CONTENTSI SI.No. PageNo. 1..Syllabus(i) 2Unit - 11 DifferentialCalculus 3.Unit 231 PartialDifferentialEquations 4.Unit - 358 MatricesandDeterminant 5.Unit - 491 IntegeralCalculus 6.Unit 5121 ProbabilityandStatistic 7.Unit- 6149 NumericalMethods ..........-.: SYLLABUS ADVANCE ENGINEERING MATHEMATICS- (AC/AA-l.l) UNIT-I: Differential Calculus:Introduction, functions of several variables, Partial Differentiation, Homogeneous function, Euler's Theorem, Total derivatives, Taylor's theorem for function of two variables, Maxima and Minima of function of the variable, Lagrange's method of undetermined multipliers. UNIT-II: PartialDifferentialEquations:Introduction,formationofpartialdifferential equations, method of separation of variables differential equation Of the first order and first degree.Application of differential equations of firstand higher degree. UNIT-III: Matrices:Definition,Propertiesofmatrix,addition,subtractionand multiplication. Inverse of a matrix, elementary transformation and theorem. Rank of a matrix, Caley Hamilton Theorem. UNIT-IV: IntegralCalculus:Integrationof substitution, bypartsandbypartialfractions. Integration of trigonometric and irrational functions.Reduction L__ ..:;, Formulae for indefInite integrals involving powers of circular functionsof x andproductsof sinxandCosx.Elementaryideasof definiteintegralsandtheir calculations.Simpsonrule .for approximateintegration.Lengthsof simplecurves. Volumes and surfaces of solids of revolution. Mean value and root- means-square value. Double and triple integrals and their simple applications . UNITV: Probability and Statistic:Concept of probabilitytlaws of probability, Binomialpoisson, Normal Distribution. The t-distribution tworking rule. Nature and purpose of Mathematical statistics, Tabular and Graphical Representation sample mean and variance. UNITVI: NUmericalMethods:Introduction,Numericalanalysisincludingsolutionof equationsgraphically,iterationalNewtonRaphson'sorsuccessivesubstitution methods- Ruleoffalseposition(RegulaFabi).Numericalintegrationand differentiation. ii ( Unit - 1 : Differential Calculus Multiple Choice TypeQuestions: Q.l.Iff=1,thenfxx+ fyy+ fzzis ~ x 2 + l + Z 2 (a)f(b)(c)(d)o 24 Q.Z. If u - f [ ~ J. then:if. +Y.f,is (a)0(b)1(c)-1(d)None of the above x+ Yauau Q.3. If u =..Jx.JY ,then x-+ Yis x+Yaxay uu (a)(b)(c)0(d)u 22 Q.4.The functionf(x, y) = x2+ l+6x +14attains the maximum value at the point (a)(-3,0)(b)(3,0)(c)(0,-3) (d)(0,3) Q.S.Letafunctionf(x, y)beacontinuousandpossessesfirstandsecondorderpartial derivativesatapoint( XI, Yl) .IfP(XI, Yl) isacriticalpointand r=fxx(XI'YI),s=fxy(x"Yl),tfyy(XI'YI),thenthepointPisapointofrelative maxima if (a)rt-s2>O,r >0 (b)rt-s2>O,r (d)rt-s2 xax +Yay=0. Hence proved. Q.S.If z is homogeneous function of degree n,show that22a2za2z2az x-+2xy-+y -=n(n-l)z.ax2 axayay2 Sol.Since z is a homogeneous function of degree n,by Euler's theorem azazx-+y-=nz...(1)axay Differentiating (1) partially with respect to 'x', we get aza2za2zaz-+x-+y--=naxax2 axayax 8 a2za2ZaZ=>x-+ y-=(n-l)- ... (2)aX2 axayax Multiplying (2) by x ,we get a2 2a2 Zz()azx-+xy--= n-l x- ... (3) ax2 axayax Differentiating (1)partially with respect to 'V, we get a2zaza2zaz x aXay +ay +Yal =nay a2za2zaz => x-+y-=(n-l)- ...(4) axayay2ay Multiplying (2) by Y,we get a2z2a2zazxy-+y -=(n-l)y- ...(5)axayay2ay Adding (3)and(5)/weget 2a2za 2 z 2z()(azazJ() 2ax-+2xy--+y -= n-lx-+ y- =n-l nz(from(l)) aax2 axayay2axoy 2za2z02Z =>X2-2 +2XYa~ .+l-a2=n(n-l)z.oxXuyy x3+lQ.6.Verify Euler's theorem for the functionz =---"-xy x3 +3 Sol. Z=Y xy It canbe easily verified that zis a homogeneous function of degree 1. By Euler's theorem, ozazx-+y-=nz=1.z=z...{1)axay Verification: 9 L x3 +y3X2 z==-+ ...{2) xyYx Differentiating (2)partially with respect to 'x', we get az2xl ... (3)ax =-Y-7 Multiplying (3) by x,we get az2X2y2 x-=--- ...{4)axyx Differentiating (2)partially with respect to 'v', we get 2yaz= x2 +... (5)ayx Multiplying (5) by Y,we get az_ ... (6)y ay - y x Adding (4) and (6),we have 22azaz2X2lx 2lxx-+=-----+-=+=---'-axayyxyxyxxy => azazx-+y=z.axay Q.7. Sol. Hence proved. Examinef(x, y) = x3 +l-3xyfor maximum and minimum values. f(x, y) = x 3 +l-3xy f=3X2_3yf=3y2x'y 3x'f=6x'f=-3-f=6y'xx'xy'yy For maxima and minima,fx=0,fy= 0 =>3x2-3y =0,3l-3x=O 10 =>X2 = Y , y2= X =>(X2t =xorX4=xor X4- x = 0or x (x 3 -1) = 0 =>x(x-l)(x2 +x+l)=O or x=O,1 Forx=0,y=(0)2:::::0and forx=l,y=(1)2 =1 The two stationary points are(0,0)and(1,1). (0,0)(1,1) 06fxx=6x -3-3fxy=-3 06fyy-6y 27 -9fxx.fyy -(fxyt Fromthe above table,we see that at the pOint(0,0), fxx.fyy _(fxy)2 0andfxx>0,:. (1,1)is a pOint of minima. The minimum value isf(1,1) =1 +1-3 =-1. Findthe extreme values of the functionf(x, y) =x 3l(1- x - y) . f(x, y) = x 3l(1- x - y) fx=x 3l(-I)+(1-x- y).3x2l=x 2l(-x+3-3x-3y) =x 2l(-4x-3y+ 3) fy=x 3 l(-1) +2yx 3 (1- x - y) = x 3Y( - Y+2- 2x - 2y ) =x 3 Y( - 2x - 3y +2 ) For extreme values,fx= 0,fy= 0 11 L :::} x2 y2 (-4x-3y+3) = 0andx3y(-2x-3Y+2)=0 :::} -4x-3y+3=0and-2x-3y+2=0 :::} 4x+3y=-3and2x+3y=-2 Solving above two equations for x and Y,we get .....\;;.....\_1_1A.'2- - ; ~::.7" x-Z,y-3""./ .(.1..1)The extreme value of the given function is~ Z' 3". ~~ Q.9.If (cx-az,cy-bz)=O,showthat ap+bq=c,wherep=-&q=axdy. Sol.let u =cx-az andv =cy -bz. Then(u,v)=O...(1) Differentiating (i) partially with respect to 'K, we get a au +a av==0 auaxavax aa :::} -(c-ap)+-(-bq)=O... (2)auav Differentiating (1) partially with respect to 'V,we get a au +a av=0 audyavdy :::}a (-ap)+ a(c-bq) =0 ... (3) auav Eliminatingaandafrom (i) and (2),we get .auav c-ap -bq1=0 -apc-bq :::} (c-ap)( c-bq)-abpq =0 12 :=} c2 -cbq-apc+abpq-abpq = 0 :=} c 2 -cbq-apc =0 :=} c ( c - bq - ap) =0 :=} ap+bq=c(.: c;t:O) Henceproved. Q.l0. Sol. () auauauIfu =uy-z,z-x,x- y,prove that -+-+-=0. axayaz leta=y-z,b=z-x,c=x-y. Thenu=u(a, b, c)...(1) Differentiating (1)partially withrespect to 'x, we get auauaaauabauac -=--+--+-axaaaxabaxacax :=} au=au (0) + au(-1) + au (1) = _ au+ au axaaabacabac ...(2) Differentiating (1)partially withrespect to 'y', we get auauaaauabauac -=--+--+-ayaaayabayacay :=} au= au (1)+ au (0)+ au (-1) =au_au ayaaabacaaac ...(3) Differentiating (1) partially withrespect to I Z', we get auauaaauabauac -=--+--+-azaaazabazacaz :=} au=au (-1) + au(1) + au (0) = _ au+ au azaaabacaaab ...(4) , " Adding (2),(3) and (4) ,we get 13 au + au +au=. axayaz Hence Proved. 0.11.ExpandeX cos yinpowers of x and y as far as terms of second degree. Sol.I(x, y) =eX cos y x=o,y=o I (x, y) eX cos y 1 . Ix (x,y) i eXcos y 1 Iy(x,y) _ex siny0 Ixx(x, y) eXcos y 1 Ixy (x, y) X . -esmy 0 Iyy(x, y) '- -_ex cosy-1 . By Taylors theorem, aa)1 (aa)2l(x,y)=/(o,O)+x-+y- 1(0,0)+- x-+y- 1(0,0)+ ...( axdy2!axdy 2 x2xyl =I(O,O)+xf(0,0)+ yly (O,O)+-Ixx (O,O)+-I (O,O)+-I (0,0)+ ...xyyy x2!2!2! 2 x2xyl=l+x(l)+ y(O)+-(l)+-(O)+-(-1)+ ... 2!2!2! +... 2! x2 l=l+x+---+... 2!2! 14 Q.12.Iff(x,y)= 2x-y ,show thatIim[Iimf(x,y)] *Iim[limf(x,y)].2x+ yx-+oy-+Oy-+Ox-+o 2x-ySol.f(x,y)=-2x+y l.H.S 2x- y]Ii[2X-OJIi(1)1Iim[Iimf (x,y)] =Iimm=m -- =m= x-+oy-+O_ x-+o [Iiy-+O2x+yx-+O2x + 0x-+O R.HS. Iim[Iimf (x,y)] =Iim[Iim 2x- y] =Iim[O- y] =lim ( -1) :::;-1 y-+Ox-+oy-+Ox-+O2x+ yx-+O0+ yx-+o l.H.S* R.H.S. Hence proved. Long Answer Type Questions: Q.l.IfZ =f (u) is a homogeneous function of x,y ofdegree n, then d2Ud2Ud2U x2dX2+ 2xy dXdy+ ldy2=g (U)( g' (U)-1) f(u) where,g ( u) =n f' (u) Sol.Since z is a homogeneous function of degree n,by Euler's theorem dZdZx-+y-=nz...(1)dXdY It is given that,z:::; f(u) . dZ:::;f'(u) dU and dZ=f'(u) dU ... (2)dXdXdydy Using (2) in (1),we have I x.f'(u) dU+y.f'(u) dU:::;nf(u)dXdY 15 Lw dudu/(u)=> x-+y-=n-dxdyf'(u) dudu => x-+y-=g(u)... (3)dxdy Differentiating (3) partially with respect to lx',we get 2d2UduJd u,dux-+-.l +y-=g (u)( dx2 dxdxdydx d2ud2uIdu => X-+y-=(g (u)-l)- :..(4)dx2 dxdydx Multiplying (4) by x ,we get 2d2u. d2udu X -2+xy-=(g I (u)-l)x.- ...(5)dxdxdydx Differentiating (3) partially with respect to 'V, we get 2 2 d u(d uduJ'du x dXdy+y dy2+ dy.1= g (U)dy d2ud2u,du => y-+X-=(g (u)-l)- ...(6)dy2 dXdydy Multiplying (6)by y ,we get 2d2ud2u,du y-+xy-=(g (u)-l)Y.- ... (7)dy2dxdydy Adding (5) and(7) ,we get 22 2 2d ud u2d u,(dUdu Jx-+2xy-+y -=(g (u)-l)x-+ydx2 dxdydy2dydy =g(u )(g'(u)-l). Hence proved. 16 .+Q.2.If u =sm1111,Prove that X76+y76 22 22a ua u2a u1(2)X-2+2xy--+y-2 =-tanutanu-ll ,axaxdydy144 .-1 + yX; JSol.u =sm1111 X76+y76 Letz =sin u =xX;+yX;xl(;+yl(;. 1xX;+Here, z isa homogeneous function of degree- as forf(x, y) =l(;l(;; 12x6+ Y6 ":-':'..,. +(AY)X;AX; + yX;)+yX;)f(Ax,AY)l(;11==A){2A){2f ()(AX)6+ ( Ay)'6Al(; (x,%+ y'%)(x,%+ y'%)=x,y. We know that if z =f(u) is a homogeneous function of degree n,then 2 2 2a ua ua uI x2 ax2 +2xy axay +l dy2=g(u)(g (u)-1)..,(i) f(u) where,g (u) =n f' (u ) 1 Here,n = 12 z = f(u ) = sin u =>f' (u) = cos u f(u)1sinu1 => g(u)=n--=---=tanu(2)f'(u)12 cosu12 Using (2) in (l},we get 17 L 22 2 2a Ua U2a U1(12)X-+2xy--+y -=-tanu -sec u-l2ax axdydy21212 2 1(secu-12J=-tanu 1212 2 1(tanU-llJ=-tanu 1212 1 2=144 tanu(tan u-ll) 22 2 -I [X+ yJ2a ua ua u sinucosuQ.3. If u=sm ,Prove thatx-2+2xy--+i-= x- yaxaxayay2 4cos3 . ,-1[x+ yJSol. u=sm x+y Letz =sin u= -.JY . 1x+yHere, z is a homogeneous function of degree- =as forf(x, y).JY;2x- y f(AX,AY)==AX+AY=A(X+y)=Alix+y=Alif(x,y). ffx-jiYWe know that if z ==f (u) isa homogeneous function of degree n,then 22 2 2a Uau2a U()('())x-+2xy-+y -=g ugu-1...(1)2ax axaydy2, f(u) where,g (u) = nf' (u) Here,n =2"1 z =f (u) =sin u 18 =>f' (u) =cos u f(u)_.!.. sin u ...(2}=> g(u)=n I'(u) - 2cosu f(u)__1 => g'(u}=n I'(u) - 2cos2 u Using (2) in (1),we get 2a2 u2a2 u2azu1 sin u (11)x-+ xy--+y -=---: --2ax axayal2 cosu ,2cos2 U =.!..sinu (1-2COS2 u) 2 cosu2coszu =.!..sinu ("-COS2UJ 2 cosu2cos2 U sinucosu - 4cos3 u Hence proved. Q.4. 3)ht(a u= ___9..If u =log (x3 +l+Z- 3xyz,prove taaxayaz(x + y +z) Sol.u=1og(x3 +l+z3 -3XYz)...(1) Differentiating (1)partially with respect to ')(, we get au1a(3333)3x2 -3yz-=.- x+y+z- xyzaxx3 +l+Z3- 3xyzaxx3 +l+Z3- 3xyz au3x2 -3yz=> ...(2)-ax =x3 + l+Z3- 3xyz Differentiating (1) partially with respect to 'V, we get au1a (3333)3l-3xz- =.- x+y+Z- xyz=--:-----''':---:--ayx3 +l+Z3- 3xyzayx3 +l+Z3- 3xyz 19 au3l-3xz => ,.. (3)ay= x3+ y3+Z3 -3xyz Differentiating (1) partially with respect to i Z', we get au1a'3z2-3xy--::-_-:-----:-__._(x3+l+ Z3- 3xyz) = azx3+ l+ Z3- 3xyzazx3+ l+Z3- 3xyz au3z2-3xy=> ... (3)'az= x3+ l+Z3 -3xyz Adding (1),(2) and (3),we get 2auauau3 ( x+ l+Z2- xy - yz - ZX)- + - +-= ---=--....,.---..,,-....,.---_..:.. axdyazx3+l+Z3- 3xyz 3 ( x2+ y2 +Z2- xy - yz - zx)3 (x+y+z}(X2+y2+Z2_xy_yZ-ZX) = x+y+z auauau3 => -+-+-=-axdyazx+ y +z Now, (aaaJ2(aaaJ,3 ax + ay + azu=ax + ay +azx+ y + z =333 - (x+y+zt(x+y+zt(x+y+zt 9 =--- 2' (x+ y+ z) Hence proved, 20 Q.5.Find the expansion for sin x sin y inpowers of x,y upto fourth order terms. Sol.Letf (x, y) = sin x sin y i x=O,y =0 0 sinxsinyf(x, y) 0 cosxsin yfx(x, y) 0 sinxcosyfy (x, y) 0 -sin xsin yfx2(x, y) cos xcos y 1fxy(x,y) -sin xsin y0fi (t,y) -cosxsiny0.f ~(x, y) 0 -sinxcos yfX2y(x, y) 0 -cosxsin yfxy2(X,y) 0 -sinxcos yfydx,y) 0 sin x sin yfx4(x, y) cosxcos Y -1fxJy (x, y) 0 sin x sin yfx2i(x,y) cos x cos y -1fxy3(X,y) 0 sinxsin yfl (x,y) 21 L By Taylor's theorem, I(x. y) =I(O,O)+;ifx (0,0)+ yly x2Ixx (0,0) +2xylxy (0,0) +llyy (0,0)J+ 2.. 3Ix2;![x 3(0,0) +3x yfx2y(0,0)+3xy2 Ixy2(0,0) +lly3 (0,0)] + [X4 IX4(0,0) +4X3Ylx3y (0,0)+6x2lIx2/(0,0) +4xy3Ixy3(0,0)+ lly4 (0,0)]+... Substituting the values fromabove table,we get sin x sin y =+x(0) +y (0) + [ x2(0) +2xy (1) +l(0)J+ ;![x3(0)+3x2y(0)+3xy2 (0)+ l(O)J+ X4(OJ+4x3y(_1)+6x2y2 (0)+4xy3 (-1)+ y4 (O)J+ ... 3sin xsin y = xy -.!. (x y +xy3 )+'" 6 a.6.Expandx2y +3y - 2 inpowers ofx -1 andy +2using Taylor's theorem. Sol.I(x, y) =x2y+3Y-2 By Taylor's theorem, 22 2 l(a+h,b+k)=/(a,b)+(h al +-.!..(h2a{+2hkaL+k2 a {)+ axOY)(a.b)2!axaXuydy(a,b) a33 3a3 3a31 (I2I2aII)- h-3+3hk:\2:\+3hk:\ +k-::;-T+......(1)3! laxoxoyoXuyoy(a,b) Here,a+h=xandh=x-l=>a=l b+k =yand k = y+2=>b=-2 22 I(x,y) Ix (x, y) Iy(x,y) Ixz(X,y) : Ixy(X,y) Ii (x,y) (x,y)

