Linear and Nonlinear Numerical Solvers - Numerical Methods - Engineering Sciences Previous Year Questions

1

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

Consider the following four FORTRAN statements
S1: X = 5 ** 3
S2: X = (-5) ** 3.0
S3: X = 5 ** -3
S4: X = 5 ** 3.0
Which of the following sets contains the set of valid statements from above?

A

{S1, S3}

B

{ S1, S4}

C

{S2, S3}

D

{S2, S4}

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2

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

The iteration scheme \( x_{n+1} = 1/(1 + x_n^2) \) converges to a real number x in the interval (0, 1) with \( x_0 = 0.5 \). The value of x correct up to 2 places of decimal is equal to

A

0.65

B

0.68

C

0.73

D

0.80

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3

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

The value of sum that will be printed by the above program is

A

369

B

361

C

303

D

261

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4

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

The equation \( g(x) = x \) is solved by Newton-Raphson iteration method, starting with an initial approximation \( x_0 \) near the simple root \( \alpha \). If \( x_{n+1} \) is the approximation to \( \alpha \) at the \( (n+1)^{th} \) iteration, then

A

\(x_{n+1} = \frac{x_n g'(x_n) - g(x_n)}{1 - g'(x_n)} \)

B

\(x_{n+1} = \frac{x_n g'(x_n) - g(x_n)}{g'(x_n) - 1} \)

C

\(x_{n+1} = g(x_n) \)

D

\(x_{n+1} = \frac{x_n g'(x_n) - g(x_n) + 2x_n}{g'(x_n) + 1} \)

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5

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

If Runge-Kutta method of order 4 is used to solve the differential equation $\frac{dy}{dx} = f(x), \; y(0)=0$ in the interval $[0,h]$ with step size $h$, then

A

$y(h) = \frac{h}{6}\left[f(0)+4f(h/2)+f(h)\right]$

B

$y(h) = \frac{h}{6}\left[f(0)+f(h)\right]$

C

$y(h) = \frac{h}{2}\left[f(0)+f(h)\right]$

D

$y(h) = \frac{h}{6}\left[f(0)+2f(h/2)+f(h)\right]$

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6

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

If a root of \( f(x) = x^2 - 2x + 1 = 0 \) is obtained by using the iterative scheme \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) with initial value \( x_0 = 0.5 \), then the convergence rate is

A

1

B

1.62

C

1.84

D

2

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7

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

Consider the system of linear simultaneous equations \( x + 10y = 5; \; y + 5z = 1; \; 10x - y + z = 0 \). On applying Gauss-Seidel method the value of x correct up to 4 decimal places is

A

0.0385

B

0.0395

C

0.0405

D

0.0410

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8

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

The iteration scheme
\(x_{n+1} = \sqrt{a} \left(1 - \frac{3a^2}{x_n^2} - \frac{3a^2}{x_n}\right)\), \(a > 0\) converges to the real number

A

√a

B

a

C

a√a

D

a^2

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9

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

Consider the following recursive function g().

Recursive integer function g(m,n) result (r)
integer:: m,n
if (n == 0) then
r = m
else if (m <= 0) then
r = n+1
else if ((n - n/2*2) == 1) then
r = g(m-1, n+1)
else r = g(m-2, n/2)
end if
end

Which value will be returned if the function g is called with 6, 6?

A

2

B

4

C

6

D

8

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10

2007 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2007

If the following function is called with x = 1

real function print_value(x)
real:: x, sum, term
integer:: i
i = 0
sum = 2.0
term = 1.0
do while (term > 0.00001)
term = x * term/(i+1)
sum = sum + term
i = i + 1
end do
print_value = sum
end

the value returned will be close to

A

log_e 2

B

log_e 3

C

1 + e

D

e

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11

2008 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2008

An approximate solution of the equation \( x^3 - 3x + 1 = 0 \) is 0.347296. Which of the following iterating functions will converge most rapidly to this root?

