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Test: Numerical Methods - Civil Engineering (CE) MCQ


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10 Questions MCQ Test Engineering Mathematics - Test: Numerical Methods

Test: Numerical Methods for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation. The Test: Numerical Methods questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Numerical Methods MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Numerical Methods below.
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Test: Numerical Methods - Question 1

Find the difference between the sum of (x + y + z) and the Trace upper triangular matrix formed by using the gauss elimination method.

Detailed Solution for Test: Numerical Methods - Question 1

Concept:
Gauss elimination method,
In this method, an augmented matrix is formed by the coefficient of x, y, & z then,
By using row transformation, it is converted into an upper triangular matrix.
Calculation:
AX = B

R2 → R2 - 2R1
R3 → R3 - R1

R3 → R3 + R2

Now, again converting this matrix into equation,
x + 3y + 2z = 5
-2y - 10z = -14
-9z  = -9
on solving 
z= 1, y = 2 & x = -3
x + y + z = -3 + 2 + 1 = 0 
Trace of upper traingular matrix = 1 - 2 - 9 = -10 
Difference = ( x + y + z ) - trace = 0- (-10 =)
Difference = 10 

Test: Numerical Methods - Question 2

If dy/dx = x - y2 and y(0) = 1, then y(0.1) correct upto two decimal places (approx.) is:

Detailed Solution for Test: Numerical Methods - Question 2

Concept:
Euler's Method to generate a numerical solution to an initial value problem of the form:
y' = f (x, y), y(x0) = y0
yn + 1 = yn + h f(xn, yn)
Calculation:
We have,
x0 =  0, y0 = 1, h = 0.1
x1 = x0 + h
For n = 0
⇒ y1 = y0 + hf(xn, yn)
⇒ y1 = 1 - 0.1 ×  (0 - 1)2
⇒ y1 = 1 - 0.1
⇒ y1 = 0. 9
For n = 1
⇒ y2 = y1 + hf(x1 - y12)
⇒ y2 = 0.9 + 0.1[0.1 - (0.9)2]
⇒ y2 = 0.9 + 0.1[0.1 - 0. 81]
⇒ y2 = 0.9 - 0.1 ×  0.71
⇒ y2 = 0.9 - 0.071
⇒ y2 = 0. 829
⇒ y2 = 0.83
∴ The approximately. two decimal place value is 0.83 

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Test: Numerical Methods - Question 3

Which order of Polynomials can best be integrated using Trapezoidal Rules?

Detailed Solution for Test: Numerical Methods - Question 3

Concept:
The following table shows the different methods of numerical integration and degree of polynomials for which they will produce results of minimum error or zero error:

From the above table, it is clear that both Trapezoidal Rule polynomials of degree ≤ 1
Alternate Method
We know,
While deriving the formula for numerical integrations f(x) is assumed as -

  • Quadratic polynomial → Simpson's 1/3 Rule
  • Cubic polynomial → Simpson's 3/8 Rule
  • Linear polynomial → Trapezoidal Rule
Test: Numerical Methods - Question 4

Considering four subintervals, the value of  by Trapezoidal rule is:

Detailed Solution for Test: Numerical Methods - Question 4

Concept:
The trapezoidal rule states that:

Calculation:

Total number of intervals = 4

Test: Numerical Methods - Question 5

The bisection method is applied to compute a zero of the function f(x) = x4 – x3 – x2 – 4 in the interval [1, 9]. The method converges to a solution after _______ iterations.

Detailed Solution for Test: Numerical Methods - Question 5

Concept:
Bisection method:
Used to find the root for a function. Root of a function f(x) = a such that f(a)= 0
Property: if a function f(x) is continuous on the interval [a…b] and sign of f(a) ≠  sign of f(b). There is a value c belongs to [a…b] such that f(c) = 0, means c is a root in between [a….b]
Note:
Bisection method cut the interval into 2 halves and check which half contains a root of the equation.
1) Suppose interval [a…b] .
2) Cut interval in the middle to find m : m = (a + b)/2
3) sign of f(m) not matches with f(a) proceed the search in the new interval.
Calculation:
The bisection method is applied to a given problem with [1, 9]
After 1 iteration

Now since f(x1) > 0, x2 replaces x1
Now, x0 = 1 and x1 = 5
And after 2nd iteration 

Now since f(x1) f(x2) > 0, x2 replaces x1 and x0 = 1 and x1 = 3 and after 3rd iteration

Now, f(x2) = f(2) = 24 – 23 – 22 – 4 = 0
So the method converges exactly to the root in 3 iterations.

