This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Bisection Method – 1”.
1. Using Bisection method find the root of cos(x) – x * ex = 0 with a = 0 and b = 1.
a) 0.617
b) 0.527
c) 0.517
d) 0.717
View Answer
Explanation: Iteration table is given as follows
| No. | a | b | c | f(a)*f(c) |
| 1 | 0 | 1 | 0.5 | 0.053 |
| 2 | 0.5 | 1 | 0.75 | -0.046 |
| 3 | 0.5 | 0.75 | 0.625 | -0.357 |
| 4 | 0.5 | 0.625 | 0.562 | -7.52*10-3 |
| 5 | 0.5 | 0.562 | 0.531 | -2.168*10-3 |
| 6 | 0.5 | 0.531 | 0.516 | 3.648*10-4 |
| 7 | 0.516 | 0.531 | 0.524 | -9.371*10-5 |
| 8 | 0.516 | 0.524 | 0.520 | -3.649*10-5 |
| 9 | 0.516 | 0.520 | 0.518 | -3.941*10-6 |
| 10 | 0.516 | 0.516 | 0.517 | 1.229*10-5 |
The difference between final values is less than 0.01 hence we stop the iterations. So one of the roots of cos(x) – x * exp(x) = 0 is approximately 0.517.
2. Find the root of x4-x-10 = 0 approximately upto 5 iterations using Bisection Method. Let a = 1.5 and b = 2.
a) 1.68
b) 1.86
c) 1.88
d) 1.66
View Answer
Explanation: Iteration table is given as follows:
| No. | a | b | c | f(a)*f(c) |
| 1 | 1.5 | 2 | 1.75 | 15.264(+ve) |
| 2 | 1.75 | 2 | 1.875 | 2.419(+ve) |
| 3 | 1.75 | 1.875 | 1.812 | 2.419(+ve) |
| 4 | 1.812 | 1.875 | 1.844 | -0.303(-ve) |
| 5 | 1.844 | 1.875 | 1.86 | -0.027(-ve) |
We stop iterations after 5 and thus obtain the approximated value of x. So one of the roots of x4-x-10 = 0 is approximately 1.86.
3. If a function is real and continuous in the region from a to b and f(a) and f(b) have opposite signs then there is no real root between a and b.
a) True
b) False
View Answer
Explanation: If a function is real and continuous in the region from a to b and f(a) and f(b) and if both have different signs then there is at least one real root between a and b. This is because the function has to cross the x-axis at least once.
4. A function is given by x – e-x = 0. Find the root between a = 0 and b = 1 by using Bisection method.
a) 0.655
b) 0.665
c) 0.565
d) 0.656
View Answer
Explanation: Iteration table is given as follows:
| No. | a | b | c | f(a)*f(c) |
| 1 | 0 | 1 | 0.5 | 0.107 |
| 2 | 0.5 | 1 | 0.75 | -0.03 |
| 3 | 0.5 | 0.75 | 0.625 | -0.00956 |
| 4 | 0.5 | 0.625 | 0.562 | 0.0007758 |
| 5 | 0.562 | 0.625 | 0.593 | -0.0003317 |
| 6 | 0.562 | 0.593 | 0.577 | -0.0001307 |
| 7 | 0.562 | 0.577 | 0.569 | -0.00002978 |
| 8 | 0.562 | 0.569 | 0.565 | 0.00002078 |
The difference between the final iterating values is less than 0.01. So one of the root of x – e-x = 0 is approximately 0.565.
5. A function f(x) is given as e-x * (x2+5x+2) + 1 = 0. Let a = 0 and b = -1. Find the root between a and b using Bisection Method.
a) -0.557
b) -0.575
c) -0.775
d) -0.0577
View Answer
Explanation: Iteration table is given as follows:
| No. | a | B | c | f(a)*f(c) |
| 1 | 0 | -1 | -0.5 | 1.763(+ve) |
| 2 | -0.5 | -1 | -0.75 | -0.89(-ve) |
| 3 | -0.5 | -0.75 | -0.625 | -0.219(-ve) |
| 4 | -0.5 | -0.625 | -0.562 | 0.076(+ve) |
| 5 | -0.562 | -0.625 | -0.593 | -0.015(-ve) |
| 6 | -0.562 | -0.593 | -0.577 | 1.733*10-3(+ve) |
The difference between the final iterating values is less than 0.01. So one of the roots of e-x * (x2-5x+2) + 1 = 0 is approximately -0.577.
