Engineering Mathematics - Numerical Methods
Exam Duration: 45 Mins Total Questions : 30
Let \({ X }^{ 2 }-117=0.\) The iterative steps for the solution using Newton-Raphson's method is given by
- (a)
\({ X }_{ k+1 }=\frac { 1 }{ 2 } \left( { X }_{ k }+\frac { 117 }{ { X }_{ k } } \right) \)
- (b)
\({ X }_{ k+1 }={ X }_{ k }-\frac { 117 }{ { X }_{ k } } \)
- (c)
\({ X }_{ k+1 }={ X }_{ k }-\frac { { X }_{ k } }{ 117 } \)
- (d)
\({ X }_{ k+1 }={ X }_{ k }-\frac { 1 }{ 2 } ({ X }_{ k }+\frac { 117 }{ { X }_{ k } } )\)
Equation \({ e }^{ x }-1=0\) is required to be solved using Newton's method with an initial guess \({ X }_{ o }=-1\) . Then after one step of Newton;s method, estimate x1 of the solution will be given by
- (a)
0.71828
- (b)
0.36784
- (c)
0.2.587
- (d)
0.000
The square root of a number N is to be obtained by applying the Newton-Raphson iteration 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)
\({ X }_{ i+1 }=\frac { 1 }{ 2 } \left( { X }^{ 2 }_{ i }+\frac { N }{ { X }^{ 2 }_{ i } } \right) \)
- (c)
\({ X }_{ i+1 }=\frac { 1 }{ 2 } \left( { X }_{ i }+\frac { N^{ 2 } }{ { X }_{ i } } \right) \)
- (d)
\({ X }_{ i+1 }=\frac { 1 }{ 2 } \left( { X }_{ i }+\frac { N }{ { X }_{ i } } \right) \)
The convergence of the bisection method is
- (a)
cubic
- (b)
quadratic
- (c)
linear
- (d)
None of these
For the differential equation \(\frac { dy }{ dx } =x-{ y }^{ 2 }\) is given that
| X | 0 | 0.2 | 0.4 | 0.6 |
| Y | 0 | 0.02 | 0.0795 | 0.1762 |
Using predictor-correction method, the y at next value of x is
- (a)
0.5114
- (b)
0.4648
- (c)
0.3046
- (d)
0.2498
Which one of the following is correct?
- (a)
Bisection method is used for iteration
- (b)
Regula-falsi method is direct method
- (c)
Secant method is direct method
- (d)
Newton-Raphson method is not iterative method
if n=3, ao=1, a1=0, a2 = -1, a3 = -11, then the root of the equation between 2 and 3 by regular-falsi method is
- (a)
2.0
- (b)
2.09
- (c)
2.9
- (d)
2.2
If \({ e }^{ o }=1,{ e }^{ 1 }=2.72,{ e }^{ 2 }=7.39,{ e }^{ 3 }=20.09\) and \({ e }^{ 4 }=54.60\) , then by Simpson's \(\frac { 1 }{ 3 } rd\) rule value of \(\int _{ 0 }^{ 4 }{ { e }^{ x } } dx\) is
- (a)
52.78
- (b)
53.87
- (c)
5.278
- (d)
5.387
Newton-Raphson method is used to compute a root of the equation \({ X }^{ 3 }-13=0\) with 3.5 as the initial value. The approximation after one iteration is
- (a)
3.575
- (b)
3.677
- (c)
3.667
- (d)
3.607
Match the following and choose the correct combination.
| Group I | Group II |
| E. Newton-Raphson method | 1. Solving non-linear |
| F. Runge-Kutta method | 2. Solving linear simultaneous equations |
| G. Simpson's Rule | 3.Solving ordinary differential equations |
| H. Gauss elimination | 4. Numerical integration |
| 5. Interpolation | |
| 6. Calculation of eigen values |
- (a)
E-6, F-1, G-5, H-3
- (b)
E-1, F-6, G-4, H-3
- (c)
E-1, F-3, G-4, H-2
- (d)
E-5, F-3, G-4, H-1
The differential equation \(\frac { dy }{ dx } =0.25{ y }^{ 2 }\) is to be solved using the backward (implicit) Euler's method with the boundary condition y= 1 at x = 0 and with step size of 1. What would be the value of y at x=1?
- (a)
1.33
- (b)
1.67
- (c)
2.00
- (d)
2.33
The table below gives values of a function F(x) obtained for values of x at intervals 0.25
| X | 0 | 0.25 | 0.5 | 0.75 | 1.00 |
| f(x) | 1 | 0.9412 | 0.8 | 0.64 | 0.50 |
The value of the integral of the function between the limit 0 to 1 using Simpson's rule is
- (a)
0.7854
- (b)
2.3562
- (c)
3.1416
- (d)
7.5000
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
The equation x3 +4x-9 = 0 needs to be numerically solved using the Netwon-Raphson method is
- (a)
\({ X }_{ k+1 }=\left( \frac { 2X^{ 3 }_{ k }+9 }{ 3X^{ 2 }_{ k }+4 } \right) .8\)
- (b)
\({ X }_{ k+1 }=\frac { 3{ x }^{ 2 }_{ k }+4 }{ 2{ x }^{ 2 }_{ k }+9 } \)
- (c)
\({ X }_{ k+1 }={ x }_{ k }-3{ x }^{ 2 }_{ k }+4\)
- (d)
\({ X }_{ k+1 }=\frac { 4{ x }^{ 2 }_{ k }+3 }{ 9{ x }^{ 2 }_{ k }+2 } \)
Given, a>0, we wish to calculate its reciprocal value \(\frac { 1 }{ a } \) by using Newton-Raphson method for f(x) = 0.
