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Maths3 Gold

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501 views122 pages

Maths3 Gold

This is helpful for the engineering students as this pdf includes all the previous year questions

Uploaded by

drstrangee700
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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FOURIER TRANSFORMS eee Qu Express the function ‘Ass Fourier integral fee P is Fie 21 0. fro aes Jing Fourier integral representa- fon, show thet Qi: Find the Fourler transform of fy tsl hea, /) ay Wekoww tae Piel = [ROEM | oo I ri 8 zi > ruta = | Mayers [fede 9 Find the Pour of Blok ficson | iar tence bacocnscats ren a) fa fortes no) a |o otherwise 04 fa~x'yae +0 UNT-1~B-TE04, Joon sisive [> even det] O a aapede = of-P oer deco [fesies =2f pasar ip and 04 f= 09) ns = ofa eocte =v) ate Qs. Find Fourier transform of she {llowing functions {i+2.toro F) ; Senco ener Smears eaey Soa Pa een tse BS eu escent Titoli is eS = Levees # Gara est 140-70 eptia [oon Pee) tte - " | Be, 2 eas a ; QS, Find Laplace Transformof®e%, no U8 { anitatorss ae s Woke Het = ees : i Shy change of ele poports, ve ave i aot nee a Beene =} F ion intel ee fem spa Aon bth sitesk Pusu ofr fs 1fto log (6+ 1) — log (6+ 3) [Putting 5 = 9) 29. Using unit step function, find jh, Laplace Transform OF) 1)? u(t) Gy sintn (9), Ane Lit yt D) = eb (6 +1~ Fj ogo Ue —all= end Kian] eta! _ Bet Fer Let = OO Hoi ta e— pm eZ Gain 42) = © Lain t) =e Lain y) = os Ans, 2.80. State second shifting theorem for Laplace transform and hence find the Laplace transform of the follow ing function: fe i>e POSH ee Ans. Second Shifting theorem: Form if fofe Fler then U0-«) wea] = ea my) Form I: Tf £G()] = Fe) then L/@)u(~o)] =e Li @4 @) Form DEHLI @)]= Fe) ft-a) toa od B= 1 ee then Lig @) =e F@) toa f= hee or Ae eu ta) 3 UO) = 0% Leloag iby te) Fal i following function {(}) =a Q31. Express the following | in terms of unit step function and find i? | Qs2. | Laplace transform: terme fet acres ae = irl execs Roe 124 =H eqn =H O47 uy et Bina =- Dud) Tug Laplace transform of beth siden a Ue ue) — Yh ue ay) 4, { Ser Tak * 0 Ans, Ift ioe botweon Gand 1, (2) = Tec tine between 1 and 2, (A seprasents mt — 9)) 9 [ce — 3) ..c4) straight line passing throvgh (i, 0) a0 so20nil shifting theorm, we Ts eqjuation te ull =<" 2 G+ 0] be © Aepipineabove result in (vege, Bee Bat DOI= DLE + 0)" 2 rhe + aya) “Line passing through (x,,.,Jand] et, [ Thus, Ho) e-1icred 1 se In terms of unit step fimetion 1 (¢ ~ a), HO =O fu 0) — ult — 1+ —1) fu @-1)—wt—2 +1 fw 25) —NuG—D-—¢— Due D6 ute G-be =H Blu 4) yy second shifting theorem, we have Lf] wea = Lp + @)] Applying above result-in (1), we at, Lif @)= 6? Dt~ 3) €*1[@42)-2] =e* (Lio) —e** (LO) lr La" Q.d8. Define Laplace Transformot Periodic function. 638 the following function in Funit step function and find its per —unre tne nh Teoy — un —MaTHMaries.n, aallifVis npc fontpeeih pind 7 1 jerind 2a, elven eon 8 Tr a meres nentn Sayeceafetone | gxtsa astaie Ric: As ee eee a Riowins nected ee paren eee Ts hb << Raw tooth wove of ad Pare + Fees bate nk period 5 (0703011) GoMart wth fei ier ee bono th ; erodes ty rein Ped lo bapender ae on a eee ‘ 28 ya righ of eve Bint co" when Treatment 1 (renee, it : 4 se Af0) fe ; ik 4 x ren sins agora z + , net fonegyeh y fe ye a f afo1= j 0p (were 7 pend ot Be ttapiand ate Heanolormofihe sinag tar no eaten sider nave tush % THe Stu) 5: Lean the graph and Bac transform: of the (allows [Ba-t astcoq When tlies between 0 nnd, T= twhich is lice y en ¢ Hes between 9 and = ce ebetrean a and Ba, 0) = | Fanction it Periodic with period 2a, Above BrAPHs Aro repeated at intorvel of a4 Thus eraph of above function ie a below. Bi ) a] t 7 we nN a, rs 2nd Part: We know that for 2 periodic function f), UpOl= aa hy oF [where T riod of the function] 1 F EMO = q_—e ei [+ Givon function has period 2a] ee ee Res ee “Fare tod * is Q.97. A periodic function of period“ defined by Rsinot0=t<" MON, E A= Putting =~ 4. we get.