Numerical Analysis Questions and Answers – Newton Raphson Method – 2

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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

Answer: a
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.
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2. The Newton Raphson method is also called as ____________
a) Tangent method
b) Secant method
c) Chord method
d) Diameter method
View Answer

Answer: a
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 x2-4=0. Considering the initial approximation at x=6 then the value of x1 is given as ____________
a) 10/3
b) 4/3
c) 7/3
d) 13/3
View Answer

Answer: a
Explanation: Iterative formula for Newton Raphson method is given by
x(1)=x(0)+\(\frac{f(x(0))}{f’x(x(0))}\).
Hence x0=6 (initial guess), f(x0)=32 and f’(x0)=12.
Substituting the values in the equation we get x1= 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

Answer: a
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.
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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

Answer: d
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) = x2-117 = 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

Answer: b
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

Answer: d
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.
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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

Answer: c
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 x0.

9. The equation f(x) is given as x3+4x+1=0. Considering the initial approximation at x=1 then the value of x1 is given as _______________
a) 1.67
b) 1.87
c) 1.86
d) 1.85
View Answer

Answer: c
Explanation: Iterative formula for Newton Raphson method is given by
x(1)=x(0)+\(\frac{f(x(0))}{f’x(x(0))}\)
Hence x0=1 (initial guess), f(x0)=6 and f’(x0)= 7.
Substituting the values in the equation we get x1=1.857 which is 1.86 rounded to 2 decimal places.

Sanfoundry Global Education & Learning Series – Numerical Methods.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn