Tamil Nadu Open University (TNOU) 2009-1st Sem M.C.A - - Numerical methods -- I - Question Paper

M.C.A. DEGREE EXAMINATION, 2009
( 1st SEMESTER )
( PAPER - I )
111. NUMERICAL METHODS
( New Regulations )
December ] [ Time : three Hours
Maximum : 100 Marks

part - A (5 × eight = 40)
ans any 5 ques..
All ques. carry equal marks.
1. obtain the positive root of x – cos x = 0 by bisection method.
2. Round - off the number 34·89126543 to 5 significant digits (correct to 2 decimal places)
and obtain the absolute and relative errors.
3. Solve the subsequent system of formula by Gauss - Jordan method :
5x – 2y + 3z = 18
x + 7y – 3z = –22
2x – y + 6z = 22.
4. Solve the subsequent system of formula by Gauss - elimination method :
2x + 3y – z = 5
4x + 4y – 3z = 3
2x – 3y + 2z = 2.
5. obtain the formula y = f(x) of lowest degree and passing through the points
(–1, –21,) ; (1, 15) ; (2, 12) ; (3, 3)
obtain also, y at x = 0.
6. The subsequent data are taken from the steam tables :
Temperature ºC : 140 150 160 170 180
Pressure kgf/cm2: 3·685 4·854 6·302 8·076 10·225
obtain the pressure at temperature t = 142º.
7. Using Euler 's method, solve numerically the formula :
y¢ = x + y,
y(0) = one for x = 0·2, 0·4.
8. obtain the value of f¢(0·5) using Stirling 's
formula from the subsequent data :
part - B (3 × 20 = 60)
ans any 3 ques..
All ques. carry equal marks.
9. (a) obtain a positive root of x ex = 2
by the method of False posit ion.
(b) obtain a positive root of 2x3 – 3x – six = 0 by Newton - Raphson method.
x : 0·35 0·40 0·45 0·50 0·55 0·60 0·65
y = f(x) : 1·521 1·506 1·488 1·467 1·444 1·418 1·389
10. (a) provided the subsequent table, obtain y(35) using
Stirling 's formula :
(b) Use Lagrange 's formula, obtain f(6) from the subsequent data :
11. Solve the subsequent system by Gauss - Siedel method :
10x – 5y – 2z = 3
4x – 10y + 3z = –3
x + 6y + 10z = –3.
12. Evaluate :? x4 dx by using
(i) Trapezoidal rule.
(ii) Simpson's rule.
x : two five seven 10 12
f(x) : 18 180 448 1210 2028
x : 20 30 40 50
y : 512 439 346 243
13. Using Milne 's method, obtain y(4·4), given,
5xy¢ + y2 – two = 0 provided
y(4) = 1,
y(4·1) = 1·0049
y(4·2) = 1·0097
and y(4·3) = 1·0143.

Earning:   Approval pending.