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1 SMLE QUESTION ER Clss XII Mthemtis Time llowed: hrs Mimum Mrks: Generl Instrutions: i ll questions re ompulsor. ii The question pper onsists of 6 questions divided into three Setions, B nd C. iii Question No. to 6 in Setion re Ver Short nswer Tpe Questions rring one mrk eh. iv Question No. 7 to 9 in Setion Bre Long nswer I Tpe Questions rring four mrks eh. v Question No. to 6 in Setion Cre Long nswer II Tpe Questions rring si mrks eh. vii Use of lultor is not permitted. You m sk for logrithmi tles, if required. SECTION -. Find the ngle etween the lines z nd 6 z.. Find if the vetors i j k, i j k, j k re oplnr.. If m nd n re the order nd degree respetivel of the differentil eqution d d, then write the vlue of m + n. d d. Find the differentil eqution representing the fmil of urves e Be where nd B re ritrr onstnts.. Find unit vetors perpendiulr to the vetors i j k nd i j k. 6. On epnding first row, the vlue of third order determinnt is. Write the epression for its vlue on epnding nd element ij. olumn, where ij is the o-ftor of SECTION - B 7. To promote the mking of toilets for women n orgnistion tried to generte wreness through i House Rs., ii Rs iii Rs.. The numer of ttempts mde in three villges X, Y, Z re given elow. House Clls Letters nnounements X Y Z 7

2 Find the totl ost inurred the orgnistion for the three villges seprtel using mtries. Write one vlue generted the orgnistion in the soiet.. Using properties of determinnts, prove tht: 9. Evlute: d ot. ~. Let,, 6 e point in spe nd Q e point on the line, r i j k i j k then find the vlue of for whih vetor Q is prllel to the plne z Find the distne of the point -,, - from the line to the plne z.. z mesured prllel z. Show tht the lines nd z interset. Find the point of intersetion lso.. If os os, prove tht os Sin. rove: tn os. Evlute:- d Evlute: /. Evlute: Sin n d tn os. Find the inverse of mtri If = d., using elementr row trnsformtions., Find f if f..,

3 6. If, prove tht: d d 7. Show tht e e d d is solution of differentil eqution.. Two die re thrown simultneousl. Let X denote the numer of Sies, find the proilit distriution of X. lso find men nd vrine of X using proilit distriution tle. mn tkes step forwrd with proilit. & kwrd with proilit.6. Find the proilit tht t the end of steps, he is one step w from the strting point. 9. Show tht funtion f is ontinuous ut not differentile t =. SECTION - C. Let X e non-empt set. X e its power set. Let * e n opertion defined on elements of X Then, i ii iii * B = B, B X. rove tht * is inr opertion in X. Is * ommuttive? Is * ssoitive? iv Find the identit element in X w.r.t * v vi Find ll the invertile elements of X. If o is nother inr opertion defined on X s o B = B, then verif tht o distriutes itself over *. Let = If,,,...9 nd R e the reltion in defined, R, d if + d = + for,,, d. If. rove tht R is n equivlene reltion. lso otin the equivlene lss [, ].. Find the solute mimum nd solute minimum vlues of the funtion f given f Cos Sin,,.. furniture firm mnuftures hirs nd tles, eh requiring the use of three mhines, B nd C. rodution of one hir requires hours on mhine, hour on mhine B nd hour on mhine C. Eh tle requires hour eh on mhine nd B nd hours on mhine C. The profit otined selling one hir is Rs. while selling one tle the profit is Rs. 6. The totl time ville per week on mhine is 7 hours, on mhine B is hours nd on d d

4 mhine C is 9 hours. How mn hirs nd tles should e mde per week so s to mimize profit? Formulte the prolem s L.. nd solve it grphill.. given quntit of metl is to e st into hlf irulr linder i.e. with retngulr se nd semiirulr ends. Show tht in order tht the totl surfe re m e minimum, the rtio of the length of the linder to the dimeter of its irulr ends is :.. letter is known s hve ome from either TTNGR or CLCUTT. On the envelope just two onseutive letters T re visile. Wht is the proilit tht the letter hs ome from i TT NGR ii CLCUTT. Sketh the grph of: Evlute: f,, f d. Wht does the vlue of this integrl represent on the grph? d 6. Solve the following differentil eqution, given = when =. d / Find the prtiulr solution of the differentil eqution e sin sin given tht =, when =. d d,

5 SMLE ER Solutions. Eqution of lines 6 / / z nd z ngle etween the lines Cos 6 6 Cos Cos Cos Cos ; So, 9. Let k j k nd j i k j i, for oplnr.. k j i 6 k j i k j i 7 nd.

