx As D > 0 4 2  k  (–1,1) ...(1) = I1 + I2 Now, ( – )2 < 2 Put x = –y in I1 and (x + 6) = t in I2  ( + )2 – 4 < 2  I1 + I2 = 0 (Q) g(0) = 1. Now, 4 (x2 – x + 1)6.g(x) = (1 + x)7. 1  x2  k2  4  2 Differentiate b.t.s w.r.t x and put x = 0, we get g'(0) = 13 4  k2  6  6k2  4 5  j  9 j1  11  4 (R) cos2  (cosine series)  k2  2 ...(2) (S) p = –3,–2,2,3 = 0 3 SECTION – IV  (1) (2) gives     1. Ans. 2 k   1,    2  2 ,1 ƒ 'x ƒ \"x  ƒ 'x 3 3 ƒ3 x 12. Ans. (C) ƒ ' x 2  1 x  C Let ƒ(x) = kx2 + 2x + k Now,  2ƒ2  k ƒ(2) < 0 2  k(5k + 4) < 0 As ƒ(1) = 1 and ƒ'(1) = 0  C  1 2  4  k  0 ...(3)  ƒ'x   1  ƒ2 x 5 ƒ x Also, ƒxƒ'x kƒ(5) < 0   1  ƒ2 x dx   dx  k(26k + 10) < 0 1001CT103516017 HS-5/13

 ƒ x  2x  x2 (As ƒ(1) = 1) Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-1 4. Ans. 3 y V(1,1) ƒ t   3 t ; t 1 t 1 ; t 1 ƒ(t) (0,0) x (0,3) ƒ(t)=3–t ƒ(t)=t+1 (2,0) 2 (0,0) (1,2) t 1  S   2x  x2 dx 0  (Put x = 2sin2) Now,  2 16S  3  x2  x  1 ; x 1,0   so,  4 2 g x    2 ; x  ,1  0,1 2. Ans. 5  x 1 ; x 1,  ƒ '(x)  4 sin x.esin 2x  cos x  1   cos x  1   n1 = 0, n2 = 3  2   2   (n1 + n2) = 3 5. Ans. 2 +–+ –+ – 0 /3 /3  /3 /3  sign of ƒ'(x)  m = 3, n = 2  (m + n) = 5 Let the first term and common difference of A.P. be a and d respectively. 3. Ans. 5 t2 1 1 x dx ...(i) 50  g t  cot1 t2  x  a2i1  50 i1 2t Also,  a + (a + 2d) + (a + 4d) +.....+(a + 98d)=50 t2 1  t2 x  a + 49d = 1 ...(1) g t  cot1 dx ...(ii) Now, 2t 1  x (Using king property) 50 j j1  (i) + (ii) gives   1 2 a2j1 2g  t   t2    dx 2t  2  j1  g t   t 12 1 = |–a1 – a3 + a5 + a7 – a9 – a11+....... 4 .......+ a93 + a95 – a97 – a99| = |–a1 – a99| = |–2(a + 49d)| = |–2(1)| = |–2|= 2 so, g 5   15  5 g 3 4  3 4 HS-6/13 1001CT103516017

Paper Code : 1001CT103516018 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 : 11 - 03 - 2017 PAPER-2 SOLUTION 1. Ans. (2) 5. Ans. (4) Sol. E= dV Sol. Wext = Uf – Ui  4kq2 2kq2 dx Ui = a  2a 2. Ans. (2) Sol. mCVCg  mSVS1 ... (i) 4kq2 2kq2 Uf =   mCVCg  mC  mS VS2 2a 2a ... (ii)  mC  mS VS2  mCVCg  mSV 'S2 ... (iii) 6. Ans. (3) 2mC VCg  V 'S2 Sol. mv cos  = 2mvx' y mS – mv sin  + 2mv = 2mv'y 3. Ans. (4) v cos  x v'x = 2 VCD/A v'y = 2v  v sin  60° 2 Sol. VA v' = v 'x2  v '2y VCD   7. Ans. (2) VCD/ A  VCD  VA 8. Ans. (4)   Sol. mv0R = mva ....(i) ....(2) VCD  VCD/ A  VA 1 mv20 1 mv2 GMm 2 2 a   VC2D  VA2  V2  2VA VCD/ A cos 60 CD/ A VCD = 2m/s 9. Ans. (1) 4. Ans. (1) Sol. 1 mv2  1 Qe 2 M 2a2  2 4 0 R Sol. MV0a = 3  1  Qe  4 0  mR2   3 V0 v= 2  8a  10. Ans. (4) 1 I2  Mg 2a  Mga  2Mga  Mga u2 u2m 22 Sol. Range = R   for  = 45° a qE  V0 = 16 u2m 3 2 1 ga E = qR Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-7/13 +91-744-5156100 [email protected] www.allen.ac.in

Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-2 1 mu2  KE 14. Ans. (1) 2 15. Ans. (1) 16. Ans. (1) E  m 2KE  2KE Sol. Pt = Pc(1 + k2/2) qR m qR Pt is total transmitted power (sidebands and carrier). 11. Ans. (2) Sol. q1  Q ' Pc is carrier power. rR k is modulation index. q1 + Q' = Q Maximum sideband power occurs when k = 1. Q '  r  1   Q  R  17. Ans. (4) Q '  QR , q1 = Qr N 45° 45° Rr Rr N Similarly Q\"  QR2 , q2  QrR R  r2 R  r2 Sol. Similarly q3  QrR2 mgcos 2N cos 45° = mg cos  R  r3 mg cos   QrR  2 N = 2 cos 45 mg sin  – 2 µkN = ma q 2    R  r 2   QR2r  q3 a = g sin  – 2kmg cos  2   R  r3 2cos 45 m q1  Qr   R  r  a = g (sin  – 2k cos  ) 18. Ans. (3) 12. Ans. (2) Sol.   qnet 0 for a charge placed at corner of cube   qnet 8 0  thus for given system 1  2  3  4  6  7  5  8 T  Sol. mg 2 0    1 O 2 0 13. Ans. (2) Sol. IE = Ii = Vi  200mV  10mA mg  cos   T Ri 20 2 IC  IE T = Tmin = mg cos  2 IL  Ii  10mA 19. Ans. (1) VL = ILR = (10 mA)(5k) = 50 V Sol. NY /2 Voltage Gain: /2 mg AV  VL  50V  250 Vi 200mV mg – NY = ma .....(i) 1001CT103516018 HS-8/13

 =  m   2   ....(ii) Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2 mg 2  2 24. Ans. (4)    Sol. S2 is open and S1 is on position-2    Rb Rc NY = 0 20. Ans. (2) x F Ra  Ra Sol. Fthrust Pb is max Pc = Pd dm P = I2R F = Fthrust = vrel dt Pb > Pa > Pc = Pd F = v d x Rb > Ra > Rc = Rd S2 is closed and S1 is on position-1 dt F = v dx  v2 I1 Rb dt 21. Ans. (2) 22. Ans. (1) V0 0 I2 Sol. 0 I Rc x V0  Ra Rd 0 dU d  Q2  1 Q2 dC P'b = V02 , P'c = V02 P'c > P'b HS-9/13 F= =    Rb RC dx dx  2C  2 C2 dx b : length of one side of plates I = I1 + I2 P'a = I2Ra C = 0 A  0 bx dC  0 b P'd = I2Rd ; 0 0 dx 0 Since Ra > Rd F  1 Q2  0 0 b P'a > P'd 2 20 b2x2 0 2   Rb   1 Q2  0  1 P'c = I22Rc = I  R b  Rc  Rc  2 0 b  x2    F    23. Ans. (4) P'c = I2  RcRb 2 1  Rc  R   b   Rc  Sol. Consider system of two disks to be short  dipole.  RcRb  2  1   Rc  Rb   Rc  2kdp      Rc  Rd dE = r3 dp = dq = dA P'c < P'd P'a > P'd > P'c > P'b 2k dp  2k A 25. Ans. (2) r3 E = r3 6  a 2 R2 E= 4 0 r3 R2  2 0 r3 Sol. 12V  b 1001CT103516018