Test: Numerical Methods - Question 6

Find the sum of the value of x, y, & z by using gauss jordan method.
3x - y + 2z = 12
x + 2y + 3z = 11
2x - 2y - z = 2

Detailed Solution for Test: Numerical Methods - Question 6

Concept:
Gauss Jordan method,
In this method, an augmented matrix is formed by the coefficient of x, y, and z. then by using row transformation, it is converted into a diagonal matrix i.e. all non-diagonal elements will be zero.
Calculation:
x + 2y + 3z = 11
2x - 2y - z = 2
3x - y + 2z = 12

AX = B

R2 → R2 - 2R1
R3 → R3 - 3R1

R3 → 6R3 + R2
R1 → 3R1 + R2

R1 → R1 + 2R3
R2 → R2 - 7R3

Now, again converting this matrix into equation,
3x = 9 ⇒ x = 3
-6y = 6 ⇒ y = 1
-z = -2 ⇒ z = 2
x + y + z = 3 + 2 + 1 = 6

Test: Numerical Methods - Question 7

In which of the following categories can we put Bisection method?

Detailed Solution for Test: Numerical Methods - Question 7

Bracketing Methods:

  • All bracketing methods always converge, whereas open methods (may sometimes diverge).
  • We must start with an initial interval [a,b], where f(a) and f(b) have opposite signs.
  • Since the graph y = f(x) of a continuous function is unbroken, it will cross the abscissa at a zero x = 'a' that lies somewhere within the interval [a,b].
  • One of the ways to test a numerical method for solving the equation f(x) = 0 is to check its performance on a polynomial whose roots are known.

Bisection method:
Used to find the root for a function. Root of a function f(x) = a such that f(a)= 0
Property: if a function f(x) is continuous on the interval [a…b] and sign of f(a) ≠ sign of f(b). There is a value c belongs to [a…b] such that f(c) = 0, means c is a root in between [a….b]
Note:
Bisection method cut the interval into 2 halves and check which half contains a root of the equation.
1) Suppose interval [a, b] .
2) Cut interval in the middle to find m : m = (a + b)/2
3) sign of f(m) not matches with f(a), proceed the search in new interval.

Test: Numerical Methods - Question 8

Consider the below data:

The value of  by Trapezoidal rule will be:

Detailed Solution for Test: Numerical Methods - Question 8

Concept:
Trapezoidal rule states that for a function y = f(x)

xn = x0 + nh, where n = Number of sub-intervals
h = step-size

For a trapezoidal rule, a number of sub-intervals must be a multiple of 1.
Calculation:

Here: x0 = 4, x1 = 3, x2 = 12, h = 1
From equation (1);

Key Points:
Apart from the trapezoidal rule, other numerical integration methods are:
Simpson’s one-third rule:
For applying this rule, the number of subintervals must be a multiple of 2.

Simpson’s three-eighths rule:
For applying this rule, the number of subintervals must be a multiple of 3.

Test: Numerical Methods - Question 9

Trapezoidal Rule gives exact value of the integral when the integrand is a

Detailed Solution for Test: Numerical Methods - Question 9

Trapezoidal rule
It integrates a linear function exactly and produces errors for polynomial functions of degree 2 or higher.

Here, the interval is divided into 'n' number of intervals (even or odd) of equal width 'h' b is the upper limit, a is the lower limit, h is the step size
It fits for a 1-degree (linear) polynomial.
According to Simpson's 1/3 rule:

It fits for 2-degree (quadratic) polynomial.
According to Simpson's 3/8 rule

It fits for 3-degree (cubic) polynomial.

Test: Numerical Methods - Question 10

If f(x) = x2, then the second order divided difference for the points x0, x1, x2 will be:

Detailed Solution for Test: Numerical Methods - Question 10

Concept:
If data points are given as a function of f, then the various order divided differences are as follows,
Zeroth-order divided difference:
f[x0] = f(x0);
First-order divided difference:

Second-order divided difference:

Calculation:
Given f(x) = x2;
Using the second-order divided difference formula, we get 

∴ the second-order divided difference of x2 is 1.

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