This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Approximation of Functions using Least Square Method”.

1. Fit the straight line to the following data.

a) y=x

b) y=x+1

c) y=2x

d) y=2x+1

View Answer

Explanation: The normal equation are:

Σy = aΣx + nb

and

Σxy = aΣx

^{2}+ bΣx

Now,

x | y | x^{2} |
xy |

1 | 1 | 1 | 1 |

2 | 2 | 4 | 4 |

3 | 3 | 9 | 9 |

4 | 4 | 16 | 16 |

5 | 5 | 25 | 25 |

Σx = 15 | Σy = 15 | Σx^{2} = 55 |
Σxy = 55 |

Substituting in the equations,

15 = 15a + 4b and 55 = 55a + 15b

Solving these two equatons, we get a=1 and b=0,

Therefore the required straight line equation is y=x.

2. Fit the straight line to the following data.

x | 0 | 5 | 10 | 15 | 20 |

y | 7 | 11 | 16 | 20 | 26 |

a) y = 0.94x + 6.6

b) y = 6.6x + 0.94

c) y = 0.04x + 5.6

d) y = 5.6x + 0.04

View Answer

Explanation: First drawing the table,

x | y | x^{2} |
xy |

0 | 7 | 0 | 0 |

5 | 11 | 25 | 55 |

10 | 16 | 100 | 160 |

15 | 20 | 225 | 300 |

20 | 26 | 400 | 520 |

Σx = 50 | Σy = 26 | Σx^{2} = 750 |
Σxy = 1035 |

The normal equation are:

Σy = aΣx + nb

and

Σxy = aΣx^{2} + bΣx.

Substituting the values, we get,

80 = 50a + 5b

1035 = 750 a + 50 b

Solving them, we get a = 0.94 and b = 6.6.

Therefore the straight line equation is y=0.94x + 6.6.

3. Fit the straight line curve to the following data.

x | 75 | 80 | 93 | 65 | 87 | 71 | 98 | 68 | 84 | 77 |

y | 82 | 78 | 86 | 72 | 91 | 80 | 95 | 72 | 89 | 74 |

a) y = 0.9288x + 7.78155

b) y = 7.78155x + 0.9288

c) y = 0.8288x + 6.78155

d) y = 6.78155x + 0.8288

View Answer

Explanation: First drawing the table,

x | y | x^{2} |
xy |

75 | 82 | 5625 | 6150 |

80 | 78 | 6400 | 6240 |

93 | 86 | 8349 | 7998 |

65 | 72 | 4225 | 4680 |

87 | 91 | 7569 | 7917 |

71 | 80 | 5041 | 5680 |

98 | 95 | 9605 | 9310 |

68 | 72 | 4624 | 4896 |

84 | 89 | 7056 | 7476 |

77 | 74 | 5929 | 5698 |

798 | 819 | 64422 | 66045 |

The normal equation are:

Σy = aΣx + nb

and

Σxy = aΣx^{2} + bΣx.

Substituting the values, we get,

819 = 798a + 10b

66045 = 64422a + 798b

Solving, we get

a = 0.9288 and b = 7.78155

Therefore, the straight line equation is :

y = 0.9288x + 7.78155.

4. Fit a second degree parabola to the following data.

x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

y | 2 | 6 | 7 | 8 | 10 | 11 | 11 | 10 | 9 |

a) y = -0.2673x^{2} + 3.5232x – 0.9286

b) y = 0.2673x^{2} + 3.5232x – 0.9286

c) y = 0.2673x^{2} + 3.5232x + 0.9286

d) y = -0.2673x^{2} + 3.5232x + 0.9286

View Answer

Explanation: Here,

x | y | x^{2} |
x^{3} |
x^{4} |
xy | x^{2}y |

1 | 2 | 1 | 1 | 1 | 2 | 2 |

2 | 6 | 4 | 8 | 16 | 12 | 24 |

3 | 7 | 9 | 27 | 81 | 21 | 63 |

4 | 8 | 16 | 64 | 256 | 32 | 128 |

5 | 10 | 25 | 125 | 625 | 50 | 250 |

6 | 11 | 36 | 216 | 1296 | 66 | 396 |

7 | 11 | 49 | 343 | 2401 | 77 | 539 |

8 | 10 | 64 | 512 | 4096 | 80 | 640 |

9 | 9 | 81 | 729 | 6561 | 81 | 729 |

45 | 74 | 285 | 2025 | 15333 | 421 | 2771 |

The normal equations are:

Σy = aΣx^{2} + bΣx + nc

Σxy = aΣx^{3} + bΣx^{2} +cΣx

Σx^{2}y = aΣx^{4} + bΣx^{3} + cΣx^{2}

Substituting the values, we get

74 = 285a + 45b + 9c

421 = 2025 a + 285 b + 45 c

2771 = 15333a + 2025 b + 285 c

Solving them, we get the second order equation which is,

y = -0.2673x^{2} + 3.5232x – 0.9286.

