Paper Code : 0000CT103116002 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE TARGET : JEE (ADVANCED) 2016 Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced TEST DATE : 12 - 02 - 2017 PAPER-1 PART-1 : MATHEMATICS 2 34 5 ANSWER KEY Q. 1 D BC A 12 13 6 7 8 9 10 SECTION-I A. A A,C B,C 5 B A,C B,C,D A,B,C A,C Q. 11 2 34 7 2 25 A. B,C,D SECTION-IV Q. 1 A. 5 SECTION-I SOLUTION 1. Ans. (A) 3. Ans. (B) (2p+ 2p + 6) (–2p – 2q + 6) < 0 n m! n 1! m 1! m n m n 1 n!m 1 n! n 1!m 1 m (p + q + 3) (p + q – 3) > 0 mn > (m – n + 1) (m – n) m2 – 3mn + m + n2 – n < 0 Also p2 < 4 p (–2,2) m2 – (3n – 1)m + n2 – n < 0 and q2 < 4 q (–2,2) Region is m 3n 1 5n2 2n 1 , 3n 1 5n2 2n 1 q 2 2 (0,3) clearly some integer must be lying in this interval let it be M(n) (–2,0) (2,0) P 3n 1 5n2 2n 1 1 M n 3n 1 5n2 2n 1 (0,3) 22 According to sandwich theorem Area = 2 1 11 1 lim M n 3 5 2 n n 2 2. Ans. (D) 4. Ans. (C) 4x–3y=0 a0 5 22 1 2 2 (h,k) 3x+4y–9=0 a1 a20 2 17 24 1 4 22 (3,0) 28 1 44 1 24 42 (0,0) a2 k 3h 4k 9 4h 3k x 4 1 3 x2 22n 1 55 Let an , x , k 9 3h 4k 4h 3k 3 1 1 n ak 1 5 5 ak k0 ak k0 lim h – 2k = 6 x – 2y = 6 n Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/16 +91-744-5156100 [email protected] www.allen.ac.in
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-1 lim a0 1 . a1 1 . a2 1 ..... an 1 7. Ans. (A,C) an a1 a2 an L.H.S. > 0 and hence R.H.S > 0 0 lim 22 2 1 . 42 4 1 . 162 16 1 .... x2 x 1 8. Ans. (B,C,D) 22 1 42 1 162 1 x2 1 x (A) lim e esin x.ntan x tan x.nsin x lim 22 1 1 x4 x2 1 3 x0 22 2 x4 1 7 x sin x.ntan x tan x .nsin x 5. Ans. (A) lim e x 1/x e x 1/ x Let a variable point on the parabola be x0 (1 + t2, 2 t) and its reflection in the given 1+1=2 line be (h,k) lim 1 ex 1 cos x sin x 1 ex 1 cos x 2 1 t2 2t 2 x0 (B) h – (1 + t)2 = k – 2t = 2 h – (1 + t2) = –1 – t2 – 2t + 2 h = 2(1 – t) x2 x x2 and k = –(t2 – 1) 1 .... 1 .... 2! 1! 2! lim 1 x0+ sin x k t 1t 1 x h 21 t t 1 2k x 1 ex 1 cos x h (C) Base is exact 1 and t 2k 1 n 1n 22n 3 h lim 6. n.n 12n 1 1 (D) n and 1 t 2 2k h h2 6 h2 9. Ans. (A,B,C) xdy – y2dy = ydx –y2dy = ydx – xdy 2h – 2k = 2 4x – 4y = x2 –dy = x A + B = 0. d y 6. Ans. (B) We have to maximize zyxzyy y x c y i.e xy3z2 where x + y + z = 1 2 4 c c 4 2 x 3. y 2. z y3 z2 . 1/6 32 x –y2 = x – 4y 6 33 z2 x = 4y – y2 = 4–(y – 2)2 (y – 2)2 = –(x – 4) 27.4 xy3z2 < 66 where x y z k focus is 15 , 2 32 4 1 equation of directrix 6 k + 3k + 2k = 1 k x 17 4x 17 0 4 x 1 y1 z1 6 23 e = 1. HS-2/16 0000CT103116002
