Math Chapter Wise

EnglishUPPSC GDC 2021

Q101.The second order partial differential equation ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ 2 2 2 2 2 2 z z z z – 2sinx – cos x – cos x = 0 x x y y y is

AParabolic for all values of x, y BParabolic in half plane x ≥ 0 CParabolic in the half plane y ≥ 0 DHyperbolic for all values of x, y

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EnglishUPPSC GDC 2021

Q102.The Fourier transform of 2 x – 2 e is

A2 s – 2 2 e π B2 s 2 2 e π C2 s 2 2 e π D2 s 2 2 e − π

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EnglishUPPSC GDC 2021

Q103.-1 2 s –1 L s – 6s +13 is equal to

A3x/e sin 2x B2x/e cos2x C() 2x e sin3x cos3x + D() 3x e cos2x sin 2x +

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EnglishUPPSC GDC 2021

Q104.The curve, for which () ∫ π/2 2 0 y' + 2xyy' dx subjected to y (0) = 1, y π 2 = 1 is extremised, is

Ay = 2 sinx + cosx By = sinx + 2 cosx Cy = sinx – cosx Dy = sinx + cosx

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EnglishUPPSC GDC 2021

Q105.Inner product of tensors Aij and Bhk will be a mixed tensor of the type

A(2, 1) B(1, 2) C(2, 2) D(1, 1)

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EnglishUPPSC GDC 2021

Q106.General solution x = z (x, y) of the partial differential equation y2 zp + x2 zq = xy2, is

A() 3 3 2 2 F x y,x z 0 − − = B() 3 3 2 2 F x y,x z 0 − − = C() 3 3 2 2 F x y,x z 0 + + = D() 3 3 2 2 F x y,x z 0 + − =

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EnglishUPPSC GDC 2021

Q107.The value of L–1 –1/s e s is

AJ0 (2t) BJ0 ( )/2 t CJ0 ( )/2 t DJ0 ( )/2t

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EnglishUPPSC GDC 2021

Q108.If φ1 and φ2 are arbitrary functions, then the solution of the partial differential equation γ = a2 t, is

A() () 1 2 z y ax y ax = φ + + φ − B() () 1 2 z y y ax = φ + φ − C() () 1 2 z y ax y = φ + + φ D() () 1 2 z x y axy = φ + φ +

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q109.An external of the functional [ ] () 3 + ∫ ' 2 y 0 I y(x) = e dx,y(0) = 0,y(2)=1, is-

AWeak minimum on x y 2 = BWeak maximum on x y 2 = CStrong maximum on x y 2 = DStrong minimum on x y 2 =

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q110.The condition that ∂ ∂ d f f – = 0 dx y' y for the functional [ ] () ∫ 2 1 x I y(x) = f x,y,y' dx to have a stationary value is-

Aonly necessary Bonly sufficient Cnecessary and sufficient Dneither necessary nor sufficient

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q111.The extremal of the functional () ∫ 1 2 0 V[y(x)] = y' +12xy dx,y(0) = 0, y(1)=1, represents-

Afamily of circles Bfamily of straight lines Ccatenary Dnodal cubic curve

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EnglishUPPSC GDC 2021 UPPSC Polytechnic Lecturer 2021 (I)

Q112.If F(x,y,y') is independent of x, then Euler's equation for the functional () ∫ 2 1 x F x,y,y' dx reduces to-

AFy'= C BF–y'Fy'=c CF+y'Fy'=C DFy=C

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q113.The Euler equation for the functional () () z ∫∫ 2 2 D x y I z x,y = + z dxdy satisfy-

AHeat equation BWave equation CLaplace equation DNone of the above

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q114.The extremals of the functional ∫ 2 1 θ 2 2 '2 θ r + r dθ where () dr r = r θ,r' =, dθ is given by-

A1 2 rsin C rcos C θ = θ + B1 2 rcos C rcos C θ = θ + C1 2 rsin C rcos2 C θ = θ + DNone of the above

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q115.The extremals of the functional () ∫ π 2 2 2 5 0 I = y'' – y + x dx, () () y 0 = 0,y' 0 = 1 π 1 y' = 4 2 is-

Ay=cosx By=sinx Cy=tanx Dy=tanx+8

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q116.An extremals to the functional () () () ∫ b 2 a V y x = 1+ y' x dx

AA catenary BThe straight line connecting (a,y(a)) and (b,y(b)) CThe family of circles DDoes not exist

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q117.The functional () ∫ π '2 2 2 0 y – y + 2xy dx with () π y 0 = 0,y = 0 2 can be extremized on the curve-

Ay=x+cosx By=x–x sin x Cy x sin x 2 π = + Dy x – sin x 2 π =

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q118.How many extremals s are possible for the extremals () ∫ 2 1 2 0 d y I y x = + y dx,y(0) = 1, dx () () () y 1 = –1,y' 0 = 1,y' 1 = 1?

AOnly me BExactly two CInfinite many DNone of the above

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EnglishUPPSC Polytechnic Lecturer 2021 (I)

Q119.The variation problem of eternizing the functional [ ] ∫ 2 2π 2 0 dy I y(x) = – y dx,y(0) = 1, dx y(2π) = 1 has-

AA unique solution BExactly two solutions CAn infinite many solutions DNo solution

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q120.The Eigen values of the homogeneous integral equation φ(x) = φ ∫ 2 1 λ xt + (t) xt dt is –

A() 2 17 265 ± B() 4 17 265 ± C() 1 17 265 2 ± D17 265/±

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