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

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

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

\begin{gathered} What\,will\,be\,the\,value\,of\,'a'\,in\,the\,given\,divide\,diffidence\,table? \hfill \\ \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D}&{3rdD.D} \\\ 1&{0.7}&{0.25}&{0.025}&{} \\\ 3&{1.2}&{0.35}&{ - 0.0625}&a \\\ 5&{1.9}&{0.1}&{}&{} \\\ 7&{2.1}&{}&{}&{} \end{array} \hfill \\\ \end{gathered}

• A) -0.0146
• B) -0.0387
• C) -0.0021
• D) -0.0245

If y(x) is approximated by a polynomial {P_n}(x) of degree n then the error is given by

• A) \varepsilon (x) = y(x)\,\, \div \,\,\,{P_n}(x)
• B) \varepsilon (x) = y(x) - {P_n}(x)
• C) \varepsilon (x) = y(x) + {P_n}(x)
• D) \varepsilon (x) = y(x)\,\, \times \,\,{P_n}(x)

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

• A) 3
• B) -2
• C) -3
• D) 2

If there are (n+2) values of y corresponding to (n+2) values of x, then we can represent the function f(x) by a polynomial of degree

• A) n+1
• B) n+2
• C) n-1
• D) n

In Romberg’s method, accuracy of Simpson and Trapezoidal rules is improved by ---------.

• A) extrapolation
• B) interpolation
• C)
• D)

\delta \,\, = \,\, - - - -

• A) {E^{\frac{1}{2}}}\,\, - \,\,{E^{ - \,\,\,\,\frac{1}{2}}}
• B) None
• C) \frac{{{E^{\frac{1}{2}}}\,\, + \,\,\,{E^{ - \,\,\,\,\frac{1}{2}}}}}{2}
• D) {E^{\frac{1}{2}}}\,\, + \,\,\,{E^{ - \,\,\,\,\frac{1}{2}}}

\Delta = - - -

• A) \frac{{E - 1}}{2}
• B) 1 - E
• C) None
• D) E\,\, - \,\,1

\begin{gathered} For\,the\,given\,four\,data\,{\text{point}}s,\,the\,{\text{degree}}\,of\,Lagrange's\,{\text{interpolation}}\,polynomial\,could\,be \hfill \\ \begin{array}{*{20}{c}} x&{0.3}&{0.7}&{0.9}&{1.0} \\\ y&{0.067}&{0.248}&{0.518}&{0.6812} \end{array} \hfill \\\ \end{gathered}

• A) three
• B) five
• C) six
• D) four

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

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