Problem Set#1
Multiple Choice Test
Chapter 03.06False-Positiion Method
COMPLETE SOLUTION SET
1. The false-position method for finding roots of nonlinear equations belongs to a class of a (an) ______method.
(A)open
(B)bracketing
(C)random
(D)graphical
Solution
The correct answer is (B).
The false-position method brackets the root with the two initial guesses.
2. The newly predicted root for false-position and secant method can be respectively given as
and
,
While the appearance of the above 2 equations look essentially identical, and both methods require two initial guesses, the major difference between the above two formulas is
(A)false-position method is not guaranteed to converge
(B)secant method is guaranteed to converge
(C)secant method requires the 2 initial guesses to satisfy
(D)false-position method requires the 2 initial guesses to satisfy
Solution
The correct answer is (D).
In order for the two initial guesses to bracket the root, must be true. In other words, the function must change signs between the two guesses for there to be a root present.
3. Given are the following nonlinear equation
two initial guesses, and , and a pre-specified relative error tolerance of 0.1%. Using the false-position method, which of the following tables is correct = predicted root)?
(A)
Iteration / / /1 / 1 / 4 / ?
2 / ? / ? / 2.939
(B)
Iteration / / /1 / 1 / 4 / ?
2 / ? / ? / 2.500
(C)
Iteration / / /1 / 1 / 4 / ?
2 / ? / ? / 1.500
(D)
Iteration / / /1 / 1 / 4 / ?
2 / ? / ? / 2.784
Solution
The correct answer is (A).
First check given function is of the form , then proceed as follows:
Iteration 1:
Initial guesses are and . Check that they satisfy the requirement.
, which is <0.
Now can estimate the root:
Use to what new roots are:
Since this value is >0, the root must lie between and . Therefore, and .
Iteration 2:
Initial guesses are and . Again, check requirement:
, which is <0.
Now find new root estimate:
Therefore, for the second iteration,
4. Given are the following nonlinear equation
two initial guesses, and , and a pre-specified relative error tolerance of 0.1%. Using the false-position method, which of the following tables is correct = predicted root, = percentage absolute relative approximate error).
(A)
Iteration / / / / || %1 / 1 / 4 / ? / ?
2 / ? / ? / ? / 11.63
(B)
Iteration / / / / || %1 / 1 / 4 / ? / ?
2 / ? / ? / ? / 6.11
(C)
Iteration / / / / || %1 / 1 / 4 / ? / ?
2 / ? / ? / ? / 5.14
(D)
Iteration / / / / || %1 / 1 / 4 / ? / ?
2 / ? / ? / ? / 4.15
Solution
The correct answer is (A).
First check given function is of the form , then proceed as follows:
Iteration 1:
Initial guesses are and . Check that they satisfy the requirement .
, which is <0.
Now can estimate the root:
Use to what new roots are:
Since this value is >0, the root must lie between and . Therefore, and .
does not exist for the first iteration because there is no previous approximation to use in the equation.
Iteration 2:
Initial guesses are and . Again, check requirement:
, which is <0.
Now find new root estimate:
Now can solve for :
5. The root of was found using false-position method with initial guesses of and , and a pre-specified relative error tolerance of%. The final converged root was found as , and the corresponding percentage absolute relative approximate error was found as . Based on the given information, the number of significant digits of the converged root that can be trusted at least are
(A)3
(B)4
(C)5
(D)6
Solution
The correct answer is (B).
Knowing , match it within the significant digit pattern:
/ Number of trusted significant digits in root0.5 / 1
0.05 / 2
0.005 / 3
0.0005 / 4
0.00005 / 5
Because not less than 5x10-5, it falls in the category above it and the corresponding number of trusted significant digits is associated with it.
6. The false-position method may have difficulty in finding the root of because
(A) is a quadratic polynomial
(B)a straight line
(C)one cannot find initial guesses and that satisfy
(D)the equation has two identical roots
Solution
The correct answer is (C).
The graph of the function demonstrates that all the function’s values are positive. Therefore, a value of can never be found, even though the root exists.