Linked Answer Questions 1 and 2.
Give a > 0, we wish to calculate its reciprocal value 1/a by using Newton Raphson Method for f(x) = 0.
1. The Newton Raphson algorithm for the function will be
(a) { x }_{ k+1 }=\frac { 1 }{ 2 } \left( { x }_{ k }+\frac { a }{ { x }_{ k } } \right)
(b) { x }_{ k+1 }=\left( { x }_{ k }+\frac { a }{ 2 } { x }_{ k }^{ 2 } \right)
(c) { x }_{ k+1 }={ 2x }_{ k }a{ x }_{ k }^{ 2 }
(d) { x }_{ k+1 }={ x }_{ k }\frac { a }{ 2 } { x }_{ k }
(2 Mark, 2005)

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2. For a = 7 and starting with x_{0} = 0.2, the first two iterations will be
(a) 0.11, 0.1299
(b) 0.12, 0.1392
(c) 0.12, 0.1416
(d) 0.13, 0.1428
(1 Mark, 2005)

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3. The following equation needs to be numerically solved using the NewtonRaphson method. x^{3}+ 4x – 9 = 0.
The iterative equation for this purpose is (k indicates the iteration level).
(a) { x }_{ k+1 }=\frac { 2{ x }_{ k }^{ 2 }+9 }{ 3{ x }_{ k }^{ 2 }+4 }
(b) { x }_{ k+1 }=\frac { 3{ x }_{ k }^{ 2 }+4 }{ 2{ x }_{ k }^{ 2 }+9 }
(c) { x }_{ k+1 }={ x }_{ k }3{ x }_{ k }^{ 2 }+4
(d) { x }_{ k+1 }=\frac { 4{ x }_{ k }^{ 2 }+3 }{ 9{ x }_{ k }^{ 2 }+2 }
(2 Mark, 2007)

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4. Given that one root of the equation x^{3} – 10x^{2} + 31x – 30 = 0 is 5, the other roots are
(a) 2 and 3
(b) 2 and 4
(c) 3 and 4
(d) 2 and 3
(2 Mark, 2007)

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5. Three values of x and y are to be lifted in a straight line form y = a + bx by the method of least squares. Given and ∑x = 6, ∑y = 21, ∑x^{2} =14 and ∑xy = 46, the value of a and b are respectively
(a) 2 and 3
(b) 1 and 2
(c) 2 and 1
(d) 3 and 2
(2 Mark, 2008)

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6. If probability density function of a random variable x is
f(x) = x^{2} for 1\le x\le 1, and
= 0 for any other value of x
Then, the percentage probability P\left( \frac { 1 }{ 3 } \le x\le \frac { 1 }{ 3 } \right) is
(a) 0.247
(b) 2.47
(c) 24.7
(d) 247
(2 Mark, 2008)

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7. In the solution of the following set of linear equations by Gauss elimination using partial pivoting 5x + y + 2z = 34; 4y – 3z = 12; and 10x – 2y + z = 4; the pivots for elimination of x and y are
(a) 10 and 4
(b) 10 and 2
(c) 5 and 4
(d) 5 and 4
(2 Mark, 2009)

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8. The table below gives values of a function F(x) obtained for values of x at intervals of 0.25.
x 
0 
0.25 
0.5 
0.75 
1.0 
F(x) 
1 
0.9412 
0.8 
0.64 
0.50 
The value of the integral of the function between the limits 0 to 1 using Simpson’s rule is
(a) 0.7854
(b) 2.3562
(c) 3.1416
(d) 7.5000
(2 Mark, 2010)

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9. The square root of a number N is to be obtained by applying the Newton Raphson Iterations to the equation x^{2} – N = 0. If I denotes the iteration index, the correct iterative Scheme will be
(a) { x }_{ i+1 }=\frac { 1 }{ 2 } \left( { x }_{ i }+\frac { N }{ { x }_{ i } } \right)
(b) { x }_{ i+1 }=\frac { 1 }{ 2 } \left( { x }_{ i }^{ 2 }+\frac { N }{ { x }_{ i }^{ 2 } } \right)
(c) { x }_{ i+1 }=\frac { 1 }{ 2 } \left( { x }_{ i }+\frac { { N }^{ 2 } }{ { x }_{ i } } \right)
(d) { x }_{ i+1 }=\frac { 1 }{ 2 } \left( { x }_{ i }\frac { { N } }{ { x }_{ i } } \right)
(1 Mark, 2011)

