Question:
How many minimum number of iterations are required to get an accuracy of \(0.001\) for the interval \([1,3]\) in the bisection method?
How many minimum number of iterations are required to get an accuracy of \(0.001\) for the interval \([1,3]\) in the bisection method?
Show Hint
Minimum iterations in the bisection method:
\[
\boxed{
n\ge
\log_2\left(\frac{b-a}{\varepsilon}\right)
}
\]
where \(\varepsilon\) is the required accuracy.
Updated On: Jul 14, 2026
- \(11\)
- \(10\)
- \(9\)
- \(8\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the bisection error formula.
The error after \(n\) iterations satisfies
\[
\frac{b-a}{2^n}\le \varepsilon.
\]
Here,
\[
a=1,\qquad
b=3,\qquad
\varepsilon=0.001.
\]
Thus,
\[
\frac{2}{2^n}\le0.001.
\]
Step 2: Find the smallest integer \(n\). This gives \[ 2^n\ge2000. \] Taking logarithm, \[ n\ge\log_2(2000)\approx10.97. \] Hence, \[ \boxed{n=11.} \] Therefore, \[ \boxed{(A)} \] is the correct answer.
Step 2: Find the smallest integer \(n\). This gives \[ 2^n\ge2000. \] Taking logarithm, \[ n\ge\log_2(2000)\approx10.97. \] Hence, \[ \boxed{n=11.} \] Therefore, \[ \boxed{(A)} \] is the correct answer.
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