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

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

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

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

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

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

----------- difference is a symmetric function of its arguments.

• A) Backward
• B) None
• C) Forward
• D) Divided

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

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

At which of the following point the derivatives or slopes the functions f(x) = x – 2 and g(x) = x + 2 may differ?

• A) does not differ for any value of ‘x’
• B) x = -2
• C) x = 2
• D) differ for every value of ‘x’

The percentage error in numerical integration is defined as

• A) = (Theoretical Value-Experiment Value)/ Experiment Value*100
• B) = (Theoretical Value-Experiment Value)* Experiment Value*100
• C) = (Theoretical Value +Experiment Value)/ Experiment Value*100
• D) = (Theoretical Value-Experiment Value)/ Theoretical Value *100

We prefer ………over the Lagrange’s interpolating method for economy of computation.

• A) Newton’s forward difference method
• B) Newton’s backward difference method
• C) Newton’s divided difference method
• D) None of the given choices

Which of the following reason(s) lead towards the numerical integration methods?

• A) All above choices are true
• B) Analytical evaluation of integral is very complicated
• C) Analytical evaluation of integral is impossible
• D) Integrand is given in tabular form

Which of the following is the Richardson’s Extrapolation limit: F3(h/8) provided that F2(h/8) = F2(h/4) = 1 ?

• A) 63
• B) -1
• C) 64
• D) 1

The idea of Richardson’s extrapolation is to combine two computed values of derivative of y using the same method but with ……… different step sizes.

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

The single definite integral of a function is called ………

• A) Area under the curve
• B) Length of the curve
• C) None of the given choices
• D) Volume of the curve

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

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

Simpson’s 3/8 rule is based on fitting ……………… points by a cubic.

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

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

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

Which of the following is the Richardson’s Extrapolation limit: F2(h/4) provided that F1(h/4) = F1(h/2) = 1 ?

• A) 1
• B) 16
• C) 15
• D) -1

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

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

In Newton-Cotes formula for finding the definite integral of a tabular function, which of the following is taken as an approximate function then find the desired integral?

• A) Exponential Function
• B) Logarithmic Function
• C) Polynomial Function
• D) Trigonometric Function

Given the following data x:1 3 8 y:2 4 9 f(3) can be found by using

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

If the area under ‘f(x) = x’ in interval [0,2] is subdivided into two equal sub-intervals of width ‘1’ with left end points, then which of the following will be the Truncation Error provided that I(definite integral) = 2 and approximate sum = 3 ?

• A) 1
• B) 3
• C) 0
• D) -1