12th (Date: 05-01-2011)ANSWERKEYPHYSICSSECTION-3PART-AQ.1 CQ.2 BQ.3 CQ.4 CQ.5 BQ.6 CQ.7 BQ.8 BQ.9 AQ.10 DQ.11 DQ.12 CQ.13 BQ.14 AQ.15 CQ.16 AQ.17 AQ.18 DQ.19 BQ.20 CQ.21 AQ.22 CQ.23 BQ.24 B,C,DQ.25 A,DQ.26 A,B,DQ.27 A,CQ.28 A,CPART-BQ.1 (A) R(B) Q(C) Q(D) QPART-CQ.1 0002Q.2 0027MATHSSECTION-1PART-AQ.1 DQ.2 DQ.3 AQ.4 DQ.5 CQ.6 AQ.7 CQ.8 BQ.9 BQ.10 AQ.11 DQ.12 CQ.13 DQ.14 BQ.15 A,B,C,DQ.16 A,B,CQ.17 A,BQ.18 A,B,C,DPART-CQ.1 0014Q.2 0003Q.3 0001Q.4 0002Q.5 0005Q.6 0002CHEMISTRYSECTION-2PART-AQ.1 BQ.2 AQ.3 BQ.4 BQ.5 AQ.6 BQ.7 DQ.8 BQ.9 CQ.10 CQ.11 BQ.12 CQ.13 AQ.14 BQ.15 BQ.16 AQ.17 DQ.18 CQ.19 A,B,C,DQ.20 A,B,DQ.21 A,BQ.22 A,CQ.23 B,C,DQ.24 A,CPART-CQ.1 0002Q.2 0508Q.3 0040Q.4 0112Q.5 1637Q.6 0320MATHEMATICSXII Page # 1PART-AQ.1[Sol. LetL =! "x 3 1x 11xx 7 e Lim #$ %($ form)! "&&'())*+% $ % $ , # -&'()*+# - #$ % $ % $ %0x 31Lim and ) 0 1 ( eex 71 Lim x 7 e Lim Asxx 11x 11xx 11xln L= ! "x 3x 7 e nLimx 11x#$ %l(UsingL-Hospital rule)ln L = ! "! " x 7 e 37 e 11Limx 11x 11x##$ % = ! "! "x 11x 11xxe 7 1 3e 7 11Lim##$ %##Hence L = 311e]Q.2[Sol. As, f (x) = 1 + ln&'()*+. . 1 x x2 + 5x3 4x4Clearly, ln &'()*+. . x 1 x2, 5x3 are odd functions, soI =! "4x114e x 4 1##/# dx= 2! "4x104e x 4 1#/# dx = 2 e20e12ex2 dx e . xdxd1010xx44- &'()*+# -&&'())*+- &'()*+/# ]Q.3[Sol. ! f is onto0 Range of tan1 (2x x2 + 1) should be 234)*+ 5 #0 ,26Range of 2x x2 + 1 should be ($, 0] ; henceD = 0 6 4 + 41 = 0, hence 1 = 1 ]Q.4[Sol. f ' (x) = 789: #; . #1 x , 21 x , 10 x 2 x 32since for the quadratic 3x2 2x + 10,D < 0 hence it is always positive6 function f (x) increases in ($, 1) and decreasingin (1, $)0 for max at x = 1f (1) < f (1+)6 1 1 + 10 5 < 2 + log2 (a2 2)log2 (a2 2) ; 7a2 2 ; 276 a2 ; 130Hence, 2 < a2 ; 1306 a = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}0 No. of integral values of a = 20 ]MATHEMATICSXII Page # 2Q.5[Sol. f (x) =/ /# #.1xx tx0t xdt e dt ef (x) = / /# #.1xt xx0t xdt e e dt e e(1, e 1)(0, e 1)21x - x = 1XYOGraph of y = f(x)in [0, 1]2( e 1) >f (x) = ex(1 ex) + ex(e ex)0 f (x)= ex + e1 x 2f ' (x) = ex e1 x = 00 x = 21f '' (x) = ex + e1 x > 0 6 minima?@A# -# -1 e ) 1 ( f1 e ) 0 ( f Maximum value&'()*+21f = ! " 1 e 2 #% Minimum value]Q.6[Sol. Fory < 0,0y = e | x | 21 = 77789B #< ##0 x if21e0 x if21exx.....(1)and for y < 0, y = 21 e | x | = 77789B #< ##0 x if e210 x if e21xx.....(2)Combining (1) and (2) we have the required area given asOln 2 n 2 lxy2 1 #2 1A = 4 /&'()*+##2 n0xdx21el = 4 &'()*+# 2 n2121l = (2 2 ln 2)Ans.]Q.7[Sol. ! " ! " ! "/#20dx x 4 f x f= 10. Let x = 2t6 dx = 2dtSo, ! " ! " D E/#10dt t 8 f t 2 f 2= 10 6! " ! " D E/#10dx x 8 f x 2 f = 5 ... (1)Given that ! " ! " ! "/#10dx x 2 f x f= 5 ... (2)From (1) + (2), we get! " ! " ! "/#10dx x 8 f x f= 10 Ans.]MATHEMATICSXII Page # 3Q.8[Sol. Let Ak = (2t, t2)0 Slope of FAk = 0 t 21 t2## = tan&'()*+F .5k20 tan Fk = 2t 1t 2# = tan(2G) (Say)F(0,1)Fk+ FkO(0,0)xyA(2t,t)k2526 G = 2kF = n 4k5 where tan G = tAlsoFAk = 2 2 2) t 2 ( ) 1 t ( . # = t2 + 1 = 1 + tan2G = sec2&'()*+ 5n 4k0 H H-$ %-$ %&&'())*+&'()*+ 5-n1 k2nn1 kkn nk4secn1Lim FAn1Lim =/&'()*+ 5102dx4xsec = 104xtan4234789 55 = 54 Ans.]Paragraph for question nos. 9 to 11[Sol.