Numerical Analysis Questions and Answers – Bisection Method – 1

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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

Answer: c
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
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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

Answer: b
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

Answer: b
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

Answer: c
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.

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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

Answer: c
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

Answer: a
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

Answer: b
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)
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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

Answer: a
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

Answer: d
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

Answer: a
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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn