Math Chapter Wise

EnglishBihar (TRE 1.0) -2023

Q01.Let φ:[0, 1] → → → ℝ be three times continuously differentiable. Suppose that the iterates defined by xn+1 = φ(xn), n ≥ 0 converge to the fixed point ξ ξ of φ. If the order of convergence is three, then.

Aφ"(ξ ξ ξ) = 0, φ" = 0 Bφ"(ξ ξ ξ) ≠ 0, φ" = 0 Cφ"(ξ ξ ξ) ≠ 0, φ" ≠ 0 DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q02.While solving the equation x2 – 3x + 1 = 0 using the Newton-Raphson method with the initial guess of a root as 1, the value of the root after one iteration is

A1.5 B0 C1 DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q03.Consider the system of equations 1 2 3 5 2 1 x 13 -2 5 2 x = -22 -1 2 8 x 14 With the initial guess of the solution [ ] T 1 2 3 (0) (0) (0) x,x,x = 1,1,1 The approximate value of the solution T 1 2 3 (1) (1) (1) x,x,x After one iteration by the gauss-Seidel method is

A[2, –4.4, 1.625]T B[2, –4, –3]T C[2, –4, 3]T DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q04.For what values of α and β, the quadrature formula ∫ 1 -1 f(x)dx = αf(-1) + f(β) is exact for all polynomials of degree ≤ 1?

Aα = 1, β = 1 Bα = –1, β = 1 Cα = 1, β = –1 DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q05.The value of the integral () ∫ 1 2 -1 I = exp x dx Using a rectangular rule is approximated as 2. Then, the approximation error |I – 2| lies in the interval.

A(2e, 3e] B2,2e 3 Ce 0, 8 DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q06.Let y be the solution of the initial value problem () () 2 d y + y = 6cos2x,y(0) = 3,y' 0 = 3,y' 0 = 1 dx Let the Laplace transform of y be F(s). Then, the value of F(1) is

A17/5 B13/5 C11/5 DMore than one of the above ENone of the above

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EnglishRPSC PGT 2022

Q07.The third divided difference of 1 x based on the points x0, x1, x2, x3, is-

A0 1 2 1 x x x − B0 1 2 3 1 x x x x − C0 1 2 1 x x x D0 1 2 1 x x x −

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EnglishRPSC PGT 2022

Q08.If interval of differencing is unity, then 10 2 3 4 ∆ (1- ax)(1- bx)(1- cx)(1- dx)is -

A1 – abcd B10 abcd C10.abcd D(abcd) / 10

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EnglishRPSC PGT 2022

Q09.Given – X: 0 π 6 π 3 π 2 f(x) = esinx: 1 1.64872 2.3632 2.71828 Then by Simpson’s 3 8 rule, value of ∫ π sinx 6 0 e dx is – (using π= 3.14159)

A3.09329 B3.0999 C3.9529 D3.09000

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EnglishRPSC PGT 2022

Q10.The interpolation formula f(x) = f(1) + x–1 C1 ∆ f(0) + x C2 ∆ 2 f(0) + x C3 ∆ 3 f(–1) + (x+1) C4 ∆ 4 f(– 2) +……………..is known as -

AGauss Forward formula BGauss Backward formula CGauss Third formula DNone of these

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EnglishRPSC PGT 2022

Q11.In solving by iteration method, the equation x2 – x – 1 = 0, to be convergent in the neighborhood of its root x = – 0.62; can be written in which of the following form?

Ax = 1– 1 x Bx = 2x–x2 +1 Cx = x 1 + Dx = x2 –1

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EnglishOPSC PGT 2017

Q12.For Simpson rule to solve a definite integral, each section of the curve is replaced by:

AA secant chord BA tangent to curve CA second degree curve DA spline arc

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EnglishOPSC PGT 2017

Q13.Gauss elimination method solved:

AA system of linear equations BA cubic equation CAn algebraic equation of degree 4 DAn integral equation

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EnglishOPSC PGT 2017

Q14.Gauss-Seidel method represents:

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

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EnglishOPSC PGT 2017

Q15.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

Q16.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

Q17.Order to convergence of secant method is approximately:

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

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EnglishUPPSC GDC 2021

Q18.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

Q19.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

Q20.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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