Question 10398 – GATE 1993
January 31, 2024GATE 1996
February 1, 2024GATE 2003
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Question 1
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Consider the following C function.
float f(float x, int y)
{
float p, s; int i;
for (s=1, p=1, i=1; i < y; i ++)
{
p*= x/i;
s+=p;
}
return s;
}
For large values of y, the return value of the function f best approximates
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Xy
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ex
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ln(1+x)
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Xx
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Question 1 Explanation:
P = P * (x/i)
S = S+P
Iteration 1:
P=1; i=1; S=1
P=x
S = 1+x
Iteration 2:
P=x; S = 1+x; i=2
P = x * x/2 = x2/2
Iteration 3:
P = x2/2; S = 1+x+x2/2; i=3
P = (x2/2)(x/3) = x3/6
S = 1 + x + x2/2 + x3/6
Continue upto n
Then f( ) returns:
S = 1 + x/1 + x2/2⋅1 + x3/3⋅2 + …
= 1 + x/1! + x2/2! + x3/3! + … + xn/n!
= ex
S = S+P
Iteration 1:
P=1; i=1; S=1
P=x
S = 1+x
Iteration 2:
P=x; S = 1+x; i=2
P = x * x/2 = x2/2
Iteration 3:
P = x2/2; S = 1+x+x2/2; i=3
P = (x2/2)(x/3) = x3/6
S = 1 + x + x2/2 + x3/6
Continue upto n
Then f( ) returns:
S = 1 + x/1 + x2/2⋅1 + x3/3⋅2 + …
= 1 + x/1! + x2/2! + x3/3! + … + xn/n!
= ex
Correct Answer: B
Question 1 Explanation:
P = P * (x/i)
S = S+P
Iteration 1:
P=1; i=1; S=1
P=x
S = 1+x
Iteration 2:
P=x; S = 1+x; i=2
P = x * x/2 = x2/2
Iteration 3:
P = x2/2; S = 1+x+x2/2; i=3
P = (x2/2)(x/3) = x3/6
S = 1 + x + x2/2 + x3/6
Continue upto n
Then f( ) returns:
S = 1 + x/1 + x2/2⋅1 + x3/3⋅2 + …
= 1 + x/1! + x2/2! + x3/3! + … + xn/n!
= ex
S = S+P
Iteration 1:
P=1; i=1; S=1
P=x
S = 1+x
Iteration 2:
P=x; S = 1+x; i=2
P = x * x/2 = x2/2
Iteration 3:
P = x2/2; S = 1+x+x2/2; i=3
P = (x2/2)(x/3) = x3/6
S = 1 + x + x2/2 + x3/6
Continue upto n
Then f( ) returns:
S = 1 + x/1 + x2/2⋅1 + x3/3⋅2 + …
= 1 + x/1! + x2/2! + x3/3! + … + xn/n!
= ex
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