PREVIOUS YEAR SOLVED PAPERS MCQ - GATE 2017 Shift 2

31 Consider a quadratic equation x² - 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b = _____.

Note – Numerical Type question

Clearly 13 = 1 x 10 + 3 and 36 = 3 x 10 + 6 ⇒ base b = 10
The quadratic equation with solutions x = 5 and x = 6 is x² - 11x + 30 = 0
According to the given condition, we have b + 3 = 11 and 3b + 6 = 30 ⇒ b = 8
Answer is 8
Alternative solution:
x² - 13x + 36 = 0 (given quadratic equation)
In bae b, 13 = 1xb¹ + 3xb⁰ = b+3 and
36 = 3x¹ + 6x⁰ = 3b + 6
So the equation becomes x² - (b + 3)x + (3b + 6) = 0
Since x = 5 is a solution
∴ 5² - (b + 3)5 + (3b + 6) = 0 ⇒ b = 8
Similarly, by putting x = 6, we get b = 8

32 Identify the language generated by the following grammar, where S is the start variable.
S → XY
X → aX∣a
Y → aYb∣ϵ

The given grammar with S as start symbol is
S → XY
X → aX∣a
Y → aYb∣ϵ
X generates atleast one 'a'. While Y generates equal no of a's and b's( including epsilon).
L = { a , ab, aab, aabb, aaabb ....}

33 The representation of the value of a 16-bit unsigned integer X in hexadecimal number system is BCA9. The representation of the value of X in octal number system is

(BCA9)₁₆ = (001 011 110 010 101 001)₂ = (136251)₈
Convert hexadecimal to the octal number system.

34 Consider the following function implemented in C:
void printxy (int x, int y) {
int *ptr;
x = 0;
ptr = &x;
y = * ptr;
* ptr = 1;
printf (“%d, %d,” x, y);
}
The output of invoking printxy (1, 1) is

void printxy (int x, int y) {
int *ptr;
x = 0; // x=0,y=1
ptr = &x; // x=0,y=1, and ptr points to x
y = * ptr; //x=0, y=0
* ptr = 1; //x=1, y=0
printf (“%d, %d,” x, y); // It prints 1 AND 0.
}

35 The maximum number of IPv4 router addresses that can be listed in the record route (RR) option field of an IPv4 header is _________.

Note – Numerical Type question

Record Route is an optional field used to record address.
In IPv4 header, 40 bytes are reserved for OPTIONS. For Record Route to stores, 1 byte is used to store type of option, 1 byte for length and 1 byte for pointer. Out of 40, 37 bytes are left.
An IP4 address takes 32 bits or 4 bytes.
We can store at most floor(37/4) = 9 router addresses.
Alternate Solution
In IPV4 frame format there are 40 bytes of option field, out of which only 38 bytes can be used, 2 bytes are reserved. Also one IPV4 address is of 4 bytes
Thus maximum IPV4 address that can be hold <= 38/4 = 9