Math Chapter Wise

EnglishUPPSC Polytechnic Lecturer 2021(II)

Q121.The solution of φ(x) = (1 + x) – φ ∫ x 0 (t)dt, (x)=1 by using the method of successive approximation is –

Aφ(x) = (1 + x) Bφ(x) = (1 + x2) Cφ(x) = x Dφ(x) = 1

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q122.The value of L–1 () 2 2 2 s s + a is –

At a cos at Bt a sin at Ct 2a sin at Dt 2a cos at

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q123.The solution of linear integral equation φ(x) = x + φ ∫ 1/2 0 (t) dt is –

Aφ(x) = x + 1 4 Bφ(x) = x2 + 1 Cφ(x) = x2 + 1 4 Dφ(x) = 3 x 1 3 4 +

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q124.Which one of the following is a Volterra integral equation?

Ay(x) = 1 x t 0 e y(t)dt − ∫ By(x) =2x2 + 0 sin x cos t y(t)dt π ∫ Cy(x) = x + x 0 xt y(t)dt ∫ Dy(x) = f(x) + b a xt y(t)dt ∫

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q125.All iterated Kernels of a symmetric Kernel is -

ASkew symmetric BSymmetric CHilbert Kernels DHermitian Kernels

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q126.The complete integral of px + qy = pq. ∂ ∂ z z where p= andq = x y is –

Aaz = 1 2 (ax + b)2 + b Bz = 1 2 (x + by)2 + a Cz = (ax + b)2 + y2 Dz = (x + ay)2 + b

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q127.The complete integral of p2 + q2 = x + y, ∂ ∂ z z where p = andq = x y is –

Az = (x + a)3/2 + (y – b)3/2 + b Bz = (x + a) 2 + (y – b)2 + b Cz = (x + a)–3/2 + (y – a)–3/2 + b Dz = 2 3 (x + a)3/2 + 2 3 (y – a)3/2 + b

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q128.Solution of the partial differential equation yzp + zxq = xy, ∂ ∂ z z where p = andq = x y is –

Af(x2 + y2, x2 – z2) = 0 Bf(x2 – y2, x2 + z2) = 0 Cf(x2 – y2, x2 – z2) = 0 Df(x2 + y2, x2 + z2) = 0 Where f is arbitrary function.

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q129.The solution of pq = k, where k is constant and p = ∂ ∂ z z,q =, x y is –

Az = a (x + y) + b Bz = ax + by + c Cz = x + ay + b Dz = ax+ k a y + c

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q130.The second order partial differential equation Rr + Ss + Tt + f(x, y, z, p, q) = 0, where R, S, T are continuous functions with continuous partial derivatives is called hyperbolic, if –

AS2/< 4RT BS2/> 4RT CS2 = 4RT DS2 = RT

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q131.The complete integral of q = 3p2, ∂ ∂ z z where p = and q = x y is –

Az = ax + b3 y + c Bz = (x + ay)2 + b Cz = (ax + y)2 + b Dz = ax + 3a2 y + b

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q132.The equation x2 (y – 1) () u y ∂ ∂ ∂ 2 2 2 u – x y –1 x ()∂ ∂ ∂ 2 u u +y y –1 + = 0 y, is hyperbolic in the region –

AEntire xy-plane except x-axis BEntire xy-plane except y-axis C2 1 (x, y) R: x 0, y < 3 ∈ ≠ D() 2 1 x, y R: x 0, –1 < y < 3 ∈ ≠

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q133.The solution of the equation ∂ ∂ ∂ 2 z =4xy +1 x y is –

Az = x2 y2 + xy + φ (x) Bz = x2 y2 + xy + φ (x) + ψ (y) Cz = x2 y2 + φ (x) + ψ (y) DNone of these

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q134.The complete integral of the equation (y – x) (qy – px) = (p – q)2, ∂ ∂ z z where p =,and q = x y is –

Az = a (x + y) – a xy c + Bz = a (x + y) + a xy c + Cz = a (x – y) + a xy c + Dz = a (x – y) – () a x y c + +

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q135.One dimensional heat (diffusion) equation, where u (x, t) is the temperature at a distance 'x' from one end at any time 't' is –

A2 2 2 2 2 u u c t x ∂ ∂ = ∂ ∂ B2 2 2 2 2 u u c 0 x y ∂ ∂ + = ∂ ∂ C2 u u c t x ∂ ∂ = ∂ ∂ D2 2 2 2 2 u u x t x ∂ ∂ = ∂ ∂

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q136.Monge's method is used is used to solve a partial differential equation of –

Asecond order Bnth/order Cfirst order DNone of these

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q137.The solution of the integral equation sinx = φ ∫ x 0 J (x – t) (t)dt is –

Acos x Bsin x CJ0 (x) DJ0 (2x)

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q138.An integral equation corresponding to the differential equation 2 d y dy + x + y = 0 dx with the initial condition y(0) = 1, y' (0) = 0 is –

Aφ(x) = – 1 – x 0 (2x t) − ∫ φ(t) dt Bφ(x) = 1 + x 0 (2x t) − ∫ φ(t) dt Cφ(x) = 1 – x 0 (x t) − ∫ φ(t) dt Dφ(x) = 1 + x 0 (x t) + ∫ φ(t) dt

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q139.The solution of the Fredholm integral equation φ(x) = 1 + φ ∫ 1 0 λ xt (t) dt is -

Aφ(x) = 1 + 3 x 2(3) λ − λ Bφ(x) = 1 + x (3) λ −λ Cφ(x) = 1 + 2 x (2) λ − λ Dφ(x) = 1 + 3 x 2 λ

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EnglishUPPSC Polytechnic Lecturer 2021(II)

Q140.The solution of the integral equation φ(x) = 1 + φ ∫ x 0 (t) dt is –

Aφ (x) = sin x Bφ (x) = cos x Cφ (x) = log x Dφ (x) = ex

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