1
GATE CE 2008
+1
-0.3
The Newton-Raphson iteration $${x_{n + 1}} = {1 \over 2}\left( {{x_n} + {R \over {{x_n}}}} \right)$$ can be used to compute
A
square of $$R$$
B
reciprocal of $$R$$
C
square root of $$R$$
D
logarithm of $$R$$
2
GATE CE 2007
+1
-0.3
The following equation needs to be numerically solved using the Newton $$-$$ Raphson method $${x^3} + 4x - 9 = 0.\,\,$$ The iterative equation for this purpose is ($$k$$ indicates the iteration level)
A
$${X_{k + 1}} = {{2X_k^3 + 9} \over {3X_k^2 + 4}}$$
B
$${X_{k + 1}} = {{3X_k^3 + 9} \over {2X_k^2 + 9}}$$
C
$${X_{k + 1}} = {X_k} - 3_k^2 + 4$$
D
$${X_{k + 1}} = {{4X_k^2 + 3} \over {9X_k^2 + 2}}$$
3
GATE CE 2007
+1
-0.3
Given that one root of the equation $$\,{x^3} - 10{x^2} + 31x - 30 = 0\,\,$$ is $$5$$ then other roots are
A
$$2$$ and $$3$$
B
$$2$$ and $$4$$
C
$$3$$ and $$4$$
D
$$-2$$ and $$-3$$
4
GATE CE 2005
+1
-0.3
Given $$a>0,$$ we wish to calculate it reciprocal value $${1 \over a}$$ by using Newton - Raphson method for $$f(x)=0.$$ The Newton - Raphson algorithm for the function will be
A
$${X_{k + 1}} = {1 \over 2}\left( {{X_k} + {a \over {{X_k}}}} \right)$$
B
$${X_{k + 1}} = {X_k} + {a \over 2}X_k^2$$
C
$${X_{k + 1}} = 2{X_k} - aX_k^2$$
D
$${X_{k + 1}} = 2{X_k} - {a \over 2}X_k^2$$
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