~9=-3B>B 2x-1 = Ge) Pn x=st 4 eking inverse Leplace Transform of both sides 208 Ue Q.48. Find the Inverse Laplace of the Fanction # Q4i Find the Inverse Laplace trans! form 1 Fete) z 1 fh Z ) irerake of Te + ¢ as perfect equars, We me ae da andisubteact (3 |] SFE Fa) oa asform of both #4 Taking inverse Laplace tr ake ada y18 Taking mverse Laplace Transform of both fara ides, (30) SUNT — B.TECH, —Uc.U.—MATHENATICS- sa) =o 1 FY) t 4 1) e+ a= e981 ERY 6 i] \ = cosaz, £1[ 3 | Q45. Find the Inverse Laplace s+ Transformation of + aaa ‘tontion Of (3 65+ 15 i = 52 Boost + dn gi Ans, E 5o' tein 3) . Qi44. Find the luverse Lapiace Transform we add and subtract | @48. Find the Inverse Laplace ‘Transform iF ~~ gt t4g+5 (8 +1 Taking Inverse Laplace Trasrtorm of bat a h| Peet place Trasnform of both laae we get, HELP) = 70), De RG) = L(s#2yP 41. Bee) & Putting above value in Gs gan) = sinntt| 1-5 —Eain ete) i Be MELT) =/O.then | Wren Fe) =ft—aut-0)] @) [FO =s Pew FS} = fit-o.ut-a)] 4) ace L shen Le ® FO} =/tt (Fare =- (rere) inet -2y¢-1) [sy x = ¢-Dut-)-3@-3)ue-3) From @ and G0) Q.49. Find the Inverse Laplace’ ‘Transform of epi e?- sera a lem di), Oa writ nd") Fy sad. @ a ‘Ana We vhall fini Gnd 6% Tween partial acco [sS it) il 1 AB, cy = e y ay ere Q.60, Find the Inverse Laptnes 9, ' L i ah ca 60.7 : eLuplace Teinstorg ne 1 ettog (+2) Putting #=0, we got, 1 = Ba) > B a Ans. Hore F(s)= og (1 +4) Prittingss=—a, we get b= Cats C= | We hav & _ Lara Qs2 Pemgiramniinabdeniece | iy a) (4) j2'| 2 re) flos 5 2a" A+ 2a] — . et, Taking inverse Laplace transform of both Q5L Find the Tavetse Laplac ¢ Transform ‘orem for Inver vansform we now that FA eM) =f ayy.¢ Ww)= fe AppIving above roante, econd shifting | Beli ste)| Pind the Inverse Laplace Transfortn Cs 2) ure UTU 2010, 2012, 5 mark ena, ES = 13) fh Sec ere form Ae { 1 ra iat : By convolution thearem 1, ©) FO) = [AC xpd Lf sia = sfsintet—2xycoea de w lity = 7 sinl2e—2aycon de (Multiplying and diving i z 5 [evinces 2x + x)+ sin 2e ay [. 2sin A.cos B= cin (A + B) + ain (A-gy F (foin(2t — x)-+ 9in(2e ~ 3x) lee fins) «ay /202% || os eh PCs, = {cost +i%c080— 20086] 2 ; ( [+ Joiners bye = 08a) { ae : L = St emt—1 = 1+ 5005021 —3)] 60. Applying convolution to find =F, 6), (0) where F, (s) = 1 = (e031 ~ cos 24) Q.6L. Using convolution theroem, find inverse Laplace Transformation of ete cey ahL" B : =a y ¥ convolution thorem, we have => IA (R, @) =cost at =f etl op A : ; 0 : © Fe) = fi fcayp sya coat Gyo 8 = oad and = By convolution theorem, we have Eh, OR O1= [ Ae@ae mae = 3) m, find fon of sinatt —*) p fener as Ty'sin(at — ax)co8ax dy pt ede f tein tar ax re0n ay ds [Multiplying and dividing by 3) 1 plsin(at ax eax) 2 gdtssin(at —ax —ax)}dx Acos B= sin (A+R) pam sin (A—B)) JL fener snes anos 62. Apply convolution theorem to eanate E” [= 7 1" IF, (s) ad 1 (P, (0) BY convolution theor 11, @) Hye) facone-oe wed O44 = t p = foontx ona Bit — xd = 1 fe coosiccon (2k 2290 a Multipiying and dividing by 21 Lf feon(te + Bt Be) oon 24 2a] [52 coe-A.cns B= cos (A + B)> cos (8 a 1 = Lfoostade 14 | conan — 2d Eee EE | fi [i fovstex-o) sin(ax +6) 08 2)» [eincAl 22) —ha(-29)] = Lreosat + 4[sin2¢+sin2y] 3 [sin @@)=—sine) 3 i 4 peoa2s + tain 2t re Q.63. Using conyelation theorem, find inverse Laplace Transform of F wmizattn thoes, we HAYS oleate -} Fanmgnr= LC BR OA ; : 2 Baar ra nina tinal) gy eae eee Olea) Hainaain(at on) aginavsin(at ~ axe {Auliplying and dividing by 2) J jeosax us a8) eo(or 4 08 iee [Bain Asin B= cos (4—2)— cos (A+ B)) 2a 1 | fsintax zal 2a [: Joostne+pyie= 1 /sin@ai—at) a = ss inat—atensai) Q64.Using convolution theorem, evaluate Mere sl [UTU 2010 10 marks.] Cre Anis, = © Piva + i) where F; () F,0= Fe +E (F, (9)] = 008 at =/, (#) [Let] and L7 (2, (6)] = e08 bt=f, ( [Let) By convolution theorem, we have, | ELF, (6) P, @) = f KGa —ayar Qe ae =f Faye +B 1s axccos bt — zx)ckx An Zeoaaxccoe(bt — bxde [Multiplying and dividing by) [castor +-ht — br) +cos(axr— + ble +B) + cos (AB) cotta Ds Lbthd +t [costa tbs —a Hf sin oD r0a)] = b= vorf te] 14F GH . fen ainee y= Bed 21 ‘| AD =i by Laplace vin transform, Ans. Replacing by x Given equation ie ae Making Leplace transform and ucing ‘convolution theorem, we havo, a =i 5 {=} = rn aig _ Tel Hien NY rt) = al fe rt 11 Taking Laplave invorss, wo got Perec Byyat ps Ge; *0= 2 | a(t) i] MATHEMATICS 1 y= Qe, Solve the ¥() the equation 5 4 1 + foeneose ~x\dx ‘Ans. Given equation is 1@O= 1+ [reyes — aay Taking 1.7. of both: aides, Hy @ ]= Hi +1) f sadease— i) = 2 Hxobteosto} 8 By Convolution theorem ade | = EF OELA 7 where F =Liy( ain) se ‘Taking the Inverse Laplace Transform of both sides, we get > pe eo | 73 [eave Gaya | ein2t| [eves “3 Bees, Ges oe! int + requitedsolution. Qt. Using Laplace Transform, find the Solittion of the initial value problem é Get bcos 3.189 0-2, @)-0 sin 24) which is the Ans, Given differential equation is ¥"*9y=6 cos 3 © Conditions are y (0) @=0 Taking Laplace transform of both sides FG), we get TON+9'% by) = 67. Joos 34] Q.72. Solve the followin, different equation b: transformation, Ds-3 =e, Dy tx Ans. ‘Taking Laplace Transforms of (1 a and Fr | 2008 2¢+ tain 3¢ > ig simultaneous y using Laplace sin 6, x)= 1,5, (UTU-2011) an 0=¢ @ D pce ai @ ae taint ~@ we get _ ied HOS Sap ae = dx Taking Laplace transform of (2), we get dy [3-9] Solving (8) and @) for = and 7, we have x 1 Sel > HEA 7=0 Fa) «09 00)=0) => aes Dy a0 sn tél) Again, taking Laplaca Transform of Gi) we Bae emt shins Loin tons) vane Ube) 4 2 Ly] = Let] 9 a Sige cost+ 2sint i cas dle beast) ~@ = ax —x(0) vay = on Neo) 8 Inverse Laplace of (6), Place | ‘oking aplace of 6), we get i wae “Thay = | a a ‘=O 1-207 x i 2) a: +2y = eal (ivy a Equation (i) equations (i) gives, s-1-2), + dy Wwe get sy= eae) { 4 Qi Solve the following simultaneous Giirential equations by Laplace Transform @ - get W)=y (0) =0. Bis: Givon equations are 0 ave =3 P -0.70-0 - Fromequation @O J Given system of equations can tere" itten a s ; +2y eet Ft4yy= 0 Be ee « Rt oy= ing Laplace Tranef? rm of (), we get = {U. — MATHEMATICS Ans. Convolution Theorem: Lf, O1=F, and Eff, (0) = Fy (0) then olf icone x)dx Proof: We know that LYON = fe" fede ia] AGI was] ; fe [Aeris ae B 30= dee") = fol 5? “hCG ade BE erties axa or = LL enCOne —xdeds a expontntial order a as.¢ >», then prov {By Changing the order of intagratio that the Laplace Transform of f (t) exists ‘ng the order of integration] for >a, fuTv-2011) Ue ° es | £4 —2at | J dt = fe" Pidat + fem Rar " Tg ya nally 2 Z, okists since Fi) is sect sontinuous in every finite mterval 0 ra ytuh= a > 8+ u(t) [eB => u=x-3 Putting above values and consemed values from table in (), we get = Se 29 EO 9) Eee) m De RL ~ x 4 6)4 G3 — G3 + bx + 63-6) ® +48 Beto) Gas ete ae 6) 2 i Fon (4) ME PUES Ain above rol 308 fay = 44 Ra) = 39 3 fhe interpolati: d Uibokasans ing polynomial fale Reeoroen She /eiven tab which satisfies the following data and hi = 1 hoe TUTU 2016, 17, 5 Marke | — aieae i 17, 6 Marks) toe » [10 : aL. - 1 ios sip Wa teas AU oh =a we Newton's backword int ynomiy ie Neston becorord tnvrpolation formals, For a Ee. i 7 3 8 | 1 10 wo Br Newton's Gregory backword interpolation formula, flor, tah) = £&) +2 Vie) — Dorp) ® Here ‘ nierval of diffexencing = 1 yalue of which fis to be calculated u oncened values from table -9 — Mes 1048-9) il +8 (¢—10))* 4 n (1), we get, 2 — 19x +90) + (2-8) (2 — 19x + 90) _ 19+ 90) + Ge — 1922+ 908) get + 1522 - 720) 2+ 242 x— 720) zi a= a 9+8(x—10) + 42-1 9x + 90) + G ~ 431 |EMATICS-I Putting concerned values from table ull 5 6 i iid ate ee sehy ene Nort divided diferenc G8) + 9 Ge For thi we HAO, whore divided = 48 + 52x—208 + 15 (9, using 8A) = “pa pando. 