Deemed University 2009 A.M.I.E.T.E Electronics & Communication Engineering NUMERICAL ANALYSIS & COMPUTER PROGRAMMING – ET (OLD SCHEME) in ELECTRONICS & COMM ENGG - Question Paper
Tuesday, 30 April 2013 04:20Web
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AMIETE – ET (OLD SCHEME)
Code: AE07 Subject: NUMERICAL ANALYSIS & COMPUTER PROGRAMMING
Time: three Hours Max. Marks: 100
NOTE: There are nine ques. in all.
· ques. one is compulsory and carries 20 marks. ans to Q. 1. must be written in the space given for it in the ans book supplied and nowhere else.
· Out of the remaining 8 ques. ans any 5 ques.. every ques. carries 16 marks.
· Any needed data not explicitly given, may be suitably presumed and said.
Q.1 select the accurate or the best option in the following: (210)
a. What is the output of the subsequent C program
#include
void main( )
{
int arr[ ]={10, 20, 36, 72, 45, 36}
int *j, *k
j = &arr[4]
k = (arr + 4)
if (j = = k)
printf("1010")
else
printf("0101")
}
(A) fault (B) 0101
(C) 1010 (D) No output
b. Consider the subsequent program
#include
void main( )
{
int x, y
scanf("%d %d", &x, &y)
fun(x, y)
}
void fun(int a, int b)
{
a = a + b
b = a – b
a = a – b
}
The above coding can be used for
(A) Addition and subtraction of 2 numbers.
(B) Exchanging the value of 2 variables
(C) Finding the Fibonacci series
(D) None of these
c. The convergence of Newton-Raphson method is
(A) linear (B) quadratic
(C) cubic (D) None of the above
d. If is the Forward Difference operator and is the shift operator, then
equal to
(A) 6x (B) 3x2
(C) 3x3 (D) None of the above
e. The value of y6 if y0 = –8, y1 = –6, y2 = 22, y3 = 148, y4 = 492, y5 = 1222 is
| Earning: Approval pending. |
