This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Iteration Method”.

1. What will be the root of the equation f(x) = 3x^{4} – 2x^{2} using iteration method?

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation: Given,

Let f(x) = 3x

^{4}-2x

^{2}

Here 3x

^{4}-2x

^{2}= 0

3x

^{4}= 2x

^{2}

x

^{4}= ((2x

^{2})/3)

x = \(\sqrt[4]{\frac{(2x^2)}{3}}\)

\(\phi(x) = \sqrt[4]{\frac{(2x^2)}{3}}\)

Here

x |
0 | 1 |

f(x) |
0 | 1 |

Here f(0) = 0

Root of the equation 3x^{4}-2x^{2} is 0.

2. What will be the root of the equation f(x) = tan(x) using iteration method?

a) -0.0573

b) -0.0572

c) 2.4587

d) -0.0571

View Answer

Explanation: Given,

Considering f(x) = tan(x) as \(\phi(x)\) = tan(x)

We can also directly input \(\phi(x)\) value like x = value format, e.g. x = cos(x)/exp(x)

x = tan(x)

\(\phi(x)\) = tan(x)

x1 = \(\phi(x0) = \phi(2)\) = -2.185

x2 = \(\phi(x1) = \phi(-2.185)\) = 1.4179

x3 = \(\phi(x2) = \phi(1.4179)\) = 6.4906

x4 = \(\phi(x3) = \phi(6.4906)\) = 0.2104

x5 = \(\phi(x4) = \phi(0.2104)\) = 0.2136

x6 = \(\phi(x5) = \phi(0.2136)\) = 0.2169

x7 = \(\phi(x6) = \phi(0.2169)\) = 0.2203

x8 = \(\phi(x7) = \phi(0.2203)\) = 0.224

x9 = \(\phi(x8) = \phi(0.224)\) = 0.2278

x10 = \(\phi(x9) = \phi(0.2278)\) = 0.2318

x11 = \(\phi(x10) = \phi(0.2318)\) = 0.2361

We can calculate in this mode

Approximate root of the equation tan(x) using iteration method is -0.0571.

3. What will be the root of the equation f(x) = 4cos(x)-x using iteration method, with the method of

f(x) = cos(x)-x*e^{x}?

a) 0.1282

b) It is divergent

c) 0.1284

d) -0.8723

View Answer

Explanation: Given,

Considering f(x) = 4cos(x)-x as \(\phi(x)\) = 4cos(x)-x

We can also directly input \(\phi(x)\) value like x = value format, e.g. x = cos(x)/exp(x)

x = 4cos(x)-x

\(\phi(x) = 4cos(x)-x\)

x1 = \(\phi(x0) = \phi(2)\) = -3.6646

x2 = \(\phi(x1) = \phi(-3.6646)\) = 0.1993

x3 = \(\phi(x2) = \phi(0.1993)\) = 3.7216

x4 = \(\phi(x3) = \phi(3.7216)\) = -7.0675

x5 = \(\phi(x4) = \phi(-7.0675)\) = 9.899

x6 = \(\phi(x5) = \phi(9.899)\) = -13.4576

x7 = \(\phi(x6) = \phi(-13.4576)\) = 15.9715

x8 = \(\phi(x7) = \phi(15.9715)\) = -19.8334

x9 = \(\phi(x8) = \phi(-19.8334)\) = 22.0487

x10 = \(\phi(9) = \phi(22.0487)\) = -26.0421

x11 = \(\phi(x10) = \phi(-26.0421)\) = 28.4991

x12 = \(\phi(x11) = \phi(28.4991)\) = -32.3985

x13 = \(\phi(x12) = \phi(-32.3985)\) = 34.6181

x14 = \(\phi(x13) = \phi(34.6181)\) = -38.6107

x15 = \(\phi(x14) = \phi(-38.6107)\) = 41.0606

x16 = \(\phi(x15) = \phi(41.0606)\) = -44.9643

x17 = \(\phi(x16) = \phi(-44.9643)\) = 47.1858

x18 = \(\phi(x17) = \phi(47.1858)\) = -51.1781

x19 = \(\phi(x18) = \phi(-51.1781)\) = 53.6248

x20 = \(\phi(x19) = \phi(53.6248)\) = -57.5304

x21 = \(\phi(x20) = \phi(-57.5304)\) = 59.7528

x22 = \(\phi(x21) = \phi(59.7528)\) = -63.745

x23 = \(\phi(x22) = \phi(-63.745)\) = 66.19

x24 = \(\phi(x23) = \phi(66.19)\) = -70.0966

We can calculate this mode

The process clue us away from the solution, so it is divergent.