20) + Ge— 7) (9 — O44) anderen and a =~ 160+ 52x +15 (2 arena An a ee ml are] Aue) | S72) [4°79 4 24 3p | paved] 8% de RO * 63x] | ae | (Putting + = 2] = 72) = @)-@y > 72, Also, (8) = 83 — 8? = AS) = 48 Again, (15) = 153 [Putting «= 8 in fy [Putting x = 15 inf ae _ = f15) = 3150 (Putting « = 15 in ff ty Newton's divided difference: es y toe J Avia) # G25) Hence, f(2)= 4, F(8) = 448, 15) = 3150 7) + @— 3) x) @~ x.) 0° fi Q.22. Construct the divided difference table for the data. T [= 05 5.0 | 181.0 28: 12 | polynomial and an approximation to 16 [30 | im) 162 bar | so Hetice find tho interpolating 8 the value of fix} Ans. Weconsteuct the fivited difforonce table by using A/a) = 2 divided difference Pealetlated between and b. table follows. 131.0 100.75 282.12 | 159.25 80 521.0 By Newton's divided difference formula wae AA = [0%5) + (x — 19) Aflay) + = 80 —% Z BE fle) “+ (x — 9) (pe — ay) (ty). 4? Pity) Pave Paitisgconcorned values fiom table, weget aA T h |= UNIT Bren — ya a 24 K-05) 2112 A) 18) 8) Spee : a 1OLsGie-2) wee a0 GMP Mase 2129) Uh Iterations a fix) = 2° +x+ 0,995, which a Slee = -0,051514(-ve) 2 : aoanial sesso vel sy = 1875 fl) =0:224608 ( v0) i : Beard cont root ots" “== between Berg by bisection method. Compe Root Hes beuween x amt ty Pere ‘Thue, after five iterations, roovor the given eqquution is 1.04878: Q.24, Pind a positive real root of - c08> @ by bisection method, correct upto 4 decimat ‘Ans. Gwen equation 19 Jeep = 8 v9) jana, ens x= © Let fe) = 3-8 Now, ((0,73) =— 0.01514 @-001937, fT) Hirst iteration: fe OEE Second iteration a Roat lies bersreent Sy ano, alue off) ed differ en =i second iteration: [2 1, = 0.795 fied =< * = _ p.ooget1 Gv) y= 0.7 Had = DEINE GN nx, a8 %y Tos betavce Root fe) Phird terakion: @ } n= 0991S Fourth iteration’ seats) 5 ira > fx) 0.001081. 8 D5 1608 Root Hes between BPs 9.294609 9*) = 131 between % and %y “ = fi) = — 0.000602 (- ve) Fourth iteration: ¥, = 0.738750 fix) =~0,000561 (- = 0,14 => fix,) = 0.001 Root ios Botwoon.s, nnd x, o7aa976 flay) = 0.00048 ion: 1 = 0.798750 M&,) = — 0.000561 (- ve % = 0 He) = 0.000486 Root lies between x, and Pith ite, %, = 0.739068 i) =— 0.00038 ve) Sixth iteration: %y = 0.73908 => fe)= — 0.000038 va) Root and, Seventh iteration: = 0.7390663 = 9.000038 (ve) = fis) =—0.000224 (+-ye) Root lies between x, and > fx) == 0,75 > % = 0.79141 SG) = 0.000094 (tve) Eighth iteration: my) = 0.7 390008 0.000038 (vi o.7agia) = fr) 1 (hve) 2 ys a x, = 0.739102 =x, and 4, are same i Root of the equ 3 0.7991 iteration method. Ans, Given equation cos £-3r+1=0 Let f(x) = coe x - 3x4 1 Now, /0) AA) =~ 1.459698 0) and /0) ae Root lies betwe Let us tak > 19° @9) = W@ p<, 1 ; [sina by iteration ion method ‘ ation method = foe) = 0.000004 6 ve) Root ltes between x, and, d 4 real root of the ect to. 8 decimal t POA places of tion conrect to 4 gaan eta ecuation ¢ 1 placer wai ritten ay of opposite signs 1 Oand 1 isin x} sl] a deg ee £0: eae = y= B.7eRaKH ‘Third iteration: Putting n = 2in (i), we get ‘89221 ourth iteration; Putting n [Putting x, in (i), wo got = Ttlog 2, (Putting, Fifth iteration: Putting n=4 (), wwe got 789221] Sus 2 3.789273 [Putting =, = 3.789275) © And A, are same upto 4 places of decimal. Root of the equ: Blaces in 3.7692. Q.27. Find a root ofthe equation x? — 5x +3=0 