1. Starting with x = 1, the solution of the equation x^{3} + x = 1, after two iterations of newton raphson’s method (up to two decimal places) is

- 0.233
- 0.686
- 0.889
- 0.614

2. Newton raphson method is to be used to find root of equation 3x – e^{x} + sinx = 0. If the initial trial value of the roots is taken as 0.333, the next approximation for the root would be

- 0.49
- 0.68
- 0.25
- 0.36

3. The root of the function f(x) = x^{3} + x – 1 obtained after first iteration on application of newton raphson scheme using an initial guess of x_{o} = 1 is

- 0.682
- 0.686
- 0.750
- 1.000

4. Solve the equation x = 10cos(x) using the newton raphson method. The initial guess is x = π/4. The value of the predicted root after the first iteration, up to second decimal, is

- 1.5639
- 1.2546
- 1.3694
- 1.1121

5. The newton raphson method is used to solve the equation f(x) = x^{3} – 5x^{2} + 6x – 8 = 0. Taking the initial guess as x = 5, the solution obtained at the end of the first iteration is

- 4.2143
- 4.2903
- 4.1489
- 4.6387

6. Newton raphson method is used to find the roots of the equation, x^{3} + 2x^{2} + 3x -1 = 0. If the initial guess is x_{o} = 1. then the value of x after 2^{nd} iteration is

- 0.3043
- 0.2689
- 0.3598
- 0.2358

7. In Newton raphson iterative method, the initial guess value (x_{ini)} is considered as zero while finding the roots of the equation: f(x) = -2 + 6x – 4x^{2} + 0.5x^{3}. The correction, Δx, to be added to x_{ini} in the first iteration is

- 1/2
- 1/3
- 1/4
- 1/6

8. In the Newton raphson method, an initial guess of xo = 2 is made and the sequence x_{o}, x_{1}, x_{2}, …….. is obtained for the function 0.75x^{3} – 2x^{2} – 2x + 4 = 0

Consider the statements

- x
_{3}= 0 - The method converges to a solution in a finite number of iterations.

Which of the following is true?

- only i
- only ii
- both i and ii
- neither i nor ii

9. The function f(x) = e^{x} – 1 is to be solved using newton raphson method. If the initial value of x_{o} is taken as 1.0, then the absolute error observed at 2^{nd} iteration is

- 0.426
- 0.152
- 0.352
- 0.248

10. When the newton raphson method is applied to solve the equation f(x) = x^{3} + 2x – 1 = 0, the solution at the end of the first iteration with the initial guess value as x_{o} = 1.2 is

- -0.82
- 0.49
- 0.705
- 1.69

11. A numerical solution of the equation $f(x)=x+\sqrt{x}-3=0$ can be obtained using newton raphson method. If the starting value is x = 2 for the iteration, the value of x that is to be used in the next step is

- 0.306
- 0.739
- 1.694
- 2.306

12. Newton Raphson method is used to compute a root of the equation x^{2} – 13 = 0 with 3.5 as the initial value. The approximation after one iteration is

- 3.575
- 3.677
- 3.667
- 3.607

13. The newton raphson iteration $x_{n+1}=\frac{1}{2}(x_n+\frac{R}{x_n})$ can be used to compute the

- square of R
- reciprocal of R
- square root of R
- logarithm of R

14. Equation e^{x} – 1 = 0 is required to be solved using newton’s method with an initial guess x_{o} = -1. Then, after one step of newton’s method, estimate x_{1} of the solution will be given by

- 0.71828
- 0.36784
- 0.20587
- 0.00000

15. The equation x^{3} – x^{2} + 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

- 2/3
- 4/3
- 1
- 3/2

16. Starting from x_{o} = 1, one step of newton raphson method in solving the equation x^{3} + 3x – 7 = 0 gives the next value (x_{1}) as

- x
_{1}= 0.5 - x
_{1}= 1.406 - x
_{1}= 1.5 - x
_{1}= 2