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

• A) 0
• B) -2
• C) -3
• D) -1

If\,any\,three\,data\,{\text{point}}s\,are\,given,\,the\,formula\,for\,Lagrange's\,{\text{interpolation}}\,polynomial\,will\,be

• A) y = f(x) = \frac{{(x - {x_1})(x - {x_2})}}{{({x_0} - {x_1})({x_0} - {x_2})}}{y_2} + \frac{{(x - {x_0})(x - {x_2})}}{{({x_1} - {x_0})({x_1} - {x_2})}}{y_1} + \frac{{(x - {x_0})(x - {x_1})}}{{({x_2} - {x_0})({x_2} - {x_1})}}{y_0}
• B) y = f(x) = \frac{{(x - {x_1})(x - {x_2})}}{{({x_1} - {x_0})({x_1} - {x_2})}}{y_0} + \frac{{(x - {x_0})(x - {x_2})}}{{({x_0} - {x_1})({x_0} - {x_2})}}{y_1} + \frac{{(x - {x_0})(x - {x_1})}}{{({x_2} - {x_0})({x_2} - {x_1})}}{y_2}
• C) y = f(x) = \frac{{({x_0} - {x_1})({x_0} - {x_2})}}{{(x - {x_1})(x - {x_2})}}{y_0} + \frac{{({x_1} - {x_0})({x_1} - {x_2})}}{{(x - {x_0})(x - {x_2})}}{y_1} + \frac{{({x_2} - {x_0})({x_2} - {x_1})}}{{(x - {x_0})(x - {x_1})}}{y_2}
• D) y = f(x) = \frac{{(x - {x_1})(x - {x_2})}}{{({x_0} - {x_1})({x_0} - {x_2})}}{y_0} + \frac{{(x - {x_0})(x - {x_2})}}{{({x_1} - {x_0})({x_1} - {x_2})}}{y_1} + \frac{{(x - {x_0})(x - {x_1})}}{{({x_2} - {x_0})({x_2} - {x_1})}}{y_2}

\begin{gathered} What\,will\,be\,the\,value\,of\,'a'\,in\,the\,given\,divide\,difference\,table? \hfill \\ \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D}&{3rdD.D} \\\ 1&{0.4}&{0.25}&{0.0375}&{ - 0.0104} \\\ 3&{0.9}&{0.4}&a&{} \\\ 5&{1.7}&{0.3}&{}&{} \\\ 7&{2.3}&{}&{}&{} \end{array} \hfill \\\ \end{gathered}

• A) -0.0109
• B) -0.025
• C) -0.0012
• D) -0.0343

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

• A) none
• B) 59
• C) 82
• D) 76

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

• A) Four
• B) Three
• C) Five
• D) Two

\[\begin{gathered} What\,will\,be\,the\,value\,of\,'a'\,in\,the\,given\,divide\,difference\,table? \hfill \\ \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D}&{3rdD.D} \\\ 1&{0.4}&{0.25}&{0.0375}&{ - 0.0104} \\\ 3&{0.9}&{0.4}&a&{} \\\ 5&{1.7}&{0.3}&{}&{} \\\ 7&{2.3}&{}&{}&{} \end{array} \hfill \\\ \end{gathered} \]

• A) -0.0109
• B) -0.0012
• C) -0.0343
• D) -0.025

If\,only\,two\,data\,{\text{point}}s\,are\,given,\,the\,formula\,for\,Lagrange's\,{\text{interpolation}}\,polynomial\,will\,be

• A) y = f(x) = \frac{{(x - {x_0})}}{{({x_1} - {x_0})}}{y_0} + \frac{{(x - {x_1})}}{{({x_0} - {x_1})}}{y_1}
• B) y = f(x) = \frac{{(x - {x_1})}}{{({x_0} - {x_1})}}{y_0} + \frac{{(x - {x_0})}}{{({x_1} - {x_0})}}{y_1}
• C) y = f(x) = \frac{{(x - {x_0})}}{{({x_0} - {x_1})}}{y_0} + \frac{{(x - {x_1})}}{{({x_1} - {x_0})}}{y_1}
• D) y = f(x) = \frac{{({x_1} - {x_0})}}{{(x - {x_0})}}{y_0} + \frac{{({x_0} - {x_1})}}{{(x - {x_1})}}{y_1}

\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) = - 0.0389 + (x - 1)(2.2) + (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) = 2.2 + (x - 1)( - 0.0389) + (x - 1)((x - 4)(0.4333)

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

\[For\,the\,given\,data\,{\text{points}}\,(1,0.3),\,(3,1),\,and\,(5,1.2)\,\,the\,divide\,difference\,table\,will\,be\,given\,as\]