Pmax = Va2b ; Pmax = 16  0.4watt Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-2 4Rab 4 10 28. Ans. (4) 12 C Vab = 9  3 = 4 volt Sol. 63 D Rab = 8 + 9 = 8 + 2 = 10 26. Ans. (1) Sol. x R AB  d y It can be worked out that the circuit diagram dEsin is a regular tetrahedron, whose edges contain 6 identical capacitors. From P symmetry consideration, irrespective of the dE  pair of points between which the source is connected, there always exists uncharged dEcos capacitor in the circuit Ey   dE cos  29. Ans. (1) Ey = 0 MR2 2 Ex   dEsin  1  20  1 K R2   2 Sol. 2   kdq  2 KR22 Ex  R2 sin  02  MR2 Ex  k0 /2 d 82k R 0  M sin2 30. Ans. (2) Sol. Rayleigh criterion : /2 When the central maximum of one source falls on the first minimum of another source, Ex  E1  k0  the sources are said to be just resolved. 2R E1  0 R 8 0 27. Ans. (3) 31. Ans. (3) Sol. mg  (0.10 kg) (10 m/s 2 )  1 N 32. Ans (2) mg = 4x01/2 33. Ans. (1) x0 = 0.0625 m 34. Ans (2) F 35. Ans.(3) mg dF H2O(l)  H2O(g) dx X  X0  ke H = U + ngRT O x0 x 10000 = U + 1 × 2 × 350 ke   dF   4  1 x 1/ 2  2  2  4  8 N/m  U = 9300 cal  dx  x0 2 x0 36. Ans.(3) ke 8 AZX  126C + 01e m 0.1  4 5 Sol. A = 12, Z = 7 2 2 2 fn  37. Ans (4) HS-10/13 1001CT103516018

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2 38. Ans. (1) 60. Ans. (4) 61. Ans. (2) 13 NH3 2 N2 + 2 H2 LHS : cos x  cosec 2x k2 Eq. (1–)  3 3 3 22 sin 2x cos x 33 13   2  3 2 tan–1(tan2) = 2 – . =  2   2  KP . Pº 62. Ans. (3) (1 ) (1 ) Solving : Given 2b = a + c  2b  1  c ...(i) a a 1      b  15  b  15  c  31  3 3 Pº  2 aa a   1 4KP   = –31. 39. Ans. (1) 40. Ans. (4) 63. Ans. (1) Solubility of MX = (4 × 10–10)1/2 21 2  ƒ(x)dx   1  x dx  7x  61/ 3 dx MX3 : Ksp = 27s4 00 1 = 27 (2 × 10–5)4  55 = 4.32 × 10–18 42 41. Ans. (4) 64. Ans. (2) 42. Ans. (4) ƒx x x x.......  x. ƒ x  ƒx  xx 43. Ans. (1) ƒ2(x) = x2ƒ(x)  ƒ(x) = x2  ƒ'(x) = 2x 44. Ans. (4)  ƒ'(3) = 6 65. Ans. (2) 45. Ans. (4) 46. Ans. (2) d  ƒx  dx  ƒ ' x  47. Ans. (1) (ƒ'(x))2 – ƒ(x)ƒ\"(x) = 0   0 48. Ans. (1) ƒ x  ƒ 'x = constant 49. Ans. (4) 50. Ans. (3)  ƒx  1  ƒ x  e2x ƒ'x 2 51. Ans. (3) The equation e2x = x2 has one solution. 52. Ans. (3) 66. Ans. (3) 53. Ans. (1) Differentiate both sides wrt 'x', 54. Ans. (2) (e – 1)exy  xdy  y   2x  ex2  y2  2x  2y dy   dx   dx  55. Ans. (3) 56. Ans. (3) e  1  dy   2  e2  dx  57. Ans. (2) 1,0 58. Ans. (2) dy  2 dx 1,0 59. Ans. (1) 1001CT103516018 HS-11/13

Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-2 67. Ans. (2) 75. Ans. (1) Put ex  t ƒ  8   8   ƒ  8   \"  16  x  x   x   x3 e et  1  tnt  e et  nt 1  ee   1  t  1  t   I dt   dt  76. Ans. (1) 68. Ans. (4) Tn  tan1  n2 3   tan 1 n  2  tan1 n  1  n 1   = 135º tan   1 2 Use : Sn = Tn 1 lim sin   cos    2 77. Ans. (4) 3  4 ƒ(x) = x3 + bx2 + cx + d, 0 < b2 < c 69. Ans. (3) ƒ'(x) = 3x2 + 2bx + c. h(x) = ƒ(g(ƒ(x)) D = 4b2 – 12c = 4(b2 – 3c) = 4((b2 – c) – 3c)<0 h'(x) = ƒ'(g(ƒ(x)).g'(ƒ(x)).ƒ'(x) h'(2) = 64. ƒ'(x) > 0  x  R. 70. Ans. (4) Hence, ƒ(x) is strictly increasing.   3k2  3k  1   1 1  78. Ans. (2)   k1 k3(k  1)3  k1  k3   13  1 Pn = e –nPn–1 P10 = e – 10P9 k e – P10 = 10e – 90P8 P9 = e – 9P8 71. Ans. (2)   1 4x6  4x3  ƒ x ƒ x  4x6 dx  4 – 9e = P10 – 90P8 e  P10  e  9P8 79. Ans. (3) 10 07 1   1 20   ƒ x  2x3 2 dx  0  ƒ x  2x3 20 i1 xi  x 2  5 0 72. Ans. (2) d y  20  dt   d y x  xi  x2  100 x  i 1  x   x new observations are 2x1,2x2,......,2x20. dy dy 2 x1  x2  ...  x20  dx Their mean = x1   2x   dy  y  n 1 x  C 20 1  x 73. Ans. (3) Now,  1 20  2x2 variance 20 (tan2x – 1)2 = 3 – [a]2 2xi i 1 Hence, 3 – [a]2 > 0  a   3, 3 1 20 xi x 2 1 20 20 4 i 1    4 100  20  [a] = –1,0,1  a  [–1,2) 80. Ans. (2) 74. Ans. (3) x C O 5m ƒ(x) = ƒ(–x)  x  R tan = 5 ...(i)  2Ax3 – 2Bx = 0 xR  A = B = 0  Bx simultaneously 30m 25m tan 2  30 x  1 2 1 ...(ii)  2 A A  0  sin     0  sin   2  from (i) and (ii) tan   2  1 2 1 3 B  0   tan   3  0  tan    3     x = 5cot = 5 3  2 Hence 2 solution only. HS-12/13 1001CT103516018

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2 81. Ans. (1) 1  1  n  1 .  1 1 Do yourself hn 1 50 100 n  82. Ans. (3) 1  2n 1  n 1  hn 1 100n 1 ƒ(1) + ƒ(1) = ƒ(1) = 5 hn 1 100 n 1  n  3 83. Ans. (4) a2hn–1 = 5000. a = 2; b = 1; c = –5 88. Ans. (2) a + 2b – c = 9 ' ' 84. Ans. (3) ( + )2 – 4 = (' + ')2 – 4''  a2 – 4b = b2 – 4a b1 + b2 = 1  b1(1 + r) = 1 b1 1  a2 – b2 = 4(b – a) 1r   1 1 2r  2 bk  1  r2 2 1  r1  r k 1  b1 1 1 (a – b) (a + b) = –4(a – b) 1  r  2  2 2  (a + b) = –4 1 2 89. Ans. (2) 85. Ans. (1) dx  x  y ƒ 1  0   dy y ƒ 2  0 x  y  c  x  2y  y2 ƒ(1) = 1 – 2a + a2 – 6a < 0 y y = 3 when x = – 3 a2  8a  1  0  a  4  86. Ans. (2) 15,4  15  ...(i)  Do yourself ƒ(2) = 4 – 4a + a2 – 6a < 0 87. Ans. (1) 50,a1,a2,....,an,100 are in A.P. a2 – 10a + 4 < 0 a2 = 50 + 2d; where d  50 a  5  21,5  21 ...(ii) n 1 a2   50  100   50n  150  50 n  3 (1)  (2)  a  5  21, 4  15   n 1  n 1 n 1  90. Ans. (3) 50, h1,h2,......,hn,100 are in H.P. 1 1 Do yourself hn 1 50   n 1d ' ; 1  1  1  100 50  where d'   100n 1 n 1 1001CT103516018 HS-13/13