by Rogula-Palsi methor corroct to four decimal places eorrectto 4 decimal and 1)= -1 (i) First iteration: Ay= 0 = (4) =3 (+ ve) % =4 > fiz,) == 1 (v9) “ Root lies between s, and 2, Hof Gy) = f(s) i il ta) Ha) f(x,) = 0.75 oti) [Putting x, = 8.772034) 1.788288] 0,75 = fx) +; Root lies between x and, replacing; by x im RLS. off) we ‘y yf iep) “ 1G)— 1G) 0.676056 0.071289 (— ve) ‘Whird iteration: 0 > ft) =3 ve) is = 0.676065 => fix.) = 0.71289 (ve) + Root lies between x, and 2, Replacing a by x,in RALS. of Gil), woe ~ BEG) =a) 1 eae i => x= O6eog04 =. fla) = —0.013847-ve) Fourth iteration: %= O= fiz) = = 0.600864 = fix) — + Root lies between x, an: Repleciag x, by x, in RHLS. of @0), weet i= 24M) FG) Fx) *, = 0.657330 = fs)=-01002629 (va) Fifth itoration; % = 0 /G,) =3 ve) %5 = 0:697380=> f(r.) =—0.04 +: Root lies between x, and x, Replacing x, by «, in RES. of @), wate 20 (ve) 0.656754 0.000497 (ye) = 1 fix) =2.177980 6+ ve) First iteration: * Root lies hewteon x aind xy y= 0=3 /tz,)=—"T Eve) . +4 Replacing x by ay i RAS of (8), we get, = 1-3 fC, 123189 (tye! ss te in (x) " Roctlies between xy and y= We) = Vika) » (9) = x= 0617660 = _ fix) = —0.000269 vo) Ninth iteration: soflay) nfl) © 1e)= fl) 0.470990 265169 (+ve) 517660 Second iteration: 4p = => fle) =-1 Eve) i = 0.470990 => /(,) = 0.265159 (+ ve) 177980 (+-ve) Root lies between «, and x, dix, Replacing x, by x in RLS. of (i), we got id ty in HS of ©), wo get Sof Cen) aah Cx) Z fe) He) “4 0.372277 > fq) = 0.028534 (+ ve) ‘Third iteration (10) fx ).029534 (+ve) Rope lies between and «,. Replocing x, ond xin ALS. of Gi), we gee Beh) =f) f( FG) ) %= 13/6, * Root lies between x, + Replacing, by andy, RES of (10), we get, = 0.861597 > _ Hy) = 0.002940 Gye) > 450511743 Fourth iteration: : - > %ig O04 Hy are saine upto 4 p| of 29 = 0 fl) =—1 Kye) decimal” a as * 7= fe) = 0.002940 (4 ve) = Root of the equation correct to 44 tween xq and x, Placesia 0.5177 by xjin RLS. of Gv), we get = Sfe)—n/e,) @29, Perform five iterations of false Position method to complete the smallest Positive root of the equation gx + sin x — = e=0, Ans. Given equation is Ox + sin x e=0 Let fi) = 9x4 sin x Now, (0) = — 1 and /(1) = 1.129169 By method of talk position, Suaflte,)~x, foe...) = et ea) Fret F=f, a) os soathd, ay es TOOL to Xy anid, ee pie ALG : {One 20113 Markay enna. fy ee ame. f + Ro i slacen ie a5g ts re a = G1) fat GYM = 2969. solve above equation by Newton's tah By Newton-Rs a Besnethod. ion by Newton's valli ot agyim, mon mmnnea indie Ue anf) = 2-31 =—15 ff) = 3'-S1= 50 Ana, Lot {UTU 2611-12) ) (4ay® = f@) and (3) ave of opposite sign + Root of f (2) lies between 2 and a, 43 ; BS = 25 be tie on Wo shall solve above equation by Newton =) ae 2 Mal’ Repheon metho approximation. me ify Newton Raphosn method, £(%,) + (8) and fil) are of opposite sign Reot of fle) lies hetween Band 4 4 Let xy = “S> = 2:5 he the initial approximation: By Newton Raplison Method See xe) ni First iteration: 9, weet Putting n= Oin (), we set +48 639456 zi i any, . (Potting) = 35] = x, = 2871000 (4) = 24) gecond iteration: Second iteration: in (i), we get, Puiting n = 1 in (i), we veh ane 3x 9.634240 [Putting x= 650488) : a7 Third iteration: ae (Pateing = 28710001 TH ing = ain we ee iteration: Paiting n= 1 in @, we ee" 2 ovale ‘Ant. We know that ot < = lee eee. by w= a0 64) Gv} ai aT ea ROD Where 8, () =< e0cb2% Absolute erar= Be “Percentage error in u=|-—| x 100% Fo] < 200% Maximum absolute error odig, the Ja | = 0.25 x 100% = 25% | g88, Prove that the relative error of a . Maximum relati eg panluct of three non zero ae OSE Firexceed the sum of the relative Ces a < intho numbers. a inet s am [UTU 2009-10] = lit a Ise Let x, and < be the threo non zen Forn decimal ae Tuskers and P he their product 4 nnaneinal Dae ae i as 3.4, | = ee Gg) Maximum relative errar< 310" Se eo show that cae 2 | ce + For 6 decimal place accuracy ; Maxinnura relative emot< 5x10° | log P= log (xy2) Slog P= log x + log y + log = erate 2 (mn) = log m+ log a ant 72x 108 Peat c oe oe gmt le] going above relation, we got, n=10 12857 aking differential, we 2°% Hence we reed 10 tormsof the exponential series 90 that the sum is cornect to 6 decimal ; places. 