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-1 10. Ans. (A,C) a 1 b det (A – I) c ad a d bc 1 x y zc d 1 Equation of L1 0 b c a b c d det A I x y zc Equation of L2 a 0 c 1001 0 ˆi ˆj kˆ 1 0 0 1 0 Direction of is 0 b c 1 0 0 1 0 a0c 1 0 0 1 4 bcˆi acˆj abkˆ . Equation of is 0110 0 –bc(x – 0) + ac(y – 0) + ab(z – c) = 0 0 1 1 0 2 0 1 1 0 2 0 1 1 0 0 x y z 1 0 13. Ans. (B,C) a bc z Distance between L1 & L2 is (13,9) 0, 0, 2c.bc, ac, ab z2 B z1 (10,6) (16,6) b2c2 a2c2 a2b2 1 2abc 4 b2c2 c2a2 a2b 1 1 1 locus is major arc of circle a2 b2 c2 64 z 13 9i 13 102 9 62 11. Ans. (B,C,D) z 13 9i 3 2 Normal ...(i) SECTION–IV y = mx – 2m – m3 ...(ii) 1. Ans. 5 and y mx km m m3 common ratio of G.P is 24 2 tan1 x whose modulus is less than 1. (i) & (ii) same 1 m 2m m3 1 m km m m3 tan1 x 24 ƒ x 2 tan1 8 4m2 1 x m2 4k 1 2 3m2 = 4k – 6 Now domain of (ƒ(x))2 + (sin–1x)2 = a is x [–1,1] m2 > 0 4k 6 0 k 3 Now, ƒ x 1 2 2 1 tan1 2 12. Ans. (A,C) 1 x Let A a b a b a c 1 0 d , then c d b 0 c d 1 ƒ x at x = –1 min 2 a2 + b2 = 1 and ƒ x at x = 1 ac + bd = 0 max 6 c2 + d2 = 1 0000CT103116002 HS-3/16
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-1 2 2 2 2 g'x ex/2 ƒ'x 1 ex/2 ƒx ƒ x 0, 4 and sin1 x 0, 4 2 minimum ƒ2(x) + (sin–1x)2 is 0 g\"x ex/2 ƒ\"x ex/2 ƒ'x 1 ex/2 ƒ x 2 4 g\"' x ex/2 ƒ\"' x 3 ex/2 ƒ \"x 3 ex/2 ƒ ' x 1 ex /2 ƒ x and maximum is 2 48 2 a 0, 2 1 ex/2 8ƒ\"'x 12ƒ\"x 6ƒ 'x ƒ x 2 8 integral values are 0,1,2,3,4 i.e. 5 4. Ans. 5 2. Ans. 2 x 1 1 1007 I dx ƒ x ƒ x tan 1 x ...(1) x1007 1 x2016 0 Replace x by x 1 , we get Put x1008 t x1007dx dt 1008 x x 1 1 1 x 1 1 1 1007 1 1 x x dt 1 t 1007 1 t 1007 dt ƒ ƒ tan I 1 x ...(2) 1008 0 1 t2 1008 0 1 1 1 2 t 1007 dt Replace x by 1 x in (1) we get t1007 I 1008 0 1 1 Now, Put t = 2z ƒ ƒ x tan 1 ...