4. What will be the root of the equation f(x) = x^{4}+2x^{3}+x^{2}+x-1 using iteration method?

a) 0.4848

b) 0.4835

c) 0.4838

d) -0.4835

View Answer

Explanation: Given,

Let f(x) = x

^{4}+2x

^{3}+x

^{2}+x-1

Here x

^{4}+2x

^{3}+x

^{2}+x-1 = 0

x

^{4}+2x

^{3}+x

^{2}+x = 1

x(x

^{3}+2x

^{2}+x+1) = 1

x= 1/(x

^{3}+2x

^{2}+x+1)

\(\phi(x)\) = 1/(x

^{3}+2x

^{2}+x+1)

Here

x |
0 | 1 |

f(x) |
-1 | 4 |

Here f(0) = -1<0 and f(1) = 4>0

Root lies between 0 and 1.

x0 = 0+12 = 0.5

x1 = \(\phi(x0) = \phi(0.5)\) = 0.4706

x2 = \(\phi(x1) = \phi(0.4706)\) = 0.4956

x3 = \(\phi(x2) = \phi(0.4956)\) = 0.4742

x4 = \(\phi(x3) = \phi(0.4742)\) = 0.4924

x5 = \(\phi(x4) = \phi(0.4924)\) = 0.4769

x6 = \(\phi(x5) = \phi(0.4769)\) = 0.4901

x7 = \(\phi(x6) = \phi(0.4901)\) = 0.4789

x8 = \(\phi(x7) = \phi(0.4789)\) = 0.4885

x9 = \(\phi(x8) = \phi(0.4885)\) = 0.4803

x10 = \(\phi(x9) = \phi(0.4803)\) = 0.4872

x11 = \(\phi(x10) = \phi(0.4872)\) = 0.4813

x12 = \(\phi(x11) = \phi(0.4813)\) = 0.4864

x13 = \(\phi(x12) = \phi(0.4864)\) = 0.482

x14 = \(\phi(x13) = \phi(0.482)\) = 0.4857

x15 = \(\phi(x14) = \phi(0.4857)\) = 0.4826

x16 = \(\phi(x15) = \phi(0.4826)\) = 0.4853

x17 = \(\phi(x16) = \phi(0.4853)\) = 0.483

x18 = \(\phi(x17) = \phi(0.483)\) = 0.4849

x19 = \(\phi(x18) = \phi(0.4849)\) = 0.4833

x20 = \(\phi(x19) = \phi(0.4833)\) = 0.4847

x21 = \(\phi(x20) = \phi(0.4847)\) = 0.4835

x22 = \(\phi(x21) = \phi(0.4835)\) = 0.4845

x23 = \(\phi(x22) = \phi(0.4845)\) = 0.4836

x24 = \(\phi(x23) = \phi(0.4836)\) = 0.4844

We can analyze in this custom

Approximate root of the equation x^{4}+2x^{3}+x^{2}+x-1 using iteration method is 0.4838.

5. What will be the root of the equation f(x) = 2^{x}-x-1.5 using iteration method?

a) 0.3602

b) 0.3607

c) -0.3602

d) -0.3607

View Answer

Explanation: Given,

Considering f(x) = 2

^{x}-x-1.5 as \(\phi(x)\) = -x+2

^{x}-1.5

We can also directly input \(\phi(x)\) value like x = value format, e.g. x= cos(x)/exp(x).

x = -x+2

^{x}-1.5

\(\phi(x)\) = -x+2

^{x}-1.5

x1 = \(\phi(x0) = \phi(2)\) = 0.5

x2 = \(\phi(x1) = \phi(0.5)\) = -0.5858

x3 = \(\phi(x2) = \phi(-0.5858)\) = -0.2479

x4 = \(\phi(x3) = \phi(-0.2479)\) = -0.41

x5 = \(\phi(x4) = \phi(-0.41)\) = -0.3374

x6 = \(\phi(x5) = \phi(-0.3374)\) = -0.3711

x7 = \(\phi(x6) = \phi(-0.3711)\) = -0.3557

x8 = \(\phi(x7) = \phi(-0.3557)\) = -0.3628

x9 = \(\phi(x8) = \phi(-0.3628)\) = -0.3595

x10 = \(\phi(x9) = \phi(-0.3595)\) = -0.361

x11 = \(\phi(x10) = \phi(-0.361)\) = -0.3604

x12 = \(\phi(x11) = \phi(-0.3604)\) = -0.3607

Approximate root of the equation 2^{x}-x-1.5 using iteration method is -0.3607.

