Question

Let \( f(x) = x^3 - x - 2 \). Using the bisection method on the interval \( [1, 2] \), how many iterations are required to approximate a root correct to two decimal places?

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

A useful shortcut for decimal precision: Every 3.32 iterations of the bisection method roughly gain one decimal digit of accuracy (because \( \log_2(10) \approx 3.32 \)).

Answer 1

The Correct Option is D

Concept:
The Bisection Method repeatedly halves an interval and selects a sub-interval in which a root must lie. The error in the \( n \)-th iteration is bounded by the width of the interval. To find a root correct to \( d \) decimal places, the maximum error must be less than \( 0.5 \times 10^{-d} \).

Step 1: Define the error tolerance.
For a root to be correct to two decimal places, we need the error \( \epsilon \) to be less than \( 0.005 \). The formula for the number of iterations \( n \) is derived from the interval width: \[ \frac{b - a}{2^n} < \text{Tolerance} \] Given \( a = 1, b = 2 \), and Tolerance \( = 0.005 \).

Step 2: Set up the inequality.
\[ \frac{2 - 1}{2^n} < 0.005 \] \[ \frac{1}{2^n} < \frac{5}{1000} \quad \Rightarrow \quad \frac{1}{2^n} < \frac{1}{200} \] Taking the reciprocal (which flips the inequality sign): \[ 2^n > 200 \]

Step 3: Find the smallest integer \( n \).
Let's check powers of 2:
• \( 2^6 = 64 \)
• \( 2^7 = 128 \)
• \( 2^8 = 256 \) Since \( 256 > 200 \), the smallest integer \( n \) that satisfies the condition is \( n = 8 \).