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 * e^{x} = 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 x^{4}-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 x^{4}-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} * (x^{2}+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} * (x^{2}-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) – e^{x} = 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) – e^{x} = 0 is approximately 0.3605.

9. A function is defined as f(x) = x^{2} – 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) = x

^{2}– 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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