Numerical Methods MCQ

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• The error in $${\left. {\frac{{\text{d}}}{{{\text{dx}}}}{\text{f}}\left( {\text{x}} \right)} \right|_{{\text{x}} = {{\text{x}}_0}}}$$   for a continuous function estimated with h = 0.03 using the central difference formula $${\left. {\frac{{\text{d}}}{{{\text{dx}}}}{\text{f}}\left( {\text{x}} \right)} \right|_{{\text{x}} = {{\text{x}}_0}}} = \frac{{{\text{f}}\left( {{{\text{x}}_0} + {\text{h}}} \right) - {\text{f}}\left( {{{\text{x}}_0} - {\text{h}}} \right)}}{{2{\text{h}}}},$$        is 2 × 10^-3. The values of x₀ and f(x₀) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h = 0.02 is approximately
  ক

  1.3 × 10^-4

  খ

  3.0 × 10^-4

  গ

  4.5 × 10^-4

  ঘ

  9.0 × 10^-4
• The integral $$\int\limits_{{{\text{x}}_1}}^{{{\text{x}}_2}} {{{\text{x}}^2}{\text{dx}}} $$  with x₂ > x₁ > 0 is evaluated analytically as well as numerically using a single application of the trapezoidal rule. If $$I$$ is the exact value of the integral obtained analytically and J is the approximate value obtained using the trapezoidal rule, which of the following statements is correct about their relationship?
  ক

  J > $$I$$

  খ

  J

  গ

  J = $$I$$

  ঘ

  Insufficient data to determine the relationship
• The estimate of $$\int\limits_{0.5}^{1.5} {\frac{{{\text{dx}}}}{{\text{x}}}} $$  obtained using Simpson's rule with three-point function evaluation exceeds the exact value by
  ক

  0.235

  খ

  0.068

  গ

  0.024

  ঘ

  0.012
• The differential equation $$\frac{{{\text{dx}}}}{{{\text{dt}}}} = \frac{{4 - {\text{x}}}}{\tau },$$   with x(0) = 0 and the constant $$\tau $$ > 0, is to be numerically integrated using the forward Euler method with a constant integration time step T. The maximum value of T such that the numerical solution of x converges is
  ক

  $$\frac{\tau }{4}$$

  খ

  $$\frac{\tau }{2}$$

  গ

  $$\tau $$

  ঘ

  $$2\tau $$
• Using Simpson's 1/3 rule for numerical integration, the consecutive points are joined by a
  ক

  line

  খ

  parabola

  গ

  polynomial with power 3

  ঘ

  polynomial with power 1/3
• In numerical integration using Simpson's rule, the approximating function in interval is a
  ক

  constant

  খ

  straight line

  গ

  cubic B-spline

  ঘ

  parabola
• The integral $$\int_1^3 {\frac{1}{{\text{x}}}{\text{dx,}}} $$   when evaluated by using Simpson's 1/3 rule on two equal subintervals each of length 1, equals
  ক

  1.000

  খ

  1.098

  গ

  1.111

  ঘ

  1.120
• Function f is known at the following points<br>
  ক

  <table class="table-style-1"> <tr> <td>  x</td> <td>0</td> <td>0.3</td> <td>0.6</td> <td>0.9</td> <td>1.2</td> <td>1.5</td> <td>1.8</td> <td>2.1</td> <td>2.4</td> <td>2.7</td> <td>3.0</td> </tr> <tr> <td>f(x)</td> <td>0</td> <td>0.09</td> <td>0.36</td> <td>0.81</td> <td>1.44</td> <td>2.25</td> <td>3.24</td> <td>4.41</td> <td>5.76</td> <td>7.29</td> <td>9.00</td> </tr></table><br>The value of $$\int_0^3 {{\text{f}}\left( {\text{x}} \right){\text{dx}}} $$   computed using the continuous at x = 3? 8.983

  খ

  9.003

  গ

  9.017

  ঘ

  9.045
• A piecewise linear function f(x) is plotted using thick solid lines in the figure below (the plot is drawn to scale).<br><img src="/images/question-image/engineering-maths/numerical-methods/1689253547-a-piecewise-linear-function-fx-is.png" title="Numerical Methods mcq question image" alt="Numerical Methods mcq question image"><br>If we use the Newton-Raphson method to find the roots of f(x) = 0 using x₀, x₁ and x₂ respectively as initial guesses, the roots obtained would be
  ক

  1.3, 0.6 and 0.6 respectively

  খ

  0.6, 0.6 and 1.3 respectively

  গ

  1.3, 1.3 and 0.6 respectively

  ঘ

  1.3, 0.6 and 1.3 respectively
• Let x² - 117 = 0. The iterative steps for the solution using Newton-Raphson's method is given by
  ক

  $${{\text{x}}_{{\text{k}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{k}}} + \frac{{117}}{{{{\text{x}}_{\text{k}}}}}} \right)$$

  খ

  $${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - \frac{{117}}{{{{\text{x}}_{\text{k}}}}}$$

  গ

  $${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - \frac{{{{\text{x}}_{\text{k}}}}}{{117}}$$

  ঘ

  $${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - \frac{1}{2}\left( {{{\text{x}}_{\text{k}}} + \frac{{117}}{{{{\text{x}}_{\text{k}}}}}} \right)$$
• The velocity v (in m/s) of a moving mass, starting from rest, given as $$\frac{{{\text{dv}}}}{{{\text{dt}}}} = {\text{v}} + {\text{t}}{\text{.}}$$   Using Euler forward difference method (also known as Cauchy-Euler method) with a step size of 0.1 s, the velocity at 0.2 s evaluate to
  ক

  0.01 m/s

  খ

  0.1 m/s

  গ

  0.2 m/s

  ঘ

  1 m/s
• The root of the function f(x) = x³ + x - 1 obtained after first iteration on application of Newton Raphson scheme using an initial guess of x₀ = 1 is
  ক

  0.682

  খ

  0.686

  গ

  0.750

  ঘ

  1.000
• Solution of the variables x₁ and x₂ for the following equations is to be obtained by employing the Newton-Raphson iterative method<br>equation (i) 10x₂ sin x₁ - 0.8 = 0<br>equation (ii) 10$${\text{x}}_2^2$$ - 10x₂ cos x₁ - 0.6 = 0<br>Assuming the initial values x₁ = 0.0 and x₂ = 1.0, the Jacobian matrix is
  ক

  \\

  খ

  </span> \\
• Numerical integration using trapezoidal rule gives the best result for a single variable function, which is
  ক

  linear

  খ

  parabolic

  গ

  logarithmic

  ঘ

  hyperbolic
• What is the value of (1525)^0.2 to 2 decimal places?
  ক

  4.33

  খ

  4.36

  গ

  4.38

  ঘ

  4.30
• The differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = 0.25{{\text{y}}^2}$$   is to be solved using the backward (implicit) Euler's method with the boundary condition y = 1 at x = 0 and with a step size of 1. What would be the value of y at x = 1?
  ক

  1.33

  খ

  1.67

  গ

  2.00

  ঘ

  2.33
• The following equation needs to be numerically solved using the Newton-Raphson method. x³ + 4x - 9 = 0. The iterative equation for this purpose is (k indicates the iteration level)
  ক

  $${{\text{x}}_{{\text{k}} + 1}} = \frac{{2{\text{x}}_{\text{k}}^3 + 9}}{{3{\text{x}}_{\text{k}}^2 + 4}}$$

  খ

  $${{\text{x}}_{{\text{k}} + 1}} = \frac{{3{\text{x}}_{\text{k}}^2 + 4}}{{2{\text{x}}_{\text{k}}^2 + 9}}$$

  গ

  $${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - 3{\text{x}}_{\text{k}}^2 + 4$$

  ঘ

  $${{\text{x}}_{{\text{k}} + 1}} = \frac{{4{\text{x}}_{\text{k}}^2 + 3}}{{9{\text{x}}_{\text{k}}^2 + 2}}$$
• With respect to the numerical evaluation of the definite integral $${\text{K}} = \int_{\text{a}}^{\text{b}} {{{\text{x}}^2}{\text{dx,}}} $$   where a and b are given, which of the following statements is/are TRUE?<br>I. The value of K obtained using the trapezoidal rule is always greater than or equal to the exact value of the definite integral.<br>II. The value of K obtained using the Simpson's rule is always equal to the exact value of the definite integral.
  ক

  I only

  খ

  II only

  গ

  Both I and II

  ঘ

  Neither I nor II
• During the numerical solution of a first order differential equation using Euler method with step size h, the local truncation error is of order of
  ক

  h²

  খ

  h³

  গ

  h⁴

  ঘ

  h⁵
• The square root of a number N is to be obtained by applying the Newton Raphson iterations to the equation x² - N = 0. If i denotes the iteration index, the correct iterative scheme will be
  ক

  $${{\text{x}}_{{\text{i}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{i}}} + \frac{{\text{N}}}{{{{\text{x}}_{\text{i}}}}}} \right)$$

  খ

  $${{\text{x}}_{{\text{i}} + 1}} = \frac{1}{2}\left( {{\text{x}}_{\text{i}}^2 + \frac{{\text{N}}}{{{\text{x}}_{\text{i}}^2}}} \right)$$

  গ

  $${{\text{x}}_{{\text{i}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{i}}} + \frac{{{{\text{N}}^2}}}{{{{\text{x}}_{\text{i}}}}}} \right)$$

  ঘ

  $${{\text{x}}_{{\text{i}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{i}}} - \frac{{\text{N}}}{{{{\text{x}}_{\text{i}}}}}} \right)$$
• Given that one root of the equation x³ - 10x² + 31x - 30 = 0 is 5, the other two roots are
  ক

  2 and 3

  খ

  2 and 4

  গ

  3 and 4

  ঘ

  -2 and -3
• The recursion relation to solve x = e^-x using Newton-Raphson method is
  ক

  $${{\text{x}}_{{\text{n}} + 1}} = {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}$$

