Practice Test: Computer Science Engineering (CSE) - 11 - Computer Science Engineering (CSE) MCQ

If we try with base ‘8’,-(53)₈-= 8 × 5 + 3 = (43)₁₀

-(22)₈-=8 × 2 + 2 = (18)₁₀

-43 × 18-= (774)₁₀

whereas,-(1276)₈ -=512×1+64 × 2+8 × 7+6 =(702)₁₀

If we try with base ‘9’,-(53)₉- = 9 × 5 + 3 =(48)₁₀

-(22)₉ - = 9 × 2 + 2 =(20)₁₀

-48 × 20 -= (960)₁₀

whereas,- (1276)_{9 -} = 729 × 1 + 81 × 2 + 9 × 7+ 6 = (960)₁₀

Converting-(4371)_{9 –}

= 4 × (729) + 3 × (81) + 7 × (9) + 1 =3223.

After rationalization,

On comparison, we get

a = 2

b = 1

a:b = 2:1

Divide the given word in 2 equal parts

M A J O R I T Y and each part is rewritten in reverse sequence

O J A M Y T I R.

Position of letters in the alphabet is 15/10/1/13 and 25/20/9/18 and position of letters in the given code PKBN/XSHQ is 16/11/2/14 and 24/19/8/17.

In comparison, you notice, the first half is obtained by adding ‘1’ and the second half is obtained by subtraction ‘1’.

• A touch screen is a display that can recognize a touch to its surface area, either with a finger or a stylus.

• Touch screens are generally used in applications like ATMs, hospitals, airline reservations, supermarkets, etc.

Conclusion:

I. C > A: True

II. D ≤ B: True

The total number of boys enrolled in Management and IT together

= 3500 × 20 + 16100 - 1500 × 18 + 12100

= 35 × 36 - 15 × 30

= 1260 - 450 = 810

If we arrange alphabets in two rows as shown below, the letters are related as follows,

Thus JP are the odd letters.

• Almeida was the first Viceroy of Portuguese in India .

•  

  Almeida took the island of Diu in his possession in AD 1509.

•  

  Albuquerque was the next Viceroy who won Goa in AD 1510.

•  

  Later on, the Portuguese also took Bassein & Daman in their possession.

Clearing means an open space in a forest, especially one cleared for cultivation.

Freeing means release from confinement or slavery.

Saving means release from confinement or slavery.

Eye-opening is a phrase meaning something that surprises you and teaches you new facts about life, people, etc.

After carefully observing the figures given in the question, it is very clear that the question figure is embedded in answer figure (A). It is shown as given below

Hence, Option A is the correct response.

L1= {ab,abab,abb,bbab,.........,bbb,bbbbbb........}

For string where after ‘ a ‘ never ‘ b ‘ appears ,language is-

L2={ε,a,b,aa,ba,,bbb,bbbb.}

Some strings are there on intersection of L1 & L2 (Like bbb, bbbb etc)

So L1 & L 2 are not complementary to each other .

In a language – no. of state in minimal NFA is n, then in worst case, maximum no. of state in equivalent DFA is 2n.

More than one dead state gets combined during minimization of DFA, so DFA contains only one dead state.

Number of 4 x 16 MUX required

= 256 / 16 + 16/16 = 16 + 1 = 17

A [i - k] < A [ i ] < A [i + k]

for each i such that k < i < n – k.

For example, the Arrays 1 4 2 6 3 7 5 8 are 2- ordered.

Now come to question: Arrays are both 2 ordered and 3 ordered. i.e.

1 2 3 4 5 6 7 8

Now if the Arrays were 1-ordered then Arrays will be 1 2 3 4 5 6 7 8 maximum number of positions that an element can be from its position=1.

No conflict in above DFA.

Consider characteristic equation of D-Flip-Flop : Q_N+1=D

Consider characteristics equation of T-Flip-flop: Q__N+1=T⊕Q_{_N}

Using equations (i), (ii) and (iii)

The number of used states = 5

∴ Modulus value of the counter = 5.

The characteristic equation of A is

|A-XI| = 0

⇒ f(x) = x3 + 6x2 + 11x + 6 + 2a

= (x + 1)(x + 2)(x + 3)+2a = 0

f(x) cannot have all 3 real roots (if any) equal

for if f(x) = (x-k)3, then comparing coefficients, we get

6 = –3k, 3k2 = 11

No such k exists

A) Thus f(x) = 0 has repeated (2) roots α, α, β

or

B) f(x) = 0 has real roots (distance) α, β, δ

Now

At x1, f(x) has relative maximum

At x2, f(x) has relative minimum

The graph of f(x) will be as below

Case A. repeated roots (α, α, β)

Case B. distinct roots

Note that the graph of f(x) cannot be like the one given below

Thus in all possible cares we have

Pattern is a rule describing the set of lexemes that can represent a particular token in the source program. For example, if the token is related then its informal description of the pattern will be < or <= or = or > or >= or <> etc.

Lexeme is a sequence of characters in the source program that is matched by the pattern for a token. For example let the token be an identifier then its sample lexemes can be pi, count, a, b etc.

- It’s used to send the content type of a response to be sent to a client, if the response has not been sent yet. The content type may include character encoding specifications.

Now for remaining n-4 vertices there is a choice whether to be included or not. So using the product rule, we get 2 × 2 × 2 × ….. (n-4) times.

So, we got the degree of the vertex having 4 elements is ⁴C₃ x 2^n-4

Q →M

Q →N

Q →P

M→R

P →O

Using BFS Algorithm Order of visiting the nodes QMNPRO or QMNPOR

A) MNOPQR → If you try to run BFS, after M, you must traverse NQR (In some order) Here P is traversed before Q, which is wrong.

B) NQMPOR → This is also not BFS. P is traversed before O!

D) QMNPOR → Incorrect. Because R needs to be traversed before O.(Because M is ahead of N in the queue.

10¹⁸_-1= 2ⁿ

Take log both side

log10(10¹⁸_-1) = log₁₀(2ⁿ)

~ log₁₀(10¹⁸) = log₁₀(2ⁿ)

18 = n log10²

n = 18/ log10²

n = 59.79

~ n = 60

(a) x ⊕ y = (xy + x′y ′)′

= (xy )′

= x ⊕ y, it is valid.

(b)

=

= Σm(1, 2, 4, 6)

x⊕(y+z)=x ̅(y+z)+x((y+z) ̅ )

=

= Σm(1, 2, 3, 4)

(x + y) ⊕ z ≠ x ⊕ (y + z)

So option (b) is invalid.

For asymptotic behaviour

h(n) = O(f(n))

g(n) = Ω(f(n))

A function having its upper bound denoted by O notation

Ω notation provides a lower bound of asymptote.

Probability of 1^st friend's selection is 3/5

Probability of 1^st friend's rejection is (1-3/5)=2/5

Probability of 2^nd friend's selection is 1/6

Probability of 2^nd friend's rejection is 1-1/6=5/6

probability of both friends get rejected is (2/5)×(5/6)=1/3

aij = i² – j²∀ i, j

∴ A3×3 is singular

∴ |A| = 0.

P₁ → 5 → 4

P₂ → 9 → 8

P₃ → 13 → 12

So P₁ + P₂ + P₃ + 1 = 25.

order never arises.

Process P₂ can execute D first then E. So option D. is correct.