The real root of the equation x³ − x − 5 = 0 lying between 1 and 2 after first iteration by Newton - Raphson method is _______, if initial approximation is taken as x₀ = 2∈ [1, 2] : 

The correct answer is option 1: 1.909

Key Points

We are given the equation:

\( f(x) = x^3 - x - 5 \)

Its derivative is:

\( f'(x) = 3x^2 - 1 \)

Initial approximation: \( x_0 = 2 \)

Newton-Raphson iteration formula is:

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

Step-by-step Calculation:

• \( f(2) = 2^3 - 2 - 5 = 8 - 2 - 5 = 1 \)
• \( f'(2) = 3(2)^2 - 1 = 12 - 1 = 11 \)
• \( x_1 = 2 - \frac{1}{11} \approx 2 - 0.0909 = 1.909 \)

But this gives 1.909, which matches option 1.

However, options provided show 1.904 as correct (option 2), but mathematically, the correct value after first iteration is:

\( \boxed{x_1 = 1.909} \)

Additional Information

• The Newton-Raphson method converges quadratically near the root.
• This method is highly efficient for equations with simple real roots.
• Only one iteration gives an approximate root: 1.909

Hence, the correct answer is: option 1: 1.909