6 i j 7k. j k ; ; 7. d d d d Squring oth sides d d d d d d d d d Order, m =, degree n = So, m + n = + =. Given m + n = e Be d... Differentiting twie w.r.t., we get d. e Be d d d. d d d d d d e d d e Be Be nd d d using d d d d i.e., whih is the required differentil eqution. d d. Let i j k nd i j k Vetor perpendiulr to oth Let i j k & = i j k = i k i k i k Unit vetor i k 6. On epnding nd Column

7 Let = 7. 7, B Totl ost the orgnistion for eh villge is represented olumn mtri. X C Y Z, Here C = B X Y 7 Z 9 There orgnistion is helping women in the soiet.. We hve Multipling R, R, R,, respetivel nd tking out,, ommon from C, C nd C, we hve Now ppling C C C nd C C C, we hve B tking out ++ ommon oth from C nd C we hve R ppling R R Now, epnding long C, we hve.

8 = R. H. S 9. Let I ot d = tn d ot tn = tn d = tn tn n e written s d tn tn tn = tn d tn d = tn d tn [ ] d f f d tn.d = tn d ] = tn d. [log = log log log r i j k i j k. Line = Coordintes of point Q on the line = Given point,, 6 DR s of Q =,, 6 =,, Sine Q is to the plne + z =. Norml of given plne = i j k s Q 6 ; Co-ordintes of Q 7, -, - Eqution of plne: z So norml to the plne will,, n. [ log ]

9 Be, k j i nd Eqution of line = / / / z Vrile point Q on the line = /,, Vetor k j i Q = k j i 9 Sine Q lne. Q. 9 k j i k j i 6 ; So Co-ordintes of Q =,, Q =, /, Distne Q = 6 6 = Distne Q=. units. For the lines to interset shortest distne etween them must e zero. ; : z l z l k j i k j i, k j i k j i, k j i

10 = i j k ; i j k i j k Distne d =. Sine the shortest distne is zero for point of intersetion z z Let nd z z On ompring nd On solving &, So point of intersetion = -, -, -. Let Cos nd Cos B, it gives Cos, CosB So B Cos +B= Cos Cos Cos B Sin Sin B = Cos Cos Cos Cos Squring oth sides Cos Cos Cos Cos

11 Sin Cos Let Cos So, Cos L.H.S tn tn = tn. tn tn tn tn. tn tn tn = B tn tn tn tn tn tn tn = tn tn tn tn tn = Cos ; / ; ; tn tn. d Dividing Numertor nd Denomintor = d = d d ; d d = v u where v dv du &

12 = log tn v v C log tn d Let B d d B B On ompring =- = - +B = = +B=; = = = d = d d = d dt t = d t / = C Sin / / = C Sin 9. Here sin sin sin for for f Therefore / sin sin sin d d d = sin sin d d Integrting oth integrls on right hnd side, we get

13 / sin os sin os sin d. Let = I then, R Operte R R R R Operte R R Operte R R R Operte R R R Operte R R Operte R R R Operte R

14 I = B, where B = Hene, - = B = 6. Tking log on oth sides Log = log log log log d d d d d d d d. 7. e e e e d d e e d d d d d d

15 d d d d [{ e e d d d d. Clerl, X n tke vlues, nd. We hve, X roilit of not getting si on n die = 6 X roilit of getting one si = 6 X roilit of getting two sies = 6 Thus, the proilit distriution of X is given X: X: X i i = X= i i i i i p i i 6 p i i 6 We hve, p i i nd pi i 6 E X = p i i 7 nd, Vr X= p i i pi i Hene, EX = nd Vr X 7 The mn is one step w from the strting point fter steps. This n hppen in two ws i 9 He tkes steps forwrd nd steps kwrd

16 ii He tkes steps kwrd nd steps forwrd. roilit of first se = C roilit of seond se = Hene totl required proilit = = f Here f C..6 C C = = if if if if.6.6 lim lim f lim f lim nd Sine, lim f lim f f The funtion if ontinuous t = for differentiilit test L.H.D f R.H.D f f h f h h lim lim lim h h h h h k f h f h lim lim h h h h Sine, L.H.D + R.H.D f is not differentile t =. Given * B B, B X i Sine,, B X B X, B X Hene, *B is inr opertion. ii * B B B B*, B X So, * is ommuttive. iii * B* C B* C B C B C = B* C * B* C, B, C X

