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vection- A 41- Cnoose the Correct opons un the following If FT is a non-aero vector then A xay = a a b> yar oo 70 ‘ x : ¥ ~ \ Name the mequalty Ja+ bl Ee 1al +] Blis a> 5 arly 7 , Cauchy Schuuar ty Thequ ality ae Trangle Tnequality cr Rolle's Thequatity a> lagrange’s Thequality. im, = . WA Vector eq oF the Une x«-5 = Ys ar3 —= ——= is ar = UL-SF- aR _~ mn av b> = TAS F SA +S + C54 33 3h) cy t 4 i a Ste 373k + p (4TH 9 Sta) Ua Se St +37 ivy 5 ; at rl Objective Function of a Unear Programming problem is ay AluraysS qwadralc Ve lures su ra] S Near cy May be Untar or Quadratic depending Ypon Problem C8 May be cubic somelime = V7 TE PCA), PCB) 8, PLANB)= Ley then PCA g)= ay & Ue & © a S a. = we | 9 WAY poate EE NAD , Relation R= {ouy xy? uunere 4 ERs “> Reflexive bet not Symmetric & SYmmettic ana transtlve but not reflexive Cc? ua Reflexive and Symmetic Nevthes vefleruve nor symmebic Or ranscuve Vay Range of function cos! is a> [omy £34 b> (0,71) ($28) fof on Wp TF ALS square matrix of order &x8 and [Al = 5, then |AdgUad| is 8s by las KS 5 a> tio ix> ore [rx-3 5 —_ | ‘ - © 8 » ther y is equal to 5 se Ss 3 a 2 br 3 erg eS y x> TE flx) = Sa? , meg is Continuous at % =O, then mrt , L=0 vawe of mm US MS 3 ary 5 b> - 5 % 5 wv] xV> TE 4 = ° at ‘then ay Ax : log x = > ay log t- x b> xe x by th a Using integration, find the area hounded by e circle xt yrs 1G wm the first quadrant Ya 0,4) Lien (arog “1 Sel Given eq” of circle xy yes 1G yrs tee? <—— yr Jey “Gt y —© . wa “he vequired area = [dx " oO “Pere oe ° a ef 4 . [* Jaret 4 0" sin (4) | 3 8 . [o-0) + $§sw'(4) ~ sin ay] y > Sxl & = ut Square uncts- 4 The volume of Spherical balloon is intreaging at the rate of aScci/s- Find the rate Of change of its Surface area at the instant when its radius is Scum Sol lel vo be the volume of Spherical balloon Given av - + 85 e¢/e at Also v= UR? | uthere Rts radius 3 av 2 Us BR*KgdR At z at as = UNE aR Ac —= Me “op. as —O - ry Unp? Ler S$ be surface area of balloon s= UnR? QS. UxnxdR xa a at o AS 2 Uxt xARx aS cre Unp™ ak B= Bay , as - xt XB xIS - 50 - 10 t Gtx § 5 aS. lo cm /g at S* Find the crtucal points of the function fcx)= ax2-axhj2e45 Sol Given foo = axF- 3x2 18x 416 (cx) = 6x7- 6x -1B +6 flex)> 6 (x7. x ~8) Feed= 6 (x. gx yx 3) flcx) = 6 Cx Cx -3) 10x 3) ) flex) = 6 Cx aiy Ce-a) Now, for cretical points , per Teoso ac =" 4,80 are critica ports We t-F+38BK be of UWtangle prea of “Urarg Le 2 Ayla x By 4. Th as paoa 3a Y 4 b&b (apy: Bia Sol We are quen, ie Jr f g gue . a) n-{ 4 | > Be 6 Woo4 6 . ves fas 3] >| 0 a uoi4o~o 5 6 AR = 9 +O+15 -araro \ UtOoto -Y+r gro pH: oer -acra pons r aA [> from ® ard ©, | NB = [ °, | | 5 | a 4 fe) 4 ys oO i) Y+toto TS ~UFATS 4 | ~@) -a (AB) = BA Hence proved “then ver fy “that S find ou) ip y tan (2%) Ox al Sol let y = tanfa x 1-3? pua x= tano be B= +taw lex) 4° tan! (a Lane ) | > Lan’ 6 d= tan! (ten A) { “tandx = 94 Ane , 4 : -_ L~ tan x Yt A-tan'cx) dy. Axl. 4 Ole \4ec2 \+x2 oP Hind ae if x34 axy ¥-3xy*> ty2= loo ax Sol Byven x34 ax y ~ 3x47 4 4? = loo anf Fervent ating hoth Sides , we qe 3x24 “el ~ 8f xy dy +947 43g dy = 0 ax ice ax dy ax? -6xy +3y7 4 = 4 Sea Yocy “3y°f ax ay 2 - [8x74 ty - 8g? an _— wo. ss {ax’- Guy + By? f Sections © \B- Syalaate \ Bx da (x a) y Cx-&) Sol lea gf - \ 3% 4 a aa (2741) Cx-B) A 4 (Rx+c) EY Cx741) pus 3x+3 - A (eA41) Cx-Qgy x- Bev = ACK) + Beorc) Ce-8) pur x= a) acaysa = A(Can) +0 6+8 = BC44) B= SA A= 8 S put = O B00) 8 = A CoH) + (BOd4c) Co-a) a= A -ac = 2€ Qo " | acc Sol 44- nara Comparing coehh, Of *2 ure ger oz AB Be- A = 8 S _ 3x78 7 BY . CRx ~e An) Cx-ad 78 x74 | “sD ={2 dx ~£ [Seat os 2 x-a S acy. y Ts Bf a ae Bx ax > [ S -3 Ss HY T+ B& toglx-al ~ Leu far ax + ( S: S at T= B toghe-a] ~ Y tog j7ary - = 9 = eg! 5 Maximise and Minimise Subject to the constaints MFO, 420 Ale q Solve the problem graphically - Az 4Uxt dy-7, m+ By £60 Ate tO mMaximige and Mutimi ge > Ux 4 ayo] Subyect to the constiarnts X+3y £ Go x+y 210 M-Y EO, £20,426 tan lay tc nity SE ; scary 2le, M4 50, 7H ze WES HR Fou pornt Gr, solving rey my dy t 66 ON yx x 60 Ux + 60 Bis 1S . Y ° Ss O C1S,18) For Pporwt Solving KEY i | © tS ~4 xr [Oo x= ilo m= Sy, o = 5 405,35) Now MM pt A(o,a0) At pl: CColo) | At pl G@Cisjis) | At pl + C5,5) A= Yloy+Qcio)- J} Az 4CIS)+ ACIs)-7 A= Yo)+9(a0)-7 | * * do-] 4.= Yo + 8y-7 A= YCS)3a05)-7 k= 60+ 30-7 Z } A=) in = Z= Oro-7 ” 80 +1o-F Go-4 re a3 i 2 = $3 A= 33 t Z is maxmum at &CIS,!S) = 83 AL mMmmmum at C Coysto) = 13 12- Solve log x DH + y= 2 logy Ox x So! The gwen arf eq” is x logx Ay y = 3. log ax O8 il ye. 8 ax ~t0gx 9 ek, pat i VY 3 ay. yoga ea ‘ax a vy ody 3x74 x CV ryt Ox)? ae ot Vox AY > (34+ 9+") Ax — * a = 3404 v7 lobe AL Br yt asx Av « aoe \o74 3 3U | Ae log Jap te 974 (I3y > 1 tan' (i \ = log lal + c NEY wr ) a tan! ( 3) : Log| >] tC —@ U3 ws Given uchen yd, x4 4 tan’ ( 4 = leq | a) tc NEY W3 J | 1 «th o +e SR 6 ®>s 1 tan! (2 W3 « = log|x)+ Tr 643 Le ee ‘3 of Solving Lt are Pre babe tty that \ ie pe «students + Problen LS given to “tyre 9 uthose ¢ hances 2 € dy, Uak ‘ eta Ye, dz alee) 16 IL LS the I> exactly one of therm solve, the Problem Vy “the preblem ts solved, Sol Ler AR c be the events that Problens US Solved by 8 tudent mil Bot mespe ctoely 6 Swen PCAye , PCA)? 9 % 4 PC = 4 R)- UW i) Q PCB ) = PCC)= a 5 PCE) SS & 6 'y P Cexactly one Of them sol the problem) PCABC or ARE oy AR c) = PCA RE) + PCARE) 4 PCA Rc) PCA) PCB) Pct) + PCA) PCAYPCC) + PCA) PCB) Pccy ~ BEE) * (HRD? HYG) Te) —~ + = + 3 = 38 Vo “0 Yo “Q6 = 18 ug - & P ( the problem is solved ) PC the Problem ic solved by atleast one of they ) 1 -Pone of ” “them solve ut) 4- PCR BT) the required proba bulity PC Mie) © PEAY pCeIn) PCA) Pen) + PCB PCy) . (oro01) (o-qay 7 “(6*00)) C6°99) + Co°%99) Cor005 ) : aes la loo009 - d90° - 2a 33 + Maas 5S Jay 33 looo00 \o 00000 \4o Prove that the fundion FIRAIR, PCM Sx 43 Y Sine -one and onto on Sel Siiven FI RaR Such “Chey feay= B+ 3 4 © fF tS one-one lek <4 tg ER St Ford = Fcxs ) > = S243 y > Sx,4+2- Soy Q ry Sx, + Sx». sy) xX, <9 Function Fits one-one @ { is on bo Lea YER be any element put f Ge) sy D> Sev 2 Ly — 3 =) Sx+a=4y S Sx = Yy-3 yS «xc = YW-3 ER LO yeRs S$ y yer 3x= Uys ce S Such that fox) = Fa) S , ‘fos de ) +3 3 ty = f(a) = 3 fis orto iS- fy) Useng area of triangle formed bY the Porn and (a, 8) is LA sq) Unit: Sal Gri ven determinants » Find the value of bh, Uf the 3,6). C4, 9) pl of “atangle AC3,6), BEY,4) CCR >a) No given area of Liangle = 1a Square unt A)-36 1a al-yud a - a4 sd 4 ~6/-4Y da [4a =o lo9 