• A) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{0.3}&{0.35}&{} \\\ 4&1&{0.1}&{ - 0.125} \\\ 6&{1.2}&{}&{} \end{array}\]
• B) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{0.3}&{0.35}&{} \\\ 4&1&{0.1}&{ - 0.225} \\\ 6&{1.2}&{}&{} \end{array}\]
• C) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{0.3}&{0.35}&{} \\\ 4&1&{0.1}&{ - 0.0625} \\\ 6&{1.2}&{}&{} \end{array}\]
• D) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{0.3}&{0.35}&{} \\\ 4&1&{0.1}&{ - 0.525} \\\ 6&{1.2}&{}&{} \end{array}\]

\[\begin{gathered} What\,will\,be\,the\,value\,of\,'a'\,in\,the\,given\,divide\,difference\,table? \hfill \\\ \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D}&{3rdD.D} \\\\ 2&{0.5}&{0.3}&a&{} \\\\ 4&{1.1}&{0.3}&{ - 0.0125}&{ - 0.0021} \\\\ 6&{1.7}&{0.25}&{}&{} \\\\ 8&{2..2}&{}&{}&{} \end{array} \hfill \\\\ \end{gathered} \]

• A) 0.0893
• B) 0.0612
• C) 0.0115
• D) 0

If any ten data points are given,the degree of Lagrange's interpolation polynomial could be

• A) eleven
• B) twelve
• C) nine
• D) ten

\[\Delta = - - - \]

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

\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&3&4&6 \\\ y&{0.067}&{0.248}&{0.518} \end{array} \hfill \\\ \end{gathered}

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

\[\begin{gathered} What\,will\,be\,the\,value\,of\,'a'\,in\,the\,given\,divide\,difference\,table? \hfill \\ \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 3&{0.4}&{}&{} \\\ 6&{0.9}&{0.1667}&{} \\\ 9&{1.7}&{0.2667}&a \end{array} \hfill \\\ \end{gathered} \]

• A) 0.0349
• B) 0.0254
• C) 0.0167
• D) 0.0211

\[For\,the\,given\,data\,{\text{points}}\,(2,0.3),\,(4,1),\,and\,(6,1.2)\,\,the\,divide\,difference\,table\,will\,be\,given\,as\]

• A) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{0.3}&{0.1}&{} \\\ 4&1&{ - 0.0625}&{0.35} \\\ 6&{1.2}&{}&{} \end{array}\]
• B) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{0.3}&{0.35}&{} \\\ 4&1&{0.1}&{ - 0.0625} \\\ 6&{1.2}&{}&{} \end{array}\]
• C) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{0.3}&{ - 0.0625}&{} \\\ 4&1&{0.1}&{0.35} \\\ 6&{1.2}&{}&{} \end{array}\]
• D) \[\begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{0.3}&{ - 0.0625}&{} \\\ 4&1&{0.35}&{0.1} \\\ 6&{1.2}&{}&{} \end{array}\]

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

• A) 2
• B) 5
• C) 0
• D) 1

\[\begin{gathered} Which\,of\,the\,following\,method\,can\,be\,used\,for\,{\text{interpolation}}\,for\,the\,given\,values\,of\,x\,and\,y? \hfill \\ \begin{array}{*{20}{c}} x&3&4&6 \\\ y&{0.067}&{0.248}&{0.518} \end{array} \hfill \\\ \end{gathered} \]

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

\[\begin{gathered} For\,the\,given\,divide\,difference\,table \hfill \\ \begin{array}{*{20}{c}} x&y&{1stD.D}&{2ndD.D} \\\ 2&{2.2}&{0.325}&{} \\\ 6&{3.5}&{0.15}&{ - 0.0219} \\\ {10}&{4.1}&{}&{} \end{array} \hfill \\ the\,Newton's\,divide\,difference\,\,{\text{interpolation}}\,formula\,will\,be \hfill \\\ \end{gathered} \]

• A) \[y = f(x) = 2.2 + (x - 6)(0.325) + (x - 12)((x - 6)( - 0.0219)\]
• B) \[y = f(x) = 2.2 + (x - 2)(0.325) + (x - 2)((x - 6)( - 0.0219)\]
• C) \[y = f(x) = 2.2 + (x - 2)((x - 6)(0.325) + (x - 2)( - 0.0219)\]
• D) \[y = f(x) = 2.2 + (x - 12)(0.325) + (x - 6)((x - 2)( - 0.0219)\]

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}&{ - 0.098} \\\ {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}&{ - 0.0219} \\\ {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}&{ - 0.065} \\\ {12}&{4.1}&{}&{} \end{array}