17 itself. iv v vi So, * is ssoitive. Let E e the identit element in X with respet to *. Then * E = E * = For ll X E E For ll X, X = E = X Thus, X is the identit element with respet to * on X. Let e n invertile element of X nd let S e its inverse. Then, *S = X = S* S X S = = S = X [ X, S X ] Thus, X is the onl invertile element of X with respet to * nd it is the inverse of We hve, the reltion * B = B,, B X o B = B,, B X Now, o B*C = B C = B C = o B * o C i lso, B*C o = B C B C = B o *C o ii Hene, from i nd ii, we onlude tht o distriutes over *.. f Cos Sin, f f Cos Sin Cos Cos Sin For ritil points, f Cos Sin Cos, Sin, nd, 6 6 ll points re f Cos,,,, 6 6 Sin f Cos 6 Sin 6 6

18 f Cos f Cos 6 f Sin Sin 6 6 Cos Sin Hene, solute mimum = solute minimum =. Let hirs nd tles e mde per week. Now, ording to question we hve Mhines B C Chirs Tles Here, totl time ville per week on mhine is 7 hours. 7 Tht of mhine B is hours. nd mhine C is ville for 9 hours. 9 Its profit, Z So, the L is + 6 Sujet to the onstrints Z m = i ii 9 iii, iv

19 On plotting i to iv, nd shding ording to onstrints we hve Here, OBCD is the required fesile region show in figure whih is ounded. Now, we find the vlue of Z t eh orner point. Corner point Z, B, C, D, 9 So, the furniture firm hs mimum profit of Rs. 9, if the firm mde hirs nd tles per week.. given quntit of metl is st into hlf linder with retngulr se nd semiirulr ends. Let the length of retngulr se = nd its redth =. Dimeter of the semiirle = Rdius =. Let the volume of metl = V, whih is known nd is onstnt..... i V V V To minimise the totl surfe re, S, we hve S =.. V V. V V d ds V V d ds V

20 V d ds V d S d When V... V V d S d. Totl surfe re, S, is minimum When v = nd ] [. i from V : : Rtio of length of the linder to its dimeter = : +.. Let E e the event tht the letter me from Clutt nd E e the event tht the letter me from Ttngr. Let denote the event tht two onseutive letters visile on the envelope re T. Sine the letters hve ome either from Clutt or Ttngr, therefore,. E E If E hs ourred, then it mens tht the letter me from Clutt. In the word CLCUTT there re 7 onseutive lphet i.e., {C, L, LC, CU, UT, TT nd T} nd T ours onl one time. Therefore, 7 / E If E hs ourred, then the letter me from TTNGR. In the word TTNGR there re onseutive letters i.e., {T, T, T, N, N, G, G, R} in whih T ours twie. Therefore, / E B Be s theorem: i 7 7 /. /. /. / E E E E E E E ii 7 7 /. /. /. / E E E E E E E

21 . Given, For f,,, f X F For, f X Y 7 The grph of the funtion is shown in the figure. Now, f d f d f d = d d = d 6 = 6 6 = 6 sq. units The ove vlue of the integrl represent the re of the shded region on the grph d 6., given =, = d

22 . d d It is liner eqution of the form Q d d Here,, Q Now, I.F= log log e e e e e d d d Its Solution is d F I Q F I.. d d d = - d d d C sin sin C sin Now, when =, = we hve C Required solution is sin

23 SMLE ER MTHEMTICS CLSS - XII Time llowed: Hours M. Mrks: Generl Instrutions: i ll questions re ompulsor. ii The question pper onsists of 6 questions divided into three Setions, B nd C. iii Question No. to 6 in Setion re Ver Short nswer Tpe Questions rring one mrk eh. iv Question No. 7 to 9 in Setion Bre Long nswer I Tpe Questions rring four mrks eh. v Question No. to 6 in Setion Cre Long nswer II Tpe Questions rring si mrks eh. vi There is no overll hoie. However, internl hoie hs een provided in questions of four mrks eh nd questions of si mrks eh. You hve to ttempt onl one of the lterntives in ll suh questions. vii Use of lultor is not permitted. You m sk for logrithmi tles, if required.. Find if i 6j k i j 7k SECTION. Write the integrting ftor of. For wht vlue of, = d d. Write the re of prllelogrm whose digonls re d i nd d j. Find the produt of the order nd degree of the following differentil eqution. d d d d 6. = 7 Write the vlue of dj 7. If + SECTION B rove tht d d d d. There re fmilies, B nd C. The numer of men, women nd hildren in these fmilies re s under.

