This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Newton Raphson Method – 1”.

1. Rate of convergence of the Newton-Raphson method is generally __________

a) Linear

b) Quadratic

c) Super-linear

d) Cubic

View Answer

Explanation: Rate of convergence of the Newton-Raphson method is generally Linear. It states that the value of root through the Newton Raphson method converges slowly.

2. The equation f(x) is given as x^{3} – x^{2} + 4x – 4 = 0. Considering the initial approximation at x=2 then the value of next approximation correct upto 2 decimal places is given as __________

a) 0.67

b) 1.33

c) 1.00

d) 1.50

View Answer

Explanation: Iterative formula for Newton Raphson method is given by

x(1)=x(0)+\(\frac{f(x(0))}{f’x(x(0))}.\)

Hence x

_{0}=2 (initial guess), f(x

_{0})=8 and f’(x

_{0})=12.

Substituting the values in the equation we get x

_{1}=1.33

3. The Newton-Raphson method of finding roots of nonlinear equations falls under the category of which of the following methods?

a) bracketing

b) open

c) random

d) graphical

View Answer

Explanation: The Newton Raphson method involves the guessing of the root. Hence it falls under open methods.

4. The Iterative formula for Newton Raphson method is given by __________

a) x_{1} = x_{0}-f(x_{0})/f’(x_{0})

b) x_{0} = x_{1}-f(x_{0})/f‘(x_{0})

c) x_{0} = x_{1}+f(x_{0})/f’(x_{0})

d) x_{1} = x_{0}+f(x_{0})/f‘(x_{0})

View Answer

Explanation: The Iterative formula for Newton Raphson method is given by x(1)=x(0)+\(\frac{f(x(0))}{f’x(x(0))}\). It depends on the initial value and converges slowly.

5. If f(x) = x^{2} – 153 = 0 then the iterative formula for Newton Raphson Method is given by __________

a) x(n+1) = 0.5 [x(n)+\(\frac{153}{x(n)}\)]

b) x(n+1) = 0.5 [x(n)-\(\frac{153}{x(n)}\)]

c) x(n+1) = [x(n)+\(\frac{153}{x(n)}\)]

d) x(n+1) = [x(n)-\(\frac{153}{x(n)}\)]

View Answer

Explanation: Consider x(n+1)=0.5 [x(n)+\(\frac{N}{x(n)}\)]

Where N=117. Hence x(n+1)=0.5 [x(n)+\(\frac{117}{x(n)}\)].

6. 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

View Answer

Explanation: If the curve f(x) is constant then the slope of the tangent drawn to the curve at an initial point is zero. Hence the value of f’(x) is zero.

7. For what values of 0 the initial guess will be equal to the next iterative values?

a) 70 degrees

b) 90 degrees

c) 100 degrees

d) 55 degrees

View Answer

Explanation: Iterative formula is given by x(1) = x(0) + \(\frac{f(x(0))}{f’x(x(0))}\). For f’(x

_{0}) at x=90 degrees approaches ∞. Then for all values of x

_{1}=x

_{0}. Hence if f(x

_{0}) = 0 then the tan0=90 degrees.

8. The equation f(x) is given as x^{2}-4=0. Considering the initial approximation at x=6 then the value of next approximation correct upto 2 decimal places is given as __________

a) 3.33

b) 1.33

c) 2.33

d) 4.33

View Answer

Explanation: Iterative formula for Newton Raphson method is given by

x(1)=x(0)+\(\frac{f(x(0))}{f’x(x(0))}\).

Hence x

_{0}=6 (initial guess), f(x

_{0})=32 and f’(x

_{0})=12.

Substituting the values in the equation we get x

_{1}=3.33.

9. At which point the iterations in the Newton Raphson method are stopped?

a) When the consecutive iterative values of x are not equal

b) When the consecutive iterative values of x differ by 2 decimal places

c) When the consecutive iterative values of x differ by 3 decimal places

d) When the consecutive iterative values of x are equal

View Answer

Explanation: When the consecutive values of iterations are equal the iterations of Newton Raphson method are stopped. This allows maximum accuracy as compared to other methods.

10. The Newton Raphson method fails if __________

a) f’(x_{0})=0

b) f’’(x_{0})=0

c) f(x_{0})=0

d) f’’’(x_{0})=0

View Answer

Explanation: When f’(x

_{0}) becomes zero then the value of f(x

_{0})/f’(x

_{0}) becomes ∞. Hence Newton Raphson method fails at f’(x

_{0})=0.

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