Math Chapter Wise

EnglishOPSC PGT 2017

Q7181.Gauss-Seidel method represents:

AA matrix inversion BAn interative procedure CAn integral evaluation DAn interpolation technique

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EnglishOPSC PGT 2017

Q7182.Newton-Raphson method is applied to solve:

AAn algebraic equation BA transcendental equation CA system of simultaneous equations DAny of these

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EnglishOPSC PGT 2017

Q7183.Runge-Kutta methods are used to solved the differential equation of:

AUpto second order BUpto order three CFirst order only DAny order

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EnglishOPSC PGT 2017

Q7184.Order to convergence of secant method is approximately:

A1.427 B1.618 C1.84 D2.0

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EnglishUPPSC GDC 2021

Q7185.Unique polynomial f(x) of degree 2 or less such that f (0) = 1, 1 f(1) = 3 and f(3) = 55 is

A1 + 6x – 8x2 B1 – 6x + 8x2 C1 + 6x + 8x2 D1 – 6x – 8x2

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EnglishUPPSC GDC 2021

Q7186.Newton-Raphson iterative formula to obtain the cube root of N is:

Ai 1 i 2 i 1 N x 2x 3 x + = − Bi 1 i 2 i 1 N x 2x 3 x + = + Ci 1 i 2 i 1 N x 2x 3 x + = + Di 1 i 2 i 1 N x 2x 3 x + = −

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EnglishTGT 2004

Q7187.In using Newton–Raphson method for finding the roots of an equation, the original curve is being replaced by -

AA line through the origin BA line joining two points on either side of x- axis CA line joining two points on either side of y- axis DA tangent to the curve

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EnglishUPPSC GDC 2021

Q7188.The order of the differential equation- 2 x x x ∆ y + 3∆y – 3y = x, is

A1 B2 C3 D4

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EnglishUPPSC GDC 2021 UKPSC Lecturer (Mains) 2020

Q7189.The rate of convergence of Newton-Raphson method is:

ACubic BFourth order CLinear DQuadratic

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EnglishUP Higher Asst. Prof. 2021

Q7190.Using Runge-Kutta method of order 4 for the following initial value problem () 2 2 dy x y,y 1 0 dx = + = the value of y(1.1) shall be

A0.110 B0.117 C0.119 D0.101

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EnglishUP Higher Asst. Prof. 2021

Q7191.Using Euler's modified method. the value of y(0.02) for the differential equation () 2 dy x y,y 0 1 dx = + = and taking h = 0.01 is.

Ay(0.02) = 1.2020 By(0.02) = 1.0250 Cy(0.02) = 1.0201 Dy(0.02) = 1.0203

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EnglishUP Higher Asst. Prof. 2021

Q7192.Solution of the initial value problem u' = –2tu2 u(0) = 1 with h = 0.2 on the interval [0, 1] is

Au(0.2)= u1= 0.8615241 Bu(0.2)= u1= 0.7615241 Cu(0.2)= u1= 0.9615241 Du(0.2)= u1= 0.5615241

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EnglishUPPSC GDC 2021

Q7193.If y (0) = 3, y (1) = 10, y(2) = 18, y(3) = 30 and y(4) = 50, then value of ∆ ∆ ∆4 y(0) is

A0 B– 1 C1 D2

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EnglishUPPSC GDC 2021

Q7194.If n is an integer, then the value of Pn (– 1), where Pn (x) denotes Legendre's polynomial, is

A0 B(– 1)n C1 Dn 1 2 n

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EnglishUPPSC GDC 2021

Q7195.Simpson's three-eighth rule of integration can be applied when the range of integration [a, b] is divided into a number of subintervals which must be multiple of

A2 B3 C4 D8

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EnglishUPPSC GDC 2021

Q7196.Which relation is NOT correct?

A() () E 1 2 E 1 + δ = µ − B2 1 4 δ µ = + C1/ 2 E µ + δ = D1/2 E δ = ∇

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EnglishUPPSC GDC 2021

Q7197.Let f1(x) = 4, f2(x) = x3 and f3 (x) = 1 + Ax + Bx2. If f3 (x) is orthogonal to f1 (x) and f2 (x) on the interval (–2, 2). Then

AA = 0, B = 1 BA = 0, B = 0 CA = 0, B = 3 4 DA = 0, B = – 3 4

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EnglishUPPSC GDC 2021

Q7198.If Pn(x) denotes Legendre's polynomial, then the value of P'n+1 (x) – xP'n(x) is equal to

A(n + 1) Pn (x) Bn2 CPn(x) DP'n(x)

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EnglishUKPSC Lecturer (Mains) 2020

Q7199.Solution of L.P.P.: Max. z = 3x1 + 4x2 Subject x1 –x2 ≤–1 –x1 + x2≤0 and x1, x2≥0 is:

Ax1 = 2, x3 = 3 Bx1 = 1, x2 = 2 Cx1 = 0, x2 = 1 DNon-existent

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EnglishUKPSC Lecturer (Mains) 2020

Q7200.For a set of two linear equations with four variables, the maximum number of basic solutions is:

A2 B4 C6 D8

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