(i) Let /1 -20dx ) x ( f (where 1 = constant)0 f '(x) = f(x) + 16 / /-1 .dx 1) x ( fdx ) x ( ' f6 ln(f(x) + 1) = x + c(where cis integration constant)6 f(x) + 1 = ec ex!f(0) = 1 6 1 + 1 - ec0 f(x) + 1 = (1 + 1)ex 6 f(x) = 1 + (1 + 1) ex!1 =/20dx ) x ( f =! "/1 . . 1 #20xdx e ) 1 (= 21 + (1 + 1) (e2 1) 6 31 = 1 (e2 1) + e2 16 1 (4 e2) = e2 1 6 1 = 22e 41 e##0 f(x) = 22e 4e 1## + x22ee 41 e1&&'())*+##.= 22e 4e 1## + x2ee 43&'()*+# = &&'())*+#. #2x 2e 4e 3 e 10 For the number of points of intersection with x-axis1 e2 + 3ex = 063ex = e2 1Clearly,y = 3exis increasing function and it will cuty = 3e2 1 at only one point.MATHEMATICSXII Page # 4(ii) g(x) = IJKB 0O x > 06 f ' (x) > f (0)6ln(1 + x) > x 1x tan1.#]Q.17[Sol.f ' (x) = x 22x 1 xx 1x1. .&&'())*+.. +ln &'()*+. . 1 x x2 1 xx2.= 1 xx2. + ln &'()*+. . 1 x x2 1 xx2.= ln&'()*+. . 1 x x2> 0 O x > 0< 0 O x < 00 h ' (x) = (1 2 f (x) + 3f2 (x)) f ' (x)= f ' (x) (1 2 f(x) + 3f2 (x))D < 06 h (x) is increasing when f is increasing & decreasing when f is decreasing0h (x) increasing in (0, $)& h (x)decreasing in (0, $) ]Q.18[Sol. See graph y = f(x) =2 | 3 x 4 x |2# . # , y = m is a horizontal line with intersection points, from whichthe x-values have different signs, only if m > 2.]MATHEMATICSXII Page # 8PART-CQ.1[Sol. f (x) = i x 21# = 1 x 4i x 22..Let (h, k) be locus of points a1, a2............a6.then h = 1 x 4x 22.,k = 1 x 412.0kh = 2x0 k = 1 k h12 2.= 2 22k hk.6 h2 + k2 k = 00 Locus is x2 + y2 y = 0centre = &'()*+21, 0radius = r = 21A hexagon inscribed in circle will have side same as radius0 Area = 6 Area of one triangle = 6 43 221&'()*+ = 83 3Hence a = 3; b = 3 and c = 86(a + b + c) = 14 Ans.]Q.2[Sol. We have, f (1) =a + b + cand f (2) = 4a 2b + cHence f (1) f ( 2) = 3(b a)So, E = a bc b a#. . = ) 2 ( f ) 1 ( f) 1 ( f 3# # = ) 1 ( f) 2 ( f13##Hence Emin. occurs whenf (2) = 0. R x 0 f(x) as negative be tcan' c, f(0) Clearly,. positive also a is b ) given ( a b and0 a so R, x 0 f(x) as 0 c) b (a f(1)= O < -6 :: = O < : . . - : NoteHenceEmin. = 3 Ans.Aliter: f (x) < 0Ox6 a > 0andb2 4ac ; 00 c < a 4b20 a + b + c < a + b + a 4b2a + b + c < a 4b ab 4 a 42 2. .since b a > 0MATHEMATICSXII Page # 90a bc b a#. . < ) a b ( a 4b ab 4 a 42 2# . . = ) a b ( a 4) b a 2 (2#. = ) a b ( a2) a b ( a 32#&'()*+ # .UsingA.M. < G.M. on3aandb a < ) a b ( a) a b ( a 3## = 3Equality holds when3a = b a0 4a = b6 b = c = 4a.Ans. ]Q.3[Sol. We havef (x) = 777789&'()*+55= # 5234789 5=,2x ; ) x (2, 0 x ; x22;g (x) = 777789&'()*+55= # 5234789 5=,2x ; ) x (2, 0 x ; x