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)
\(x_{ k+1 }=\left( { x }_{ k }+\frac { a }{ 2 } { x }^{ 2 }_{ k } \right) \)
- (c)
\(x_{ k+1 }=2{ x }_{ k }-{ ax }^{ 2 }_{ k }\)
- (d)
\(x_{ k+1 }={ x }_{ k }-\frac { a }{ 2 } { x }^{ 2 }_{ k }\)
The double (repeated) root of
\(4{ x }^{ 3 }-8{ x }^{ 2 }-3x+9=0\)
by Newton-Raphson method will be
- (a)
1.4
- (b)
1.5
- (c)
1.6
- (d)
1.55
Using trapezoidal rule and the table given below, \(\int _{ 4 }^{ 5.2 }{ In\quad x\quad dx } \) will be
| x | 4 | 4.2 | 4.4 | 4.6 | 4.8 | 5 | 5.2 |
| In x | 1.39 | 1.44 | 1.48 | 1.53 | 1.57 | 1.61 | 1.65 |
- (a)
1.8277
- (b)
1.9284
- (c)
1.6424
- (d)
0.98795
Using trapezoidal rule, taking 10 equal interval \(\int _{ 0 }^{ \pi /2 }{ sinx\quad dx } \) will be
- (a)
1.902
- (b)
1.941
- (c)
1.888
- (d)
0.99795
The Newton-Raphson iteration
\({ x }_{ n+1 }=\frac { { x }_{ n } }{ 2 } +\frac { 3 }{ 2{ x }_{ n } }\)
can be used to solve the equation
- (a)
X2 = 3
- (b)
X3 = 3
- (c)
X2 = 2
- (d)
X3 = 2
A real root of equation \({ x }^{ 3 }+{ x }^{ 2 }-1=0 \) by iteration (method of successive approximation) method is
- (a)
0.7548765
- (b)
0.7548756
- (c)
0.7548776
- (d)
0.7548764
Using bisection method, the negative root of \({ x }^{ 3 }-4x+9=0\) correct to three decimal places is
- (a)
-2.506
- (b)
-2.706
- (c)
-2.406
- (d)
None of these
A real root of the equation x-cos x = 0 by the method false position correct to four decimal places is
- (a)
0.7391
- (b)
0.7439
- (c)
0.7347
- (d)
None of these
Using Newton-Raphson method, a root correct to three decimal places of the equation \({ e }^{ x }=1+2x \) is
- (a)
1.256
- (b)
1.255
- (c)
1.286
- (d)
None of these
Using Newton-Raphson method, a root correct to three decimal places of the equation
sin x = 1-x is
- (a)
0.511
- (b)
0.500
- (c)
0.555
- (d)
None of these
Using bisection method the negative of x3 - x + 11 = 0 is
- (a)
-2.3736
- (b)
-2.3838
- (c)
-2.3737
- (d)
None of these
Which of the following statements applies to the bisection method used for finding roots of functions?
- (a)
coverage within a few iteration
- (b)
guaranteed to work for all continuous functions
- (c)
it is faster than the Newton-Raphson method
- (d)
requires that there will be no error in determining the sign of the function
Newton-Raphson iteration formula for finding \(\sqrt [ 3 ]{ C } \) where C > 0, is
- (a)
\({ x }_{ n+1 }=\frac { 2{ x }^{ 3 }_{ n }+\sqrt [ 3 ]{ C } }{ 3{ x }^{ 2 }_{ n } } \)
- (b)
\({ x }_{ n+1 }=\frac { 2{ x }^{ 3 }_{ n }-\sqrt [ 3 ]{ C } }{ 3{ x }^{ 2 }_{ n } } \)
- (c)
\({ x }_{ n+1 }=\frac { 2{ x }^{ 2 }_{ n }+C }{ 3{ x }^{ 2 }_{ n } } \)
- (d)
\({ x }_{ n+1 }=\frac { 2{ x }^{ 2 }_{ n }-C }{ 3{ x }^{ 2 }_{ n } } \)
The iteration formula to find square root of a positive real number b using the Newton-Raphson method is
- (a)
\({ x }_{ k+1 }=\frac { 3({ x }_{ k }+b) }{ 2{ x }_{ k } } \)
- (b)
\({ x }_{ k+1 }=\frac { ({ x^{ 2 } }_{ k }+b) }{ 2{ x }_{ k } } \)
- (c)
\({ x }_{ k+1 }=\frac { { x }_{ k }-2{ x }_{ k } }{ ({ x^{ 2 } }_{ k }+b) } \)
- (d)
None of these
Using Newton-Raphson method, a root correct to three decimal places of the equation x3 - 3x -5 = 0 is
- (a)
2.275
- (b)
2.279
- (c)
2.222
- (d)
None of these
A real root of equation x3 -5x -7 = 0 by the method of false position correct to three decimal places is
- (a)
2.7472
- (b)
2.0844
- (c)
2.0774
- (d)
None of these
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