  খ

  $${{\text{x}}_{{\text{n}} + 1}} = {{\text{x}}_{\text{n}}} - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}$$

  গ

  $${{\text{x}}_{{\text{n}} + 1}} = \left( {1 + {{\text{x}}_{\text{n}}}} \right)\frac{{{{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}{{1 + {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}$$

  ঘ

  $${{\text{x}}_{{\text{n}} + 1}} = \frac{{{\text{x}}_{\text{n}}^2 - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}\left( {1 + {{\text{x}}_{\text{n}}}} \right) - 1}}{{{{\text{x}}_{\text{n}}} - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}$$
• P(0, 3), Q(0.5, 4) and R(1, 5) are three points on the curve defined by f(x). Numerical integration is carried out using both Trapezoidal rule and Simpson's rule within limits x = 0 and x = 1 for the curve. The difference between the two results will be
  ক

  0

  খ

  0.25

  গ

  0.5

  ঘ

  1
• The following table lists an n^th order polynomial f(x) = a_n xⁿ + a_{n - 1} x^{n - 1} + , .... a₁x + a₀ and the forward difference evaluated at equally spaced values of x. The order of the polynomial is<br>
  ক

  <table class="table-style-1"> <tr> <td>  x</td> <td>   f(x)</td> <td>    Δf</td> <td>  Δ²f</td> <td>  Δ³f</td> </tr> <tr> <td>-0.4</td> <td>1.7648</td> <td>-0.2965</td> <td>0.089</td> <td>-0.03</td> </tr> <tr> <td>-0.3</td> <td>1.4683</td> <td>-0.2075</td> <td>0.059</td> <td>-0.0228</td> </tr> <tr> <td>-0,2</td> <td>1.2608</td> <td>-0.1485</td> <td>0.0362</td> <td>-0.0156</td> </tr> <tr> <td>-0.1</td> <td>1.1123</td> <td>-0.1123</td> <td>0.0206</td> <td>-0.0084</td> </tr> <tr> <td>  0</td> <td>    1</td> <td>-0.0917</td> <td>0.0122</td> <td>-0.0012</td> </tr> <tr> <td>0.1</td> <td>0.9083</td> <td>-0.0795</td> <td>0.011</td> <td>0.006</td> </tr> <tr> <td>0.2</td> <td>0.8288</td> <td>-0.0685</td> <td>0.017</td> <td>0.0132</td> </tr></table> 1

  খ

  2

  গ

  3

  ঘ

  4
• A calculator has accuracy up to 8 digits after decimal place. The value of $$\int\limits_0^{2\pi } {\sin {\text{x dx}}} $$   when evaluated using this calculator by trapezoidal method with 8 equal intervals, to 5 significant digits is
  ক

  0.00000

  খ

  1.0000

  গ

  0.00500

  ঘ

  0.00025
• Equation e^x - 1 = 0 is required to be solved using Newton's method with an initial guess x₀ = -1. Then, after one step of Newton's method, estimate x₁ of the solution will be given by
  ক

  0.71828

  খ

  0.36784

  গ

  0.20587

  ঘ

  0.00000
• The iteration step in order to solve for the cube roots of a given number N using the Newton-Raphson's method is
  ক

  $${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} + \frac{1}{3}\left( {{\text{N}} - {\text{x}}_{\text{k}}^3} \right)$$

  খ

  $${{\text{x}}_{{\text{k}} + 1}} = \frac{1}{3}\left( {2{{\text{x}}_{\text{k}}} + \frac{{\text{N}}}{{{\text{x}}_{\text{k}}^2}}} \right)$$

  গ

  $${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - \frac{1}{3}\left( {{\text{N}} - {\text{x}}_{\text{k}}^3} \right)$$

  ঘ

  $${{\text{x}}_{{\text{k}} + 1}} = \frac{1}{3}\left( {2{{\text{x}}_{\text{k}}} - \frac{{\text{N}}}{{{\text{x}}_{\text{k}}^2}}} \right)$$
• The equation x³ - x² + 4x - 4 = 0 is to be solved using the Newton-Raphson method. If x = 2 is taken as the initial approximation of the solution, then the next approximation using this method will be
  ক

  $$\frac{2}{3}$$

  খ

  $$\frac{4}{3}$$

  গ

  1

  ঘ

  $$\frac{3}{2}$$
• Starting from x₀ = 1, one step of Newton-Raphson method in solving the equation x³ + 3x - 7 = 0 gives the next value (x₁) as
  ক

  x₁ = 0.5

  খ

  x₁ = 1.406

  গ

  x₁ = 1.5

  ঘ

  x₁ = 2
• A 2^nd degree polynomial, f(x) has values of 1, 4 and 15 at x = 0, 1 and 2, respectively. The integral $$\int\limits_0^2 {{\text{f}}\left( {\text{x}} \right){\text{dx}}} $$   is to be estimated by applying the trapezoidal rule to this data. What is the error (defined as "true value - approximate value") in the estimate?
  ক

  $$ - \frac{4}{3}$$

  খ

  $$ - \frac{2}{3}$$

  গ

  0

  ঘ

  $$\frac{2}{3}$$