Ixyz(x,y) Ii (x,y) x2y+3y-2 2xy x2+3 2y 2x 0 0 2 0 0 x=l,y =-2 1 -10 -4 4 -4 2 0 0 2 0 0 substituting the values from above table in(l),we get x2y+3y - 2=-10+( (x-l)( -4) +(y +2)(4))+((X_l)2 (-4)+ 2( x-l)(y + 2)(2)+( y+ 2)2 (0)) (0)+3(x-l)2 (y+2)(2)+3(x-l)(y+2)2 (0)+(y+2)3 (0))+ ... x2y +3y - 2 =-IO-4(x-l) +4(y + 2)-2(x-I)2 + 2(x-I)(y +2) +(x_I)2 (y + 2) +... Q.7.Find the extreme values ofI(x, y) = x3+3xy2 -l5x2-l5l +72x .Also find the points of relative maxima andminima. Sol.I(x, y) =x3+3xy2 -l5x2-l5l +72x Ix= 3x2+3l-30x+72 Iy=6xy-30y 23 L fxx=6x-30,fxy =6y,fyy =6x-30 For extreme values,fx= 0 ,fy= 0 Now, fx= 0gives 3x2+3y2 -30x+72 =0...(1) Also,fy=0gives 6xy-30y=0 ~6y(x-5) = 0 ~y=0,x=5...(2) substituting the value of y from (2) in(l),we get 3x2-30x+72=0 ~x2-lOx+24=0 ~x=4,x=6 :.The extreme values are(4,0),(5,0)and (6,0). (4,0)(5,0)(6,0) fxx=6x-30-6 06 fxy= 6y 000 fyy=6x-30-6 06 fxxfyy-(fxy t 36036 From the table,we see thatf xx.fyy- (fxy )2=0for the point(5,0). :.The point(5,0)isneither a point of maxinia nor a point of minima. 24 At the point(4,0), f:o:< 0, :. (4,0) is a point of maxima. At the point(6,0) If:o:> 0, :. (6,0) isa point of minima. 3 b33a cQ.8.If u =-2+-2+2" ,wherex + y +z =1 .prove that the stationary value of uis given by xyz abc x=,y=,z=--a+b+ca+b+ca+b+c 3 b33a cSol.f=-+-+x2 ll x+y+z=l=>x+y+z-l=O Takeg = x+ y +z-1. LetF=f+Ag 3 b33a c=> F =-+-+-+A(X+ y+z-l)2x lZ2 2a3 2b3 2c3 =>F.=--3+A;Fy=--3+A;Fz=--3+A;F,t =x+y+z-l xYz For stationary values, Fx= 0, Fy= 0, Fz = 0, F,t= 0 ForFx=0, 3 2x3 2a+A= 0=>1= a ...(1)- 3/1,3X ForFy=0, 32_L3 2b+A=0=>1- b3 ...(2)- 3/1,Y ForF =0 Iz ~ -3__2l2c+A= 0=>')- c ... (3)- 3/1,3Z 25 L ForFA,=0, x+y+z-l=Ox+y+z=l... (4) From (1),(2) and (3),we get 33Z3 -=-3Y--=k(say) aX b -3 ... (5) Using (5) in (4),we get

(a +b +c}=l __l ... (6)(a +b + substituting (6)in (5),we have abc x=,y=,Z=--.(a+b+c)(a+b+c)(a+b+c) Hence proved. Q.9.Divide 24 into three parts such that the continued product of the first, the square of the second and the cube of the third may bemaximum. Sol.Let x,y,z be the three numbers such thatx +y +z = 24. We need to find the maximum value of xy2Z3given thatx + y +z = 24. TakeJ= xy2Z3andg =x+y+z-24 LetF=J+Ag F =xy2Z3+A(X+ y+ z-24) Fx=y2Z3+A; Fy=2XYZ3+A; F =3xy2Z2+A; FA,=x + y +Z- 24z 26 For stationary values, Fx=O,Fy=O,Fz = 0, FA=0 ForFx=0, lz3+ A=0 => A=_y2Z3 ...(1) ForFy=0 , 2XYZ3 +,1,=0 => A=-2xyz3 ...(2) ForFz =0, 3xy2Z2 +,1,=0 => A=-3xy2z2 ...(3) ForFA=0, x+ y+z-24=O=>x+y+z=24... (4) From (1) and(2),we get y2z3=2XYZ3 => y=2x...(5) From (1) and (3),we get y2Z3= 3xy2Z2 =>z=3x... (6) Using (5) and (6) in (4),we get x+2x+3x=24=>x=4 x=4,y=8,z=12. Q.l0.Using Lagrange's method of multipliers ,find the shortest distance of the point (1,2,2) to the spherex2+ l+ Z2= 36. Sol.LetP ( x, y, z)be any point on the given sphere. The distance between the points(x, y, z) and (1,2,2) is d=J(X-I)2 +(y_2)2 +(Z_2)2 27 L LetJ (x, y, z) =(x_l)2 +(y - 2)2 +( z- 2)2 And g(x, y, z) =x2+ y2+Z2- 36 . LetF=J+Ag ~F =(x_l)2 +(y_2)2 +(Z-2)2 +A(X2 + y2+Z2 -36) Fx=2(x-l)+Ax=(2+A)x-2 Fy=2(y - 2) +AY =(2 +A) y - 4 =2{z -2)+AZ =(2+A)Z-4 Fz FA,=x2+ l+Z2 -36 For stationary values, Fx=O,Fy =O,Fz =O,FA,=0 2 Now,Fx=0~... (1)x=2+A 4Fy =0~...(2)y=2+A 4 F=0~...(3)z Z=2+A FA,=0~x2+l+Z2=36...(4) substituting the values of x,y, and z from (1),(2) and (3) in (4),we get 2)2(4)2(4)2- +- +- =36( 2+A2+A2+A ~4+16+16=36(2+A)2 ~36=36{2+A)2 28 ~ (2+Af =1 ~2+2=1or 2+2=-1 ~2=-1and2=-3 For2= -1,x =2, Y = 4, z = 4 . For2=-3, x=-2,y=-4,z=-4. :.we obtain two points(2,4,4)and(-2,-4,-4). The distance of the point(1,2,2)fromthe point(2,4,4)is 3 and the distance of the point(1,2,2)fromthe point(-2,-4,-4) is9. Hence the shortest distance of the point(1, 2, 2)fromthe given sphere is3. Q.ll.A rectangular box ,which is open at the top ,has a capacity of 256 cubic meter, Determine the dimensions of the box such that the least material isrequired for the constructions of the box.use Lagrange's method of multipliers to obtain the solution. Sol.Let x,y,z be the length ,breadth and height of the box. Itis given that,volume of box is 256 cubic meter. ~xyz =256 Letg = xyz - 256 Let 5 be the surface area of the box. Then S =xy+2yz+2xz LetF =S+2j. ~ F =xy+2yz+2xz+2(xyz-256) ~ Fx=y + 2z + 2yz ; Fy= x + 2z + 2xz ; Fz = 2 y + 2x + 2X)' ; FA,=X)'Z - 256 For stationary values, Fx=0, Fy=0, Fz =0, FA,=0 Now,Fx=0~y+2z+2yz=0ory+2z=-2yz... (1) 29 L Fy=0=>X+2Z+AXZ=0orX+2Z=-AXZ...(2) Fz =0 =>2y+2x+AXY =0or2y+2x=-AXY...(3) FA= 0=>xyZ 256 = 0orxyZ = 256... (4) Multiplying (1)by x ,we get xy +2xz = -AxyZ = -256 A... (5) Multiplying (2) by Y,we get xy+2YZ=-AXYZ=-256A...(6) Multiplying (3) by Z ,we get 2zy +2xz = -AxyZ = -256 A...(7) From (5) and(6), xy+2xz = xy+2yz=>x= y From (6) and (7), xy+2yz = 2yz+2xz .=>y = 2z Substituting the values of x and y in (4), y.y.[ ~ ] = 256 =>y=8 =>x=8,z=4 Hence length =8m, breadth=8m and height=4m. 30 Unit - 2 : Partial Differential Equations Multiple Choice Type Questions: Q.l.If the unknown function in a differential equation depends on more than one independent variable then the differential equation is said to be (a)ordinary differential equation (b)partial differential equation (c)homogeneous differential equation (d)none of above 'I d'ff'I.azaz,Q.2.The order 0 f t he partlaIerentlaequationy - - x - = ZISayax (a)1(b)2(c)a(d)None of the above 2 2 'I d'ff'I,azaz3az,Q.3.The order 0 f t he partlaIerentlaequation-2- -2+- = xyIS dyaxax "'.:.': yZ_yz =cz ...(2) From (1) and (2), xZ+l=f{l-yz}. a.6.Solvep+3q=5z+tan(y-3x). Sol.p+3q=5z+tan(y-3x) This is the Lagral(lge's linear equation. Its auxiliary equations are dxdydz "1="35z+tan(y-3x) From first tworatio's ,we have dxdy-=13 Integrating both sides,we get 1 =>x=-y+c3 => y-3x=c1 ... (1) From first and last ratio's,we have dxdz "1- 5z+tan(y-3x) dxdz=>-=-15z+tanc1 Integrating both sides,we get 1 x= -log( 5 z+ tan C1)+ d 5 =>5x = log(5z +tancl)+d =>5x-d =log(5z+tancl)ore 5x-d =5z+tanc\ 38 5x e =C2=>SZ+tanc1 5x e => =C2 ...(2)Sz+tan(y-3x) Form (1) and (2},we have 5x e f(y-3x) =Sz+tan(y-3x) Q.7.Solvep(1+q)=qz. Sol.p(1+q)=qZ Lett=x+ay atat =>-=1-=a...(1)ax()y azazatdz p=-=--=axataxdt azazatdz q=-=--=a()yat()ydt (1) reduces to dZ(dZ)dz- 1+a- =a-z dtdtdt dz =>1+a-=az dt dzaz-1 adz=dt=> -=-- => dtaaz-1 Integrating both sides,we get log ( az -1) =t+c =>log ( az -1) =x + ay + C 39 L Q.8.Solve(D3-6D2D' +12DDfl -8D'3)Z = o,whereD = ~ , D ' = ~ . axdy Sol.(D3-6D2D' +12DD'2 -8D'3)z =0 PuttingD= mandD' = lithe auxiliary equation is given by m3-6m2+12m-8=0 =>(m-2)(m2-4m+4) =0 =>(m-2f =0 =>m=2,2,2 The solution is given by Z=h(y +2x) +xft (y +2x) +x2h(y +2x) . a2a2Zzx+2yQ.9.Solve----=eax2ay2. Sol.The given equation canbe written as 2y a,a(D2_ D'2) z =ex+whereD=ax' D=dy The auxiliary equation is m2 -1=0=>m=l Complementary function isgiven by G.F=h(Y+x)+ 12(Y-X) Particular integral is given by 12__ex+2yP.I. _D'2D =1ex+2y= _.!..eX+2Y e-223 :.The complete solution is 2YZ=C.F.+ P.I.= h(y+x) +12 (y_x)_.!..eX+ . 3 40 d2d2d2Zzz.Q.I0.Solve-2-2--:i:+-2 =smx. dXdXvydy Sol.The given equation canbe written as d,d(D2 - 2DD' +DI2 ) Z =sin x whereD= dX,D=dy The auxiliary equation is m2-2m+1=0 =>(m_1)2=0orm=1,1 Complementary function is given by C.F. =.t; (y+x)+x f2 (y +x) Particular integral is given by P.l.=21,. ~sin x D-2DD +D 1.. 22smx=-smx (-1)-2(0)+(0) :.Thecomplete solution is z=C.F.+ P.I.=.t; (y +X)+Xf2 (y+x)-sinx. Q.ll.Eliminate the arbitrary constants a and b fromZ =aebt sin bx . Sol.z=aebt sinbx...(1) Differentiating (1) partially with respect to lX, we get dZbdX=abetcosbx ... (2) Differentiating (2) partially with respect to 'x, we get d2 -dXZ 2 =-ab2l t sinbx... (3) Differentiating (1)partially with respect to It', we get dZ=abebt sinbx ... (4)dt Differentiating (4)partially with respect to It', we get 41 'l_ 2 az = ab2ebt sin bx ... (5)at2 From(3)and (5),we get a2za2z ax2 =- at2 a 2z + a 2z =0. ~ax2 at2 Q.12.Form the partial differential equation fromj(x +y +z, x2 +l+) =0 . Sol.Letu = x + y +z, v = x2 +l+Z2 Thenj(u, v) =0... (1) auazauazavazavaz Now-=1+- -=1+- -=2x+2z- -=2y+2z, axax ' dydy, axax ' aydy Differentiating (1)partially with respect to 'x', we get ajau + aj av=0 auaxavax aj (az)aj (az)~ - 1+- +- 2x+2z- =0~ ..(2)auaxavax Differentiating (1)partially with respect to 'V, we get ajau + aj av=0 audyavay aj (az \Jaj (az J~ - 1+- +- 2y+2z- =0...(3)auayavay Eliminatingajandajfrom (2)and(3),we get auav azazl1+- 2x+2z-, axaxl=o azaz1+- 2y+2zayay l+p2x+2zPI=0azaz ~ whereP =-and q =:l..l+q2y+2zqaxv)' 42 ::::}(1 + p)( y +qz) - (1 +q)( x +zp) = 0 ::::}(y-z)p+(z-x)q=x-y. d2Z2d2 Z2 Q.13.Solvedx2- ady2= X . Sol.The given equation canbe written as d,d(D2 _ a2D/2 )z = x2 whereD= dx ,D =dy The auxiliary equation is m2 _a2 =0 ::::}m=a Complementary function is given by C.F. =.r. (y+ax)+ f2 (y-ax) Particular integral is given by 12P.I.=22. ~x D-a D =1 12X2 = -1( 1-a2_D/2 J-1 2112 D' (I-a' ~ ,JD'D'x =D' (I+a' ~ ,Jx' =_1(x2+a2_1 (0))=_1x2=1.(x3 J=X4 D2\.D2D2D312 The complete solution is z =C.F.+ P.l. =.r. (y+a x)+ f2 (y-ax)-..!..x4. 12 Long Answer Type Questions: Q.l.Find the general solution of x( Z2-l) dz+y(X2 - Z2)dz=z(l- X2). dxdy Sol.We have xZ ( 2-y2)dZ-+y(2x-Z2)dZ(22)-=Z Y-xdxdy The auxiliary equations are 43 dxdydz ...(1)x( Z2_y2)y(X2 _Z2) x(l-x2 ) Using multipliers x,Y,z,we get "< r m~~~ + ~ + ~~ + ~ + ~ x(z2-l)y(;-l)x(l-;);(l-l)+l(;_Z2)+;(l-;)o xdx+ ydy+ zdz = 0 Integratingboth sides,we get x2 + l+ Z2=c1 ... (2) (1) canbe rewritten as dxdy dz dxdydzdxdydz-+-+- -+-+y xYzxYzx .z 22(Z2_y2)(X2_Z2)(l_X2)( Z2-l)+ ( x - Z2 ) +(y2_x )o dxdydz => -+-+-=0 xyz Integratingboth sides,we get log x +log y +log z = logc2 =>xyz =c2 ... (3) 2 xyz =f (x+ l+Z2 ) Q.2.Solvepx (z - 2y2 )=(z - qy)( z -l-2x3 ) Sol.The given equation can be rewritten as px( z - 2l )+qy (z -l-2x3 )=Z ( z -l-2x3 ) The auxiliary equation is dx dydz ...{1)x(z-2l)y( z-l-2x3 )z ( Z- y2 - 2x3 ) From last two ratio's of (l),we get dydz -=yZ Integrating,we get 44 log y = log z + log c1 or Y=C1 Z ...(2) From first and third ratio's,we get dxdz x(z-2l)z(z-l-2x3) dxdz => 2x (z - 2l) - z ( z - Cl Z2- 2x3) => Zdx-CI2Z2 dx-2x3dx = xdz-2cl2Xzdz 2dxdxdz-zdx22xzz-z+2xdx=O=> 2 Cj Ax Integrating,we get 2 Z2Z2 --C1 -+x =C2 ...(3) xx From (2)and (3),we get L =f[!:"_CI 2+x 2J. ZXx Q.3.Solve(X3+3xy2)p+(y3 +3x2y)q = 2(X2 +l}z. 2Sol.We have(X3+3xy2) p+(l +3x2y)q =2( x+ l)z The auxiliary equation is dx_dy dz -:----;:-...(1) x3+3xy2- l+3x2y 2(X2 + l)z 112 Using multiplier's-,-,--,we get xyz 112112-dx+-dy- dz-dx+-dy--dzdxdydzxy zxz = == x3+3xl y3+3x2y 2(X2+ l)zx2+3l + l+3x2-4x2-4l0 112 => -dx+-dy--dz =0 xyz Integrating ,we get logx+logy2logz=logcj 45 l .xy=C::::::> 21 ... (2) Z From last two ratio's of (l),we get dx_dy_dx+dy_d(x+ y) ...(3) x3 +3.xy2- l+3x2Y - x3 +3.xy2 +y3+3x2Y - (x + yl dxdydx-dyd(x-y)Also,=--::----=----'::----::- ... (4) x3+3.xy2l+3x2yx3+3.xy2-l-3x2y(X_y)3 From (3)and (4), d(x+y)d(X-y) 3=3 (x+y)(x-y) Integrating,we get 1_ 1c +2(X+y)2- 2(X_y)22 11 -22=C2 ... (5) (x+y)(x-y) From (2)and(S),we have .xy(17= 1(x+ y)2 2 2 21 J (x- y)2 aza zaz 2x .Q.4.Solve-2-3-,:\-+2-2 =e +3y +sm(x+2y).ax(JXdyay Sol.The given equation canbe written as 2x a,a(D2-3DD' +2D'2)Z =e +3y +sin(x+2y) whereD = ax' D= dy . The auxiliary equation is m2-3m+2=0 ::::::>(m-l)(m-2) =0orm =1,2 Complementary function is given by .-... C.F. =.h (y+ x)+ 12 (y+2x) Particular integral is given by 46 P.I.=21 ,f2[e2X+3Y +sin(x+2y)]D-3DD +2D 1e2x+3y+1sin (x +2y) D2 - 3DD' + 2D'2D2 - 3DD' + 2D12 Put D =2, D' =3in firsttermand D2=_12, DD' =-2, D'2=_22in sec ond 1e2x+3y+1sin (x+ 2) 22-3.2.3+2.32 (-12)-3(-2)+2(-22)y. =!e2x + 3Y -.!.sin(x+ 2y) 43 :.The complete solution is z = C.F.+P.I. =h (y+x)+ 12 (Y+2x)+!e2X+3Y -.!.sin(x+2Y)43 Q.S.Solve(D3- 7DD'2 - 6D'3 ) Z =cos ( x +2 y) . Sol.(D3-7DD'2 -6Df3 ) z=cos(x+2y) The auxiliary equation is m3-7m-6=O =>(m+1)(m+2)(m-3)=O =>m = -1,-2,3 Complementary function is given by C.F. =h (y - x) +12 (y - 2x) +13 (y +3x) Particular integral Is given by P.l.=1D3_ 7DDI2 _ 6Df3cos (x + 2y) PutD2=-e,DD'=-2,D'2 =_22 11 D(-1)-7D(-22 )-6D'(_22) cos(x+2y) =27D+24D,cos(x+2y) 1D =3 9D2+8DD,cos(x+2y) 47 =1D1 3 9( -1)+8(-2) cos(x+2y) =- 75 cos(x+2y) Hence the complete solution is 1 Z= h(y-X)+ 12 (y-2x)+ 13 (y+3x)--cos(x+2y).75 a3za2zQ.6.Solve-3- 2-2- =2e2x +3x/.axaxay Sol.The given equation canbe written as a,a(D3- 2D2D') z =2e2x +3x/ whereD= ax,D= ay The auxiliary equation is m3 -2m2 =0 =>m2 (m-2)=0orm=0,0,2 Complementary function is given by C.F.= h(y)+XI2 (y)+ 13 (y+ 2x) Particular integral is given by 122x3.1 2=>=e+xyD3 -2D2D'D3 -2D2D' 12 122x+3D')XY2'-22'(0)eD'(1-2D" ')-11 2x 1D2 =-e+3- 1-2- xy4D3 ( D 11(( +... 4D3DD=_e2x +3- 1+2-+D' 2-D')2Jxy2 11(D'(D')2J=4"e2x +3 D3xy2+21)(xy2)+21)(xy2)+ ... 12 1 D2 1 2x '2=-e+3- xy+2-(2xy)+4-2(XY)+... J 3 (4DDD 48 1 2x +3-1(221())2x+... =-e 3 xy+2xy+4-24DD =_e2x +3- 1( xy2 +2x2y+4.-3 J 1x 4D3 3 12x1 (12 22X4J=-e+3- -x y+-x3 y+4D2233 12x1 (13214X5 J=-e+3- -x y+-x y+4Dl6615 6 J 1 2x +3(11x 4243090 =_e -x4l+_x5y+_ The complete solution is 6 J (1 4l+_x5y+_ Z =f, (y)+XI2 (y)+ 13(y+2x)+_e1 2X +3-' x1x. a4243090 2za2za2zQ.7. Solve-2+-a--6-2 =ycosx. axaxyay Sol.We have a2za2za2z-+---6-=ycosx ax2 axayal a,a(D2+DD' -6D'2) z =ycosx whereD=- D= ax'ay The auxiliary equation is m2 +m-6=0 =>(m+3)(m-2) =0orm = 2,-3 Complementary function is given by C.F.= f, (y+2x)+ 12 (y-3x) Particular integral isgivenby =>P.l.=1D2 +DD' -6DI?ycosx 49 L_ 1 =(D-2D')(D+3D') .ycosx = D_12D'f(c+3x)cosxdx(Puty=c+3x) =1,[(c+3x)sinx+3cosx] =1,[ysinx+3cosx]D-2DD-2D = f[(c-2x)sinx+3cosxJdx(PutY =c - 2x ) = ( c- 2x) ( - cos x) - 2 sin x + 3 sin x = - y cos x + sin x Hence the complete solution is z =it (y +2x)+ 12 (y -3x)+sinx- ycosx. Q.S.Using method of separation of variables/find the solution of the equation auau.4-3xh0- +u =- If u =ewent=. axat auau Sol.-+u=..(1)axat Letu =X ( x ) T (t ) ...(2) where Xis a function of x only and Tis a function of t only be a solution of the given equation. Putting the value of u in(l)/we get a(XT)a(XT) ax+XT=at Tax +XT= xaT::::::> axat dX +XT= XdT::::::> dxdt ::::::>T x' +XTXT' ::::::>x' + 1 = ~ =k(say)X Now, x' +1= k ::::::> x' =k-1 XX Integrating both sides,we get 50 - (k-l)x(3)log X = ( k -1) x +log a => X-ae.., T' Now,-=k T Integrating both sides,we get logT =kt+loga => T =aekt ...(4) From (2),(3) and (4),we get u =ab e(k-l)x ekt =abe(k-l)x+kt ...(5) Given:u=4e-3x whent=O. Using this in (5), 4e-3x = abe(k-l)X Comparing the constant term and exponent of e on both sides,we have ab=4c-1=-3 =>ab=4.c=-2 Hence the solution is u =4e-3x-2t dud2u.(0).Q.9. Solveat = dx2 Ifux.= sm l[X d2duuSol... (1)at= dx2 letu = X(x)T(t)...(2) where X is a function of x only and T is a function of tonlybe a solution of the givenPutting the value of u in (l),we get d(XT)d2 (XT) dt- dx2 XdT= Td2X=> dtdx2 d2XXdT =T=> dx2dt 51 L_ :::::>XT'=TX" x" _T' =k(say):::::>-X-T There arises three cases: Case-l:k =0 "T'X--=0 :::::>-X-T :::::>X" =T' =0 Integrating ,we have X=ax +b,T =c,where a,b,c are any arbitrary constants :::::>u=(ax+b)c It is given that, u =sin ttxatt=0 . This is not true here. Therefore this case isnot possible. Case-2: k= a2 X"T'2 :::::>-=-=aXT X"2T'2 :::::>-=a-=a X 'T 1d2 X _21 dT2 2 :::::>----a--=aor d x =a2 XdT =a2dt dx2 ,TXdx2 'Tdt Integrating-dT =a 2dt ,we getT IlogT =a2t+c1 :::::> T =bea2 d2 x 2Now,we shall solve-2- =aX dx This canbe written as(D2- a2) X=0 It's auxiliary equation ism2 - a2 = O. :::::>m=a,-a 52 :. Solution is x =aeax +peax :. u =(aeax +peax )bea2t It is given that, U=sin trxatt=0 . This isnot true here. Therefore this case is not possible. Case-3:k =_a2 => X"T'2 -=-=-a XT => X"2 --=-a X T'2 -=-a'T => 1d2 X__..~ _2 Xdx2--a ..!...dT =_a2 'Tdt or d 2 X--a2X dx2 -, dT =-a2dtT .dT2dIntegrating- =-at,we getT logT =-a2 t+C1=> T =be-a2 t d2x 2Now,we shall solve-2- =_a X dx 2This canbe written as(D2+a )X=0 2 +a It's auxiliary equation ism 2 = 0. =>m =ai,-ai :. Solution is X=acos ax+p sin ax U=(acosax+ psinax)be-a2/... (3) It is given that, U=sin trxatt=0 . =>sinJrx =(acosax+ pSinax)b =abcosax+ pbsinax Comparing the terms onboth sides,we get ab =0, pb =1, a =Jr 53 Hence the solution is u = sinJrx e-a2t Q.l0.The vibrations of an elastic string is governed by the partialdifferential equation a 2 ~= a 2 ~.The length ofthe string isJrand the ends are fixed.The inial velocity is zero atax aand the initial deflection isu(x,O) = 2( sinx+sin 3x) .Find the deflectionu (x,t) of the vibrating string fort>0 . 2a2 uu Sol.... (1)at2 =ax2 Letu=X ( x)T (t )... (2) where X is a function of x only andT is a function of t only be a solution of the given equation. Putting the value of u in (1"we get a2 (XT) _ a2 (XT) at2 - ax2 a2TTa2 X => X-=-at2 ax2 d2Td2X => X-=Tdt2 dx2 => XT" =T X" X" _T" =_a2 (say)Let XT X"T'2 => _=_a2 -=-a X'T 1d2X 2 1 d2T2 22 ---=-aor dX= _a2XdT-a2T=> Xdx2=_a , Tdt2 dx2 'dt2 d2x 2Now,we shall solve-2- =_a X dx This canbe written as(D2 +a2)X=0 2 +a It's auxiliary equation ism2 O. 54 =>m=ai,-ai :. Solution is x = a l cos ax + Asin ax Similarly,T = azcos at + pz sin at -1u =(a l cos ax + PI sinax)(azcos at + pz sin at)... (3) Puttingx = 0, u = 0in (3),we get o= a l (azcos at + pz sin at) =>at =0 Putting the value of al in (3),we get u = PI sin ax(azcos at + pz sin at)...(4) Puttingx =1(, U=0in (4),we get o= Asin a1((azcos at + pz sin at) => o= Asin a1(orsinatr=O =>a1l =n1lora =n, n is any integer Now,(4) reduces to U= PI sin nx( lXzcos nt + pz sin nt)...(5) Differentiating (5) w.r.t. tt',we get du= PI sin nx(-nazsin nt + npz cos nt )dt du Put-=Oandt=O. dt O=nPIPzsinnx=>pz =0 (5) reduces to u = azA sin nxcos nt... (6) Given:u(x,O) = 2(sin x+sin3x) =>u = azPI sin nx= 2(sin x + sin 3x) =>azA sin nx = 4sin 2xcos x 55 '~aill =4cosxandn=2 substituting these values in (6),we get the required solution as u = 4cosxsin 2xcos2t. Q.ll.A rod of length I with insulated sides is initially at a uniform temperature u.lts ends are -- ..suddenly cooled to0Candare kept at that temperature. Prove that the temperature 2 2 2 00c1r nt .()..b()~ b.ntrx--/2functIonUx,tISgivenyUx,t= ~nSlD--.e n=1I wherebis determined from the equationU0=! bsin ntrx . nn n=1I Sol.let the equation for the conduction of heat be au2a2u -=c- ... (1)ax2at letU=X( x) T (t )... (2) where X is a function of x only and T is a function of tonly be a solution of the above equation. Putting the value of uin (l),we get a(XT)2a2 (XT)----'---'- =C-----'-....:....atax2 XaT=Tc2 a 2 x ~ atax2 1d2T1 d2X ~ c2Tdt2 =X dx2 1d2T1 d2X2 let--=---=-a c2Tdt2 Xdx2 1d2 X1dT2d 2XdT22T 2 --=-aor2X-=-a cX dx2=_a dx2=-a,dt'c2Tdt Solving these equations ,we get b_a 2 c 2 tX =acosax+psinax,T -- e ~ U= (acosax+ psinax)be-a2c2t ... (3) Puttingx =O,u=0in(3),we get _a2c2tO=ab e~a=O 56 (3) reduces to .22 U= /ibsinaxe-a ct ... (4) Puttingx = l,u = 0in (4),we have o=/ibsinal.e-a2c2t =::} sin al = Oor al = nli ,n is aninteger nli =::} a=1 Hence (6)reduces to 2 2c2n;2n2tc n; n2t --2- --2- nlixnlixU=RbeIs1O--=b eIS1O-jJ1n1 This equation satisfies the given conditions for allintegral values of n. Hence takingn =1,2,3, ..... Ithe most general solution nis U(xt) - ~ b.nlix ,- ...nS1O--.e n=lI By initial conditionsu =U0whent =0 Substituting this in above solution,we get .. U- "b.nlix0- ...n S1O-n=lI 57 L Unit - 3 : Matrices and Determinant Multiple Choice Type Questions: Q.l.The rank of the matrix(10Jis o-1 (a)1(b)2(c)0(d)3 Q.2. - [12-1]The characteristic roots of the matrix043are o02 (a)1,-1,0(b)1,2,4(c)1,2,2(d)0,1,2 Q.3. Q.4. Q.5. If A is of order 5 x 3and Bis of order 3X 4then the order of (AB Yis (a)4x5(b)4x4(c)5x5(d) If A and B are two matrices of order 3 andAB =O,then (a)A=O (b)B0 (c)detA = OordetB = 0but A and B need not be zero matrices (d)det A =det B =0 ifA andBare not zero matrices The rank of the matrix246is equal to[ 1 2 3] 369 5x4 Q.6. (a)3(b)2(c)1(d)o Which of the following is not true ,in general (a)A+BB+A(b) A+(B+C)=(A+B)+C (c)A(BC)=(AB)C(d)AB=BA Q.7.Which of the following is true (a) (ABt =B-1A-1 (b) AadjA =JAIl (c)ladjA! =!A!n-l (d)All of the above 58 Q.B.IfAAt=I ,then the matrix A is called (a)Idempotent(b)Orthogonal (c)Symmetric(d)Skew symmetric Q.9.If (a+b J,the values of a and b are respectively5 (a)a=2,b=4(b)a=4,b=2 (c)a =2,b= 40ra =4,b=2(d)None of the above x-23 Q.l0.If I=O,the value of x is xx+2 (a)-lor4(b)-20r4 (c)-lor2(d)-20r2 Q.ll.Ifare eigenvalues of a matrix A of order two, the eigenvalues of transpose of A are (a)

