Q.30 Consider a nonlinear algebraic equation, 𝑥𝑙𝑛𝑥 + 𝑥 − 1 = 0. Using the NewtonRaphson method, with the initial guess of 𝑥₀ = 3, the value of 𝑥 after one
iteration (rounded off to one decimal place) is __________.

Newton-Raphson Method: Solve x ln x + x − 1 = 0

The Newton-Raphson method applied to the nonlinear equation
x ln x + x − 1 = 0 with initial guess x₀ = 3
yields x₁ = 1.3 after one iteration, rounded to one decimal place.

Newton-Raphson Formula

The Newton-Raphson iteration is:

x_n+1 = x_n − f(x_n) / f'(x_n)

For the given function:

• f(x) = x ln x + x − 1
• f'(x) = ln x + 2

Step-by-Step Calculation

At x₀ = 3:

• ln 3 ≈ 1.0986
• f(3) = 3(1.0986) + 3 − 1 = 5.2958
• f'(3) = 1.0986 + 2 = 3.0986

Thus:

x₁ = 3 − (5.2958 / 3.0986) ≈ 3 − 1.709 = 1.291 ≈ 1.3

Why This Answer?

GATE BT 2022 requires only one Newton step from x₀ = 3,
rounded to one decimal place.

Actual root lies near 0.5, but more iterations are needed to reach it.