(3) 1 x 1 x 1 1/2 1007 2dz 2 z 21007 1007 1007 I 1008 0 1 z (1)- (2) + (3), we get 2ƒ x tan 1 x tan 1 1 tan 1 x 1 22015 1/ 2 1 z 1007 dz x z1007 1 x 1008 0 Also, 1/2 2ƒ 1 x tan 1 1 x tan 1 1 tan 1 x N 22015 2 x1007 1 x 1007 dx x 0 x 1 22005 1/ 2 adding we get 1z 1007 dz z1007 1008 0 2 ƒ x ƒ 1 x x 0,1 N = 2016 = 25 × 32 × 71 222 Total 5 divisors of form (4n + 2) (n N) 3 5. Ans. 7 2 Let P denotes the chances of single bacteria Now, N 1 ƒ x dx 1 1 ƒ x ƒ 1 x dx 3 to die, then P 1 1 .P.P 1 P.P.P 2 8 42 4 P3 + 2P2– 4P + 1 = 0 (P – 1) (P2 + 3P 0 0 3. Ans. 2 – 1) = 0 Let g(x) = ex/2ƒ(x) P 3 13 2 then g(x) = 0 has at least 5 distinct zero and using rolle's theorem g\"'(x) = 0 has at least 13 3 5 13 two distinct zero. 1 P 1 2 2 HS-4/16 0000CT103116002
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-1 PART-2 : PHYSICS 4 ANSWER KEY C Q. 1 2 3 5 6 7 8 9 10 4 B A,D B,D A,C B,C A,C SECTION-I A. B B B 2 5 Q. 11 12 13 4 A. A,B A,C B,C SECTION-IV Q. 1 2 3 A. 1 3 2 SOLUTION SECTION-I mH mL gh n1 n2 CV 2 1. Ans. (B) T Sol. p(r3)5/3 = c pr5 = c 131 4 10 1 103 p c 1 2 3 25 2 r5 23 dp 5c dr 127 102 2.54 102C dT r6 dT 2 3 25 2 1 dp r5 5c k 5k 23 p dT c r6 r 4. Ans. (C) 2. Ans. (B) Sol. 360° 200 div. Sol. 11 1 18 360 15° 200 div. v 30 20 11 1 3 2 v 20 30 60 LC 1nm 200 v = – 60 m1 60 2 x 10 1 mm 30 200 1 1 1 F 1 100 103 = 5 mN v 20 30 20 1 11 23 v = –60 5. Ans. (B) v 60 Sol. 1st RORI & 2nd RORI 30 20 or 1st ROVI or 2nd VORI VI for 1st lens is real object for 2nd. So 2nd m2 60 3 situation is not possible. 20 6. Ans. (A,D) m1m2 = –6 3. Ans. (B) Sol. L Sol. iB h Bv q iR 0 H c Q = 0 = U + W iB mdv = n1CvT + n2CvT dt –mLgh + mHgh 0000CT103116002 HS-5/16
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-1 Bq = mv After short Bv mv mR dv – 2CV 2C 2CV Bc B dt 3 3 – 2CV 3 a=0 C Bc 2CV m B22c m 3 v Bc B V mv Qflow 2CV B 3 q 7. Ans. (B,D) W CV V CV2 Sol. 33 9. Ans. (B,C) + e = Case-1 : v % 0.1 100 0.3% Sol. dH bT T0 1 4 v 33 dT 3 e 50 51 ve = 3 × 10–3 × 2 × 0.5 × × 80 4 2 + e = = 2.4 × 10–11 cal/sec 2 Case-2 = 2(2 – 1) Case-2 : v % 0.1 100 0.2% dH 3 103 1 T 20 × 2 v 50.8 dt v = f e 76 75.6 0.2 ve 2k 100 T 2 n 2 v 2 1 Case-3 : v % 0.1 100 0.33% 3(T – 20) = 2.8 103 100 T v v 30.4 2 1 0.7 3T – 60 = 400 – 4T e 2 31 e 46.0 46.8 ve 7T = 460 2 2 Case-4 : v % 0.1 100 0.32% T 460 65.7C v 31.5 7 e 48.2 48.0 0.1cm dH ' 3 103 460 20 2 2 7 dt 8. Ans. (A,C) dH 2.4 101 dt 1 1 11 1 1 460 140 320 Sol. 40 7 2C 2C C C Ceq 280 1 1 2 2 Ceq C 10. Ans. (A,C) 2C 3 kQ 900 Sol. r – CV CV –C3V CV kQ 3 3 3 r2 –C3V 90 +CV CV r = 10 m 3 3 – CV 3 900 10 Q 9 109 1C HS-6/16 0000CT103116002