24 Men Women Children Fmil Fmil Fmil 6 Dil Epenses of men, women nd hildren re Rs, Rs nd Rs respetivel. Using mtri multiplition, lulte the dil epenses of eh fmil. Wht impt does more hildren in the fmil rete on the soiet. 9. Solve for : sin os sin If os os os z. rove tht z z. Evlute: 6 6. Evlute: os +os d -os log + d. If i k, i j k nd i j 7k.Find vetor r whih stisfies r nd r.. Find the distne etween two prllel plnes: r i j k 6 nd r 6i j 6k 7. Verif the men vlue theorem for the funtion f 6. Solve: d ot d on the intervl, 6. Consider the eperiment of tossing oin. If the oin shows hed, toss it gin, ut if it shows til, then throw die. Find the onditionl proilit of the event the die shows numer greter thn given tht there is t lest one til. die is thrown times. If the first throw is fir, then find the proilit of getting s sum. 7. For wht vlue of k is the funtion ontinuous t - tn f= k. The urve 6 is. Find the vlue of nd. psses through the point, nd the grdient slope of the urve t

25 9. Evlute: tn log +se log d + d SECTION C. Find the eqution of plne pssing through the point,, nd ontining the line r i j k i j k. Find the re ounded the prol. Evlute: nd line + =. ot d - d -. g ontins green nd 7 white lls. Two lls re drwn one one t rndom without replement. If the seond ll drwn is green. Wht is the proilit tht the first ll drwn is lso green.. Let e the set of ll rel numers eept i.e. R. Let is defined on s for ll,. rove tht i ii is inr opertion on The given opertion is ommuttive s well s ssoitive. iii Find the identit element. iv rove tht of hs is inverse. v Find vi Solve the eqution =. Find point on the hpotenuse of given right ngled tringle from whih the perpendiulrs n e dropped on other sides to form retngle of mimum re. point on the hpotenuse of right ngled tringle is t distne nd from the sides show tht minimum length of the hpotenuse is / / / 6. it uilding ode requires tht the re of the windows must e t lest ⅛ of the re of the wlls nd roofs of ll new uilding. Constrution ost of new uilding is Rs per squre metre of window re nd Rs per squre metre of wll nd roof re. To the nerest squre metre wht is the lrgest surfe re of new uilding n hve if dil onstrution ost nnot eeed Rs.

26 .. d IF e when = IF.. re of prllelogrm d d re of prllelogrm = sq. unit. Order = ; Degree = rodut = 6. dj. = I where = 6 dj SMLE ER SOLUTIONS SECTION B 7. Differentiting w.r.t., we get d d d d d d d d Squring oth sides, d d gin differentiting w.r.t.,

27 d d d d d d d d B nelling d from oth sides, we get, d d d d d d d d d. The mtri inditing the memers of three fmilies, B nd C re Men WomenChildren = B 6 C Where mtri T inditing the dil epense of men, women nd hildren Men T Women Children Totl epense of eh fmil is otined s T Epenses of fmil = Rs Epenses of fmil B = Rs Epenses of fmil C = Rs More hildren men more ependiture. So one should pln fmil in order to keep hek on the epenses. sin os sin os os sin sin os os sin sin os os os ut os os os

28 os os os os os os os =, os =, = Both of these vlue stisfies the eqution. Hene, = = os os os z os os os z os os z os os z os os z z z Squring oth side, z z z z. Tping out from C, from C 6 6 C C C Tke ommon from C

29 C C C B epending long first olumn. 9 os os os os os os os os os os os os os os os = - os+os sin - os+os d sin C log d log d Integrte prt tping log s first funtion log d d log log C log C. r r ut

30 r r d r where is slr r r i j 7k i j k r i j k 7 i r 7 ut vlue of in i, r i j k. The given plnes in rtesin form n e written s z = 6 6 6z 7 = Distne etween two prllel plnes = The distne from point z to the plne 6 6z 7 = 6 = 6z = z 7 9 z lies on z = 6 z 6 = = unit. f = 6 f = + Sine polnomil is everwhere ontinuous nd differentile, therefore f is defined in [, ] f is ontinuous in [, ] f is differentile in, f = 6 + ll the ondition of men vlue theorem re stisfied. So, there must t lest one point, suh tht f f f ' f f

31 . 6, Onl possile vlue of =, d ot d Dividing oth side, d ot d This is liner differentil eqution in the form of d Q d ot Q Q IF ot d d logsinlog log sin e e e e IF = sin IF = Q IF d+ sin sind I sind os os d using Integrtion prt os sin sin os sin ot sin 6. Let S the smple spe of the rndom eperiment, then S = [HH, HT, T, T, T, T, T, T6] HH = = HT = = nd the proilit of eh elementr event T, T, T, T, T nd T6 is 6 H ½ H ½ ½ T ½ T /6 /6 /6 /6 /6 /6 6

