Given the following data x:1 2 5 y:1 4 10 Value of 1st order divided difference f[2 , 5] is

• A) 2
• B) 1
• C) 0
• D) -2

Given the following data x: 1 3 -7 f(x): 3 -6 -2 Which formula is useful in finding the interpolating polynomial?

• A) Newton’s backward difference formula
• B) None of the given choices
• C) Newton’s forward difference formula
• D) Lagrange’s interpolation formula

x: 1 3 7 10 f(x): -3 10 3 13 Which formula is useful in finding the interpolating polynomial?

• A) Lagrange’s interpolation formula
• B) (viii) None of the given choices
• C) Newton’s backward difference formula
• D) Newton’s forward difference formula

Trapezoidal and Simpson’s integrations are just a linear combination of values of the
given function at different values of the …………variable.

• A) Dependent
• B) Independent
• C) Arbitrary
• D) None of the given choices

If f(x) = 2{x^3} - 5{x^2} + 9x - 6 , then its------derivative is zero for all x.

• A) 2nd
• B) 4th
• C) 5th
• D) 3rd

\begin{gathered} Which\,of\,the\,following\,method\,can\,be\,used\,for\,{\text{interpolation}}\,for\,the\,given\,values\,of\,x\,and\,y? \hfill \\ \begin{array}{*{20}{c}} x&{0.3}&{0.7}&{0.9} \\\ y&{0.067}&{0.248}&{0.518} \end{array} \hfill \\\ \end{gathered}

• A) Newton’s interpolation formula
• B) Newton’s forward difference formula
• C) Lagrange’s interpolation formula
• D) Newton’s backward difference formula

For\,the\,given\,data\,{\text{points}}\,(4,1.3),\,(8,1.5),\,and\,(12,1.9)\,\,the\,divide\,difference\,table\,will\,be\,given\,as

• A) \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 4&{1.3}&{0.05}&{} \\\ 8&{1.5}&{0.1}&{0.0062} \\\ {12}&{1.9}&{}&{} \end{array}
• B) \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 4&{1.3}&{0.0062}&{} \\\ 8&{1.5}&{0.1}&{0.05} \\\ {12}&{1.9}&{}&{} \end{array}
• C) \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 4&{1.3}&{0.1}&{} \\\ 8&{1.5}&{0.35}&{0.0062} \\\ {12}&{1.9}&{}&{} \end{array}
• D) \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 4&{1.3}&{0.1}&{} \\\ 8&{1.5}&{0.0062}&{0.05} \\\ {12}&{1.9}&{}&{} \end{array}

\[For\,the\,given\,data\,{\text{points}}\,({x_{0,}}{y_0}),\,({x_1}{y_1}),\,({x_2}{y_2}),\,and\,({x_{3,}}{y_3})\,\,the\,first - order\,divide\,difference\,will\,be\,given\,as\]

• A) \[y[{y_0},{y_1},{y_2}]\]
• B) \[y[{x_0},{x_1}]\]
• C) \[y[{x_0}]\]
• D) \[y[{x_0},{x_1},{x_2}]\]

Given the following data x: 0 1 2 4 y: 1 1 2 6 Value of 1st order divided difference f[2,4] is

• A) 2
• B) 0
• C) 1
• D) -2

For\,the\,given\,data\,{\text{points}}\,(1, - 3),\,(2,0),\,and\,(3,15),\,the\,zero - order\,divide\,difference\,will\,be\,

• A) 0
• B) -1
• C) -3
• D) -2

For\,the\,given\,data\,{\text{points}}\,(4,45),\,(5,104),\,and\,(6,190),\,the\,{\text{first}} - order\,divide\,difference\,will\,be\,

• A) 82
• B) 59
• C) 76
• D) none

In Lagrange’s interpolation, for the given five points we can represent the function f (x) by a polynomial of degree

• A) 4
• B) 3
• C) 5
• D) 6

The\,first\,divide\,difference\,y[{x_0},{x_1}]\,can\,be\,given\,as\,

• A) \frac{{\nabla {y_1}}}{h}
• B) \frac{{{y_1} - {y_0}}}{{{x_1} - {x_0}}}
• C) \frac{{\Delta {y_0}}}{h}
• D) All

Let \[f(x,y) = 2{x^3} + 6{y^3} + 9xy\] For x=0, 1,2,3,4, and y=0, 1,2,3,4 Then computing the value of f(2.5,3.5) is an example of .......

• A) Interpolation in three dimensions
• B) Interpolation in two dimensions
• C) Interpolation in one dimension
• D) Interpolation in four dimensions

\[\begin{gathered} For\,the\,given\,divide\,difference\,table \hfill \\ \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 1&{2.2}&{0.4333}&{} \\\ 4&{3.5}&{0.2}&{ - 0.0389} \\\ 7&{4.1}&{}&{} \end{array} \hfill \\ the\,Newton's\,divide\,difference\,\,{\text{interpolation}}\,formula\,will\,be \hfill \\\ \end{gathered} \]

• A) \[y = f(x) = 2.2 + (x - 1)( - 0.0389) + (x - 1)((x - 4)(0.4333)\]
• B) \[y = f(x) = - 0.0389 + (x - 1)(0.4333) + (x - 1)((x - 4)(2.2)\]
• C) \[y = f(x) = 2.2 + (x - 1)(0.4333) + (x - 1)((x - 4)( - 0.0389)\]
• D) \[y = f(x) = - 0.0389 + (x - 1)(2.2) + (x - 1)((x - 4)(0.4333)\]

For\,the\,given\,data\,{\text{points}}\,(1, - 3),\,(2,0),\,and\,(3,15),\,the\,first - order\,divide\,difference\,will\,be\,

• A) -2
• B) 3
• C) 2
• D) -3

\[For\,the\,given\,data\,{\text{points}}\,(4,2.2),\,(8,3.5),\,and\,(12,4.1)\,\,the\,divide\,difference\,table\,will\,be\,given\,as\]

• A) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 4&{2.2}&{0.325}&{} \\\ 8&{3.5}&{0.15}&{ - 0.065} \\\ {12}&{4.1}&{}&{} \end{array}\]
• B) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 4&{2.2}&{0.325}&{} \\\ 8&{3.5}&{0.15}&{ - 0.0108} \\\ {12}&{4.1}&{}&{} \end{array}\]
• C) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 4&{2.2}&{0.325}&{} \\\ 8&{3.5}&{0.15}&{ - 0.098} \\\ {12}&{4.1}&{}&{} \end{array}\]
• D) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 4&{2.2}&{0.325}&{} \\\ 8&{3.5}&{0.15}&{ - 0.0219} \\\ {12}&{4.1}&{}&{} \end{array}\]

Lagrange’s interpolation formula is used when the values of the independent variable are

• A) Equally spaced
• B) None
• C) Constant
• D) Not equally spaced

To evaluate a definite integral of tabular function f(x), piecewise quardratic approximation led to ---------

• A) Trapezoidal Method
• B) Simpson’s Rule
• C)
• D)

\[For\,the\,given\,data\,{\text{points}}\,(2,5),\,(4,7),\,and\,(6,9),\,the\,{\text{first}} - order\,divide\,difference\,will\,be\,\]

• A) 1
• B) 0
• C) 5
• D) 2