GATE CS 2014 Set 2 Q.56 (CS Q.46): Calculus

In the Newton-Raphson method, an initial guess of $x_0 = 2$ is made and the sequence $x_0, x_1, x_2 \dots$ is obtained for the function

0.75x^3 - 2x^2 - 2x + 4 = 0

Consider the statements

(I) $x_3 = 0$ .

(II) The method converges to a solution in a finite number of iterations.

Which of the following is TRUE?

Only I

Only II

Both I and II

Neither I nor II

Official answer and solution

Solution

Apply the Newton-Raphson formula iteratively starting from the initial guess x0 = 2.

Newton-Raphson Iteration Formula:
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$
Function and Derivative:
$f(x) = 0.75x^3 - 2x^2 - 2x + 4,$ $f'(x) = 2.25x^2 - 4x - 2.$
Iteration 1 ( $n = 0, x_0 = 2$ ):
$f(2) = 0.75(8) - 2(4) - 2(2) + 4 = 6 - 8 - 4 + 4 = -2.$ $f'(2) = 2.25(4) - 4(2) - 2 = 9 - 8 - 2 = -1.$ $x_1 = 2 - \frac{-2}{-1} = 2 - 2 = 0.$
Iteration 2 ( $n = 1, x_1 = 0$ ):
$f(0) = 4.$ $f'(0) = -2.$ $x_2 = 0 - \frac{4}{-2} = 0 - (-2) = 2.$
Iteration 3 ( $n = 2, x_2 = 2$ ):
$x_3 = x_1 = 0.$
Analysis of Statements:
- Statement (I): $x_3 = 0$ . This is TRUE.
- Statement (II): The sequence endlessly alternates between $2$ and $0$ ( $2, 0, 2, 0, \dots$ ). Neither $2$ nor $0$ is a root ( $f(2) = -2 \ne 0$ and $f(0) = 4 \ne 0$ ). Therefore, the method oscillates indefinitely in a 2-cycle and never converges to a solution. Statement (II) is FALSE.
Conclusion:
Only Statement I is TRUE (Option A).

Final answer

Common trap

Assuming Newton-Raphson always converges; poor initial guesses can land in periodic limit cycles.

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