(b)

(c)

(d)can't be determined [14 Q.12.For what value of Ii.. ,the matrixIi..3is singular. 64 (a)4(b)7(c)-7(d)6 Q.13.The product of anymxn matrixA = (aij) by any scalar c is written ascA = (a) (c+aiJ (b) (c-aij ) (c) (caij ) (d)None of the above Q.14.A matrixA =(aij ) is symmetric matrix if " ( ( 3z3w=-l+z+w2w+3 Equating the corresponding entries of the above matrices,we get 3x= x+4=:>2x=4 orx=2...(1) 3y=x+y+6=:>2y=x+6...{2) 3z =-1+ z+w=:>2z=w-1...(3) 3w=2w+3=:>w=3...{4) Substituting the value of x from (1)in {2),we get 2y=2+6ory=4 Substituting the value of w from (4) in (3),we get 2z = 3-1orz=l x =2, y =4, z =1, w=3 . [1-23Jr02JQ.2.If A=23-1andB=012.Verifythat (ABY=BtAt. -312120 [1-23r02JSol.AB=23-1012 -312120 [1+0+30-2+62-4+0 H4 4 -2J =2+0-10+3-24+6+0=,I110 -3+0+20+1+4-6+2+0-15-4 [41 -IJ=:>(AB)' =415 ...{1) -210-4 63 l_ (I0If2-3JNow,Bt At = 012-231 2203-12 (1+0+32+0-1-3+0+2J:.::;; = 0-2+60+3-20+1+4 2-4+04+6+0-6+2+0 (4I -IJ=415 ... (2) -210-4 From (1) and {2),we get (ABr = BtAt 3411 2 4. 3 60.3.Find the rank of the matrix A=I -1-264 1-12-3 3411 2436Sol. A=I -1-264 1-12-3 Applying the row operation, R4~ Rl'we have 1-12-3 2436 -1-264 3411 Applying the row operation, ~ ~ ~ - 2 ~, R3~ ~ +R ,R4 ~ R4- 3R,Jj 1-1 06 o-3 07 2 -1 8 -5 -3 12 1 10 64 Applying the row operation, R3-t ~ +!R.z, R4-t R4_2R.z ,we have 26 1-12-3 06-112 ~ 0015/27 00-23/6--4 Applying the row operation, R4-t R4+ 23 R3 , we have 45 1-12-3 06-112 0015/27 000-19/45 Rank=Numbers of non zero rows = 4 1111 3-21Q.4.Find the rank of the matrixA = I 1 20-32 3303 1111 A =113-21Sol. 20-32 3303 Apply the row operation, R2-t R2 - R}, R3-t R3- 2RI ' R4-t R4- 3R1 1111 02-30 o-2-50 ,00-30 Apply the row operation, R3-t ~ +R.z 1111 o 2-30 o 0-80 0.0-30 " 65 Apply the row operation, R4 R4-R3 8 1111 02-30 00-80 0000 Rank=Numbers of non zero rows = 3. Q.S.Find the value ofx such that[1x1][ ..= O. 1532Jx Sol.[1x1][ =0 1532x 132 [1x1]25(j[[ 1532 [1+2x+15'3+5x+3 -1 21=0 x

(2x+16).I+(5x+6).2+(x+4).x=0 2x+16+lOx+12+x2+4x=0 x 2+16x+28=0 x 2+14x+2x+28=0 x(x+14)+2(x+14)=0 (x+2)(x+14)=0 x=-2,-14. 66 4-21J 2t 7.L Q.6.Find the inverse of the matrixA =733. z;J[201 j2J]

4-21J Sol.A=733[201 ..,.....,.,.'- 1)(2t) 4-2CIAI=ltf3=4(3-0)+2,(7-6)+1(0-6)=12+2-6=8 201 If M jj denotes the minor of ajj ,then the cofactor of ajj is C;J=t-ItJ

2 II_1-2 1 =_9 -I"30 311 =3,c21 =_1-01=2,c31 - 331CllC1273114-2114-2c13=120 = -6 ,C23= - 20=-4, C33= 73 1=26 Now, adjA = (cij r[32-9JadjA=-12-5 -6-426 Since, A-I = adjA [3 2A-I =-12-5. -6-426 67 L 2 1 3]Q.7.If A=201,verify thatA(adjA)=IAII3 = (adjA)A.(-456 Sol.We have 2 1 3]A=201(-456 213 ~IAI=12011=2(-5)-1(12+4)+-3(10)=-10-16+30=4 -456 If M jj denotes the minor of aij ,then the cofactor of aij is jCij=(-lt Mij =-5,= 9,C31 =1 C1l C21 =-16,C22 =24, C12 C32 =4 =10,C23 =-14,C33 =-2 cl3 Now, adjA=(Cijr (-591] adjA=16244 10-14-2 (211-5 9I)(40OJ[1 0]A (adjA) =20116244=040=4010 -45610-14-2004"001 ~ A (adjA) =413 =IAII3 Similarly ,it canbeproved that (adjA)A =IAII3' 68 [20 1JQ.8. Letf(x)=x2 -5x+6.Findf(A)ifA=213. 1-10 [20 1JSol. A=213 1-10 f(A)=A2 -5A+61 ~ f(A)= [22 0131J -5[22 0131J +6[10 010OJ 1-101-10001 ~f(A)= (22 013If2 0131J-5(22 0131J+6[10 010OJ 1-101-101-10001 [4+0+10+0-12+0+0J[100 5J [60 OJ ~f(A)=4+2+30+1-32+3+0- 10515+060 2-2+00-1+01-3+05-50006 (5-12J[100 5J(60 OJ~f(A)=9-25- 10515+060 o-1-25-50006 [5-10+6-1+0+02-5+0 J ~f(A)= 9-10+0-2-5+65-15+0 0-5+0-1+5+0-2+0+6 [I'-3J -1 ~f(A)=-1-1-10. -544 69 0.9.Prove that the product of two matrices 22 COS0cos Osin OJd(COSt/J cost/Jsint/JJis zero when0 and t/J2A.( cos0 sin 0sin2 0ancost/Jsint/J sm." differ by anodd multiple of f( . 2 Sol.Consider, COS 2 0cos 0sin OJ [cos 2 t/J [ cos 0sin 0sin2 0cost/J sin t/J =(cos2 Ocos2 t/J+ cos osin Ocost/Jsint/J cosOsinOcos2 t/J+sin2 Ocost/Jsint/J =[cos0cost/J (cos 0cost/J + sin 0sin t/J) cost/J sin 0(cos 0cos t/J + sin 0sin t/J) "':': ...,.cos t/J sin t/J] ..sin2 t/J cos2 Ocost/Jsint/J+cosOsinOsin2 t/J] cosOsinOcost/Jsint/J+sin2 Osin2 t/J cos 0sin t/J (cos0cost/J + sin 0sin t/J)J sin 0sin t/J (cos0cos t/J + sin 0sin t/J) =[cos0cos t/J cos (0- t/J)cos 0sin t/J cos (0- t/J)J ...(1)cost/JsinOcos(O-t/J)sin Osint/Jcos(O-t/J) It is given that0- t/J= ( 2n +1) f( . 2 => cos (O-t/J) = 0....(2) Using (2) in (l),we get 22 COS0cos 0sin OJ [cost/Jcos t/J sin t/J] [ cosOsinOsin2 0cost/Jsint/Jsin2 t/J= cosocost/J(O)cososint/J(O)J=(OOJ = O. ( cost/JsinO(O)sinOsint/J(O)00 Hence proved. -tanaJ(cos2a-sin2a]()0.10. Proveth.U,thenI +A =I - A. osin 2acos 2a Sol.A-0