02854 Q.38. Find the number of terms of the 7498041 series (te) log,| = such that is in) ane = |: the value of log 1.2 is correct to seven. eS ine ymnal places. Also compute the value depicna J correct to seven decimal Places: peter ol stal | 1d ie, relative error of product docanotexceed Frelative errors in the number Ans. We have Maximum relative error at male For ndecimal places accuracy, @neni Maximum relative error <> 10 For 7 decimal places accuracy, syst : pe? Beauleeror= 5 Maximum relative error <> 10 F Maximum abscluteesoe= 5, = t D [Replai Maximum Yelative error 10 > @n pn Oy (ake alle i a ns = n+ 1125 4x 10% = Solving above relation, we get = Hence we need 3 term series so that tho to'7 decimal places he alue of log 12 is e low, We shall calculate yetue cf, We have to compute log (1.2). boom log We write the given (asx) To calculate log 1:2 qa} series to 3 terma, i INTEGRATION. DIFFERENTIATION, S10. N & SOLUTION OF ODE exo gut 1) and” (1.1) from the following table 1.0 table: 16 3 Sissi) | = a We Enow that FM + wl) Puttingabor re wausansn ered values ow) (0.298) (0.018) e (0.080) of aiff hich dl Kh =Interval + uh =Value Teeloilsted. table eT ble: ‘We lenow that Qu 1 Butt y> f(s.) wh) = Aposis.)> a fe PUA = EOL) * HEROES 74 f(,) + | a) at Here, k=interyal of ifferencing = 1 x, =Lagt valuo of'x=5 x, Huh at whieh dovivative i to " vilated 3 abun + 5tuti) Putting above value and suitable values irom in (I), we get he distance covered by an athlete for 60 metre race is given in the following |, —U-.U.— MATHEMATICS I 1oj-—uniriy— B.TECH : = “Time tees) | Distance (metray l Det ec. correct to two deci [UTU 2010-21 ns. Representing time by x and A (a, given table can be written as [x [o [0 | We know that speed = 3 sieo and fic time. Distance is tzken ae f(s) and +. Desired speed As 6 is near ¢ f@) at la value of alo ate 76), we con: table and then caleulate 716) by =8=f"( + 5 Marks) ita where table uct hackward difibeence using flor igonparing éiven integral with a. fie wat 5 Lage Now, we make the following table. 0.266667 0.250000 Also, Exact value = 0.287682 {Bron = | Exact value — Caleulated value! = [0.287682 — 0.287683| = 0.000001 Evaluation with 8 strips: Pa sy ‘Given integral = f aa dx faub-intervals 2 =6 ee ing the giveP integral wil ped 125, Ney, omals the folowing yakenm | 2 2.195 2.250 275 o285714 ‘onisee2 o25g0E8 9.250000 Lt, +34) + 204* 904% Fou) 40, Fg ag + M te n integral= 510, +35) #203 +9 +9@ + 4G Fypt Ie F YD [0.583388 + 2(0,860073) +4(1.150223)] (6.904971) = 0.287682 ‘Also Bxact value = 0.287682 Error = [Exact value ~ calculated value! =|0,287682— 0.2876821 =0 Q.11. Using Simpson's 3/8th rul integration, le on. — dx * cag evaluate Ty subintervals, Now, fw N Ans, Given intoyn We shal! divide thy givon interval into 6 Q. 1 I fds 0 [74]—-UNITIV—B.TECH. — UT: Number of eub intervals =6 Upper limit 1 integral 5) t dx fow we make the following Simson's = rule, 1102 560 intervals. Ans, MATHEMATICS-I ee 3 2 Jey = 1.966071 , Find the value of the integral (ot u | Tea? by Simpson's rule, taking yp tegral 7= [0 : integral 7= [/—#% Hore, No, of aubintervale n = 12 Comparing given integral with "" Fn\de, we get fay= i" Ha), weiget fad = = Lower limit = 0 1 33 1.000000 0.993103 0000 |g 0.746114 0.692308 0.640000 0.590164 0.543396 ©.50¢000, Yt Mn) #20n + Joby, Anas Ht Ot nt yt Per 2 HY FH) t6 oF = he __Now, we make the following table Now, f@) =o sin 10x x0 = lower limit = 0 f= Number of subinservals = s+ wh = upper limit = 2x a » 0 1" 1 i 1.85041 2 7.403019. 