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

Answer
Answer. b

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

Answer
Answer. d

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

Answer
Answer. c

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

Answer
Answer. a

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

Answer
Answer. b

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

Answer
Answer. a

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

Answer
Answer. b

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

Answer
Answer. a

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

Answer
Answer. d

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

Answer
Answer. c

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

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

Answer
Answer. c

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

Answer
Answer. d

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

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

Answer
Answer. c

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

Answer
Answer. a

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

Answer
Answer. b

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

 

Answer
Answer. c

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