( tana L.H.S. 70 I+A=(10)+(0 o1tana -tanaJ=(1 0tana -tanaJ 1 R.H.S. (~ s 2a- sin 2aJ{I _ A) = ( c ~ s 2a- sin 2a)(.1 sm2acos2asm2acos2a-tana ( COS 2a+tan a sin 2atanacos2a- sin 2aJ =sin 2a- tan a cos 2atan a sin 2a+cos 2a tan aJ 1 2.2(sina)2'sina)().cosa-sma+-- smacosa-- 2cos2a-I -2smacosa(cosacosa = 2smacosa-. -_.-) sina)2.2 2(sina)(2cos2a-I -- smacosa+cosa-sma(cos acos a =(cos2a-sin2a+2sin2 a2sinacosa-tana-2sinacosa1 2sin acosa-2sin acosa+tana2sin2 a+cos2 a-sin2a) ooS2a+sin2 a-tanaJ = ( tan acos2a+sin2 a =(1-tana)tan a1 :.L.H.S=R.HS. Hence proved. 22JQ.ll.Verify that the matrixA = 1:.(~1-2is orthogonal. 3-2 2-1 Sol.A matrix A is called orthogonal if AAt=At A =I. Consider, (12 2J[(12 2JJAAt= ~21-2~21-2 -22-1-22-1 1(12 2r 2 -2J=- 21-2212 9-22-12-21 71 l. (1+4+42+2-4-2+4-2)(900)(100)=i2+2-44+1+4-4+2+2=i090=010 -2+4-2-4+2+24+4+1009001 AAt =1=> Now, consider [ (12 2)]f2 2)AtA = -}-}21-221-2 -22-1-22-1 1(12 -2f2 2)=- 21221-2 92-21-22-1 [+4+42+2-42-4+2)(900)(100)=i 2+2-44+1+44-2-2=i090=010 2-4+24-2-24+4+1009001 AtA=I .. AAt=AtA=I => Hence A is an orthogonal matrix. Q.12.Find the characteristic equation and characteristic roots of the matrix 1 2 -2)A= 111.(13-1 Sol.The characteristic equation of a matrix a is given byIA- .MI = 0 . (12-2)A= 111 13-1 (12-2)(100)(H2-2)-A- Al =111- A010=11- A1 ., 13-100113-I-A 72 I-A2-2 =>IA-AII=11I-A 1\=_..1,3+..1,2+4..1,-4 13-I-A The characteristic equation is given by _..1,3+..1,2 +4..1,-4=0 .;;:, => ..1,3-..1,2-4..1,+4=0 =>(..1,-1)(..1,2-4)=0 =>(..1,-1)(..1,+2)(..1,-2)=0 =>..1,=1,2,-2 The characteristic roots are1, 2 and 2 . I:,.} Q.13. [ COSX If f(x) =sin x o -sinx cos x 0 OJ[COSY 0,&fy =sin y 10 -siny cos y 0 OJ 0then show that 1 f(x)f(y) = f(x+ y) .Hence show thatf(xt = f( -x). Sol.We have COSX-sinx0J[COS y-sin yOJ [f(x)f(y)=sinx o cos x 0 0 1 siny 0 cosy 0 0 1 COS x cos y-sinxsin y-cosxsin y-sinxcos yOJ [ =sinxcosy+cosxsiny-sinxsiny+cosxcosy0 001 COS(x+ y)-sin(x+y)OJ ( =sin(x+y)cos(x+y)0 001 f(x)f(y)=f(x+y)..(1) Puttingy = -x in (1) f(x)f (-x) = f(x-x) = f(O)...(2) 73 (=0-smO0)(I0OJ Now,1(0)=sinOcosO0=0L0=13,...(3) o01001 From (2) and (3), l(x)/(-x)=13 Hence1(-x) is the inverse of 1(x) .i.e.,1(xt = 1(-x). Long Answer Type Questions: 2 0 -lJQ.l.Find the inverse of the matrix5lOusing elementary row operations only. (o13 Sol.Consider the identity, A=IA 2 0 -lJ[10 OJ510=010A(o13001 Applying the row operationRz~ Rz- -5 RIon the matrix on the right as well as 2 on the pre-factor on the right,we have 200-Ilr 10 011OIA~=-0; 0101 Applying the row operationR3~ ~ - Rzon the matrix on the right as well as on the pre-factor on the right, we have 20 01 o0 -1 5 2 5 2 :1 100 5 = 1OIA 2 5 -11 2 74 Applying the row operationRl--? Rl+2R3 , ~--? ~ - 5~on the matrix on the right as well as on the pre-factor on the-right,we have 200 010 ' < ~ : 1I 00 2 = ') 6-22 -156 -5 111 5 -11 2 1 Applying the row operation~--? - R1, ~ --? 2 ~on the matrix on the right as well as 2 on the pre-factor on the right,we have 10OJ[3-11 J o10= -156-5A [ o015-22 From above identity,the inverse of A is -11 J6-5.A-'=[+ -22 2432 3652 Q.2.Find the inverse of the matrixA = I . Sol.Consider the identity, 2432 3652 252-3 451414 = 2 52-3 451414 " A=IA 10o0 01 o0IA o010 0o01 1 Applying the row operationRl--? - R\on the matrix on the right as well as on the pre 2 factor on the right, we have 75 12~1 2 36521= 252-3 451414 1000 o100IA o010 0001 Applying the row operation1S-+ 1S - 3Rt , ~ ----t~ - 2Rt, R4-+ R4 - 4R,onthe matrix on the right as well ason the pre-factor on the right,we have 12 3 2 1 o0 1 -2 -11= o1-1-5 o-3810 ~ 2 000 3 2 -1 1 0 0 1 OIA 0 -2001 Applying the row operation1S~ ~ on the matrix on the right as well as on the prefactor on the right,we have 12 3 2 1 o1-1-5 o0 1 2 -1 o-3810 ~000 2 -101O'A = ~100 2 -2001 Applying the row operationR,----tR, - 21S , ~----t2 ~ , R4----tR4 +31Son the matrix on the right as well as on the pre-factor on the right,we have o-2010711I5 2 o1-51 =-1 -10 1 IAo01-2 -32oo o05-5 -5031 7 Applying the row operationR\----t~ -- - ~ ,1S ----t 1S +~ ,R4 ----tR4 - 5~on the matrix 2 on the right as well ason the pre-factor on the right,we have 76 13-7-20 10018 01o-7-421 = OrA -3200 o01-2 o00510-1031 1 Applying the row operationR4-7 - R4on the matrix on the right aswell as on the pre5 factor on the right, we have 13-7-2010018 -4210010-7 = -320 oIA 001-2 31o0012-2 55 Applying the row operationRI-7 RJ -18R4 , R.z-7 R.z+7R4, ~ -7~ +2R4on the matrix on the right as well as on the pre-factor on the right,we have 1000 0100 0010 0001 6418-2329 55 26710-12 = 55 IA 621-2 55 312-255 From above identity, the inverse of A is -2329 6418 55 10-12 267 A-I =1 55 1-2 62 55 2-2 31-55 77 l (12-IJ0.3.Verify(A+B)2 =A2 +AB+BA+B2forthe matricesA=203 o12 (3-11)andB= 002. 4-32 (12-IJ(3-1IJ(41 OJSol.We haveA+ B=203+002=205 o124-324-24 (41 0J(41 0)=> (A+B)2=(A+B)(A+B)= 205205 4-244-24 (184 5J => (A+B)2 =28-820 28-46 Now, (12-IJ(12-1)(513)A2=203203=274 012012227 (12-r -11)(-12 3)AB =203002=18-118 o124-328-66 (3-1T2-1)(17-4)BA =002203=024 4-32012-210-9 (3-1If -1IJ(13~3JB2=002002=8-64 . ., , 4-324-3220-102 78 5 0 2 4 7 -4)+[13 2278-66-210-920-102A'+AB+BA+B' =[2 7 4 13J +C18-112 3 8J+r 8 34J [184 5) =28-820 286 -4 Hence(A+Bl =A2 +AB+BA+B2. -2-4)Q.4. Express the matrixA= [34as the sum of a symmetric and a skew -2-3 symmetric matrix. Sol.We have 2-11 ) At=-23-2.[ -44-3 NowA =P +Q Iwhere P is a symmetric matrix and Q, a skew symmetric matrix are given as P='!'(A+At )and Q=.!.(A-A1)22 l[ 2 -2-4)[2-11J][4-3-3) -234+-23-2 -362 -4-2-3-44-3-32-6 [[ 2 -2-4)[2-11)][0-1-5) -234- -23-2 106. -4-2-3-44-35-60 Verification: 4-3-3)pI-362=P-[-32-6 P is a symmetric matrix. 79 0-1-5]Qt=.!. (A - AI ) =.!.106 225 (-60 Q is a skew symmetric matrix. ~ Q.S.IfA=( 12],expressA6 -4A5 +8A4-12A3 +14A2as a linear polynomial in A. -13 Sol.The characteristic equation of A is 1-12 3-11=0,-1 12 -41+5=0. By division algorithm, we have 16 -415 +814-1213 +1412=(12 -41+5)(14 +312-1)-41+5...(1) From (l),we have A6 -4A5 +8A4-12A3 +14A2 =(A2 -4A+5)(A4 +3A2-1)-4A+5 Since by Cayley-Hamilton theorem, A2 -4A+51 =0 Therefore, the right hand side of (2)equals-4A+51 . HenceA6 -4A5 +8A4-12A3 +14A2=-4A+51 Q.6.Using Cramer's rule, solve the system of equations x+y+z=1 1.x+2y+3z=k 12 x+ 22 y+32 z =k2 Sol.We have 111 D=ll231=(1-2)(2-3)(3-1)=2 321222 SinceD':f:.0 ,by Cramers rule solution is given by DDD x=;,y=;,z=; 80 (2-k)(3-k)2(1-k)(k-3)(1-k)(2-k) x=,y=,z=22.2 (2-k)(3-k)(1-k)(2-k)Orx=,y=(1-k)(k-3),z= Q.7.Solve the system of equations x+ y+z =7,x+2y+3z =16,x+3y+4z =22. Sol.The given equations may be written asAX = B,where A=(:i Then [: Applying elementary row operationsRz Rz - R), R3- R) ,we have l81 Applying elementary row operations~ ~ ~ - 2 ~ ,we have 1 IIJ._(XJ( 7Jo12y=9(o0-1_z-3 Thus we have reduced the co-efficient matrix A to triangular form and the system of equations get reduced to x+ y+z = 7,y+2z =9,-z =-3 =>z =3,y =9-2z =3and x= 7- y =1. Q.8.Using matrices ,solve the system of equations x+3y+2z =0,2x- y+3z =0,3x-5y+4z =0,x+17y+4z =0. Sol.The given equations canbe written as 13 3-5 2-1 ~m=117 0 0 0 0 Applying elementary row operations~ ~ ~ - 2R1, R3~ R3 - 3, R4~ R4- R1, we have 13 0-7-3200n o-14 -:~=~ 014 Applying elementary row operationsR4~ R4 +~ ,we have 132 0-7-3 0-14-2 000 m= 0 0 0 0 / Applying elementa ry row operationsR3~ ~ - 2 ~ , we have 82 I 132 0-7-3 004 000 [ ~ } =0 0 0 0 / :. The system of equations reduces to x+3y+2z =0,-7y-3z =0,4z =0 =>x=O,y =O,z=O Hencex = 0, y = 0, z= 0is the only solution. Q.9.Find for what values of k the set of equations 2x-3y+6z-5t =3, y-4z+t = 1,4x-5y+8z-9t = k Has (1)no solution (2) infinite number of solutions. Sol.The given system of equations canbe written asAX = B 2-36-5][3]o1-41X=1[4-58-9k The augmented matrix is given by 2-36-5 . 3][A:B]=01-41 .1 [ 4-58-9k Applying elementary row operationsR3~ R3- 2R1 ,we have 2-36-5 [A:B]01-41 [o1-41k ~ J Applying elementary row operationsR3~ ~ - ~ ,we have 2-36-53] [A:B]01-41.1 [o000:k-7 83 L (i)There is no solution ifrankA '# rank [A: B] Ifk-7 '#0ork '# 7,thenrankA =2, rank [A:B] =3 =>The given system has no solution if k '# 7 (ii)There are infinite number of solutions if rankA = rank [A+B] . We can easily see thatrankA = 2 andrank [A: B] = 2 will be equal to 2 if k-7=0ork =7. Fork = 7 ,the given system of equations reduces to 2-36 -5][3] x3 1 -41 Y=1[o000Z0 t This canalso be written as 2x-3y+6z-5t =3,y-4z+t =1... (1) Asr =rankA =rank [A: B] =2and the no. of equations, n =4,:. to obtain the solutionn - r = 2 variables must be taken as constants. So ,letz=kl ' t=k2 ...(2) From(1) and(2),we get x =3+3kl +2k2 ' Y =1+4kl - k2 , Z =kl ' t=k2 . This is the required solution. Q.10.Determine ,without actually attempting to solve, whether the following system is consistent or not. x-3y+ z =-1,2x+ y-4z =-1,6x-7y+8z =7. Sol.The given equations may be written as .AX = B . 1-31](-1]21-4X= -1(6-787 84 The augmented matrix is given by 1-31: -1][A:B]=21-4-1 [ 6787 Applying elementary row operations~ ~ ~ - 2Rl ~ ~ ~ - 6Rl'we have 1-31-1][A:B]07-61 [ o112:13 Applyi ng elementary row operationsR3~ R3-.!..!. ~ Iwe have 7 1-31-1 [A:B]107-61 8080o0 77 Applying elementary row operations~ ~ ~ R3/we have 80 [1-31 : -1][A:B]07-6:1 o01:1 1-311-3-1 =7 :;I!:Oand We have107-607 11=7:;1!:O. 001001 rankA = rank [A:B] = 3 Hence the given system of equations is consistent. Q.ll.Show that only value of Afor which the following system of linear equations has non zero solution is 6 and then solve the equations. Sol.The given equation canbe written as ~ (I-A)X+2Y+3Z=0} 3x+(I-A) y+2z =0...(1) 2x+3y+(I-A)Z = 0 85 s The system of equationsAX = 0has a non zero solution if the rank of the coefficient matrix A is less than 3 i.e.,IAI = O. I-A23 ... I 3 I-A 21=0 23I-A I =>(6- ..1)(..12 +3..1+3) =0 -3H => A=60r ..1=-2 Hence 6 is the only real value of Afor which the given system of equations has a non zero solution. Substituting..1=6in (l),we get (7 Applying elementary row operations -t+,we have (-:5-2-3z0 Applying elementary row operationsR2-t+R"R3-t+ R, ,we have 5 ;_ [XJ-[OJ55Y- 0 o00z0 => -5x+2y+3z=0 1919--y+-z=O55 O.z=O Equation (3) is true for all values of z,solet z =k , k be any number ...(1) ... (2) ...(3) 86 Putting the value of z in (1) and(2),we have -5x+2y=-3kandy=k =>x=k,y=k,z=k This is the required solution. ~ [1-23J Q.12.Verify thatA(BC)=(AB)C for the matricesA=23-1' -312 [102J[11 1J B = 012andC =12-3. 1202-13 [1-23f0 2J[1-0+30-2+62-4+0 J Sol.AB =23-1012=2+0-10+3-24+6-0 -312120-3+0+20+1+4-6+2+0 [44-2J=>AB=1110 -15-4 Consider [44-2f1 1J [4+4-44+8+24-12-6J (AB)C =111012-3=1+1+201+2-101-3+30 -15-42-13-1+5-8-1+10+4-1-15-12 [414-14J =>(AB)C =22-728 -413-28 Now (102f1 1J [+0+41+0-21+0+6J BC= 01212-3= 0+1+40+2-20-3+6 1,202-131+2+01+4+01-6+0 [5-17J ','=>BC= 503 35-5 87 Consider 7 J(5-10+9-1+0+157-6-15J-23](5-1 3-1503=10+15-3-2+0-514+9+5 -31235-5-15+5+63+0+10-21+3-10 414-14J =>A(BC) =22-728 [ -413-28 Hence A(BC)=(AB)C. = 1IJQ.13.Verify Cayley-hamilton theorem for the matrixA10. 12 Sol.The characteristic equation of a matrix a is given byIA - 211=O. 2-211 => I 0 1-2o1=0 112-2 =>23_522+72-3=0. According to Cayley-hamilton theorem A3-5A2+7A-3I=O Verification: [2IIf IIJ[544JA2=010010=010 112112445 A3[544fIIJ(141313J=010010=010 445112131314 [141313J[544J[2IIJ(10 OJA3-5A2+7A-3I=010-5010+701o -3010 13131444511200 1 88 14-25+14-313-20+7+0 13-20+7+0J =0+0+0+01-5+7-30-0+0-0 [ 13-20+7+013-20+7-014-25+14-3 000 Hence verified. r' o2]Obtain the characteristic equation of the matrix A = [ 21and hence o3 Calculate its inverse. Sol.The characteristic equation of a matrix a is given by IA- All =0 . I-A02 => 2-A1=0I 0 203-A =>-A3 +6A2-7A-2=0. =>A3 -6A? +7A+2=0 According to Cayley-hamilton theorem A3 -6A2+7A+21=O...(1) Multiplying (1) byA-I ,we have A-IA3 -6A-1A2+7A-1A+2A-1I = A-tO =>A2-6Al +7+2A-1 = 0