5.021384 sf 4 10,919630 { 5 | 24.736527 fr 5 : @ By a ae ; i i i asa pytagratntanne 2 ya7.690082) ee pean tM ¥s boy to, ah me) Loy) + 20,4 Diatnnes qe Soren Mtr Hayy — Merv, y= Lower fimit = 0. aust 25) nl = upper limit = =e ios estonee moved by train = 624.875 nanoele me ine velocities of « car which starts = x tah = 35 20+ Oh= & from rest (running on a straight 1 a intervals of 2 minutes aa a 2 ee ‘are'given, eo % 6 2 "= 30 3 By sinpéon’s rl feet (ermine) F(x) 23 sh, (=)a 4 30 |. [PM acte= "Sy tp) +2 tte Pa) + 36, +55 ta bI>* Yet Pio? Iu M 6 aT Ee fer va a ‘4 Float 90) * iy) + 30, +32 7% FY 1 yeror2en® 221 30+18+7)) x 285 = 3.562500 Distanice coveredl = 80 m wide. The depth ‘y ’ from one bank 562500 kms. ee ao cinvers tim os anv divide i. QUT Aaives is — oe sotthe river at'a distance‘ Bax tate becomes {fs given by the following table: \ given table be s : blown —— = To {10 }20 80) 40/50 [s0\70|5°| jy ual 83 Tine Velocity be z @ Ges) | &) (kovbx) | __— \y 0 )4 (ene 12(15 | 14] 0 0 do Vind the approximate area ‘of crosssection 1 vy | of river using Simpson's ane rule. oy ‘Ans. The required area= foe Comparing the above mtegralwith (= F(xjdx, we get y=) Lower lint 4} nl =uppor Limit o-rncde) = 80 [eh width of 10 n= We can write the given table as follows: y=) 0 7m ‘ Hi 9 ' i ; w | 3 ul .: (By Seapsens <4 cule, 5 Te =F Un +3) +20. uty, 2 de Hog ho toll = Roguired area = 5 10) +54) txt +4 ty ¥, typ) _10 > (2+ 2188) +4086) = 710 Thine, required avees= 710m? Q.18. State and prove Trapezdidal rule for numerical Integration. Ans, Trapesoidal Rule: According w thisrale, Proof: Ifyis apoly hy Newton Cove's quadratu +5)t 20, yy tat DT fal of degree n, then fovea Putting n=1 in above expression we get, 1 = nf Jo + AN, (7a) —UNIT-V— B.TECH, ~ UU. — MATHEMATICS HL (>. higher differences arg. Now we shall ealaulate Ay), ef ayy= (B= teas wee > N=) fmyey 4 ke Putting above valuein @),we get ll a 1 fi vde = M90 a ge) sell h similay ‘Adding all above welation, we get (0? sdx= "toy 1 19, State and prove Simpson's std rule 1 Ans, Simpson's +d rule; According to thisrak: +3) +2 Q20.Ste y+ 4G, +94 Proofiliy polynomial of degree n, ther by Newion Cole's quadrature formula ans Six os ca ae i n(2n—3) ZO url | Pro Newt Putting n= 2in aboye expression, we t [Psa 27 [5,22 \ [. Higher difference arent 2h | Ys ES) ae wallcalewlate wh a is am Ay = (E—Dyy ie = ye 21% f: Also, E— yy =e 2E+ Wy bors vation n(), woe t= eMA) Be BHC Ny aj Ildingall above relations, we get by Hy, toy ae sgt, Prook:lfyisa polynomi Powton Cote's quadratu in above expression, we get Now we Als [is @4 byte gp Ay, Bait + tab? bry 1B By 95 By = ya Putting above valuas in(p, wo get Je yas a = By, + 120) -9) 4180, — Oq- Sy, +3 2 Diy tay, age Sinilanty, ("Sots = 34 yg + 3), 35, +yd d 508 3h [ite = Moy oay omy ery ‘Adding all above relations, we get Oo $y) FBO, 1 Qi, Use the Runge Kutta fourth order Rethod to find the value ofy when x= 1. ion that y = | when x = 0 and that [UTU 2011-12, 6 Marks} [x¥%904)=' sadipesteae dl ene equation with h=0.2, {OTU 2018-14 ‘Ans, Given differential equation is fe i yay) 39, He Bot Ai ae Nee yaw FED Fag, NI5= 1 condition is 90-4) = 0.41 Also, ne d: a Se) comparing (with GY =fery).wege, | 2 a DAvp 99) = 0.5/10,1) = 0.5 fey)= ery Also, 2 [Given] First Step: J=050.20, bf Cy 99) 0.2, (0.4, 0.41) = 0.18 fee = Bf (ug +h 19 * hg) = D505, 225806 1 gM 2h 2h, +) 339922 = =H Second Step: = 6.20.5, 0.51)= 0.200998 450.59 =06 eB G+ hs 9g + = lef, 4) =0.5/(0.5, 1.889922) = 0.27 (0.6, 0.610998) 0.228249 a ‘Thus y= hfe + Sy oy a (0.75, 1.46047) = 0.159717 ho Gk 4 2ky + 2hy + hy ane = 0.200848 : 58 => 9, =¥,+k= 0610348 = 0.5/(0,75,1.419780) =0.1543, = ffl, +h, 9 + hy) = 0.57 1,1.494 bee ee Z 0.099080 ae! = ¥y = 0.610348, h i= hf Gey 94) 0.2/ (0.6, 0.610348) Second Step: ‘Thus, (hy + 2k, + 2h, + He,) = 0.220032 ms 169242 = +h 1499164 k wad, as 490164 i : THeriee valio cfel for e=1 in 1.499164 Ane. £02 (071, O7aCaeS = 0.238358 Q.22. Use Runge-Kutta formula of fourth h ky ws + hts} 4 hy) 8, 084944). 