If f(x) = x2 – 153 = 0 then the iterative formula for Newton Raphson Method is given by __________ b) x(n+1) = 0.5 c) x(n+1) = d) x(n+1) =

If f(x) = x2-117 = 0 then the iterative formula for Newton Raphson Method is given by ____________ b) x(n+1)=0.5 c) x(n+1)=0.5 d) x(n+1)=0.25

A

x(n+1)=0.25

B

x(n)+$\frac{166}{x(n)}$

C

x(n)+$\frac{166}{x(n)}$

D

x(n)+$\frac{166}{x(n)}$

Let x² - 117 = 0. The iterative steps for the solution using Newton-Raphson's method is given by

A

$${{\text{x}}_{{\text{k}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{k}}} + \frac{{117}}{{{{\text{x}}_{\text{k}}}}}} \right)$$

B

$${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - \frac{{117}}{{{{\text{x}}_{\text{k}}}}}$$

C

$${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - \frac{{{{\text{x}}_{\text{k}}}}}{{117}}$$

D

$${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - \frac{1}{2}\left( {{{\text{x}}_{\text{k}}} + \frac{{117}}{{{{\text{x}}_{\text{k}}}}}} \right)$$

The following equation needs to be numerically solved using the Newton-Raphson method. x³ + 4x - 9 = 0. The iterative equation for this purpose is (k indicates the iteration level)

A

$${{\text{x}}_{{\text{k}} + 1}} = \frac{{2{\text{x}}_{\text{k}}^3 + 9}}{{3{\text{x}}_{\text{k}}^2 + 4}}$$

B

$${{\text{x}}_{{\text{k}} + 1}} = \frac{{3{\text{x}}_{\text{k}}^2 + 4}}{{2{\text{x}}_{\text{k}}^2 + 9}}$$

C

$${{\text{x}}_{{\text{k}} + 1}} = {{\text{x}}_{\text{k}}} - 3{\text{x}}_{\text{k}}^2 + 4$$

D

$${{\text{x}}_{{\text{k}} + 1}} = \frac{{4{\text{x}}_{\text{k}}^2 + 3}}{{9{\text{x}}_{\text{k}}^2 + 2}}$$

Solution of the variables x₁ and x₂ for the following equations is to be obtained by employing the Newton-Raphson iterative method
equation (i) 10x₂ sin x₁ - 0.8 = 0
equation (ii) 10$${\text{x}}_2^2$$ - 10x₂ cos x₁ - 0.6 = 0
Assuming the initial values x₁ = 0.0 and x₂ = 1.0, the Jacobian matrix is

A

\\

B

\\

The square root of a number N is to be obtained by applying the Newton Raphson iterations to the equation x² - N = 0. If i denotes the iteration index, the correct iterative scheme will be

A

$${{\text{x}}_{{\text{i}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{i}}} + \frac{{\text{N}}}{{{{\text{x}}_{\text{i}}}}}} \right)$$

B

$${{\text{x}}_{{\text{i}} + 1}} = \frac{1}{2}\left( {{\text{x}}_{\text{i}}^2 + \frac{{\text{N}}}{{{\text{x}}_{\text{i}}^2}}} \right)$$

C

$${{\text{x}}_{{\text{i}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{i}}} + \frac{{{{\text{N}}^2}}}{{{{\text{x}}_{\text{i}}}}}} \right)$$

D

$${{\text{x}}_{{\text{i}} + 1}} = \frac{1}{2}\left( {{{\text{x}}_{\text{i}}} - \frac{{\text{N}}}{{{{\text{x}}_{\text{i}}}}}} \right)$$

In Newton Raphson method f’(x) for a given point is given by the formula ____________

A

y/x’

B

y’/x

C

y/x

D

y’/x’

The equation x³ - x² + 4x - 4 = 0 is to be solved using the Newton-Raphson method. If x = 2 is taken as the initial approximation of the solution, then the next approximation using this method will be

A

$$\frac{2}{3}$$

B

$$\frac{4}{3}$$

C

1

D

$$\frac{3}{2}$$

In Newton Raphson method if the curve f(x) is constant then __________

A

f(x)=0

B

f’(x)=c

C

f’’(x)=0

D

f’(x)=0

In Newton Raphson method if the curve f f(x) is constant then __________

A

f’’(x)=0

B

f(x)=0

C

f’(x)=0

D

f’(x)=c

Secant Method is slower than Newton Raphson Method.

A

True

B

False

If f(x) = x2 – 153 = 0 then the iterative formula for Newton Raphson Method is given by __________ b) x(n+1) = 0.5 c) x(n+1) = d) x(n+1) =

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