0 Required area = 2 &&'())*+# . #/ /5 212102dx ) x x ( dx ) x x (= 2 &&'())*+&&'())*+#5#&&'())*+#5. &'()*+#218 3124 31212 3= 2&&'())*+.5#5318 242 3 = 324 122 3.5#5Pa53 + b52 + c(Given)So, a=121 ,b=41 #,c=32Hence,(| a | +| b | +| c |)=3241121.#.=3241121. .=128 3 1 . .=1212=1. Ans.]Q.4[Sol. We have , f(x) =x cos 4 x sin2 4.x sin 4 x cos2 4.= x sin 4 4 x sin2 4# . x cos 4 4 x cos2 4# . = 2 sin2x (2 cos2x)= cos2x sin2x = cos 2xGiven,g(sin 2t) = sin t + cos t6 ! " ! "2t 2 sin g = 1 + sin 2tMATHEMATICSXII Page # 106 g (sin 2t) =t 2 sin 1.So, g(x) =x 1. , 1 ; x ; 1Now,! " ) x ( f g= g(cos 2x) =x 2 cos 1.= x cos2O x = R.0 Range of ! " ) x ( f gis D E 2 , 06 (a2 + b2) = 2Ans.]Q.5[Sol. LetAF = x = DEandAE = y = DFAs QCABis QCEDSo,ABDECACE- 6cxby b-#6y = b &'()*+#cx1(HereBC = a, AC = b and AB = c)Now, area of parallelogramAFDE= S = (AF) (EM) = xy sin A 6S = x b &'()*+#cx1 sin A (Note : sin A isfixed) ........(1)Now, differentiating both sides of equation (1) with respect to x, we getcbdxdS- (c 2x) sin A=062cx -Also,0cb 2dxs d2cx22B#-234-So, S is maximum whenx =2cNow,Smax =41 bcsin AA= &'()*+A sin bc2121 = ! " ) ABC ( area21Q = |1 3 91 2 41 1 1|41## = 420 =5(square units.) Ans.]Q.6[Sol. We have7789== #-I x ; 0I x ;21} x {) x ( fClearly,f(x) is periodic with period = 1.Now, /240dx ) x ( P= " "240240 I IIx ) x ( P dx ) x ( P 1) P . B . I (-/ /240dx ) x ( ' P x= 24 P(24) /240dx ) x ( ' P xAs, P(24) = /240dt ) t ( f= /10dt ) t ( f 24= /&'()*+#10dt21t 24= 0MATHEMATICSXII Page # 11and/ / / / /&'()*+# . .&'()*+# .&'()*+# - -24232402110240dx247x x ......... dx23x x dx21x x dx ) x ( f x dx ) x ( ' P x/- , -240212124 ) x ( ' P x &'()*+121is integral definite everyof value AsSo,/240dx ) x ( P=2 0#= 2.Ans.]XII Page # 1CHEMISTRYPART-AQ.2[Sol. Correct option (A) as all other Nitrogen lonepairs in the given structure are participating in resonance sotheir Nucleophilic power is decreased. ]Q.5[Sol. (A)O CN|| |Me C CH Me ! ! !" " " # "4 2SO H . aq ) acid keto (COOH|Me C CH Me||O! $! ! ! 2CO ! %"# " O||Me C MeCH2! ! P is ketone so will not give Fehling test, NaHCO3 test or cerric ammonium test. It will give ...... test withNaHSO3 and idoform test.]Q.6[Sol. HSO4 ! H+ + SO421020.01x x0.01 + x102 = x 01 . 0) x 01 . 0 ( x!&104 102 x = 102 x + x2x2 + 2 102x 104 = 0x = 210 4 10 4 10 24 4 2 ! ! !' & ' ( ' != 210 2 2 10 22 2 ! !' ( ' != 210 2 2 22 !' & ![H+] = 210 ) 1 2 (!' !pH = 2 log ) 1 2 ( !]Q.8[Sol. In osazone formation only C1 and C2 carbon react remaining carbon configuration will be same. ]Q.13[Sol. A ) H C * C CH2 CH2 CH3 most acidic among C5H8 alkyne 2 + HB ) H3C C * C CH2 CH3 most stable alkyne 5 + HC )alkadieneC C C C C H *NaNH2! C C C C * 3NH Na C &,!]Q.14[Sol. 5 + H ]XII Page # 2CHEMISTRYQ.15[Sol. Number of C atom are more ]Q.16[Sol. N2 (g) + 3H2 (g) ! 