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-1 (8 – x)2 + (7 – y)2 = 102 SECTION-IV 3 106 9 109 106 1. Ans. 1 r2 Esurface = qvB 90– B O r2 = 3 × 10–3 u r 30cm 11. Ans. (A,B) Sol. h mg E1 E2 Sol. E1 E2 qvB sin = mg 0 qvB cos mv2 h tan P E1 E2 2 qvB mv2 R1 h sin 12. Ans. (A,C) Sol. f – mg sin 37 = ma mg mv2 h f = ma + mg sin µmg cos a 8 – 6 = 2m/s2 n=1 µmg cos + mg sin = mamax 2. Ans. 3 amax = 14 m/s2 13. Ans. (B,C) A ++ Sol. –– O Sol. X mg eE eV h 90° V mgh X’ e IXX ' IXX ' 1 2m r2 Pd V iR V V 3V 3mgh 4 4 4 4e 3. Ans. 2 IXX ' 1 mr2 Sol. e kQ 1.44keV charge is constant 4 R B kQ 1440V R 7 41–mR2 Q 1440 V Ne 8 9 109 X’ A 4R/3 x0 N 1440 1019 102 9 109 1.6 X N = N0 (1 – e–t) 1012 = 4 × 1012 (1 – e–t) x 4R sin 4R 3 1 m 3 3 32 8 et 1 2 half lives 2 hrs 4 1 1 15 IC 4 1 12 1 64 64 Nd = 1012 15 49 NS 1 1012 N0 IAB 64 1 64 = 1kg m2 3 4 2 half lifes. 0000CT103116002 HS-7/16
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-1 4. Ans. 2 60 2 6 180 i 6 11 11 11 180 0 61—10 11 3 i 6 180 48 132 12 11 11 Sol. 6 60 0.5H 11 11 5 i = 2A 5. Ans. 4 63 iT 60 50.1n2 Sol. 40 2 60 m 2mv 11 1 0.5 10 2 v e 5 12 1 6 A v 5g 11 2 11 200 50l l4m PART-3 : CHEMISTRY 4 ANSWER KEY A 567 Q. 1 2 3 A A,C,D A,C,D 8 9 10 A. C C B 4 A,B A,B,C,D A,B,C,D SECTION-I 2 Q. 11 12 13 5 A. A,B,C,D D A,B,C,D 4 Q. 1 2 3 SECTION-IV A. 3 1 2 SOLUTION SECTION-I (C) [HA] = 10–5 M 1. Ans.(C) Ka = 10–2 M Weak acid will dissociate completely 2. Ans.(C) H2 2H+ + 2e– [H+] = 10–5 M 2CuBr + 2e– 2Cu(s) + 2Br– (D) H2O H+ + OH– 0.06 log[H ]2[Br ]2 x C+x Ecell = 0.6 = E0cell – 2 0.6 = E0 + 0.48 10–14 = (x) (C + x) cell 0.12 = E0 = E0 + E0 = E0 10–14 = 2 x2 cell oxidation reduction reduction 3. Ans. (B) x = 1 107 C 2 Sol. CaC2 + N2 CaCN2 + C Mixture of CaCN2 and carbon is known as [NaOH] = 1 10–7 nitrolim. It is used as a fertiliser. 2 4. Ans. (A) [OH–] = C + x = 2 × 10–7 Sol. FeSO4 absorb NO gas and forms brown ring. 5. Ans. (A) [H+] = 1 107 2 6. Ans. (A,C,D) (A) [HCl] = 0.1 M 7. Ans. (A,C,D) [H+] = 0.1 M 8. Ans. (A, B) (B) [H2SO4] = C Sol. HNO2 + NH2–C–NH2 N2 + CO2 + H2O HSO4– H+ + O C-x C + x (x) (y) (z) [H+] = C + x SO42– + Ka2 Mg + N2 Mg3N2 H2O NH3 CuO N2 x (x) (w) (p) (x) HS-8/16 0000CT103116002
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-1 Ca(OH)2 + CO2 CaCO3 SECTION-IV 1. Ans. (3) (y) Milky solution NH3 + K2HgI4/OH HgO .Hg NH2 1 RH 1 1 I 90nm 1 (p) Brown ppt. 9. Ans. (A, B, C, D) 1 1 z2 1 1 54nm 90 4 16 Sol. [Ag(NH3)2]+ sp hybridisation (linear) 90 4 16 80 z2 [Ag(CN)2]– sp hybridisation (linear) 12 54 9 [Au(CN)2]– sp hybridisation (linear) ICl2 sp3d hybridisation (linear) z=3 10. Ans. (A, B, C, D) 2. Ans.(1) Sol. This is 4n + 2 series, therefore nuclides whose mass no equal to 4n + 2 may form, only possible nuclide is U234 Ni + 4CO 50°–60° Ni(CO)4 200°–230°C 3. Ans. (2) (Impure) Tetrahedral (pure) Sol. M(AB)3 n and Ma4b2 n have two Maximum 5 atoms are in one plane in geometrical isomer Ni(CO)4 4. Ans. (2) 11. Ans. (A,B,C,D) 5. Ans. (4) 12. Ans. (D) 13. Ans. (A,B,C,D) 0000CT103116002 HS-9/16
Paper Code : 0000CT103116003 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE TARGET : JEE (ADVANCED) 2016 Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced TEST DATE : 12 - 02 - 2017 PAPER-2 PART-1 : MATHEMATICS ANSWER KEY Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. D C B A C C B,D A,C A,B A,B Q. 11 12 13 14 15 16 17 18 A. A,D A,B,C A,C A,C B D A A SOLUTION SECTION-I Equation of plane passing through A 1. Ans. (D) & perpendicular to BC is x – 2y = 0. H lies on this plane P(acos,bsin) 2(9 4) 0 = 2 = 9 – 2 = 5 (–a,0) A' A (a,0) 3. Ans. (B) 1 x2 2x 2 x2 1 I= dx Q(acos,–bsin) R x2 2x 2 0 y b sin (x a) {Equation of PA} 1 x2 2x 2 ( x 1) .(x 1) a(cos 1) x2 2x 0 2 y b cot (x a) 1 d (x 1) x2 2x 2 dx = 2 2 a2 0 dx y b tan (x a) {Equation of QA'} 4. Ans. (A) a2 B Eliminating ; y2 b2 (x2 a2 ) a2 x2 y2 1 A 'C' be the origin and C and a2 b2 CA a CB b Let 2. Ans. (C) be the position vectors of A and B {x cx} x a (a b) A(0,0,9/2) b) 1 1 b k || µ a (a 2 a 3 cx (9/4,0,9/4) EH 1 k; k B C 23 (0,9,0) (9/2,0,0) F 1 k k 1 5k k 6 CE r (0,9,0) (1,4,1) 32 2 6 5 3 H (,9 4,) cx a b 1 55 HS-10/16 Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-2 5. Ans. (C) 7. Ans. (B,D) Bi : i rupee coin C(x,y) & 4 – i paisa coin 0i4 A(0,0) B(a,0) 1 tan A y tan B y P(Bi ) 5 x ax E :- Two randomly drawn tan B 2x tan A tan B 2 coin both found to be rupee coin. 2y 2 4 x 1 2 C2 y y 5 4 C2 4x2 ax 1 y2 P 136 1 B2 / E 1 2 C2 3 C2 4 C2 5 4 C2 4 C2 4 C2 4x 1 y2 4x2 a x y2 = 4ax 1 P(B2 / E) 10 8. Ans. (A,C) Similarly P( B3 / E) 3 10 (–a,0) A' s A(a,0) P(B4 / E) 6 10 P(next drawn coin is rupee coin) P1 y2 = –4(a – ae)(x – a) ......