# Newton Raphson Method MCQ

1. Starting with x = 1, the solution of the equation x3 + x = 1, after two iterations of newton raphson’s method (up to two decimal places) is

1. 0.233
2. 0.686
3. 0.889
4. 0.614

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

1. 0.49
2. 0.68
3. 0.25
4. 0.36

3. The root of the function f(x) = x3 + x – 1 obtained after first iteration on application of newton raphson scheme using an initial guess of xo = 1 is

1. 0.682
2. 0.686
3. 0.750
4. 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. 1.5639
2. 1.2546
3. 1.3694
4. 1.1121

5. The newton raphson method is used to solve the equation f(x) = x3 – 5x2 + 6x – 8 = 0. Taking the initial guess as x = 5, the solution obtained at the end of the first iteration is

1. 4.2143
2. 4.2903
3. 4.1489
4. 4.6387

6. Newton raphson method is used to find the roots of the equation, x3 + 2x2 + 3x -1 = 0. If the initial guess is xo = 1. then the value of x after 2nd iteration is

1. 0.3043
2. 0.2689
3. 0.3598
4. 0.2358

7. In Newton raphson iterative method, the initial guess value (xini) is considered as zero while finding the roots of the equation: f(x) = -2 + 6x – 4x2 + 0.5x3. The correction, Δx, to be added to xini in the first iteration is

1. 1/2
2. 1/3
3. 1/4
4. 1/6

8. In the Newton raphson method, an initial guess of xo = 2 is made and the sequence xo, x1, x2, …….. is obtained for the function 0.75x3 – 2x2 – 2x + 4 = 0

Consider the statements

1. x3 = 0
2. The method converges to a solution in a finite number of iterations.

Which of the following is true?

1. only i
2. only ii
3. both i and ii
4. neither i nor ii

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

1. 0.426
2. 0.152
3. 0.352
4. 0.248

10. When the newton raphson method is applied to solve the equation f(x) = x3 + 2x – 1 = 0, the solution at the end of the first iteration with the initial guess value as xo = 1.2 is

1. -0.82
2. 0.49
3. 0.705
4. 1.69

11. A numerical solution of the equation 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

1. 0.306
2. 0.739
3. 1.694
4. 2.306

12. Newton Raphson method is used to compute a root of the equation x2 – 13 = 0 with 3.5 as the initial value. The approximation after one iteration is

1. 3.575
2. 3.677
3. 3.667
4. 3.607

13. The newton raphson iteration can be used to compute the

1. square of R
2. reciprocal of R
3. square root of R
4. logarithm of R

14. Equation ex – 1 = 0 is required to be solved using newton’s method with an initial guess xo = -1. Then, after one step of newton’s method, estimate x1 of the solution will be given by

1. 0.71828
2. 0.36784
3. 0.20587
4. 0.00000

15. The equation x3 – x2 + 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

1. 2/3
2. 4/3
3. 1
4. 3/2

16. Starting from xo = 1, one step of newton raphson method in solving the equation x3 + 3x – 7 = 0 gives the next value (x1) as

1. x1 = 0.5
2. x1 = 1.406
3. x1 = 1.5
4. x1 = 2

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