5. The normal equations for a straight line y = ax + b are:

a) Σy = aΣx + nb and Σxy = aΣx^{2} + bΣx

b) Σxy = aΣx + nb and Σy = aΣx^{2} + bΣx

c) Σy = aΣx + nb and Σxy = aΣx^{2} + bΣxy

d) Σy = aΣx + nb and Σx2y = aΣx^{2} + bΣx

View Answer

Answer: a

Explanation: Let the sum of residues be E.

E = Σd_{i}^{2} = Σ(y_{i} – (ax_{i}+b))^{2}

Here (frac{∂E}{∂a})=0 and (frac{∂E}{∂b})=0

Solving these two equations, we get the normal equations as

Σy = aΣx + nb and Σxy = aΣx^{2} + bΣx.

6.The normal equations for a second degree parabola y = ax^{2} + bx + c are Σy = aΣx^{2} + bΣx + nc, Σxy = aΣx^{3} + bΣx^{2} + cΣx and Σx^{2}y = aΣx^{4} + bΣx^{3} + cΣx^{2}.. Is it true or false?

a) True

b) False

View Answer

Answer: a

Explanation: The second order parabola is given by

y = ax^{2} + bx +c.

Let the sum of residues be E.

E = Σd_{i}^{2} = Σ(y_{i} – (ax_{i}^{2} + bxi +c))^{2}.

Here (frac{∂E}{∂a})= 0, (frac{∂E}{∂b})= 0 and (frac{∂E}{∂c})= 0.

Solving these three equations, we get the normal equations as

Σy = aΣx^{2} + bΣx + nc, Σxy = aΣx^{3} + bΣx^{2} + cΣx and Σx^{2}y = aΣx^{4} + bΣx^{3} +cΣx^{2}.

7. If the equation y = ae^{bx} can be written in linear form Y=A + BX, what are Y, X, A, B?

a) Y = logy, A = loga, B=b and X=x

b) Y = y, A = a, B=b and X=x

c) Y = y, A = a, B=logb and X=logx

d) Y = logy, A = a, B=logb and X=x

View Answer

Answer: a

Explanation: The equation is

y = ae^{bx}.

Taking log to the base e on both sides,

we get logy = loga + bx.

Which can be replaced as Y=A+BX,

where Y = logy, A = loga, B = b and X = x.

8. If the equation y=ab^{x} can be written in linear form Y=A+BX, what are Y, X, A, B?

a) Y=logy, X=x, A=loga and B=logb

b) Y=y, A=a, B=b and X=x

c) Y=y, A=a, B=logb and X=logx

d) Y=logy, A=a, B=logb and X=x

View Answer

Answer: a

Explanation: The given curve is y=ab^{x}.

Taking log on bothe the sides, we get,

logy = loga + x logb.

This can be written in the format of Y=A+BX

where

Y = logy, X = x, A = loga and B = logb.

9. If the equation y=ax^{b} can be written in the linear form Y=A+BX, what are Y, X, A, B?

a) Y=logy, A=loga, B=b and X=logx

b) Y=y, A=a, B=b and X=x

c) Y=y, A=a, B=logb and X=logx

d) Y=logy, A=a, B=logb and X=x

View Answer

Answer: a

Explanation: The given curve is y=ax^{b}.

Taking log on bothe the sides, we get,

logy = loga + blogx.

This can be written as Y=A+BX,

where

Y=logy, A=loga, B=b and X=logx.

10. The parameter E which we use for least square method is called as ____________

a) Sum of residues

b) Residues

c) Error

d) Sum of errors

View Answer

Answer: a

Explanation: E is given by

E = Σdi^{2} = Σ(yi – f(xi))^{2}.

Where the term inside the summation is called as residues and the sum is said to be a sum of residues.

Therefore, E is said to be the sum of residues.

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