**Q.1 - Q.25 carry one mark each.**

1.
Let p, q, r, s represent the following
propositions.

p: x Ïµ {8, 9, 10, 11, 12}

q: x is a composite number

r: x is a perfect square

s: x is a prime number

The integer x≥2 which satisfies ¬((p ⇒q)∧(¬r∨¬s))
is ..............

Answer: 11

__Explanation:__
p⇒q
= {8,9,10,12}

¬r = {8,10,11,12}

¬s = {8,9,10,12}

¬r∨¬s
= {8,9,10,11,12}

(p⇒q)∧(¬r∨¬s) = {8,9,10,12}

¬((p⇒q)∧(¬r∨¬s)) = 11

2.
Let a

_{n}be the number of n-bit strings that do NOT contain two consecutive 1s. Which one of the following is the recurrence relation for a_{n }?
(A) a

_{n}= a_{n-1}+ 2a_{n-2}(B) a_{n}= a_{n-1}+ a_{n-2}
(C) a

_{n}= 2a_{n-1}+ a_{n-2}(D) a_{n}= 2a_{n-1}+ 2a_{n-2}
Answer: B

__Explanation:__
For n = 1, number of strings = 2 (0, 1)

For n = 2, number of strings = 3 (00, 01, 10)

For n = 3, number of strings = 5 (000, 001,
010, 100, 101)

For n = 4, number of strings = 8 (0000, 0001,
0010, 0100, 1000, 0101, 1010, 1001) ...
This series follows Fibonacci series and the recurrence relation for it
is a

_{n}=a_{n−1}+a_{n−2}.
3.

Answer: 1

__Explanation:__
Put x = y + 4. So, the equation becomes

It is equal to 1. (By Property of Limits on
sine)

4.
A probability density function on the interval
[a, 1] is given by 1/x

^{2}and outside this interval the value of the function is zero. The value of a is .................
Answer: 0.5

__Explanation:___{a}∫

^{1}(1/x

^{2}) = [-1/x]

_{a}

^{1}= -1+1/a

This is
equal to 1.

So, -1+1/a =
1 => a=0.5

5. Two
eigen values of a 3 x 3 real matrix P are (2 + √-1) and 3. The determinant of P
is ...............

Answer: 15

__Explanation:__
For a 3×3
matrix, characteristic equation will be cubic, so we will have 3 roots.

If one eigen
value is complex, the other eigen value has to be its conjugate.

So, the
eigen values of the matrix will be (2 + √-1), (2 - √-1) and 3.

Determinant
is the product of all eigen values.

So, (2 + √-1)*
(2 - √-1)*(3) = (4-(-1))*(3) = (5)*(3) = 15.

6.
Consider the Boolean operator # with the
following properties:

x#0 = x, x#1 = x’, x#x = 0 and x#x’ = 1. Then
x#y is equivalent to

(A) xy’ + x’y (B)
xy’ + x’y’

(C) x’y + xy (D)
xy + x’y’

Answer: A

__Explanation:__
Here, the
operator # represents XOR. So answer is A.

7. The
16-bit 2’s complement representation of an integer is 1111 1111 1111 0101; its
decimal representation is .................

Answer: -11.0

__Explanation:__
1] 111 1111
1111 0101, first bit is sign bit.

Its 2’s
complement is 1] 000 0000 0000 1011 = -11

8.
We want to design a synchronous counter that
counts the sequence 0-1-0-2-0-3 and then repeats. The minimum number of J-K
flip-flops required to implement this counter is ...............

Answer: 3 or 4

9.
A processor can support a maximum memory of 4
GB, where the memory is word-addressable (a word consists of two bytes). The
size of the address bus of the processor is at least ............. bits.

Answer: 31

__Explanation:__
Maximum
memory = 4 GB = 2

^{32}Bytes.
Since 1 word
= 2 bytes; total number of words=2

^{32}/2=2^{31}
So to
address these words we need minimum 31 bits.

10.
A queue is implemented using an array such
that ENQUEUE and DEQUEUE operations are performed efficiently. Which one of the
following statements is CORRECT (n refers to the number of items in the queue)?

(A) Both operations can be performed in O(1)
time

(B) At most one operation can be performed in
O(1) time but the worst case time for the other operation will be Î©(n)

(C) The worst case time complexity for both
operations will be Î©(n)

(D) Worst case time complexity for both
operations will be Î©(log n)

Answer: A

__Explanation:__
By using
circular array both operations can be performed in O(1) time.

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