GATE 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)
\({ X }_{ k+1 }={ X }_{ k }-\frac { 1 }{ 2 } ({ X }_{ k }+\frac { 117 }{ { X }_{ k } } )\)
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)
\({ X }_{ i+1 }=\frac { 1 }{ 2 } \left( { X }_{ i }+\frac { N }{ { X }_{ i } } \right) \)
Gauss-Seidel iterative method can be used for solving a set of
- (a)
linear differential equation only
- (b)
linear algebraic equations only
- (c)
Both linear and non-linear algebraic equations
- (d)
Both linear and non-linear differential equations
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
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
Match the items in Columns I and II using the codes given below the columns.
Column I | Column II |
(P) Gauss-Seidel method | (1) Interpolation |
(Q) forward Newton method | (2) Non-linear differential equation |
(R) Runge-Kutta method | (3) Numerical integration |
(S) Trapezoidal rule | (4) Linear algebraic equations |
- (a)
P Q R S 1 2 3 4 - (b)
P Q R S 2 3 4 1 - (c)
P Q R S 3 4 2 1 - (d)
P Q R S 4 1 2 3
The relation to solve x = e -x using Newton-Raphson method is
- (a)
\({ X }_{ n+1 }={ e }^{ { -X }_{ n } }\)
- (b)
\({ X }_{ n+1 }={ X }_{ n }{ -e }^{ { -X }_{ n } }\)
- (c)
\({ X }_{ n+1 }=(1+x_{ n })\frac { { e }^{ { -X }_{ n } } }{ 1+{ e }^{ { -X }_{ n } } } \)
- (d)
\({ X }_{ n+1 }=\frac { { X }^{ 2 }_{ n }-e^{ { -x }_{ n } }(1+{ X }_{ n })-1 }{ { X }_{ n }-{ e }^{ { -X }_{ n } } } \)
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 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)
\(x_{ k+1 }={ x }_{ k }-\frac { a }{ 2 } { x }^{ 2 }_{ k }\)
A numerical solution of the equation \(f(x)=x+\sqrt { x-3 } \) can be obtained using Newton-Raphson method. If the starting value of x = 2 for ten iterations, the value of x that is to be used in the next step
- (a)
0.306
- (b)
0.739
- (c)
1.694
- (d)
2.306
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, 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
If y1 = 4, y3 = 12, y4 = 19 and yx = 7, then x will be
- (a)
1.42
- (b)
1.68
- (c)
1.86
- (d)
1.98
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
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
A root of the equation x3-x-11 = 0 correct to four decimals using bisection method, is
- (a)
2.3737
- (b)
2.3838
- (c)
2.3736
- (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
The Newton-Raphson method is used to find the root of the equation x2 - 2 = 0, if iterations are started from -1, the iterations will be
- (a)
converged to -1
- (b)
converged to \(\sqrt { 2 } \)
- (c)
converged to - \(\sqrt { 2 } \)
- (d)
not converged
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