2800.3, 1.290 Shy = O.N Haas wl fe +h, 9, 2/00.4, 1.976848) 168805 2? Runge-Kutta fourth orde As ene ive the following differential we tees = Ge +N, + Dh, hy) = k= 0.179208 Jy + e= 1.875268 1,196 and y= => 0.2) = 1196 and 40.4) = 1.875268 é 24. Given the initial value problem G- =1+y',10)=0, Find y(0.6) by Runge Kutta, method taking b= ones [UTU 2010-11,5 Marks} (a O.%y= 1 Ans. Given equation is DWevant tocaloulare y ai-y=O0.2and0., S Aa Comparing the given equation with ? = fle yO) = Yo We eet, 0,y)=0 =02 = flay Yo) =0:2 (0, = 02 ho KY Flite geo 2f(0.1, 0.1) = i ‘a7 12)= 0.18931 ee | = 0.3/00.2, 1.196712) habit go ».2/(0.1, 0.101) 202040 Fh Bly + Bs + ha) = 9:198 i 20 With h = 0.2 on interval [0,64] i fourth order Runge-Kutta meine’ Thun, aE (hy + 2h 2h thy) Compare with the exact solution a gosze2 ‘Ans. Given differential equation je = yy-yp-bh = 0.202707 ae ; en ae 0) Socondd step: cbs - | 0.202707. A\= 0. Condition is (0 chain 02707) Comparing (1), with 5° = 655) wo go fee) = ay", Also, %)=0,99=1,h=0.2 SOARS O88) =0.218827 by Range Kuffe fourth order formu, SN te Mya) ifs ez) = 0.2 C 2a 442) 0.2f0.5, 0.812121) =0.219484 bie,» 2 Bile, +h, 9, +%,) 0.2/0.4.0.422191) =0.235649 20 +0.1, 1+ 0)=0.940.1,1) ~ 0.040000 > ye=3, += 0120789 Third stop: 422789, 2710.4, 0, = 0.2/0.1, 0.08 Hf (ig + Ba y+ hy) 20.2, 0.981584) [+99 + hy = 0.96584) =~0.073972 iz 0.235750 iam eye), = 02/05, IR 9) == 1 AY 5 (hb, + 2k, 20.8, 0.552021) = 9 et 2 260945 . Mie +h, 36+ by) = 0.2506, Thi aoe 01683734) = 0 203499 ae 1 Ae Ths, hae 4 2h, ae ey *= 0.261345 = Pike.y,) | 299s * k= 0.684134 = 0270.2, 0.961533) OS) = 0.684134 Ang, = 0.073964 Solve the following intitial value watg Pigvatyttd 1.524 4(1.92) 2.50 =125+ 5 oe, 9 (4) = 1.64 the set oftabulated values and =0. equation 5xy’+y° -» (45) using Milne’s method. a 05 1.004914 0483 1.009663 46694 1.014256 045176 1.018701 043739. dom: Using the prediccor formula, 4h ay, + Offa + 2h) ax = 1.004914 + [2¢.046694) ~ 045176 + 2(.043739)] = 1.023006 Using the corrector formula, we obtain h I= It EUet Mee = 1.014256 + a [045 (043739) + 042376] = 1.023006 Since the corrector has given the same result as the predictor, we can accept the value, Ha the corrector y, been different, 7, would have b: recomputed and corrector formula reapplied OG _ rastraightlineto the following data: ; 66] 73[ 60 | 67 [65 [65] 67] Qo viva straight line of the formy=ax+ 70 | 68 | 67 | 68 | @1 | bt the following data: t dpe atthe straight line to be fitted is x || yratbe @ me Normalequations are 5 By =ma t bE 3 ed Shy ~ eet bis? “4 ) where ms the number of points given | 6 | pot. Now, we make th 5 pee Ea + 6 4889 sin ee) 3B 72, 4806 | Ans. Given straight line is 4 7 70 suo | ax+b 7 fa 4850 Normal equations are is Sy =arxtmb 67 ey SExy = als + bre 6 67 ee Here, m= No. of points given =8. aS a | 18 Now, wemake the following table 67 64 fe Ses a | io 1 d | 12 4 a4 | 18 el) ta | Ao 16 10.0 Putting above values and 7 | se a equations given, 40 = | Ba +5460 86 9 bea + 373146 91 64 28 Solving above 9 30.5 204 [aaa Putting the values from table, normal i a8 = 39.545455 equationshecomes, ae 30.5 2040+ 36> = 194 desired Solving above equations, we get, 4.118095, =—1 199499

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