 # Let f x z x 0 f t dt then f is a continuous function

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LetF(x) =Zx0f(t)dt, thenFis a continuous function.F(x+y) =Zx+y0f(t)dt, let(x+y)z=t, sot= 0z= 0, t=x+yz= 1. It is enough to considerx >0, becausef(x+ (-x)) =f(0) = 0 =f(x) +f(-x), sofis an odd function ofx.F(x+y)=Z10f(xz+yz)(x+y)dzF(x+y)x+y=Z10f(xz)dz+Z10f(yz)dz( Letxz=u, yz=v)=Zx0f(u)dux+Zy0f(v)dvy=F(x)x+F(y)yThus the functionG(x) =F(x)xis continuous and satisfiesG(x+y) =G(x) +G(y).SoG(x) =xG(1), andF(x) =x2G(1). Differentiating both sides,f(x) =F0(x) = 2xG(1), sof(x) =cx, which proves the result.Question 2(b)Show that the area between the cissoidx=asin2t, y=asin3tcostand itsasymptote is3πa24.Solution.y2=a2sin6tcos2t=x3a(1-sin2t)=x3a-x.Thusy2(a-x) =x3, sox=ais an asymp-tote.RequiredArea=2Za0y dx.Sub-stitutingforyandxusingtheparametricequations,theareabe-comes2Zπ20asin3tcost2asintcost dt=4a2Zπ20sin4t dt= 4a23·14·2π2=3πa24as re-quired.OAx=aδxy3
Question 2(c)Show thatZZxm-1yn-1dxdyover the positive quadrant of the ellipsex2a2+y2b2= 1isambn4Γ(m2)Γ(n2)Γ(m2+n2+ 1).Solution.PutX=ax, Y=by,(x,y)(X,Y)=ab. ThusI=ZZX0,Y0X2+Y21am-1Xm-1bn-1Yn-1ab dX dYPutX=rcosθ, Y=rsinθ(X, Y)(r, θ)=cosθ-rsinθsinθrcosθ=r. ThenI=ambnZπ20Z10rm+n-2sinm-1θcosn-1θ r dr dθ=ambnm+nZπ20sinm-1θcosn-1θ dθ=ambnm+nΓ(n-1+12)Γ(m-1+12)2Γ(n-12+m-12+ 1)=ambn4m+n2Γ(n2)Γ(m2)Γ(n2+m2)=ambn4Γ(n2)Γ(m2)Γ(n2+m2+ 1)Here we usedxΓ(x) = Γ(x+ 1) andZπ20sinmθcosnθ dθ=Γ(n+12)Γ(m+12)2Γ(n2+m2+ 1)Paper IIQuestion 3(a)Find the shortest distance from the origin to the hyperbolax2+8xy+7y2=225, z= 0.Solution.We have to minimizef(x, y) =x2+y2subject to the constraintx2+8xy+7y2-225 = 0. LetF(x, y) =x2+y2+λ(x2+8xy+7y2-225) whereλis Lagrange’s undeterminedmultiplier. For extreme values∂F∂x=2x+ 2λx+ 8λy= 0∂F∂y=2y+ 8λx+ 14λy= 04
From the second equation we getx=-y-7λy4λ.Substituting the value ofxin the firstequation, we get-(1 +λ)(1 + 7λ)y2λ+ 8λy= 0 or 16λ2-(1 + 7λ)(1 +λ) = 0, sincey6= 0 —note thaty= 0x= 0 which is not possible becausex2+ 8xy+y2= 225. Thus we get9λ2-8λ-1 = 0, soλ= 1,-19.We shall now show thatλ= 1 is not possible.λ= 14x+ 8y= 0x=-2y4y2-16y2+ 7y2= 225, which is not possible, thusλ6= 1.λ=-192x-29x-89y= 016x-8y= 0y= 2x. Thusx2+ 16x2+ 28y2= 225x=±5, y=±25. Thus stationary points are (5,25),(-5,-25).F(x, y)=x2+y2-19(x2+ 8xy+ 7y2-225)=19h8x2+ 2y2-8xy+ 225i=19h2(2x-y)2+ 225iwhich is minimized when 2x=y. Thus the shortest distance is25 = 5.Question 3(b)Show that the double integralZZRx-y(x+y)3dx dydoes not exist overR=[0,1 : 0,1].Solution.For a fixedx6= 0, the functionx-y(x+y)3is a bounded function ofyandZ10x-y(x+y)3dy=Z10-x+y(x+y)3+2x(x+y)3dy=1x+y+2x-2(x+y)210=11 +x-x(1 +x)2=1(1 +x)2NowZ10dx(1 +x)2= lim0+Z1(1 +x)-2dx= lim0+-12+11 +=12ThusZ10Z10x-y(x+y)3dy dx=12.Similarly, ify6= 0,Z10x-y(x+y)3dx=Z10x+y(x+y)3-2y(x+y)3dx=-1x+y-2y-2(x+y)210=-1(1 +y)25
ThusZ10Z10x-y(x+y)3dx dy=-12, asZ10dy(1 +y)2=12as above.ThusZ10Z10x-y(x+y)3dy dx6=Z10Z10x-y(x+y)3dx dyshowing that the double integraldoes not exist, because if the double integral exists and the two repeated integrals exist,these have to be equal. The reason is that the functionx-y(x+y)3is not bounded in the squareR[0,1 : 0,1].

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Term
Fall
Professor
John S
Tags
Professor of Mathematics, Panjab University, Dr Sunder Lal, UPSC Civil Services Main
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