6. Use Bisection Method to find out the root of x – sinx – 0.5 = 0 between 1 and 2.
a) 1.497
b) 1.947
c) 1.479
d) 1.974
View Answer
Explanation: Iteration table is given as follows:
| No. | a | b | c | f(a)*f(b) |
| 1 | 1 | 2 | 1.5 | -8.554*10-4(-) |
| 2 | 1 | 1.5 | 1.25 | 0.068(+) |
| 3 | 1.25 | 1.5 | 1.375 | 0.021(+) |
| 4 | 1.375 | 1.5 | 1.437 | 5.679*10-3(+) |
| 5 | 1.437 | 1.5 | 1.469 | 1.42*10-3(+) |
| 6 | 1.469 | 1.5 | 1.485 | 3.042*10-4(+) |
| 7 | 1.485 | 1.5 | 1.493 | 5.023*10-5(+) |
| 8 | 1.493 | 1.5 | 1.497 | 2.947*10-6(+) |
Since the difference between b and c is less than 0.01 we stop the iterations. So one of the roots of x-sinx – 0.5 = 0 is approximately 1.497.
7. Find the approximated value of x till 4 iterations for e-x = 3 log(x) using Bisection Method.
a) 1.197
b) 1.187
c) 1.167
d) 1.176
View Answer
Explanation: Iteration table is given as follows.
| No. | a | b | c | f(a)*f(b) |
| 1 | 0.5 | 1.5 | 1 | 0.555(+ve) |
| 2 | 1 | 1.5 | 1.25 | -1.555*10-3(-ve) |
| 3 | 1 | 1.25 | 1.125 | 0.063(+ve) |
| 4 | 1.125 | 1.25 | 1.187 | 0.014(+ve) |
Hence we stop the iterations after 4. Therefore the approximated value of x is 1.187.
8. Find a root of f (x) = 3x + sin(x) – ex = 0. Use 6 iterations to find the approximate value of x.
a) 0.3605
b) 0.3650
c) 0.3615
d) 0.3655
View Answer
Explanation: The iteration table is given as follows:
| No. | a | b | c | f(a)*f(b) |
| 1 | 0 | 0.5 | 0.25 | 0.287(+ve) |
| 2 | 0.25 | 0.5 | 0.393 | -0.015(-ve) |
| 3 | 0.65 | 0.393 | 0.34 | 9.69(+ve) |
| 4 | 0.34 | 0.393 | 0.367 | -7.81(-ve) |
| 5 | 0.34 | 0.367 | 0.354 | 8.9(+ve) |
| 6 | 0.354 | 0.367 | 0.3605 | -3.1(-ve) |
We have to find the value of x approximated to 6 iterations. So one of the roots of 3x + sin(x) – ex = 0 is approximately 0.3605.
9. A function is defined as f(x) = x2 – 3. Between the interval [1,2] find the root of the function by Bisection Method.
a) 1.7334
b) 1.7364
c) 1.7354
d) 1.7344
View Answer
Explanation: f(x) = x2 – 3.
| a | b | f(a) | f(b) | c = (a+b)/2 | f(c) | Substitution | new b-a |
| 1 | 2 | -2 | 1 | 1.5 | -0.75 | a=c | 0.5 |
| 1.5 | 2 | -0.75 | 1 | 1.75 | 0.062 | b=c | 0.25 |
| 1.5 | 1.75 | -0.75 | 0.0625 | 1.625 | -0.359 | a=c | 0.125 |
| 1.625 | 1.75 | -0.75 | 0.0625 | 1.6875 | -0.1523 | a=c | 0.0625 |
| 1.625 | 1.75 | -0.3594 | 0.0625 | 1.6875 | -0.1523 | a=c | 0.0625 |
| 1.6875 | 1.75 | -0.1523 | 0.0625 | 1.7188 | -0.0457 | a=c | 0.0313 |
| 1.7188 | 1.75 | -0.0457 | 0.0625 | 1.7344 | 0.0081 | b=c | 0.0156 |
| 1.7266 | 1.7344 | -0.0457 | 0.0081 | 1.7266 | -0.0189 | a=c | 0.0078 |
Thus, with the seventh iteration, we note that the final interval, [1.7266, 1.7344], has a width less than 0.01 and |f(1.7344)| < 0.01. Hence we chose b = 1.7344 to be our approximation of the root.
10. The Bisection method is also known as ___________________
a) Binary Chopping
b) Quaternary Chopping
c) Tri region Chopping
d) Hex region Chopping
View Answer
Explanation: Bisection Method is also called as Binary Chopping. It is a method involving the division of the interval into half.
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