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 x3 – x2 + 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 x0=2 (initial guess), f(x0)=8 and f’(x0)=12.
Substituting the values in the equation we get x1=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) x1 = x0-f(x0)/f’(x0)
b) x0 = x1-f(x0)/f‘(x0)
c) x0 = x1+f(x0)/f’(x0)
d) x1 = x0+f(x0)/f‘(x0)
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) = x2 – 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’(x0) at x=90 degrees approaches ∞. Then for all values of x1=x0. Hence if f(x0) = 0 then the tan0=90 degrees.
8. The equation f(x) is given as x2-4=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 x0=6 (initial guess), f(x0)=32 and f’(x0)=12.
Substituting the values in the equation we get x1=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’(x0)=0
b) f’’(x0)=0
c) f(x0)=0
d) f’’’(x0)=0
View Answer
Explanation: When f’(x0) becomes zero then the value of f(x0)/f’(x0) becomes ∞. Hence Newton Raphson method fails at f’(x0)=0.
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