(1) 1 2 3 11 P2 y2 = 4(a + ae)(x + a) ......(2) 10 4 10 4 8 Solving (1 – e)(a – x) = (1 + e)(x + a) 6. Ans. (C) x(2) = a(1 – e – 1 – e) x = –ae Put in (1) y2 = –4a(1 – e)(–ae – a) x 1 1/x y2 = 4a2(1 – e2) = 4b2 x F x ƒ t dt xF x ƒ t dt P (–ae, 2b) Q = (–ae, –2b) 0 0 PQ = 4b Differentiating Let 1 nT E x y ' 2a(1 e) m1 a (1 e) y b lim xF 1 lim 1 nT E y' 2a(1 e) m2 a (1 e) y b ƒ tdt x0 x n nT E 0 a2 nT nT E m1m2 = b2 (1 e2 ) 1 nT 1 0 t dt t dt nT ƒ lim E ƒ 9. Ans. (A,B) n Let A : – Person owns Sedan n T 1 E B : – Person owns SUV E : – Person keeps driver ƒ tdt ƒ t dt lim n nT E 0 nT E 0 4 6 3 10 10 7 P(A) 10 , P(B) , P(E / AB) , P(E / AB) 10 , 1 T ƒ t dt 1 6 2 P(E/ AB) 9 T 3 10 0 0000CT103116003 HS-11/16
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-2 P(E) P(AB)P(E / AB) Now ƒ'(x) = 2xn2 – 2x P(AB)P(E/ AB) P(AB)P(E/ AB) ƒ''(x) = 2xn22 – 2 ƒ'''(x) = 2xn32 366 74 7 349 10 10 10 10 10 10 10 10 10 ƒ''(x) = 0 has exactly one real roots ƒ(x) = 0 has exactly 3 distinct real roots 206 14. Ans. (A,C) P(E) /2 /2 500 100 sin d 100. cos / 2 100 99.cos d PAB AB/ E 243 108 76 0 1000 1000 103 00 412 /2 1000 100 sin d 0 10. Ans. (A,B) /2 100 ƒ(x+ƒ(y)) = ƒ(x) + x + ƒ(x – y) ...(1) 99 cos d Put y = 0 (ƒ(0) = say) 0 ƒ(x + ) = 2ƒ(x) + x ....(2) /2 x = – in (2) = 2ƒ(–) + 100 (10099 sin 100 cos )d 0 ƒ(–) = /2 Put y = – in (1) ƒ(x + ) = ƒ(x) + x + ƒ(x + ) (100 sin )' d ƒ(x) = –x 0 11. Ans. (A,D) (a + b) = 20ei 100 . 1 (a2 + b2) = 16ei 2 100 a3 b3 1 a b 3 a2 b2 a b2 2 Solution for Question 15 to 16 Let be positive real root a(3 + ) + b2 + (4 + 1) = 0 ....(1) 10ei 48ei 400ei2 ( a2 b2 )min 4 (3 )2 4 3520 a3 b3 4480 {(1) represents line & a2 b2 is distance 12. Ans. (A,B,C) of point on line from origin} (ƒ'(x) – ƒ(x)) g2(x) = g(x) + g'(x) (g'(x) + g(x)) ƒ2(x) = ƒ(x) – ƒ'(x) 8 22 1 2 1 2 (g(x) g '(x))ƒ2 (x)g2 (x) g(x) g'(x) 6 34 2 2 g(x) + g'(x) = 0 log g(x) = –x + c g(x) = e–x (a2 + b2)min = g(x) = (1 2)ex g(0) 1 2 2 1 3 Similarly ƒ(x) = ( 2 1)ex 2 13. Ans. (A,C) 1 1 3 2 1 2 1 2 2 2 ƒ(x) = 2x – 1 – x2 (a2 b2 )min 1 4 5 1 3 ƒ(0) 0 ƒ(x) 0 24 ƒ(1) 0 has atleast for equality 2 = 1 ƒ(2) 0 3 distinct 15. Ans. (B) ƒ(5) 0 real roots 16. Ans. (D) HS-12/16 0000CT103116003
