At which of the following points the Minimum value of 2nd derivative of function
f(x) = -(2/x) in the interval:[1,4] exits?

• A) At x=2
• B) At x=3
• C) At x=4
• D) At x=1

Simpson’s rule is a numerical method that approximates the value of a definite integral by using …………polynomials.

• A) Quadratic
• B) None of the given choices
• C) Linear
• D) Cubic

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

• A) 2nd
• B) 5th
• C) 4th
• D) 3rd

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

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

• A) Simpson’s 3/8 Rule
• B) Trapezoidal Method
• C) Romberg’s Method
• D) Simpson’s 1/3 Rule

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

While employing Trapezoidal and Simpson Rules to evaluate the double integral numerically, by using Trapezoidal and Simpson rule with respect to -------- variable/variables at time

• A) both
• B) single
• C)
• D)

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

• A) 2
• B) 1
• C) 5
• D) 0

Given the following data x:4 5 7 10 y:46 102 294 346 Value of 1st order divided difference f[5 , 7] is

• A) 94
• B) 96
• C) 91
• D) 92

\begin{gathered} For\,the\,giev\,two\,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}&{} \\\ y&{0.067}&{0.248}&{} \end{array} \hfill \\\ \end{gathered}

• A) one
• B) two
• C) four
• D) three

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

• 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