Home Explore 10- Solution Report (10)

# 10- Solution Report (10)

## Description: Solution Report (10)

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 11  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  k0  ak  k0    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.ntan x tan x.nsin x  lim 22  1 1 x4  x2 1  3 x0 22  2  x4 1 7 x   sin x.ntan x tan x .nsin x 5. Ans. (A) lim  e x 1/x  e x 1/ x  Let a variable point on the parabola be x0 (1 + t2, 2 t) and its reflection in the given 1+1=2 line be (h,k) lim 1  ex  1  cos x sin x 1  ex  1  cos x  2 1  t2  2t  2 x0  (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  x0+ sin x k t 1t 1 x  h  21  t  t 1  2k x 1  ex  1  cos x h (C) Base is exact 1 and t  2k 1 n 1n  22n  3 h lim 6. n.n 12n 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 y1 z1 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 zc d 1 Equation of L1  0  b  c a b c d det A  I x y zc 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  102  9  62 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 tan1 x whose modulus is less than 1. (i) & (ii) same  1  m  2m  m3 1 m km  m  m3 tan1 x 24 ƒ  x    2 tan1 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 tan1   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 ƒx ƒ x  0, 4  and sin1 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 1z 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 103 p  c  1  2  3  25  2 r5 23 dp 5c dr  127 102  2.54  102C 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 103 = 5 mN v 20 30 20 1 11  23  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. iB h   Bv  q  iR  0 H c Q = 0 = U + W iB  mdv = n1CvT + n2CvT 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 Bv mv mR dv – 2CV 2C 2CV Bc B dt 3 3     – 2CV 3 a=0 C  Bc 2CV m B22c  m 3  v   Bc  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  bT  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 103 1    T  20 × 2 v 50.8 dt  v = f e  76  75.6  0.2  ve  2k  100  T 2 n 2 v    2  1  Case-3 : v %  0.1 100  0.33% 3(T – 20) = 2.8 103 100  T v v 30.4 2  1 0.7 3T – 60 = 400 – 4T e  2  31 e  46.0  46.8  ve 7T = 460 2 2 Case-4 : v %  0.1 100  0.32% T  460  65.7C v 31.5 7 e  48.2  48.0  0.1cm dH ' 3  103    460  20   2 2  7  dt  8. Ans. (A,C) dH 2.4 101   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  1C 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 106 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  102 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  et  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 11 11 5 i = 2A 5. Ans. 4 63 iT  60    50.1n2  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 l4m 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 107  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 107 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 position vectors 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 0i4 A(0,0) B(a,0) 1 tan A  y tan B  y P(Bi )  5 x ax 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 ax   1 y2  P  136 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 11 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 tdt x0 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 tdt  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) = 2xn2 – 2x  P(AB)P(E/ AB)  P(AB)P(E/ AB) ƒ''(x) = 2xn22 – 2 ƒ'''(x) = 2xn32  366  74 7  349 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 PAB  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  (10099 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  b2  2 Solution for Question 15 to 16 Let  be positive real root  a(3 + ) + b2 + (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  22  1  2  1 2  (g(x)  g '(x))ƒ2 (x)g2 (x)  g(x)  g'(x) 6  34  2  2  g(x) + g'(x) = 0  log g(x) = –x + c    g(x) = e–x (a2 + b2)min =     g(x) = (1  2)ex  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 44 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 82 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 103  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  vcb n (2) G 4 vca 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 01 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  1.52 10 103  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  8R2S  U 0 2  4 0 R Q2  8R2S  U Sol. 8 0 R dU 0 = amplitude dR = 0  same.  R  Q2 U = oscillation energy  same 8 0 16S P  P0  4S  2 1 1 R 2 0 2 2 U  mVC2  IC2 4S Q2  1 mVC2  1 IC VC2 = P0 + R  2 0 162R4 2 2 R2 = P0 + 4S Q2  1282S 0  P0 R  322 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 + I00 = 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 I2 i (1 + R) = 2 2 2 i 2  1 mV 2  1 Iv 2 1R 2 2 R2 16. Ans. (A) 4R  1 v2  m  I  2  R2  Sol. P = i2R  1  R2 L2 dP  0  4(1 + R)2 – 4R × 2(1 + R) = 0 dR  2I  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 alkaline 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 1r 17. Ans. (B)  32  r T3 = T2 18. Ans. (D)  1  HS-16/16 0000CT103116003