Numerical Methods MCQs Question and Answer in Engineering Maths

1

What is the cube root of 1468 to 3 decimal places?

A) 11.340 B) 11.353 C) 11.365 D) 11.382

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Answer: Option C

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2

Torque exerted on a flywheel over a cycle is listed in the table. Flywheel energy (in J per unit cycle) using Simpson's rule is

Angle (degree)	0	60	120	180	240	300	360
Torque (Nm)	0	1066	-323	0	323	-355	0

A) 542 B) 993 C) 1444 D) 1986

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Answer: Option B

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3

Roots of the algebraic equation x³ + x² + x + 1 = 0 are

A) (+1, +j, -j) B) (+1, -1, +1) C) (0, 0, 0) D) (-1, +j. -j)

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Answer: Option D

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4

In the Newton-Raphson method, an initial guess of x₀ = 2 is made and the sequence x₀, x₁, x₂ ... is obtained for the function 0.75x³ - 2x² - 2x + 4 = 0
Consider the statements
I. x₃ = 0.
II. The method converges to a solution in a finite number of iterations.
Which of the following is TRUE?

A) Only I B) Only II C) Both I and II D) Neither I nor II

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Answer: Option A

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5

While numerically solving the differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} + 2{\text{x}}{{\text{y}}^2} = 0,\,{\text{y}}\left( 0 \right) = 1$$ using Euler's predictor-corrector (improved Euler-Cauchy) with a step size of 0.2, the value of y after the first step is

A) 1.00 B) 1.03 C) 0.97 D) 0.96

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Answer: Option D

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6

The Newton-Raphson iteration $${{\text{x}}_{{\text{n}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{n}}} + \frac{{\text{R}}}{{{{\text{x}}_{\text{n}}}}}} \right)$$ can be used to compute the

A) square of R B) reciprocal of R C) square root of R D) logarithm of R

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Answer: Option C

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7

When the Newton-Raphson method is applied to solve the equation f(x) = x³ + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x₀ = 1.2 is

A) -0.82 B) 0.49 C) 0.705 D) 1.69

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Answer: Option C

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8

The matrix \[\left[ {\text{A}} \right] = \left[ {\begin{array}{*{20}{c}} 2&1 \\ 4&{ - 1} \end{array}} \right]\] is decomposed into a product of a lower triangular matrix x[L] and an upper triangular matrix [U]. The properly decomposed [L] and [U] matrices respectively are

A) \[\left[ {\begin{array}{*{20}{c}} 1&0 \\ 4&{ - 1} \end{array}} \right]{\text{ and }}\left[ {\begin{array}{*{20}{c}} 1&1 \\ 0&{ - 2} \end{array}} \right]\] B) \[\left[ {\begin{array}{*{20}{c}} 2&0 \\ 4&{ - 1} \end{array}} \right]{\text{ and }}\left[ {\begin{array}{*{20}{c}} 1&1 \\ 0&1 \end{array}} \right]\] C) \[\left[ {\begin{array}{*{20}{c}} 1&0 \\ 4&1 \end{array}} \right]{\text{ and }}\left[ {\begin{array}{*{20}{c}} 2&1 \\ 0&{ - 1} \end{array}} \right]\] D) \[\left[ {\begin{array}{*{20}{c}} 2&0 \\ 4&{ - 3} \end{array}} \right]{\text{ and }}\left[ {\begin{array}{*{20}{c}} 1&{0.5} \\ 0&1 \end{array}} \right]\]

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Answer: Option D

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9

Consider a differential equation $$\frac{{{\text{dy}}\left( {\text{x}} \right)}}{{{\text{dx}}}} - {\text{y}}\left( {\text{x}} \right) = {\text{x}}$$ with the initial condition y(0) = 0. Using Euler's first order method with a step size of 0.1, the value of y(0.3) is

A) 0.01 B) 0.031 C) 0.0631 D) 0.1

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Answer: Option B

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10

Consider an ordinary differential equation $$\frac{{{\text{dx}}}}{{{\text{dt}}}} = 4{\text{t}} + 4.$$ If x = x₀ at t = 0, the increment in x calculated using Runge-Kutta fourth order multistep method with a step size of Δt = 0.2 is

A) 0.22 B) 0.44 C) 0.66 D) 0.88

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Answer: Option D

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