2NH3 (g)0.32.540.3 0.42.540.183 0.182=0.12 =2Kc =100) 2 ( 12 . 036 . 0 36 . 03 '''= 13.5]Q.17[Sol. HCl + NH3 "# NH4Cl0.5 0.36 0.14 0.36(HCl) = M 28 . 02 / 114 . 0-]Q.18[Sol.1412 . 02228PPrr2222NHNH' - ' -]Q.20[Sol.2ALA is not found during ester hydrolysis as H2O is a poor nucleophile ]Q.21[Sol. (A) Sucrose + H2O "# Glucose + FructoseK =./0123!!44t0r rr rlogt303 . 2; K = ./0123! !! !20 1530 15log30303 . 2K = ./01233545log30303 . 2K = 9.212 103t1/2 = K2 log 303 . 2 ' = 310 212 . 93 . 0 303 . 2!'' = 75 min ]XII Page # 3CHEMISTRYQ.23[Sol. (A)OOHHN > OOHHN>OOHNH(B)O! a.]PHYSICSXII Page # 4Q.16[Sol. Displacement wave suffers a phase change of ! at the closed end and no phase change at open - end. Asa result displacement node is formed at closed end and antinode at open end.]Q.17[Sol. Let x = 0, t = 0, ymax = 4 103y = 2ymax2 103 = 6 2 3610 4 V 10 6 . 110 4 . 6- --? ? ) ??V = 20 m/s ]Paragraphforquestionnos.18to20[Sol.(11) v = ty@@v = 0.05 360 cos 12 x cos 360 tat t = 0, x = 0, v = 18 m/s.(12) If string vibration in n loopsL2n6An = 1,2,3....(13) C =25012360K6 6B]Q.21[Sol. A1V1 = A2V24cm2 V = 1 mm2 2m/s400 mm2 V = 1mm2 2m/sV = 4002 = 420mm/s = 5 mm/s ]Q.22[Sol. Height of spring = V/A = 5cmWork done = ,KE of water + gain in PE of water + gain in PE of piston= 21(V*)v2 + m1gh1 + m2gh2= 21[(20 106)1000]22 + [20 103 10 2.5 102] + [0.1 10 0.05]= 0.04 + 0.005 + 0.05= 0.095 J ]PHYSICSXII Page # 5Q.24[Sol. Net downward force~ ~~~~ ~~~ ~~~~~WW + W1~~force = W1 + W W( W2 = W1]Q.25[Sol. a = g tan +T = 2! effgL = 2!+ cos / gL = 2!gcos L +]Q.26[Sol. x = 2t2 4tv = dtdx = 4t 4a = dtdv = 4 (constant) hence (D) correctv = 0 = 4t 4 ( t = 1 sec. hence (B) correctfor t > 2 sec. , x & v are both positive hence (A) correct ]Q.27[Sol. distance travelled by wave in one second w.r.t. wall = v + v0wave number = 1/A = v/f/ Number of waves striking wall per second = ) / v (v v0f)]Q.28[Sol. y = A cos [(kx Bt)(A) here k = 0.5!, B = 200!v = kB = !!5 . 0200 = 400 cm/sec(B) ,C = A! 2,x = 42! x 50 = 25! = 12 2! + ! = !(C) 2A = 2 cm(D) y = 2 cos [!(0.5 100 200 100)] = 2 (not zeroPHYSICSXII Page # 6PART-BQ.1[Sol. In A : Same water level implieswtt of fluid displaced is the same as cut of object hence both backets have equal weight.In B,C: Mass of water in both buckets is equal and B has additional mass of solid object hence B isheavies.In D : Same water level and object sinks *0 > *B i.e. some volume of *B is replaced by same volumeof *0 mass increases.]PART-CQ.1[Sol.T = yA . ,+ = 40N = 0.1C = ET = 20 m/s ( A = fC = 1mString vibrates in I overtoneequatio of amplitudeA = a sin x2A!x1 = 0.75 mx2 = 0.125 m2112 / 1) 75 . 0 2 sin() 125 . 0 2 sin(AA216 6? !? !6221AA""#$%%&' = 2 Ans. ]Q.2[Sol.11 T!?E 4 . 0 4320? = 8T = 27 N ]