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-2 17. Ans.(A) x2 4(2 2) a, b, c, d h x2 4 Let O be the origin & be P.Vs of A,B,C,D respectively. O 2 2 2 h 44 2 x h . O' (a × b) ×(b × c) 2 2 2 22 2 cos 4 2 1 B2 . 2 2 2 1 22 18. Ans. (A) O v 1 82 2 1 3 1 2x2 8 45° xx 2 2x2 16 2 1 3 2x2 2x2 8 x2(2 2) 8 A B 22 PART-2 : PHYSICS ANSWER KEY Q. 1 2 34 5 6 78 9 10 A B DA D A B,C B,C A,B,D A,D SECTION-I A. 11 12 13 14 15 16 17 18 Q. A,D A,C,D B,C A,C C A HS-13/16 BA A. 4 SOLUTION SECTION-I 4. Ans. (A) 4 1. Ans. (A) Sol. (1) G 4 Sol. Fy max = T sin sin tan y x y = A sin kx sin t y 4 R xmax Ak Fy max = TAk i V 4 10 2 103 2 32 R 1.6 12 N ig i 8 2V 40 12 8 4 R 3 3 2. Ans. (B) 0.2 2V b a 20 3R Sol. O S 4 4 vsound=v–c f vcb n (2) G 4 vca VR 4 3. Ans. (D) 2 Sol. h h 1 mv m 3RT MT G 4 M 4 1 M2T2 4 400 VR 2 M1T1 2 300 0000CT103116003
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-2 V 8. Ans. (B,C) 24 i 10 R 4 iG = i Sol. M 10 V4 B 24 R 10 10 2V = 2.4 + R MB sin = mg sin × R 2V = 4+ 0.6R B mg iR 4 + 0.6 R = 2.4 + R 9. Ans. (A,B,D) 1.6 = 0.4 R R = 4 5. Ans. (D) Sol. Beat frequency : f1 – f2 = ± 5 f1 = f2 ± 5 Sol. V kQ 15 103 Q 1.5 103 f1 = f2 + 5 or f1 = f2 – 5 0.1 k f1 = 596 Hz or f1 = 586 Hz V ' kq k Q q 10 103 0.1 R q 103 ,R 0.5 103 5cm Intensity : I I0 cos2 1 2 t 10 103 2 k 6. Ans. (A) I0 cos2 2 f1 f2 t 2 Sol. 27 20 I0 cos2 5 dI d cos3 cos2 9 I0 n cos3 4 2 I0 as increase, dI decrease 10. Ans. (A,D) 7. Ans. (B,C) Sol. 37° 2 3 Sol. T 2 R3 1.5 GM 5 v = A 37° A1 > A2 v1 > v2 11. Ans. (A,D) Sol. 11 C 151 B 01 e v 6 4 3 Q = (MC – MB – 2me)C2 31 4 = 933.6 keV fraction full = 12. Ans. (A,C,D) 555 20 % empty Sol. Force in both springs is always same. N = wt. of 1cm column of water kx1 = 2kx2 13. Ans. (B,C) = 1 1.52 10 103 9 N 1600 4 HS-14/16 0000CT103116003
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-2 14. Ans. (A,C) 17. Ans. (B) Sol. Q2 8R2S U 0 2 4 0 R Q2 8R2S U Sol. 8 0 R dU 0 = amplitude dR = 0 same. R Q2 U = oscillation energy same 8 0 16S P P0 4S 2 1 1 R 2 0 2 2 U mVC2 IC2 4S Q2 1 mVC2 1 IC VC2 = P0 + R 2 0 162R4 2 2 R2 = P0 + 4S Q2 1282S 0 P0 R 322 0 Q2 R 1 VC2 m IC 2 R2 15. Ans. (C) Sol. e– I VC i 18. Ans. (A) Sol. At bottom, sphere is