Difference between revisions of "GATE2010 q2"

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Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is
 
Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is
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(A) 3.575
 
(A) 3.575
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(B) 3.676
 
(B) 3.676
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(C) 3.667
 
(C) 3.667
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<b>(D) </b>3.607
 
<b>(D) </b>3.607
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
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$$x_{n+1} = 3.5 - \frac{-0.75}{7} = 3.607$$
 
$$x_{n+1} = 3.5 - \frac{-0.75}{7} = 3.607$$
 
So option <b>(D)</b> is correct.
 
So option <b>(D)</b> is correct.
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[[Category: GATE2010]]
 
[[Category: GATE2010]]
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[[Category: Calculus questions from GATE]]

Latest revision as of 11:49, 15 July 2014

Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is

(A) 3.575

(B) 3.676

(C) 3.667

(D) 3.607

Solution by Happy Mittal

We know that, according to Newton-Raphson method, $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ In this question $f(x) = x^2-13$, and so $f'(x) = 2x$.
Here current $x_n = 3.5$, $f(x_n) = f(3.5) = -0.75$, and $f'(x_n) = f'(3.5) = 7$. So $$x_{n+1} = 3.5 - \frac{-0.75}{7} = 3.607$$ So option (D) is correct.



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Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is
(A) 3.575   (B) 3.676   (C) 3.667   (D) 3.607

Solution by Happy Mittal[edit]

We know that, according to Newton-Raphson method, $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ In this question $f(x) = x^2-13$, and so $f'(x) = 2x$.
Here current $x_n = 3.5$, $f(x_n) = f(3.5) = -0.75$, and $f'(x_n) = f'(3.5) = 7$. So $$x_{n+1} = 3.5 - \frac{-0.75}{7} = 3.607$$ So option (D) is correct.