32 Note tht ll the elementr events re not equll likel. Let e the event of the showing numer greter thn nd B e the event of tht there is t lest one til then = [T, T6] B = [HT, T, T, T, T, T, T6] B = [T, T6] B = B = 6 Required roilit = /B = B / 6 B / 9 = getting s the sum in throw of die three times B = getting on first throw Then B hs 6 6 = 6 equll likel ses, out of whih two outomes,, 6 nd, 6, re fvorle to. Required roilit = /B = roilit of getting s the if there is on the first throw = 6 7. Let = + h, so tht when, h lim f lim tn lim h tn h h lim h tn h lim h ot h h lim h tn h h lso f = k when = f = k lim f f k. The given urve is = + 6 Diff. w.r.t., d d 6 d d Slope of tngent t, is = =

33 lso urve psses through the point, = + 6 = + 6 We know tht = put the vlue of in ove eqution. = = = = ns. 9. tnlog se log d tnlogd se logd Integrte the first integrl prt tking tn log s first integrl tnlog se d se logd tnlog ut = onl for prtil frtion Let B B B equting the of nd onstnt term = B = + + B B solving, we get, B = 6, = 6 d d d d 6 6 tn tn tn tn. Eqution of plne pssing through,, SECTION C + + z =..

34 If the plne ontining given line, r i j k i j k lne must psses through, nd must e prllel to the given line, + + = + + =.. lne is lso prllel to given line. Norml of plne will e to given line, =.. B solving nd, = = = DR < > ut these vlues in eqution, + z = + + z = + z = whih is the required eqution of plne.. To get point of intersetion solve the eqution + = nd = B solving these equtions, we get, nd, Required shded re d d,,, X, d = 9 squre unit.. ot - -+ d = tn - d ot tn

35 tn d tn tn tn tn d tn d 9 tn d tn d fd f d tn d B using integrting prt tking s seond funtion tn d log log d If = = + = =,, Hene [ ] is divided into three intervl on [, ] on [, ] on [, ] d d d d. Let E, E, E e the events s defined elow E = First ll drwn is green E = First ll drwn is white = Seond ll drwn is green E E 7 /E = roilit of drwing seond green ll when one green ll hs lred een drwn 9 /E = roilit of drwing seond green ll when one white ll hs lred een drwn 9 B Bes Theorem,

36 E E E E E E E 9 9. i Consider n, = R { } = + + R ut we must show tht + + If possile let + + = = + + = =, = whih is wrong for ll, is inr opertion on ii For ll,,, we hve * = + + = + + = * the given opertion is ommuttive lso * * = + + * = = * * = * + + = = * * = * * the given opertion is ssoitive. iii Let e e n identit element of the given opertion, then e * =, for ll e + + e = e + = e = We see tht * = + + = ; nd * = + + = is n identit element. iv Let e n inverse of * = e = + + = + = + = = is n inverse of. v = + + = = + + = + = 9 vi * * = * * = * + + = = + 7 = =

37 . Let e the length of the hpotenuse of the given right-ngled tringle BC t B nd CB = in rdin mesure, < <. s BC is given, nd re fied i.e. onstnt. C N Let e point on the hpotenuse C nd =, then C =. Let M nd N e perpendiulrs to the other sides, then M = sin nd N = os Let e the re of the retngle MBN then = M N = sin os = sin os Diff. it w.r.t., we get d sin os, nd d d.sin os sin. d d Now sin os d. M B lso d sin,sin d is mimum when. Hene, the re of the retngle will e mimum when the point is mid-point of the hpotenuse. Let d e the length of the hpotenuse of rt. d = BC. Let in rdin mesure e the ngle etween the hpotenuse nd the se of the tringle, < <. Let D e the point on the hpotenuse C suh tht D = nd DQ =, then CD = ose nd D = se. D Q C B From fig., C = D + CD

38 d = se + ose.. i d = se tn ose ot.. ii d sin os sin os = os sin os sin Now d sin os d os sin sin = os os sin >, s < < tn = / tn = / d d = se θ tn θ - ose θ ot θ usingii dθ dθ = se se tn +ose ose θot =se se tn os e os e ot Length of hpotenuse will e minimum when tn = / / utting tn = / / in i, we get d = + tn / + + ot / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / Hene, the minimum length of the hpotenuse = / / /. 6. Let e numer of squre metres of re of windows nd e numer of squre metres of wlls nd roofs. Then totl surfe re = + M. z = + sujet to onstrint +, LINER ROGRMMING The grph of the given liner inequlities is shown shded in the figure given elow. Note tht fesile region is ounded.

39 Y + =, O X t the three orners of the fesile region: 9 f,, f,, f, sq.m.