rolling. i v0 Vemit > VCollector 0 V = –iR hC I suddenly increases. e e VStop V0 R 1240 2 2 Li = mV0R + I00 = mV0R + I0 310 mR I0 i Lafter = mvR + I = V0 R 2A I R i = V mR 2V V V about point of contact is zero. i=2+V v i = 2 – iR K= 1 mV2 1 I2 i (1 + R) = 2 2 2 i 2 1 mV 2 1 Iv 2 1R 2 2 R2 16. Ans. (A) 4R 1 v2 m I 2 R2 Sol. P = i2R 1 R2 L2 dP 0 4(1 + R)2 – 4R × 2(1 + R) = 0 dR 2I mR2 4 4R 8R K R 1 HS-15/16 0000CT103116003
ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Advanced)/12-02-2017/PAPER-2 PART-3 : CHEMISTRY ANSWER KEY Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. C D A B D C C,D B A,B,C A,B,C Q. 11 12 13 14 15 16 17 18 A. B,C,D A,B,C B,D C B A B D SOLUTION SECTION-I 2 /3 1. Ans.(C) T2 = (32) 5/3 T3 2. Ans.(D) = (32)–2/5 T3 = 1 1 Balanced reaction are 4 T3 = 4 T1 2Cu2+ + 4I– 2CuI + I2 T4 = 32–2/5 T5 = 2–2/5 T3 = T2 = T6 4IO3– + 20I– 12I2 + 12H2O Qtotal = Cp × 3 (T2 –T1) 2Cu(IO3)2 + 24 KI 2CuI + 13I2 + 12H2O evolved I2 will react with Na2S2O3 5 R 3 3 300 = 45 × 75 = –3375 cal. equivalents of I2 evolved = eq. of hypo 2 4 nI2 × 2 = 52 × 0.1 × 1 1 nI2 = 2.6 Ssys = 0 + 1 × Rln (32)3 Ssys = –15R ln2 nCu(IO3)2 = 2 2.6 13 2 2.6 10. Ans. (A, B, C) [Cu(IO3)2] = 13 20 = 2 × 10–2 Sol. Statement (D) is incorrect because there are 3. Ans(A) gerade as well as ungerade orbitals exist. 1000 ml solution contains 10 mol C2H5OH 11. Ans. (B, C, D) 1000 gm solution contains 460 gm C2H5OH Sol. 'X' B3N3H6 WH2O = 540 gm = 30 mol H2O 12. Ans. (A, B, C) PT = 1 40 3 20 25mmHg Sol. Metal is Ag. 40 Ag+ + dil. HCl AgCl 4. Ans. (B) (X) Sol. (A) antibonding gerade (X) is insoluble in dil.HNO but soluble in NH (B) bonding moleuclar orbital gerade 33 (C) antibonding molecular orbital ungerade solution, cyanide solution and Hyposolution. 5. Ans. (D) 13. Ans. (B,D) 6. Ans. (C) 14. Ans. (C) 7. Ans. (C,D) 15. Ans. (B) 8. Ans. (B) 9. Ans. (A,B,C) Sol. Only O is released finally. 2 Sol. Q12 = Cp(T2–T1) 2MNO4– + 10I + 16H+ SI2 + 2Mn+2 + 8H2O Q34 = Cp(T4–T2) Q56 = Cp(T6–T5) I +O alkaline IO – + O T1 = T3 = T5 = T7 = 298K 23 32 for adiabatic step = 2 – 3 16. Ans.(A) Sol. P + IO H PO + I 43 34 2 P31–r T3r P21–r T2r white 1r 17. Ans. (B) 32 r T3 = T2 18. Ans. (D) 1 HS-16/16 0000CT103116003
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