1
GATE EE 2013
MCQ (Single Correct Answer)
+1
-0.3
When the Newton-Raphson method is applied to solve the equation $$\,\,f\left( x \right) = {x^3} + 2x - 1 = 0,\,\,$$ the solution at the end of the first iteration with the initial value as $${x_0} = 1.2$$ is
A
$$-0.82$$
B
$$0.49$$
C
$$0.705$$
D
$$1.69$$
2
GATE EE 2009
MCQ (Single Correct Answer)
+1
-0.3
Let $$\,{x^2} - 117 = 0.\,\,$$ The iterative steps for the solution using Newton -Raphson's method is given by
A
$${x_{k + 1}} = {1 \over 2}\left( {{x_k} + {{117} \over {{x_k}}}} \right)$$
B
$${x_{k + 1}} = {x_k} - {{117} \over {{x_k}}}$$
C
$${x_{k + 1}} = {x_k} - {{{x_k}} \over {117}}$$
D
$${x_{k + 1}} = {x_k} - {1 \over 2}\left( {{x_k} + {{117} \over {{x_k}}}} \right)$$
3
GATE EE 2008
MCQ (Single Correct Answer)
+1
-0.3
Equation $${e^x} - 1 = 0\,\,$$ is required to be solved using Newton's method with an initial guess $$\,\,{x_0} = - 1.\,\,$$ Then after one step of Newton's method estimate $${x_1}$$ of the solution will be given by
A
$$0.71828$$
B
$$0.36784$$
C
$$0.20587$$
D
$$0.0000$$
4
GATE EE 1993
Fill in the Blanks
+1
-0
Given the differential equation $${y^1} = x - y$$ with initial condition $$y(0)=0.$$ The value of $$y(0.1)$$ calculated numerically upto the third place of decimal by the $${2^{nd}}$$ order Runge-Kutta method with step size $$h=0.1$$ is
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