Question

According to Newton-Raphson method, the value of \(\sqrt{12}\) up to three places of decimal will be

• A. 3.463
• B. 3.462
• C. 3.467
• D. None of these

Hint:

Use Newton-Raphson for fast square root approximation: \(x_{n+1} = \frac{1}{2}(x_n + \frac{N}{x_n})\).

Answer 1

The Correct Option is D

The problem involves finding the value of \(\sqrt{12}\) using the Newton-Raphson method. This method is an iterative numerical technique used to approximate roots of real-valued functions. The formula for the Newton-Raphson method is:

\(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\)

Here, we want to find \(\sqrt{12}\), hence we can set the equation as:

\(f(x) = x^2 - 12\)

The derivative of this function is:

\(f'(x) = 2x\)

To use the Newton-Raphson method, let's start with an initial guess. Since \(\sqrt{12}\) is slightly more than 3, we can take the initial guess (x₀) as 3.5 (or closer to 3.46 as this is a quadratic problem).

1. Initial guess: \(x_0 = 3.5\)
2. Apply the Newton-Raphson formula:
3. Compute:
4. Second approximation (using x₁):
5. Compute:

Repeating another iteration will show that the values converge towards approximately 3.4641. Thus, \(\sqrt{12} \approx 3.464\) up to three decimal places.

Conclusion: The value \(3.464\) does not match any of the provided options. Hence, the correct answer is "None of these".