This set of Numerical Analysis Multiple Choice Questions & Answers focuses on “Newton Raphson Method – 2”.

1. The value of y’/x’ in terms of the angle 0 is given by _____________

a) tanθ

b) secθ

c) cotθ

d) cosecθ

View Answer

Explanation: The value of derivative of a function f(x) is given as f’(x)=y’/x’. In terms of theta tangent is the ration of opposite side to adjacent side hence y’/x’=tanθ.

2. The Newton Raphson method is also called as ____________

a) Tangent method

b) Secant method

c) Chord method

d) Diameter method

View Answer

Explanation: Newton Raphson method is also known as Tangent Method. It is carried out by drawing a tangent to the curve at the point of initial guess.

3. The equation f(x) is given as x^{2}-4=0. Considering the initial approximation at x=6 then the value of x_{1} is given as ____________

a) 10/3

b) 4/3

c) 7/3

d) 13/3

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}= 10/3.

4. For decreasing the number of iterations in Newton Raphson method:

a) The value of f’(x) must be increased

b) The value of f’’(x) must be decreased

c) The value of f’(x) must be decreased

d) The value of f’’(x) must be increased

View Answer

Explanation: Iterative formula is given by

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

Hence if f’(x) decreases the value of next iteration approaches the initial one. This decreases the number of iterations in finding out next iterative value.

5. 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’

View Answer

Explanation: The derivative of a function f(x) is given as f’(x) = y’/x’. It is defined as the slope of the tangent drawn at the point of assumed initial value to the curve.

6. If f(x) = x^{2}-117 = 0 then the iterative formula for Newton Raphson Method is given by ____________

a) x(n+1)=0.25 [x(n)+\(\frac{166}{x(n)}\)]

b) x(n+1)=0.5 [x(n)+\(\frac{166}{x(n)}\)]

c) x(n+1)=0.5 [x(n)-\(\frac{166}{x(n)}\)]

d) x(n+1)=0.25 [x(n)-\(\frac{166}{x(n)}\)]

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)+\(\frac{117}{x(n)}\)].

7. The points where the Newton Raphson method fails are called?

a) floating

b) continuous

c) non-stationary

d) stationary

View Answer

Explanation: The points where the function f(x) approaches infinity are called as Stationary points. At stationary points Newton Raphson fails and hence it remains undefined for Stationary points.

8. The convergence of which of the following method depends on initial assumed value?

a) False position

b) Gauss Seidel method

c) Newton Raphson method

d) Euler method

View Answer

Explanation: The Newton Raphson method the approximation value is found out by :

x(1)=x(0)+\(\frac{f(x(0))}{f’x(x(0))}\). Hence it depends on the initial value x

_{0}.

9. The equation f(x) is given as x^{3}+4x+1=0. Considering the initial approximation at x=1 then the value of x_{1} is given as _______________

a) 1.6712

b) 0.1856

c) 0.1429

d) 1.8523

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}=1 (initial guess), f(x

_{0})=6 and f’(x

_{0})= 7.

Substituting the values in the equation we get x

_{1}=0.1429.

**Sanfoundry Global Education & Learning Series – Numerical Methods.**

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