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10. The estimate of \int _{ 0.5 }^{ 1.5 }{ \frac { dx }{ x } } obtained using Simpson’s rule with threepoint function evaluation exceeds the exact value by
(a) 0.235
(b) 0.068
(c) 0.024
(d) 0.012
(2 Mark, 2011)

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11. The magnitude as the error (correct to two decimal places) in the estimation of following integral using Simpson 1/3 rule. Take the step length as 1 ______.
\int _{ 0 }^{ 4 }{ \left( { x }^{ 4 }+10 \right) }dx(2 Mark, 2013)

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12. The error in \frac { d }{ dx } f(x){  }_{ x={ x }_{ o } } for a continuous function estimated with h = 0.03 using the central difference formula \frac { d }{ dx } f(x){  }_{ x={ x }_{ o } }=\frac { f\left( { x }_{ o }+h \right) f\left( { x }_{ o }h \right) }{ 2h } , is 2 10^{3}. The values of x_{0} and f(x_{0}) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h = 0.02 is approximately
(a) 1.3\times { 10 }^{ 4 }
(b) 3.0\times { 10 }^{ 4 }
(c) 4.5\times { 10 }^{ 4 }
(d) 9.0\times { 10 }^{ 4 }
(2 Mark, 2011)

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13. The integral \int _{ { x }_{ 1 } }^{ { x }_{ 2 } }{ { x }^{ 2 }dx } dx with { x }_{ 2 }>{ x }_{ 1 }>0 is evaluated analytically as well as numerically using a single application of the trapezoidal rule. If I is the exact value of the integral obtained analytically and J is the approximate value obtained using the trapezoidal rule, which of the following statements is correct about their relationship?
(a) J > I
(b) J < I
(c) J = I
(d) Insufficient data to determine the relationship
(1 Mark, 2015[1])

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14. The quadratic equation x^{2} – 4x + 4 = 0 is to be solved numerically, starting with the initial guess x_{0} = 3. The NewtonRaphson method is applied once to get a new estimate and then the Secant method is applied once using the initial guess and new estimate. The estimated value of the root after the application of the Secant method is _______.
(1 Mark, 2015[2])

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15. In NewtonRaphson iterative method, the initial guess value (x_{ini}) is considered as zero while finding the roots of the equation: f(x) = 2 + 6x – 4x^{2} + 0.5x^{3}, The correction, \Delta x to be added to x_{ini} in the first iteration is ______.
(2 Mark, 2015[1])

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16. For stepsize, \Delta x = 0.4, the value of following integral using Simpson’s 1/3 rule is _____.
\int _{ 0 }^{ 0.8 }{ \left( 0.2+25x200{ x }^{ 2 }+675{ x }^{ 3 }{ 900x }^{ 4 }{ +400x }^{ 5 } \right) } dx(2 Mark, 2015[2])

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17. NewtonRaphson method is to be used to find foot of equation 3x – e^{x}+ sinx = 0. If the initial trail value of the roots is taken as 0.333, the next approximation for the root would be _____.
(1 Mark, 2016[1])

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18. Consider the equation \frac { du }{ dt } =3{ t }^{ 2 }+1 with u = 0 at t = 0. This is numerically solved by using the forward Euler method with a step size. \Delta t=2. The absolute error in the solution in the end of the first time step is _____.
(2 Mark, 2017[1])

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19. The quadratic equation 2x^{2 }– 3x + 3 = 0 is to be solved numerically starting with an initial guess as x_{0 }= 2. The new estimate of x after the first iteration using NewtonRaphson method is ____.
(1 Mark, 2018[2])

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