# Numerical Analysis Questions and Answers – Bisection Method – 1

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

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

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

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.
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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

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

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

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

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

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

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

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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