A

\(x_{n+1} = \frac{1}{3 - x_n^2} \)

B

\(x_{n+1} = \frac{1}{3}(x_n^3 + 1) \)

C

\(x_{n+1} = \frac{1}{5}(x_n^3 + 2x_n + 1) \)

D

\(x_{n+1} = \frac{1}{27}(10x_n^3 - 3x_n + 10) \)

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12

2008 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2008

A continuous function f(x) defined in the interval [a,b] is such that f(a)f(b)<0. A possible number of simple roots of the equation f(x)=0 in this interval is

A

0

B

1

C

2

D

4

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13

2008 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2008

The set of simultaneous equations \[ 4x - y = 15 \] \[ x + 5y = 9 \] is to be solved using Jacobi’s iterative method. Starting with the initial values \( x = 2, y = 2, \) the values of \( x \) and \( y \) after two iterations are, respectively,

A

4.25, 0.95

B

4.25, 1.4

C

4.1, 0.95

D

4.1, 1.4

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14

2008 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2008

A real root of the equation \(x^3 - 2x - 5 = 0\) lies between \(x = 2\) and \(x = 3\). The location of the root obtained after the second iteration using the method of false position is

A

2.081

B

2.061

C

2.059

D

2.041

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15

2008 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2008

Consider the following part of a Fortran 90 function ```fortran INTEGER FUNCTION RESULT(X) INTEGER :: X VALUE = 1 DO IF ( X == 0 ) EXIT TERM = MOD (X, 10) VALUE = VALUE*TERM X = X/10 END DO RESULT = VALUE END FUNCTION RESULT ``` If the above function is called with an integer \(X = 123\), the value returned by the function will be

A

0

B

6

C

9

D

321

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16

2008 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2008

A portion of a Fortran 90 program is reproduced below:
PROGRAM CHECK_CYCLE
    DO I = 1, 10, 2
        IF (MOD(I, 3) == 0 ) CYCLE
        PRINT *, I
    END DO
END PROGRAM CHECK_CYCLE
The result displayed by the program is

A

1
5
7

B

1
3
5

C

1
3
7

D

3
5
7

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17

2008 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2008

A Fortran-90 subroutine for Gauss-Siedel Method to solve a set of N simultaneous equations [A][X]=[C] is given below.
```fortran
SUBROUTINE SIEDEL(A, C, X, N, IMAX)
REAL :: SUM
REAL, DIMENSION(N,N) :: A
REAL, DIMENSION(N) :: C, X
DO K = 1, IMAX
DO I = 1, N
SUM = 0.0
DO J = 1, N
IF ( I /= J ) THEN
SUM = SUM+A(I,J)*X(J)
ENDIF
ENDDO
*****
ENDDO
ENDDO
END SUBROUTINE SIEDEL
```
The missing statement in the program, indicated by *****, is

A

X(I) = C(I) + SUM

B

X(I) = C(I) - SUM

C

X(I) = (C(I) + SUM)/A(I,I)

D

X(I) = (C(I) - SUM)/A(I,I)

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18

2008 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2008

The equation f(x) = x^3 - 1.8x^2 - 1.35x + 2.7 = 0 is known to have a multiple root in the interval [1,2]. Starting with an initial guess x_0 = 1.0 in modified Newton-Raphson method, the root, correct up to three decimal places, is

A

1.500

B

1.200

C

1.578

D

1.495

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19

2009 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2009

The root of ax + b = 0 (a, b constants), can be found by the Newton-Raphson method with a minimum of

A

1 iteration

B

2 iterations

C

3 iterations

D

an undeterminable number of iterations

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20

2011 · Engineering Sciences · Numerical Methods · Linear and Nonlinear Numerical Solvers

Engineering Sciences (XE) 2011

Suppose \(x_n\) is the \(n\)-th iterated value while finding the positive square root of 7 by the Newton-Raphson method with a positive initial guess \(x_0\) (\(\neq \sqrt{7}\)). If \(e_n = \sqrt{7} - x_n\) for \(n \geq 1\), then

A

\(e_{n+1} = \frac{e_n}{2x_n^2}\)

B

\(e_{n+1} = \frac{-\sqrt{7} e_n^2}{2x_n}\)

C

\(e_{n+1} = \frac{e_n^2}{\sqrt{7}}\)

D

\(e_{n+1} = \frac{-e_n^2}{2x_n}\)

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