-I1237=>A=--A+A--I... (2) 23 Now [102f02][508]A2=021021=245...(3) 2032038013 89 06 {0z zz -{ -{ {= I-V B=a2_b2 222 b2x a xxdx=dx- dx JJJ(x2+a2 )(x2+b2)a2_b2 (x2+a2)a2_b2 (x2+b2) 22 1b-I (x)---tana -1 (x)-1 tan- +c2 b22a - aaa - b2bb -a tan--I (x) b_I(X)2-2tan- +c. J - a2 _b2 a a-bb 1dx. Evaluate'3x2 +4x +5 We have 245)(45 2 2)3x2 +4x+5=3( x+-x+- =3x 2 +-x+---+333399 =3[(x+%)' +~ J=3((x+j)' +[ ~ J ' ) substituting this value in the given integralJwe have 1dx __ 1 J1dx f"3X'+4x+5-.J3tx+%)' +[ ~ J 1 .nh-1 [X+2/3]=.JjSl.J1i/3+c =-SI1.nh-1 (3X+--2)+c . .Jj.J1i 101 J x+2Q.8. EvaluateJdx . x2+3x+l 2Sol.let1 =Jjx+dx..(1) let)iaralpbetwotohstantssuchthar +3x+l)+,udx X+2=A(2x+3)+,u. Comparing the coefficients of x and constant terms on both sides, we have 111_1..A='2''-- 2 11 x+2=-(2x+3)+- ...(2)22 Using (2) in (1},we get 11 -(2x+3)+- 1(2x+3)11 1= J22 dx =- Jdx+- Jdx +3x+l.2 +3x+l2+3x+l =11 +/2 ...(3) 1 J(2x+3)1 J.1 where,11=- Idxand 12=- Jdx. 2...x2+3x+l2x2+3x+l Consider 1(2x+3)dx 11='2 Putt =x2+3x+1 dt=(2x+3)dx 211=1.. J=.Jx +3x+l+c12'It 102 Now, 1, = ~JJi:3x+1(1+ m-l) JSinmxcosn xdx= n+ln+ln+l m+n).cosn+1xsinm-1xm-l._m :::::::>-- Jsmmxcosn xdx= - +-- Jcosn xsm2xdx( n+ln+ln+l n+!m-IXm-1Jnm-2dxcosxsm+ __ cosxsmx.:::::::>JSinm XCOSn xdx = m+nm+n Q.3.EvaluateJcosm xcosnxdx,mand nbeing positive integers. Sol.We have Jcosm xcosnxdx Integrating by parts,we get mIJ cosmxcos nxdx = cos x sin nx + mJ cosm- X sin x sin nxdx nn We know that cos (n -1) x =cos nxcos x + sin nxsin x :::::::>sinnxsinx =cos(n-l)x-cosnxcosx mJcosmxcosnxdx = cos x sin nx + mJcosm -! x(cos(n-l}x-cosnxcosx)dx nn m sin nxmJ-ImJ=cos x--+- cosm xcos(n-l)xdx-- cosmxcosnxdx nnn 109 . (m) I m sin nxm I m =>cosm xcosnxdx =cos x--+- cos 1 xcos(n-1)xdx nnn m+n) Isinnxm I m =>-n- cosm xcosnxdx=cosm x--+-;cos 1 xcos(n-l)xdx(n I sinnxmm-1 I=>COSm xcosnxdx =COSm X--+-- COS xcos(n-l)xdxm+nm+n Q.4. /x+2 EvaluateIV2x+"3 x Sol. /x+2We haveIV2x+"3 x x+22 Put --=t 2x+3 2 =>x= 3t-2 12t =>dx=dt (1- 2t2 ) A/x+2 1dx - 2 It1- 2t2 tdt IV2x+"3-;-- '3t2 -2" (1-2t2 f 22t t=-2I(2\{2\dt=2I- .dt __ ."1 - 2IdtIdt- - (2t2 -1) -4(3t2_ 2)(Using partial fractions) 110 1(t-*) (1) v2t+-Ji ___(Jit-l)(.J3t-Ji) 1-log log (r:::;+C v2v2t+l3v3t+v2 J dxQ.5.Evaluate. (x+2)../x2 +3x+4 111-2t Sol.Putx+2=- =>dx=--dt andx=-t2tt dxdt --Jf(X+2)N+3X+4 .1dt dt=- =-J../2t2 -t+1v2t 2--+-22 1dt=--JF-Rl1 1 Jit2 __+_+___ 2244 =- h 1 f '(t-H dt ":J t-1.-14[ 1]=- hsIDh":+c 1._1[4t-l]=- Jismh.fi+C 111 14(_1)=_x+2-1 v247l+c - ----= __Sinh-;[-2=X J2J7(2+x) 4a Q.6.Show thatthe length ofthe loop of the curve3a/ = x( x- a)2is-,J3 Sol.The curve is symmetrical about x-axis and it passes through the origin. Its point of intersection with the x-axis isA (a, 0) . y o Thelength of the loop=2r . 2)2.4aWehave3ay=x(x-aISJ3 y_1r=>- Favx(x-a) => :=k(..fx+ 1+(dy)2=1+ (3x-a)2= (3x+a)2=> dx12ax12ax 112 (3x+a) dx The length of the loop= 2 r~ 1artaJ =5a ~ l 3 . . J x+..Jxdx ------------_.... 1 =5a(2a3/2 +2a.al/2 ) 4a =.J3' 22b2Q.7.Theellipseb2 x+a2i=aisdividedintotwopartsbythelinex= a andthe 2 smallerpartisrotatedthroughfourrightanglesaboutthisline.Provethatthevolume 2generated isJmb( %.J3 - ; ) . Sol. The shaded portion isrotated about the linex = E:. which isparallel to y-axis.2 Thepoints of intersection of the linex = a with the given ellipse is given by 2 2a2b2b2 -+a2 i=a 4 113 => y=..[3b2 is 2" J3b( aV=2127&x- a)2 dy where x 2 =-r(b2-l)Jy=!)2 b J3b( 2 a2)2=2..(: 7&x-ax+4 dy J3b(a2 aZa2 )2=27&"(: dy a2 a2 2J3b a2 y(b2 -l)b -1Y J3bjl"2J3b = 27&-2 b 2 y-- +-sm--2 +-IYI2b 3b22b40 r o0 =27&[ab2 2 (..[3 b2_3..[3b2 _7&)+ ..[3 aZb]224442238 -: -Q.8.Findthesurfaceofrevolutionofthesolidgeneratedbyrevolviongthe parabola yZ=4axbounded by its latus rectum about x-axis. Sol.Required surface area is s=We have yZ=4ax dyJ"d => dx=.rx S =r27&J"d.rxJl +:dx 114 arcofthe ...~~ ' . f =4f(fa r.Ja +xdx =4f(fa ~ l ( a +xt121: 8f(r[(2)3/23/2J=-"aa-a3 =*f(a2 [2.J2 -1]. Q.9.Changing the order of integration of r re-xy sin nxdx dy ,show thatrsin nx dx =f(. x2 Sol.We haver re-XYsinnxdx dy = rdy re-XYsinnxdx...(1) First we shall find the integral re-xy sinnx dx . -xy-xy Je-XYsinnxdx = -sinnx-=--- J--=--.ncosnxdx yy -xy .enJ-xydx=-Slllnx--+- ecosnx yy e-XYn [e-XYJe-XY]=-sinnx--+- ---cosnx- ---.nsinnxdx yyyy -xy2 ndx=-slllnx--..:..-e.e- ~ -,cosnx--nJ- ~ .e -,smnx yll n2)_..e-xyn_=>1+-2 Jexyslllnxdx=-Slllnx----2exycosnx( yyy =>Je-XYsinnxdx=2y2e-XYsinnx- 2n_e-XYcosnx n+yn+y 115 -xy Ie-XY sinnxdx=:2(-ysinnx-e-XY cosnx)...(2) n+y Using (2) in (l),we get rre-XYsinnxdxdy= rdY[:-xy2n+y0 = rdY[O+2n2]= r[ 2n2]dY n+yn+y

f(=Itan-I Yo=2" ...(3) On changing the order of integration,we get rre-XYsinnxdxdy=rsinnxdxre-XYdy = r = rsin nxdx(-0+1) o =rsinnxdx . (4) From (3) and (4),we get sinnxrdx- f(-x- -2"' Q.I0.EvaluateIJI( x2+y2 +z2)dxdydzR denotes the region bounded by R X=0, y =0, z =0,and x +y+ z =a ,a >. Sol.We have x+y+z=az=a-x-y Upper limit ofz =a-x- y. Onx-yplanex+y+z=a becomesx+y=ay=a-x. Upper limit ofy =a - x 116 Upper limit ofx =a JIJ( X2+l+ z2)dxdydZ R = rr-x r-X-Y(x2+l+z2)dxdydz XY""', =rdxr-x dyr- - (x2+l+z2)dz = r dx r-x dy r->:-y (X2+y2+Z2)dz =' dx r dy(x'z+ /z+ ;r-' =' dx r dy(x' (a-x- y)+ y' (a-x- y)+y)' J 2 (a-x-y)3)= rdx rX2(a - x) - x Y + ( a- x) l-l +3dy( 234(4)a-x =dxx2(a-x)y-x2L+(a-x)L_L+a-x- y) .23412r ( o =rdx(x2(a_x)2 _x2(a_x)2+(a-x) (a_x)3(a-xt +(a-x-(a-x)f] 23412 ='( (a-x)' +r= ' Ma'x' -2a.'+x' +2 51=Iia - a: + _(a;;) a =: -: + + = . o Q.11.Find the volume bounded by the cylinderx2 +l=4 and the planesy +Z = 3and z=O. 117 -xy =>Ie-XYsinnxdx=:2(-ysinnx-e-XYcosnx)...(2) n+y Using (2) in (l),we get rre-XYsinnxdxdy =rdY[:-xy2(-YSinnx-ncosnx)]OC n+y0 = rdY[O+2n2]= r[ 2n2]dY n+yn+y

tr=ltan-1 Yo=2" ...(3) On changing the order of integration,we get rre-xy sin nxdxdy =rsin nxdx re-xy dy =r=rsinnxdx{-O+l) o =rsinnxdx .. (4) From (3) and(4),we get rsinnx dx= tr . x2 Q.l0.EvaluateIII( x2+l+Z2) dx dy dzR denotes the region bounded by R X =0, y =0, z =0,and x + y+z =a, a >. Sol.We have x+y+z=a=> z=a-x-y Upper limit ofz = a-x- y. Onx-yplanex+y+z=a becomesx+y=a=>y=a-x. Upper limit ofy =a - x 116 2Sol.We havex+ l=4=>y=.J4-x2 y+z=3 => z=3-y z=o. :.zvaries from 0 to3-y,yvariesfromy=-.J4-x2 toy=.J4-x2 On the x-y plane,x2 + l= 4,:. x varies from-2 to 2. fPr'-Y :.Required volume=12 dx iMdy.bdz =f 2dx=t dxC; (3- y)dy '2IL, 2'M =Ldx3y- 21_M =l,,+J4-x'+3J4-x' += 6 r.J4- x 2dx =x2

2222-2 = 6[2 sin-l %- 2 sin-l= 12[ + = 12K. 0.12.EvaluateJJ( x2 + l)dx dy throughout the area enclosed by the curves y =4x, x +y =3, y = 0, y = 2 . Sol.let OC representsy =4x ;ABx+ y =3 ;OAy = 0and CBy =2. The given integral is to be evaluated over the area of the trapezium OCBA. AreaOCBA consists of area OCD,areaDCBE,andarea EBA. The co-ordinates of C,Dand Bare 2). (1,2) and (3,0) respectively. 118 y2 8(1,2) o E(I,O)A(3,O) JJ(XZ+ l)dxdy=JJ (XZ+ l)dxdy+JJ(XZ+ l)dxdy+JJ(XZ+ l)dxdy OCDDCBEEBA =1'Z rX(xZ+l)dydx+ tzl(x2+l )dydx+ f r-X(x2+l)dydx = I, +12+13(say) Now,I, = rnx'+y')dydx= r(x'y+ dx = r(x'(4x)+(4;)']dx = r/276 x3dx=761x4\1/2 =.!2. 048 .b334 I, = !,f(hy')dydx= L(x'y+dx = t,(x'(2)+

