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 4m2 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 m2 4 m2' ' 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 11 = 22 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 m2 m2 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 32 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 = IS 12. Ans. (D) 510 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} = 8002 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 t10 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 13 pH = 2 2 10. Ans.(B) Hence, at 1 6 = 23 =8 bt 2 2. Ans. 2 At equivalence point, M1V1 M2V2 , 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 2sin1 x ; 1 x 1 22 ; 2 ; 20 1 x 1 2 sin1 x 22 Also, ak 960 = 1 x 1 k 1 2 8. Ans. (B,C,D) 2 sin1 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 tan1 1 n1 n1 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 tan1 n 1 tan 1 n From Rolle's theorem, ƒ(x) = 0 has n1 4 4 atmost two distinct real roots. Now, ƒ(0) = –1 < 0 tan1 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 ; x0 ; x0 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 3dx sin x 62 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 j1 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, ƒxƒ'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 cot1 t2 x a2i1 50 i1 2t Also, a + (a + 2d) + (a + 4d) +.....+(a + 98d)=50 t2 1 t2 x a + 49d = 1 ...(1) g t cot1 dx ...(ii) Now, 2t 1 x (Using king property) 50 j j1 (i) + (ii) gives 1 2 a2j1 2g t t2 dx 2t 2 j1 g t t 12 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 2a2 2 4 0 R Sol. MV0a = 3 1 Qe 4 0 mR2 3 V0 v= 2 8a 10. Ans. (4) 1 I2 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 2KE 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° Rr Rr N Similarly Q\" QR2 , q2 QrR R r2 R r2 Sol. Similarly q3 QrR2 mgcos 2N cos 45° = mg cos R r3 mg cos QrR 2 N = 2 cos 45 mg sin – 2 µkN = ma q 2 R r 2 QR2r q3 a = g sin – 2kmg cos 2 R r3 2cos 45 m q1 Qr R r a = g (sin – 2k 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. 63 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 R2 2 Sol. 2 kdq 2 KR22 Ex R2 sin 02 MR2 Ex k0 /2 d 82k 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 k0 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 k2 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 61/ 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) ƒx x x x....... x. ƒ x ƒx 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 ƒx 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) ƒx 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 e2 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 tnt e et nt 1 ee 1 t 1 t I dt dt 76. Ans. (1) 68. Ans. (4) Tn tan1 n2 3 tan 1 n 2 tan1 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) k1 k3(k 1)3 k1 k3 13 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 i1 xi x 2 5 0 72. Ans. (2) d y 20 dt d y x xi x2 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 2x2 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 xR 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 2n 1 n 1 hn 1 100n 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) 1r 1 1 2r 2 bk 1 r2 2 1 r1 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 1d ' ; 1 1 1 100 50 where d' 100n 1 n 1 1001CT103516018 HS-13/13
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