1
GATE IN 2014
+1
-0.3
The iteration step in order to solve for the cube roots of a given number $$'N'$$ using the Newton-Raphson's method is
A
$${x_{k + 1}} = {x_k} + {1 \over 3}\left( {N - x_k^3} \right)$$
B
$${x_{k + 1}} = {1 \over 3}\left( {2{x_k} + {N \over {x_k^2}}} \right)$$
C
$${x_{k + 1}} = {x_k} - {1 \over 3}\left( {N - x_k^3} \right)$$
D
$${x_{k + 1}} = {1 \over 3}\left( {2{x_k} - {N \over {x_k^2}}} \right)$$
2
GATE IN 2013
+1
-0.3
While numerically solving the differential equation $$\,{{dy} \over {dx}} + 2x{y^2} = 0,y\left( 0 \right) = 1\,\,$$ using Euler's predictor corrector (improved Euler- Cauchy) method with a step size of $$0.2,$$ the value of $$y$$ after the first step is
A
$$1.00$$
B
$$1.03$$
C
$$0.97$$
D
$$0.96$$
3
GATE IN 2008
+1
-0.3
It is known that two roots of the non-linear equation $$\,{x^3} - 6{x^2} + 11x - 6 = 0\,\,$$ are $$1$$ and $$3.$$ The third root will be
A
$$j$$
B
$$-j$$
C
$$2$$
D
$$4$$
4
GATE IN 2007
+1
-0.3
Identity the Newton $$-$$ Raphson iteration scheme for the finding the square root of $$2$$
A
$${x_{n + 1}} = {1 \over 2}\left( {{x_n} + {2 \over {{x_n}}}} \right)$$
B
$${x_{n + 1}} = {1 \over 2}\left( {{x_n} - {2 \over {{x_n}}}} \right)$$
C
$${x_{n + 1}} = {2 \over {{x_n}}}$$
D
$${x_{n + 1}} = \sqrt {2 + {x_n}}$$
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