Page 215 - Engineering Mathematics Workbook_Final

Numerical Methods
 0.2 (sin x − log e ) x dx with 12 sub- (d) an infinite number of solutions
1.4

intervals of equal length, is equal to 28. Solution of the variables x and x
(round off to 3 places of decimal). 1 2
for the following equations is to be
26. A piecewise linear function f(x) is obtained by employing the Newton-

plotted using thick solid lines in the Raphson interactive method
figure (the plot is drawn to scale).
=
x
equation (i) 10 sin x − 0.8 0
2 1
equation (ii)
10x − 10 cos x − 0.6 0
=
2
x
1
2
2
Assuming the initial values x = 0.0
1
and x = 1.0, the Jacobian matrix is
2
 10 − 0.8   10 0 
(a)     (b)    
  0 − 0.6    0 10 

 0 − 0.8   10 0 
(c)     (d)    
If we use the Newton-Raphson   10 − 0.6    10 − 10 

method to find the roots of f(x) = 0
−
using x , x and x respectively as 29. The differential equation dx = 1 x
0
1
2
initial guesses, the roots obtained dt 
would be is discretised using Euler’s numerical
integration method with a time step
(a) 1.3, 0.6 and 0.6 respectively  T  0. What is the maximum

(b) 0.6, 0.6 and 1.3 respectively permissible value of T to ensure
stability of the solution of the
(c) 1.3, 0.6 and 1.3 respectively corresponding discrete time equation?


(d) 1.3, 0.6 and 1.3 respectively (a) 1 (b) / 2

[CS. GATE-2003] (c)  (d) 2

27. In the interval 0, the equation x = du
30. Consider the equation = 3t + 2 1
cos x has dt
with u = 0 at t = 0. This is
(a) No solution
numerically solved by using the

(b) exactly one solution forward Euler method with a step
size, t = 2. The absolute error in
(c) exactly two solution

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