3 J3-X 13=f rX(x2 + l)dydx= f( x2y+0dx 119 \ = r(x'y+ dx=r(x' (3-x)+ Jdx = r[3X' -x' + -:I 22 3 II( x2 + l)dx dy =19 + 23 + 22 =463 4812348' 120 l \ I Unit - 5 : Probability and Statistic Multiple Choice Type Questions: Q.l.Two dice,one is green and one is blue are thrown. What is the probability that both the dice show 5 is 1151 (a)(b)(c)(d)18363612 Q.2.A number is chosen from the first 90 natural numbers.The probability that the number chosen is a multiple of 5 or 15 is 11 . (a) .!.(b)! (c)(d)58610 Q.3.A problem in mathematics is given to three students A,Band C whose chances of solving it are respectively.!.,!and.!.. The probability that the problem is solved is 234 1131 (a)(b)(c)(d)24424 Q.4.If A,B,C are three independent events,thenP( AnBn C) is (a)P(A)+P(B)+P(C)(b)P(A)P(B)P(C) (c)P(AUBUC)(d)None ofthe above Q.S.If A andAIare complementary events in a sample space S,then (a)P{A)+P(AI)=O(b)P{A}+P(AI)=1 (c)P(A)-P(AI)=O(d)P(A)-P(AI) =1 Q.6.In tossing a fair dice,the probability of getting an odd numberor a number less than 6 is (a)1(b)3.(c)1(d)1 3326 Q.7.The probability that a non-leap year will contain 53 Sundays is 1(b)2(c)3(d)4(a) 7777 Q.8.The probability that at least one of the events A andB occurring is 0.8 and the probability that both the events occur simultaneously is 0.25.The probabilityP(A) +P(AI) is (a)1.65(b)1.05(c)1.85(d)0.95 121 4 Q.9.The mean and variance of Binomial distribution are4and - respectively.The value of n is 3 (a)12(b)4(c)5(d)6 Q.l0.The probability of having at least one tail in 4 throws with a coin is 1511 (a)(d)116(b)16(c)"4 Q.ll.let X be a Poisson random variable, suchthatP ( X =0) =P (X=1). The standard deviation of X is (a)4(b)2(c)J2(d)1 Q.12.A manufacturer of steel blades found 5% of its blade defective.He sells blades in packets each containing 5 blades.The probability that a packet contains one defective blade is (a)0.25e-O25 (b)0.5(c)e-O25 (d)0.25 e-X x ~ O Q.13.Given a probability density functionf (x) ='.The probabilityP(1 S;xS;2) is{ o,xO) =>E(X)=J=l. Now,P(X xo) =0.3 ::::;..0.5-P(11 xn+l=X n _nxn- 5 ... (1)3x2n- 2 ChoosingXo=2, we obtainI (xo)=-1 andf' (xo) =10. Puttingn =0 in (1), we obtain Xl=2 - ( - 1 ~ )=2.1 153 Now, f(Xl) =(2.1)3 - 2(2.1) -5 = 0.061, and f = 3(2.1/ -2 = 11.23. Hence X= 2.1- 0.061=2.094568. 211.23 Q.3.Finda realroot of the equationX= e -x using the Newton- Raphsonmethod. Sol.We haveX = e-x :::;>xe x = 1 Letf( X )= xe x -1 = 0 :::;>f' ( x) = ( X + 1) eX Then f(xn ) xn+l= xn- f! (x n ) = xn LetXo==1. Xl= e + 1 = 0.6839397 2e XeX.+exn n :::;>x2 +1 (Xl += 0.5774545 .. Proceeding in this way. we obtain =0.5672297andx4 =0.5671433. Q.4.Solve graphically the equationx-I =sin x. Sol.x-I ==sinx X2ex+ 1 n (xn+ 1) eX. We take two equationsy = X -1andy = sin x. Let us find out the abscissaof the point of intersection of the liney = X -1 andthe curvey ==sin Xand give a rough estimate of the root. 154 x For the straight liney = x -1, we have the table: x3Jr 14~ 1 y=x-11.4~ o For the curvey =sin x twe have the following table: x0Jrl4Jr 123Jrl4Jr Y =sinx 0 ,0.711.000.710 On the same axest and with the same scale constructing the graphs ofy =x -1 and y =sin x twe getx =1.95radians approximately. 155 2Q.5.Determine the root ofX4+x3-7x - x +5 = 0which lies between 2 and3 correct to three decimal places using Newton -Raphson method. Sol.We haveI(x) =X4+X3 -7x2 -x+5 =0 =>1 (2) =16 + 8 - 28 - 2 + 5 =-1 and1 (3) =81 +27 -63-3+5 =+47. Since1 (2) 1 (3) < O,the root lies between 2 and3. Now,f'(x) =4x3+3x2 -14x-l TakingXi=2asfirst approximate root,we have I(Xi)=2- 1(2)x2 =Xi f'(Xi)1'(2) (-1)1 => x= 2--= 2+- = 2.067[..' 1 '(2) =32+12-28-1 =15J 21515 and x=x_I (x2 )=2.067 _ _1--,-(2_.0_67--,-) 32f'(x2 )1'(2.067) = 2.067 - (-0.0028) = 2.067 +0.0001519 = 2.0671519. 18.422 Q.6.Write the Newton-Raphson procedure for findingifN, where N is a realnumber. Use it to find3 J18correct to 2 decimals, assuming 2.5as the initial approximation. 33Sol.Letx = ifN => x = Norx - N = 0 Letl(x)=x3-N=0. =>I'(x) =3x2 By Newton-Raphson Method,x +1=X_1 (xn ) nnf'(xn) X323+N2 n- N _xn" n =0 1,, .... =>xn+1=xn ~ - 3x ' nn Let N =18. We shall find the root of 1(x) = x3 -18 = 0 ,taking 2.5as the initial approximation. So, takeXo= 2.5 . 156 2X03 + 18_2(2.5f + 18=2.62667=> .x;=3X02- 3x(2.5f 2.x;3+ 182(2.62667)3 + 18 =2.620755 . => 2x =3x/=3(2.62667)2_ Q.7.Find anapproximate value of the root of the equationX3+x-I =0nearx =1, using the method of falseposition two times. Sol.We havef(x) =X3+ x-I =0 f(I)=I+I-I=+1 f(0.5) =(0.5)3 +(0.5) -1 = -0.375 Sincef(0.5) f(1) < O,the root lies between a.Sand1. LetXl=0.5andx2 =1 .x;f (x2)-xd (.x;) X3= f(x2)- f(x1) x= 0.5f(I)-lf(0.5)=0.5(1)-1(-0.375) =064 => 3f(I)- f(0.5)1+0.375. Nowf(0.64) = -D.0979andf(1)= 1 :. Root lies between.64 and1. So ,letXl= 0.64,x2 = 1 0.64f (1) -1.f (0.64)0.64(1) -1.(-0.0979) X==0.672 3f(l)f(0.64)1+0.0979 157 Q.8.From the following table of values of x and Y,obtaindy forx =2 . ax x:2345 y:82764125 501. The difference table is :x Y 28 19 32718 376 46424 61 5125 I I We know that, dY ]1 (121314J [ axx=Xo=hYoYoYo+... Here, h = 1, Xo=2, Yo= 8 . [ dY ]1(11)- =- 19--.18+-.6=19-9+2=12. axx=2123 Q.9.From the following table of values of x and y,obtaindy forx0.30 . ax Ix 0.100.150.200.250.30 Iy 0.10030.15110.20270.25530.3093 Sol.The difference table is x 0.10 0.15 y 0.1003 0.1511 0.0508 0.0516 ,

0.0008 -

0.0002

158 0.20270.0010 0.200.0002 0.05260.0004 0.25530.0014 0.25 ! 0.0540 0.300.3093 i We know that, [dY]=!"(VYn +!..V2Yn +'!'V3Yn+ !..V4Yn+ ... ) dxh234 X=X n Here, h = O.OS,xn= 0.30, Yn= 0.3093 [ dY ]=_1_(0.OS40+!..(0.0014)+.!.(0.0004)+!..(0.0002)) dxx=O.300.05234 = 1.097666. Q.l0.EvaluateI= ~ ~ , correct to three decimalplaces using trapezoidalrule with .bl+x h=O.S. Sol.Forh =0.5:the values of x andyare tabulated below: !X Y I 0.0 1.0000 I0.5 0.6667 1 1 . 00.5000 Trapezoidalrule gives Cydx= ~ [ Y o+2(Yl + Yz+ ... + Yn-l)+ Yn] ~1=!..[1.0000+2(0.6667) + 0.5] = 0.70835. 4 159 Long Answer Type Questions: 3Q.l.Find a real root of the equationf(x) = x - 2x- 5 = O.using method of False position. 3Sol.We havef (x) =x - 2x - 5 =O. ~f(2) =-1 andf(3) =If) . Hencea root lies between2 and 3. let.x;= 2andx2 = 3 The regula falsi iteration formula is .x;f (X2) - xzf (.x;) X3f(x2)-f(.x;) 2(16)-3(-1) = 35= 2.058823529. ~ X3=16-(-1)17 Now,f(X3) =-0.390799917 1it follows that the root lies between 2.08126366 and 3.0. Hence, we have 2.08126366(16)-3-(0.147204057) = 2.089639211. X3=16.147204057 Proceeding in this way, we obtain successively: X4 =2.092739575,Xs=2.09388371, X6= 2.094305452,X;= 2.094460846, ... The correct value is 2.0945"'1 so thatis correct to five significant figures.x7 Q.2.Find a root of the equationX sin x + cos x=0 using Newton Raphsonmethod. Sol.We have f (x) =xsinx+cosx ~f'(x) =xcosx. The iteration formula is therefore xn sin xn+ cos xnxn+1=xnxn cosxn WithXo= 1[, the successive iterates are given below 160 n.xnf(xll ) Xn+l 03.1416-1.02.8233 12.8233-0.06622.7986 22.7986-0.00062.7984 --:3 2.79840.02.7984 Q.3.Find, from the following table, the areabounded by the curves and the x-axis from x = 7.47tox = 7.52 . x 7.47 .7.48 7.49 f(x) 1.93 1.95 1.98 x 7.50 7.51 7.52 f(x) 2.01 2.03 2.06 Sol.We know that r.52(rrf.52Required Area=fx)dx=ydx.47.47 We are givenh=O.Ol, x 7.47 7.48 7.49 Y 1.93 1.95 1.98 x 7.50 7.51 7.52 Y 2.01 2.03 2.06 The trapezoidal rule gives In Y d x = ~ [ Y o + 2(Yl + Y2+ ... + YII-J)+ YIIJ "52hJ=>ydx=-[Yo+2(Yl+Y2+"'+Y4)+Y5r.472 = 0.01 [1.93+2(1.95+ 1.98+2.01+ 2.03)+2.06J 2 161 =0.0996 Area= 0.0996 . Q.4.Derive Simpson's 1/3- rule using the method of undetermined coefficients. Sol.We assume the formula +--,: fhydx=a-IY-I +aoYo +a1Yl' ...(1) where the coefficientsa_I' aoandal haveto be determined. For this, we assume that formula (1) is exact whenY(x) = 1, x andx2 Putting thereforeY(x) = 1, x andx2 successively in (i), we obtain the relations and a_I+aO+a1 =fh dx =2h, -a_I +al =fhXdx=O a_I+a1 ="32 h. Solving (2),(3) and(4)fora_pao and~ ,we obtain 24h a_I="3=alandao=3' Hence formula (1) takes the form fhYdx= ~ (Y-l +4yo +YI)' which is the Simpson's 1/3- rule. Q.S.By using Newton-Raphson's method, find the root of x = 2correct to three places of decimal. Sol.Letl(x}=x4 -x-1O=0. ::}f' (x) = 4x3 -I, 1(2)=16-2-10=4 1'(2}=32-1=31 TakeXo=2. ByNewton- Raphson's method 162 ... (2) ... (3) ...(4) X4- x -10 = 0 , which is near to =xo- I(xo) =2- 1(2)f'(xo)1'(2)31 1 (1.871) = (1.871)4 -1.871-10 = 12.25 -1.871-10 = 0.379 1'(1.870) = 4(1.871)3 -1-= 4x6.5497 -1 = 25.1988 1(1.871)0.379 x2 =1.871- () =1.871- =1.871-0.0150=1.856I' 1.87125.1988 1 (1.856) = (1.856t -(1.856) -10 = 11.8662-11.856 = 0.0102 1'(1.856) = 4(1.856)3 -1 = 4x6.4623-1 = 24.8492 x=1.856- 1(1.856)=1.856- 0.0102=1.856-0.00041=1.85559 31'(1.856)24.8492 1 (1.85559) = (1.85559t -1.85559-10 = 11.855572-11.85559 = 0.00013 1'(1.85559) = 4(1.85559)3 -1 = 4x6.389193927 -1 = 24.55677571 x= 1.855591(1.85559)= 1.855590.00013 41 '(1.85559)24.55677571 = 1.85559 - 0.00000529 = 1.85558471. 2 Q.6.From the following table of values of x and y,obtaindy andd;, atx =1.2 . dxdx x1.01.21.41.61.82.02.2 Y 2.71833.3201 --------4.0552 -4.95306.04967.38919.0250 Sol.Thedifference table is xY

