An approximate solution of the equation x3 - 17 = 0 is to be obtained using the
Newton-Raphson method. If the initial guess is xo = 2 , the value at the end of the
first iteration is x1 = _____ (rounded off to two decimal places).
Official Answer Key:
2.75
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.44)**
**1. Intuitive Physical Setup & Given Parameters**:
We want to find an approximate solution to:
$f(x) = x^3 - 17 = 0$
using the **Newton-Raphson method**.
- Initial guess: $x_0 = 2$.
We need to find the value $x_1$ at the end of the first iteration (rounded off to two decimal places).
**2. Step-by-Step Mathematical Derivation**:
1. **Newton-Raphson Iteration Formula**:
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
2. **Derivative of $f(x)$**:
$f(x) = x^3 - 17 \implies f'(x) = 3x^2$
3. **Evaluate at $x_0 = 2$**:
$f(x_0) = 2^3 - 17 = 8 - 17 = -9$
$f'(x_0) = 3(2^2) = 3 \times 4 = 12$
4. **Compute $x_1$**:
$x_1 = 2 - \frac{-9}{12} = 2 + \frac{9}{12} = 2 + \frac{3}{4} = 2 + 0.75 = 2.75$
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $x_1 = 2.75$.
- **Rounding Requirement**: Two decimal places $\implies \mathbf{2.75}$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **2.75** (Acceptable Range: **2.73 to 2.77**)
- **Exam Takeaway**: One step of Newton-Raphson: $x_1 = x_0 - \frac{x_0^3 - 17}{3x_0^2} = 2 - \frac{-9}{12} = 2.75$.
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