\[\begin{gathered} For\,the\,following\,data \hfill \\ \begin{array}{*{20}{c}} x&1&2&5&7&8 \\\ y&{ - 5}&{10}&{20}&{22}&{24} \end{array} \hfill \\ the\,polynomial\,of\,the\,Lagrangens\,interpolation\,could\,be \hfill \\\ \end{gathered} \]

• A) $ - \frac{{1121}}{{80}}{x^5} + \frac{{41}}{7}{x^3} - \frac{{163}}{{40}}{x^2} + \frac{{19}}{{81}}x - \frac{3}{{10}}$
• B) $\frac{4}{{41}}{x^7} - \frac{{43}}{7}{x^2} + \frac{{65}}{{28}}x - \frac{{186}}{5}$
• C) $ - \frac{{11}}{{280}}{x^4} + \frac{{419}}{{420}}{x^3} - \frac{{7603}}{{840}}{x^2} + \frac{{15019}}{{420}}x - \frac{{98}}{3}$
• D) ${x^6} - \frac{1}{{56}}{x^5} + \frac{{47}}{5}{x^4} - \frac{{67}}{{90}}x + \frac{2}{5}$

To evaluate numerically a double integral over a rectangular region bounded by the lines x = a, x =b, y = c, y = d we shall employ either trapezoidal rule or Simpson’s rule, repeatedly with respect to ………variable at a time.

• A) Three
• B) One
• C) None of the given choices
• D) Two

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

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

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

• A) 5
• B) 2
• C) 0
• D) 1

\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) y = f(x) = \frac{{(x - 0.7)(x - 0.9)}}{{(0.3 - 0.7)(0.3 - 0.9)}}(0.067) + \frac{{(x - 0.3)(x - 0.9)}}{{(0.7 - 0.3)(0.7 - 0.9)}}(0.248) + \frac{{(x - 0.3)(x - 0.7)}}{{(0.9 - 0.3)(0.9 - 0.7)}}(0.518)
• B) y = f(x) = \frac{{(x - 0.7)(x - 0.9)}}{{(0.3 - 0.7)(0.3 - 0.9)}}(0.518) + \frac{{(x - 0.3)(x - 0.9)}}{{(0.7 - 0.3)(0.7 - 0.9)}}(0.248) + \frac{{(x - 0.3)(x - 0.7)}}{{(0.9 - 0.3)(0.9 - 0.7)}}(0.067)
• C) y = f(x) = \frac{{(x - 0.7)(x - 0.9)}}{{(0.3 - 0.7)(0.3 - 0.9)}}(0.067) + \frac{{(x - 0.3)(x - 0.9)}}{{(0.7 - 0.3)(0.7 - 0.9)}}(0.518) + \frac{{(x - 0.3)(x - 0.7)}}{{(0.9 - 0.3)(0.9 - 0.7)}}(0.248)
• D) y = f(x) = \frac{{(x - 0.7)(x - 0.9)}}{{(0.3 - 0.7)(0.3 - 0.9)}}(0.248) + \frac{{(x - 0.3)(x - 0.9)}}{{(0.7 - 0.3)(0.7 - 0.9)}}(0.067) + \frac{{(x - 0.3)(x - 0.7)}}{{(0.9 - 0.3)(0.9 - 0.7)}}(0.518)

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

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

• A) 35
• B) 42
• C) none
• D) 46

x: 1 3 7 f(x): 1 4 9 f(3) Can be found using

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

\begin{gathered} For\,the\,following\,data \hfill \\ \begin{array}{*{20}{c}} x&1&2&5&7&8 \\\ y&{ - 5}&{10}&{20}&{22}&{24} \end{array} \hfill \\ the\,polynomial\,of\,the\,Lagrangens\,interpolation\,could\,be \hfill \\\ \end{gathered}

• A) {x^6} - \frac{1}{{56}}{x^5} + \frac{{47}}{5}{x^4} - \frac{{67}}{{90}}x + \frac{2}{5}
• B) \frac{4}{{41}}{x^7} - \frac{{43}}{7}{x^2} + \frac{{65}}{{28}}x - \frac{{186}}{5}
• C) - \frac{{11}}{{280}}{x^4} + \frac{{419}}{{420}}{x^3} - \frac{{7603}}{{840}}{x^2} + \frac{{15019}}{{420}}x - \frac{{98}}{3}
• D) - \frac{{1121}}{{80}}{x^5} + \frac{{41}}{7}{x^3} - \frac{{163}}{{40}}{x^2} + \frac{{19}}{{81}}x - \frac{3}{{10}}

Given the following data x:0 1 4 8 y:1 1 8 16 Value of 1st order divided difference f[4,8] is

• A) 2
• B) 8
• C) 4
• D) 6

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

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

Trapezoidal rule of integration of a definite integral is of…………

• A) O(h3)
• B) None of the given choices
• C) O(h2)
• D) O(h4)

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

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

What will be the value of first order divided difference f[1,5]for the following data x:0 1 5 y:2 1 5

• A) 2
• B) 0
• C) 3
• D) 1

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 four dimensions
• D) Interpolation in one dimension

We can improve the accuracy of trapezoidal and Simpson’s rules using ……

• A) Simpson’s 1/3 rule
• B) None of the given choices
• C) Simpson’s 3/8 rule
• D) Richardson’s extrapolation method

In Lagrange’s interpolation, for n values of y corresponding to n values of x, we can represent the function f (x) by a polynomial of degree

• A) n+1
• B) n-1
• C) n
• D) n+2

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}&{ - 0.0625}&{} \\\ 4&1&{0.1}&{0.35} \\\ 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}&{ - 0.0625}&{} \\\ 4&1&{0.35}&{0.1} \\\ 6&{1.2}&{}&{} \end{array}

Newton’s divided difference interpolation formula is used when the values of the independent variable are

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

In double integration, we keep one variable say x fixed and ……………

• A) Varying the other variable y
• B) Reliable the other variable y
• C)
• D)