1.
Consider a sequence F

_{00}defined as:
Then what shall be the set of values of the
sequence F

_{00}?
(1) (1, 110, 1200)

(2) (1, 110, 600, 1200)

(3) (1, 2, 55, 110, 600, 1200)

(4) (1, 55, 110, 600, 1200)

Answer: 1

__Explanation:__
We have
given, F

_{00}(0) = 1, F_{00}(1) = 1
F

_{00}(2) = (10*F_{00}(1) + 100)/F_{00}(0) = 110
F

_{00}(3) = (10*F_{00}(2) + 100)/F_{00}(1) = 1200
F

_{00}(4) = (10*F_{00}(3) + 100)/F_{00}(2) = 110
Since the
values repeats after the first three values, the set of values of F

2.
Match the following:_{00}will be (1,110,1200).**List-I List-II**

a. Absurd i.
Clearly impossible being

contrary to some evident truth.

b. Ambiguous ii.
Capable of more than one

interpretation or meaning.

c. Axiom iii.
An assertion that is accepted

and used without a proof.

d. Conjecture iv. An opinion Preferably based

on some experience or wisdom.

**Codes:**

a b c d

(1) i ii iii
iv

(2) i iii iv
ii

(3) ii iii iv
i

(4) ii i iii
iv

Answer: 1

3.
The functions mapping R into R are defined
as:

f(x) = x

^{3}-4x, g(x)=1/(x^{2}+1) and h(x)=x^{4}
Then find the value of the following
composite functions:

hog(x) and hogof(x)

(1) (x

^{2}+1)^{4}and [(x^{3}-4x)^{2}+1]^{4}
(2) (x

^{2}+1)^{4}and [(x^{3}-4x)^{2}+1]^{- 4}
(3) (x

^{2}+1)^{- 4}and [(x^{3}-4x)^{2}+1]^{4}
(4) (x

^{2}+1)^{‑ 4}and [(x^{3}-4x)^{2}+1]^{- 4}
Answer: 4

__Explanation:__
hog(x) = h(1/(x

^{2}+1))
= [(1/(x

^{2}+1))]^{4}= (x^{2}+1)^{- 4}
hogof(x) = hog(x

^{3}-4x)
= hog(x

^{3}-4x)
= [(x

4.
How many multiples of 6 are there between the
following pairs of numbers?^{3}-4x)^{2}+1]^{- 4}[since hog(x) = (x^{2}+1)^{- 4}]
0 and 100 and -6 and 34

(1) 16 and 6

(2) 17 and 6

(3) 17 and 7

(4) 16 and 7

Answer: 3

__Explanation:__
Number of
multiples of 6 between 1 and 100 = 100/6 = 16

Since the
range starts from zero, we need to take zero too. [zero is a multiple of every
integer (except zero itself)].

So, answer =
16+1 = 17

Number of
multiples of 6 between 1 and 34 = 34/6 = 5

Since the
range is -6 to 34, we need to take -6 and zero.

So, answer =
5+2 = 7

5.
Consider a Hamiltonian Graph G with no loops
or parallel edges and with |V(G)|=n≥3. Then which of the following is true?
(1) deg(v) ≥ n/2 for each vertex v.

(2) |E(G)| ≥ 1/2(n-1)(n-2)+2

(3) deg(v)+deg(w) ≥ n whenever v and w are
not connected by an edge.

(4) All of the above

Answer: 4

__Explanation:__**Dirac's theorem:**A simple graph with n vertices (n ≥ 3) is Hamiltonian if every vertex

has degree n/2 or greater.

**Ore's theorem:**deg(v) + deg(w) ≥ n for every pair of distinct non-adjacent vertices v

and w of G (*), then G is Hamiltonian.

6. In propositional logic if (P→Q)˄(R→S) and (P˅R) are two premises such that

Y is the premise:

(1) P˅R

(2) P˅S

(3) Q˅R

(4) Q˅S

Answer: 4

7.
ECL is the fastest of all logic families.
High Speed in ECL is possible because transistors are used in difference
amplifier configuration, in which they are never driven into ...............

(1) Race condition

(2) Saturation

(3) Delay

(4) High impedance

Answer: 2

__Explanation:__
Emitter-coupled
logic (ECL) is the fastest of all logic families and therefore is used in

applications where very high speed is essential. High speeds have become
possible in

ECL because the transistors are used in difference amplifier
configuration, in which

they are never driven into saturation and thereby the
storage time is eliminated.

8.
A binary 3-bit down counter uses J-K
flip-flops, FF_{i}with inputs J

_{i}, K

_{i}and outputs Q

_{i}, i=0,1,2 respectively. The minimized expression for the input from following, is

I. J

_{0}=K_{0}=0
II. J

_{0}=K_{0}=1
III. J

_{1}=K_{1}=Q_{0}
IV. J

_{1}=K_{1}=Q'_{0}
V. J

_{2}=K_{2}=Q_{1}Q_{0}
Vl. J

_{2}=K_{2}=Q'_{1}Q'_{0}
(1) I, Ill, V

(2) I, IV, VI

(3) Il, III, V

(4) Il, IV, Vl

Answer: 4

9.
Convert the octal number 0.4051 into its
equivalent decimal number.

(1) 0.5100098

(2) 0.2096

(3) 0.52

(4) 0.4192

Answer: 1

__Explanation:__
(0.4051)

_{8 }= 4x8^{-1}+0x8^{-2}+5x8^{-3}+1x8^{-4}
= 0.5100098

10.
The hexadecimal equivalent of the octal
number 2357 is:
(1) 2EE

(2) 2FF

(3) 4EF

(4) 4FE

Answer: 3

__Explanation:__
(2357)

= 010 011 101 111

Now we can regroup the bits into groups of 4 and convert to hexadecimal.

= 0100 1110 1111

= 4EF

_{8}can be converted into binary just digit by digit.= 010 011 101 111

Now we can regroup the bits into groups of 4 and convert to hexadecimal.

= 0100 1110 1111

= 4EF

## 2 Comments

So fast in publishing the answer within the couple of days after written the

ReplyDeleteexam. Thanks a lot very much.

Question 4 answer is 1) 16 and 6

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