1.02.7183 0.6018 1.23.32010.1333 0.73510.0294 1.44.05520.16270.0067 163 1.64.9530 0.8978 0.1988 0.0361 0.0080 0.0013 0.0001 1.8 2.0 6.0496 7.3891 1.0966 1.3395 1.6359 0.2429 0.2964 0.0441 0.0535 0.0094 0.0014 2.29.0250 Here, h = 0.2,xo = 1.2, Yo= 3.3201 Now, dY]1 (121314)[ dxx=Xo=hYoYoYo+... dY ]1(1111)- =_.0.7351--(0.1627)+-(0.0361)--0.0080+-(0.0014)[ dx=1.20.22345 =3.3205 Also, -..I_4 2 11d]1( 23 [ dx'=. - h' =2.0576 => 1(2.0576) = -0.0579 : ..now root lies between 2.1 and 2.0576. LetXl=2.0576;x2=2.1 By Regula falsimethod, next approximation to the root is given by (2.0576) 1 (2.1) -(2.1) 1 (2.0576) =1(2.1)-1(2.0576) _ (2.0576)(0.739)-2.1(-0.0579) - 0.739-(-0.0579) => =2.0607 . =>1(2.0607) =-0.0028 =>Root lies between 2.0607 and 2.1. I By Regulafalsimethod, next approximation to the root is given by 167 --(2.0607) 1 (2.1)-(2.1) 1 (2.0607) =1(2.1)- 1(2.0607) _(2.0576) (0.739) - 2.1( -0.0028) - 0.739-(-0.0028) => =2.0609 The root of the given equation correct to three decimal places is 2.0609. Q.ll.Evaluate"3 eXdxusing Simpson's rule. Sol.Divide the interval(1.1.3) into 6 equal parts of widthh =0.05. The tabulated values of x and yare x 1.00 1.05 y=e x 2.718 2.8577 1.103.0042 1.15 1.20 1.25 1.30 3.1582 3.3201 3.4903 3.6693 By Simpson's rule, C +4(Yl + Ys+ ... + Yn-l)+2(Y2+ Y4+ ... + Yn-2)+ YnJ "3 eXdx = [2.718+4(2.8577 +3.1582+3.49031) + 2(3.0042+ 3.3201) + 3.6693J =0.9510167. 168 1.30 Q.12.Evaluatetan x dxusing Trapezoidal rule by takingh = 0.05 . .10 Sol.The tabulated values of x and yare X ~ . 0.10 0.15 0.20 y=tan x 0.1003 0.1511 0.2027 0.25 0.30 0.2553 0.3093 Trapezoidal rule gives rXnydx= h[yo +2(Yl + Y2+ ... + Yn-l) + Yn]JXo2 =>f).3Otanxdx = h [Yo +2(YI + Y2+ Y3) + Y4] 1..b.102 30 0.05[()]=>tanxdx=- 0.1003+20.1511+0.2027+0.2553+0.3093 .102 =0.040695 169 Old QuestionPaper Advanced Engineering Mathematics (AC/AA 1.1) Date: 22It, July) 2008- .. Time: 2:00-PMto 5:00-PM. Max. Marks: 100 Assume any missing data If necessary. MiM'V... Choose the correct or best alternative In the following. Each question carries two marks. 1.If Z= f (x+ct) + g (x-ct),then (a) Zit=Zxx (c) Zit-_ C 2 Zxx _2 (b) Zt=Zx (d) Zxx- CZtt -1X+ Y113)''''" .2.IfU=Slfi [1131121/2Ithenx-+ '" y- IS equal to x+yax~ 11(a)-cotu(c)--tanu 126 11 (b)tanu(d)-cotu 126 22 3.The partial differential equation of the equation2z = x2+ 4- is ab dzdzdzdz z=-+- z=x-+y(a)(c)dXdydxdy d2Zdzdz z=--2z=xa-+ Y(b) dXdy (d) xdy 4.The solution of(y - Z) P+ ( Z- x) q =x- Y is (a)f(x+ y + Z)= xyz(c)f{x 2 +l+Z2, x2lz2)= 0 2(b)f{x2 +l+Z2)= xyz(d)f(x+ y+ Z)= x + l+ Z2 f.l-10]5.The rank of matrix0f.l-1is 2, forf.lequal to -10f.lf (a)any row number(c)1 (b)3(d)2 170 IIi-x 6.The value ofJ J Jxdzdx dyis olo (a)4/35(c)8/35 (b)3/35(d)6/35 7.IfAistheareaunderthecurve}'=sin x,abovex-axissuchthaLO::;; X::;;1l' 12,then the area under the curve y = sin2x,0::;; x ::;; 1l'/2is (a)A(c)Al2 (b)2A(d)1+A 8.Thefigureboundedby one arc of thewave y=sin x and x-axisis revolvedabout x-axis. The volume of solid of revolution is (a) It (c)21l' 2 21t(b)(d)1t2 2 9.AproblemofstatisticsisgiventothreestudentsA,BandC wherechargesof solvingit are.!..,~ ,and.!.., respectively. The probability that the problem will be solved is 244 329 (a)- (c)3232 13 (b) - (d)324 10.If(J' = 2,x =5,the equation of normal distribution is 1_(.>:-5)_ (x_5)2 (a)f(x)=-e8(c)f(x)=lte8 It (X_5)218 f(x) = 2e--8(b)(d)f(x) = 2J21t e iilllt-"':1 Choose TruelFalse In the following. Each question carries one mark. 1.Charpit'smethodisusefultosolveapartialdifferentialequationoffirstorderandhigher degree 2.The property of normal curve is that the curve is symmetrical about y-axis. The mean,median and mode coincide at the origin. 171 1 3.The modulus of each eigen value of matrix is any scalar. 4.r!(x)dx= O=> !(x) =0 5.Simpson's rule gives more accuracy than trapezoidal rule . .6;iI"'6 Answer any five questions out of eight questions. Each question carries five marks If A and B are non singular matrices of the same order then prove that (ABtl = B-1A-I 2.Show by means of an example that in matrices AB=O does not necessarily mean that either A=O or B=O,where0 stands for the null matrix. 3.Using the method of iteration, find a realroot of the equation 2x-Iog10 x = 7 which lies between 3.5 and 4 4.Expand!(x, y) =eX log, (1 + y) inaTaylor'sseriesintheneighborhoodof origin,retaining terms upto second degree inx andy 5.Find the maximum or minimum values of x3l(1-x-y),x:;t:O,y:;t:O,x+y:;t:l 6.Apply Newton-Raphson method to obtain a root of the equation x3 -3x-5=0, starting withXo =3. 7.A random variable x has the following probability distribution: x:01234567 p(x):0k2k2k3k2 7 (i)find k(Ii)Evaluate p(x 1 o Show that In+ n(n -1)1._2= n ( ~)"-1 7.ifXXYYZZ= c,show that at x = y = z (fzdXdy=-(xlogexr! 181 L_ 8.Find by double integration, the smallest of the areas bounded by the circle x2 +1 =9 and the straight line x+y=3 182 Old Question Paper Advanced Engineering Mathematics (AC/AA 1.1) Date: 18111,Dec., 2009Time: 2:00 PM to 5:00 PMMax. Marks: 100 Assume any missing data if necessary. gigltm., Choose the correct or best alternative In the following. Each question carries two marks. 3 x+ l. (b)au au1.Let u (x, y ) =+xtan 1- ,( x, y)(0,0) thenx- + dyequalsx+yx~ (a)(b)2u x3+3 (33)-1Y(c)Y+ 2x tan-1 L (d) 2x+y+2xtan~x+yxx+y C22 au=C2 V 2U, =K,This equation is called atup (a)Cold equation(b)Warm equation (c)Heat equation(d)None 3.TherankofamatrixAisthe...............numberoflinearlyindependentrow vectors of A. (a)Minimum(b)Maximum (c)Two(d)None 46 96]4.IfA=60- 220,This is called[5-8110 (a)Augmentedmatrix(b)Segmented matrix (c)Linear matrix(d)None 5.JJJ dxdydz ,This is called (a)Area(b)Volume (c)Perimeter(d)A n ~ l e 183 L 6.In rolling a fair die once, what is the probability of obtaining a 5 or a 6? (a)1/2(b)1/3 (c)1/4(d)1/5 7.X=-n1 t X J ~ = ~ n l(Xl+X2+ ..... +Xn) This equation represents J ~ l (a)Median(b)Mean (c)Average(d)None ~ 8.The meanf.l,(mu) is defined byf.l=J.if(x)dxfor (a) (c) Discrete distribution Both(a)and (b) (b) (d) Continuous distribution None of the above 9.D sin =cos,D2Sin = (a)-cos (c)sin (b) (d) -sin tan 10.If number ofrows and columns of matrix are equal, then matrix is called (a)Rectangular matrix(b)Square matrix (c)Vector(d)None of the above MUIt.,,,:) Choose TrueIFalse In the fOllowing. Each question carries one mark. 1.Derivative of a function of several variables is the ordinary derivative withrespect to one of the variables when all the rest are hold constant is called Integral. 2.Oneofthesimplestmethodforfindingtherootofanonlinearequationisinterval halving (also known as bisection). 3.A square matrix A = [akais called Hermitian Matrix if A = A. 4.If matrix A has no inverse, it is called singular matrix. 5.If events A and B are such that (AnB) =P(A)P(B),they are calledPermutation. 184 44n.uG Answer any five questions out of eight questions. Each question carries five marks. I r 1.Find _au ar andau ae if u = ercos (Jcos-(r sinO). 3-4-12 2. 1 Find the rank of the matrixA =I 5 7 -2 3 5 1 4 9-377 3.A bagcontains four whiteand two black balls anda secondbag contains three balls of each colour.A bagis selected at random,and a ballis then drawn atrandom from the bag chosen. What is the probability that the ball drawn is white? 4.EvaluateJJ~ xy - y2dy dx.Wheres isatrianglewithvertices(0,0),(10,1)and s (1,1). 5.ThreemachinesI,1\andIIImanufacturerespectively0.4,0.5and0.1ofthetotal production.Thepercentageofdefectiveitems producedbyI,1\andIIIis2,4 and6 percentrespectively.Anitemischosenatrandom,whatistheprobabilitythat itis effective? 6.Find the arithmetic mean for the following distribution: 3-34]7.If A =2- 34,Find two non singular matrices P and Q such that [o-11 PAQ =I,Hence find A-1 8.Find the positive root of the equationxex =1by Newton Raphson method. 185 maiN"'] Answer any five questions out of eight questions. Each question carries marks. 22 2 .01[x+ y]P.t.h.... t2dU..2..d..U2 ..d.U-sinu cos 2u1. If.u= smIe ,iovea.x:-2+.xy--tY2=3 VX+-..JYdXdXdydY4cosu 2.Evaluate by changing the order of integration JJ.x:2 dxdy,where Ais aregionin the A first quadrant by the hyperbola.xy =16and the linesy =x, y =0andx = 8. 3.Solve, with the help of matrices, the simultaneous equations: x+y+z=3;x+2y+3z=4;x+4y+9z6 4.The two regression equations of the variablesxandyare x=19.13-0.87 yandy=11.64-0.50x. Find (i) Mean ofXl s;(ii) Mean ofyl s;(iii) The correlation coefficient between x andy 5.Therearetwogroupsofsubjectsoneofwhichconsistsof5scienceand3 engineering subjectsand the other consistsof 3 science and 5engineering subjects. Anunbiaseddieiscast.If number 3or number 5turnsup,asubjectisselectedat random fromthe group,otherwise the subject is selectedatrandomfromthesecond group.Find the probability that an engineering subject is selected ultimately. 6.Find the value of Median from the following data: I No. of days for510152025303540 which absent (less than) No.of students29224465582634644,650653 7.Usemethod (Elementary matrices method) to compute the inverse of the matrix A =[3-3412-34 o-11 J 6dx 8.EvaluateJ--2by using o 1+x 0)Trapezoidal rule(ii)Simpson's.!..3 rule (iii)Simpson'srule by dividing the intelVal in six equal parts. 8 186 ula uuestion Paper Advanced Engineering Mathematics (AC/AA 1.1) Date: 24th, June 2010Time: 2:00 PMto 5:00 PMMax. Marks: 100 Assume any missing data if necessary. !iNdit.".. Choose the correct or best alternative In the following. Each question carries two marks. 1.A matrix is a ............ array of numbers (or functions) enclosed in brackets. (a)Rectangular(b)Triangular (c)Circular(d)None of the above 2.In rolling a fair dice once, what is the probability of obtaining even number? (a)1/3(b)1/4 (c)1/2(d)1/6 3.The mean,u (mu) is defined by.u = 2>j!(Xj)for (a)Discrete distribution(b)Continuous distribution (c)Both (a) & (b)(d)None of the above 4.The product of anymXnmatrixA = lajkJby any scalar c (number c)is written as cAthen themx nmatrix obtained by multiplying each entry of c by A (a)cA=lc+ajkJ(b)cA=lc-ajkJ (c)cA=lcxajkJ(d)None of the above 5.A functionf(x, y)is homogeneous of degree n in a regionR if and only if (a)f(2x, y) = 2f(x, y)(b)f(Ax, liy) =lin f(x, y) (c)f(x,2y) = 2f(x, y).(d)None of the above 33 6.The value of limitLtx2- y,is (x,y) ~ (1,.1)x- Y (a)0(b)1/2 (c)3/2(d)1 187 L ...~ ":,..;7.A transformation that transforms a function into another function is called (a)Operator(b)Separator (c)Denoter(d)None of the above 2In2Ir28. S=-L (x. -X)=-l(xI-X)+",+(xn X)2 ] n "-I j=rJn ~ 1........... Above equation represents (a)Standard Deviation(b)Variance (c)Mean(d)None of the above 9.For mutually exclusive events A andB (a)P(AUB)=P(A)xP(B)(b)P(AUB)=P(A)-P(B) (c)P(A UB) =P(A) + P(B)(d)None of the above 1O.1'1'events A and B are such that P(A fl B) = P(A)P(B) , they are called (a)Independent events(b)Dependent events (c)Discrete events(d)None of theabove man."':1 Choose TrueJFalse In the following. Each question carries one mark. 1.A POE is an equationinvolving one or more partial derivatives of a function. 2.An ODE together withaninitial condition is called aninitial value problem. 3.A matrix is a rectangular array of numbers (or functions) enclosed in brackets. 4.If the number of rowsand columns of matrix are equal, then it is called Square Matrix. 5.Numerical differentiation formulae can be developed by fitting approximating functions (e.g.,polynomials)toasetofdiscretedataanddifferentiatingtheapproximating function. 188 ClDmD1Ift Answer any five questions out of eight questions Each question carries five marks 3 -1X+ Y3 auau.1.If u=Jan,pr.ove thatx.-.- .....+ y.-.. -.::: sm2u x-y~~ 22.Find the area between the parabolasl=4ax andx=4ay 3.Anurncontains10blackand10whiteballs.Findtheprobabilityofdrawingtwo balls of the same colour. 4.If z be a homogeneous functionof degree n,show that a2 a2 a2 2zz2z x.-+2xy.--+y .-=n(n-1)zax2 axayay2 5.EvaluateIIxydxdyover the regionin the positive quadrant x + y$;1. -23-1-1 6. 1 Find the rank of matrixI 3 -1 1 -2 3 -4 -2 630-7 7.ThreemachinesI,IIandIIImanufacturerespectively0.4,0.5and0.1ofthetotal production.Thepercentageof defectiveitemsproducedbyI,IIandIIIis 2,4and6 percentrespectively.Anitemischosenatrandom,whatistheprobabilitythatitis defective? 8.Using the method of False position, find the realroot of the equationeX=xup to three decimal places. 189 @lUll.'''.] Answer any five questions out of eight questions. Each question carries ten marks. 22 21.If U-= sin-I [x + y]prove that2 au2au -fY 'x.ax2 +xy axCJy+ Y tr 21r4"a 2.EvaluateJJJr2Sin8drd8d9 000 2432 6523.Find the inverse of the matrixA =I3 252-3 451414 2 au sin ucos 2u u 4.Find the regression line of y on x for the following data: x13468 12445YIIIII Estimate the value of y, when x =10. 95.If u = log(x3 =t- y3 + Z3- 3xyz), show that (0-00)'u = + - + ax()yaz(x+y+ 6.Solve, with the help of matrices, the simultaneous equations x+y+z=3,x+2y+3z=4,x+4y+9z=6 12-1][3-11]7.IfA = 203; B = 002 [ o 124-32 Verify the result(A + BY = A 2 + AB + BA + B2. ~ 8.Evaluate!.Jcose. deby using Simpson's!rd rule. o3 190 x- LhArea under standard normal curve from 0 to oz ~ g - ~ . Table x ~ 0' 0.000.010.020.030.040.050.060.070.080.09 0.00.00000.00400.00800.01200.01600.01990.02390.02790.03190.0359 0.10.03980.04380.04780.05170.05570.05960.06360.06750.07140.0753 0.20.07930.08320.08710.09100.09480.09870.10260.10640.11030.1141 ~:..r 0.30.11790.12170.12550.12930.13310.13680.14060.14430.14800.1517 0.40.15540.15910.16280.16640.17000.17360.17720.18080.18440.1879 0.50.19150.19500.19850.20190.20540.20880.21230.21570.21900.2224 0.60.22570.22910.23240.23570.23890.24220.24540.24860.25170.2549 0.70.25800.26110.26420.26730.27040.27340.27640.27940.28230.2852 0.80.28810.29100.29390.29670.29950.30230.30510.30780.31060.3133 0.90.31590.31860.32120.32380.32640.32890.33150.33400.33650.3389 1.00.34130.34380.34610.34850.35080.35310.35540.35770.35990.3621 1.10.36430.36650.36860.37080.37290.37490.37700.37900.38100.3830 1.20.38490.38690.38880.39070.39250.39440.39620.39800.39970.4015 1.30.40320.40490.40660.40820.40990.41150.41310.41470.41620.4177 1.40.41920.42070.42220.42360.42510.42650.42790.42920.43060.4319 1.50.43320.43450.43570.43700.43820.43940.44060.44180.44290.4441 1.60.44520.44630.44740.44840.44950.45050.45150.45250.45350.4545 191 1.70.45540.45640.45730.45820.45910.45990.46080.46160.46250.4633 1.80.46410.46490.46560.46640.46710.46780.46860.46930.46990.4706 1.90.47130.47190.47260.47320.47380.47440.47500.47560.47610.4767 2J)0 ~ ~ ~ l t 4 n 80:4783---0:4788--0:4793--0.479&---0:4803- -0.4808----0:-4812- -0:41317 --_.2.10.48210.48260.48300.48340.48380.48420.48460.48500.48540.4857 2.20.48610.48640.48680.48710.48750.48780.48810.48840.48870.4890 2.30.48930.48960.48980.49010.49040.49060.49090.49110.49130.4916 2.40.49180.49200.49220.49250.49270.49290.49310.49320.49340.4936 2.50.49380.49400.49410.4943.0.49450.49460.49480.49490.49510.4952 2.60.49530.49550.49560.49570.49590.49600.49610.49620.49630.4964 2.70.49650.49660.49670.49680.49690.49700.49710.49720.49730.4974 2.80.49740.49750.49760.49n0.49770.49780.49790.49790.49800.4981 2.90.49810.49820.49820.49830.49840.49840.49850.49850.49860.4986 3.00.49870.49870.49870.49880.49880.49890.49890.49890.49900.4990 192 -- ----------- ------------------- - - - - - ~ - - - - ---