6. What will be the root of the equation f(x) = x^{3}-x-1 using iteration method, using the method

x = \(\phi(x)\) = cos(x)/e^{x}?

a) -0.3595

b) 0.3595

c) It is divergent

d) -0.3604

View Answer

Explanation: Given,

x = x

^{3}-x-1

\(\phi(x)\) = x

^{3}-x-1

x1 = \(\phi(x0) = \phi(2) \) = 5

x2 = \(\phi(x1) = \phi(5) \) = 119

x3 = \(\phi(x2) = \phi(119) \) = 1685039

The method lead us away from the solution, so it is divergent.

n | x0 | x1 = \(\phi(x0)\) | Update | Difference x1-x0 |
---|---|---|---|---|

2 | 2 | 5 | x0 = x1 | 3 |

3 | 5 | 119 | x0 = x1 | 114 |

4 | 119 | 1685039 | x0 = x1 | 1684920 |

7. What will be the root of the equation f(x) = sin(2)(x) + 1/tan(x) using iteration method?

a) 1.8568

b) 1.9564

c) -1.9568

d) It is divergent

View Answer

Explanation: Given,

Considering f(x) = sin2(x) + 1/tan(x) as \(\phi(x)\) = sin2(x) + 1/tan(x)

We can also directly input \(\phi(x)\) value like x = value format, e.g. x = cos(x)/exp(x)

x = sin2(x) + 1/tan(x)

\(\phi(x)\) = sin2(x) + 1/tan(x)

x1 = \(\phi(x0) = \phi(2)\) = 2.9257

x2 = \(\phi(x1) = \phi(2.9257)\) = 3.9018

x3 = \(\phi(x2) = \phi(3.9018)\) = 4.8678

x4 = \(\phi(x3) = \phi(4.8678)\) = 5.7945

x5 = \(\phi(x4) = \phi(5.7945)\) = 6.6688

x6 = \(\phi(x5) = \phi(6.6688)\) = 7.4859

x7 = \(\phi(x6) = \phi(7.4859)\) = 8.2449

x8 = \(\phi(x7) = \phi(8.2449)\) = 8.9472

x9 = \(\phi(x8) = \phi(8.9472)\) = 9.5951

x10 = \(\phi(x9) = \phi(9.5951)\) = 10.1918

x11 = \(\phi(x10) = \phi(10.1918)\) = 10.7403

x12 = \(\phi(x11) = \phi(10.7403)\) = 11.2441

x13 = \(\phi(x12) = \phi(11.2441)\) = 11.7063

x14 = \(\phi(x13) = \phi(11.7063)\) = 12.1301

x15 = \(\phi(x14) = \phi(12.1301)\) = 12.5184

x16 = \(\phi(x15) = \phi(12.5184)\) = 12.8741

x17 = \(\phi(x16) = \phi(12.8741)\) = 13.1996

x18 = \(\phi(x17) = \phi(13.1996)\) = 13.4976

x19 = \(\phi(x18) = \phi(13.4976)\) = 13.7702

x20 = \(\phi(x19) = \phi(13.7702)\) = 14.0195

x21 = \(\phi(x20) = \phi(14.0195)\) = 14.2475

x22 = \(\phi(x21) = \phi(14.2475)\) = 14.4559

x23 = \(\phi(x22) = \phi(14.4559)\) = 14.6464

x24 = \(\phi(x23) = \phi(14.6464)\) = 14.8206

The method lead us away from the solution, so it is divergent.

8. What will be the root of the equation f(x) = sin3(x) + x^{3} – exp^{(2x)} using iteration method?

a) It is divergent

b) 0.4811

c) -0.4814

d) -0.4825

View Answer

Explanation: Given,

Considering f(x) = sin3(x) + x

^{3}– exp

^{(2x)}as \(\phi(x)\) = x

^{3}+ sin3(x) – e

^{(2x)}

We can also directly input \(\phi(x)\) value like x = value format, e.g. x = cos(x)/exp(x)

x = x^{3} + sin3(x) – e^{(2x)}

\(\phi(x)\) = x^{3} + sin3(x) – e^{(2x)}

x1 = \(\phi(x0) = \phi(2)\) = -46.3159

x2 = \(\phi(x1) = \phi(-46.3159)\) = -99361.7367

The method chief us away from the solution, so it is divergent.

n | x0 | x1 = \(\phi(x0)\) | Update | Difference x1-x0 |
---|---|---|---|---|

2 | 2 | -46.3159 | x0 = x1 | 48.3159 |

3 | -46.3159 | -99361.7367 | x0 = x1 | 99315.4208 |

9. Iteration method is used for the purpose of approximating the solutions.

a) False

b) True

View Answer

Explanation: Iteration method is used for the calculation of approximate solution of problems which are connected with the differential equation. As the name suggests the solution can be found by the iteration procedure for a closer accuracy.

10. When should one stop the computation of iteration in the iteration method?

a) Until |f(xi)-f(xi-1)| \(\approx\) 0

b) Until |f(xi)-f(xi-1)| < 0

c) Until |f(xi)-f(xi-1)| > 0

d) Until |f(xi)-f(xi)| < 0

View Answer

Explanation: Firstly, we need to write the equation x = \(\phi(x)\). Then, find points a and b such that a < b and f(a).f(b) < 0. Later, if f(a) is more closer to 0 than f(b), then x0 = a else x0 = b.

Lastly, x1 = \(\phi(x0)\)

x2 = \(\phi(x1)\)

x3 = \(\phi(x2)\)

…

Repeat until |f(xi)-f(xi-1)| \(\approx\) 0.

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