GATE CS 2015 Set 3

The sentence suggests that students are so narrowly focused on the syllabus that they ignore anything not directly part of it. "Extraneous" means irrelevant or unrelated to the subject being dealt with, which fits the context of closing one's mind to things outside the core requirements.(A) related - This would mean they close their minds to things that are part of the exam, which is the opposite of the intended meaning.
(C) outside - While similar in meaning, "extraneous" is the more precise academic term for irrelevant information in this context.
(D) useful - This would imply they ignore helpful things, but the focus is specifically on the relationship to the exam requirements.

The relationship in the given pair is Demographic Group : Specialist Doctor. A Pediatrician is a doctor who specializes in the medical care of children.(A) Adult : Orthopaedist - Incorrect. Orthopaedists treat bones and muscles in patients of all ages, not specifically adults.
(B) Females : Gynaecologist - Correct. A Gynaecologist is a doctor who specializes in the health of the female reproductive system.
(C) Kidney : Nephrologist - Incorrect. This is an Organ : Specialist relationship, not a demographic group.
(D) Skin : Dermatologist - Incorrect. This is also an Organ : Specialist relationship.

The correct words to fill the blanks are determined by grammar and context:Therefore, option (C) is the only grammatically correct choice.

The pattern in the given coding is to add 3 to the alphabetical position of each letter:For ROAD to URDG:

• R (18) + 3 = U (21)
• O (15) + 3 = R (18)
• A (1) + 3 = D (4)
• D (4) + 3 = G (7)

Applying the same pattern to SWAN:

• S (19) + 3 = V (22)
• W (23) + 3 = Z (26)
• A (1) + 3 = D (4)
• N (14) + 3 = Q (17)

So, SWAN is written as VZDQ.The correct option is (B).

A linear function can be represented in the form $f(x) = mx + c$, where $m$ is the slope and $c$ is the y-intercept.Given two points:Subtract Equation 1 from Equation 2:
$(3m + c) - (-2m + c) = 39 - 29$
$3m + c + 2m - c = 10$
$5m = 10$
$m = 2$Substitute the value of $m$ into Equation 1:
$-2(2) + c = 29$
$-4 + c = 29$
$c = 29 + 4$
$c = 33$So, the linear function is $f(x) = 2x + 33$.Now, find the value of the function at $x = 5$:
$f(5) = 2(5) + 33$
$f(5) = 10 + 33$
$f(5) = 43$The correct option is (C).

Let P be the statement "Alexander had conquered Persia."
Let I be the statement "Alexander turned his attention towards India."The given sentence is "Alexander turned his attention towards India, since he had conquered Persia." This implies that conquering Persia was the reason or a necessary condition for Alexander to turn his attention towards India. In logical terms, this can be interpreted as: If Alexander turned his attention towards India, then he must have conquered Persia. So, I → P.Now let's analyze the options:(A) "Alexander would not have turned his attention towards India had he not conquered Persia." This statement can be written as $\neg P \rightarrow \neg I$. This is the contrapositive of I → P. Since the contrapositive of a conditional statement is logically equivalent to the original statement, if I → P is true, then $\neg P \rightarrow \neg I$ is also true. This is a valid inference based on the interpretation of "since" implying a necessary condition.(B) "Alexander was not ready to rest on his laurels, and wanted to march to India." This statement describes Alexander's motivation or state of mind, which cannot be directly inferred from the given sentence. The sentence only states a causal link between two events.(C) "Alexander was completely in control of his army and could command it to move towards India." This statement makes an assumption about Alexander's control over his army, which is not mentioned or implied in the original sentence.(D) "Since Alexander's kingdom extended to Indian borders after the conquest of Persia, he was keen to move further." This statement introduces new information (kingdom extended to Indian borders) and an interpretation of his keenness, neither of which is directly inferable from the original sentence.Therefore, option (A) is the only logically valid inference.The final answer is $\boxed{A}$

The passage states that the batsman lacks the requisite temperament despite having technical skills, which implies he needs to improve his temperament (supporting statement ii). It also mentions he threw away his wicket after laying a strong foundation, meaning he failed to convert good starts (supporting statement iii). Statement (i) is incorrect because success eludes him. Statement (iv) is incorrect because he already possesses technical skills; temperament is the issue. Thus, (ii) and (iii) are valid.

We calculate the total (Exports + Imports) for each year and find the percentage increase from the previous year.2000: $50 + 40 = 90$
2001: $60 + 50 = 110$. Increase = $\frac{110-90}{90} \times 100 \approx 22.2\%$
2002: $70 + 60 = 130$. Increase = $\frac{130-110}{110} \times 100 \approx 18.2\%$
2003: $60 + 70 = 130$. Increase = $0\%$
2004: $70 + 80 = 150$. Increase = $\frac{150-130}{130} \times 100 \approx 15.4\%$
2005: $70 + 90 = 160$. Increase = $\frac{160-150}{150} \times 100 \approx 6.7\%$
2006: $100 + 120 = 220$. Increase = $\frac{220-160}{160} \times 100 = \frac{60}{160} \times 100 = 37.5\%$
2007: $110 + 120 = 230$. Increase = $\frac{230-220}{220} \times 100 \approx 4.5\%$The highest percentage increase is in 2006.

The plot shows a graph that lies in the 4th quadrant (positive x, negative y) and possibly the positive y-axis. The curve passes through the point $(2, -1)$ as indicated by the dashed lines.Let's test the point $(2, -1)$ in the options:
(A) $x = -1 - |-1| = -1 - 1 = -2 \neq 2$
(B) $x = -(-1 - |-1|) = -(-1 - 1) = -(-2) = 2$. This matches.
(C) $x = -1 + |-1| = -1 + 1 = 0 \neq 2$
(D) $x = -(-1 + |-1|) = -(0) = 0 \neq 2$Only option (B) is consistent with the point $(2, -1)$. The function $x = -(y - |y|)$ simplifies to $x = 0$ for $y \ge 0$ (positive y-axis) and $x = -2y$ for $y < 0$ (line in 4th quadrant), which matches the visual representation.

There are 6 members (P, Q, R, S, T, U) and 5 portfolios (Home, Power, Defense, Telecom, Finance). This means one member will not be assigned a portfolio.Let's analyze the constraints:From constraint 3, S must be assigned a portfolio (either Power or Telecom). Therefore, from constraint 1, U must be the member who does not get a portfolio.So, the five members assigned portfolios are P, Q, R, S, T.Now let's re-evaluate the constraints with U being unassigned:

• Members assigned: P, Q, R, S, T.
• Constraint 2: R must get Home or Finance (since R is assigned).
• Constraint 3: S and Q get Power and Telecom (one each).
• Constraint 4: Since P is assigned, T must also be assigned (which is consistent).

Let's check the options:(A) P-Home, Q-Power, R-Defense, S-Telecom, T-Finance

• R gets Defense. This violates constraint 2 (R must get Home or Finance). So (A) is incorrect.

(B) R-Home, S-Power, P-Defense, Q-Telecom, T-Finance

• U is unassigned (consistent).
• R gets Home (consistent with constraint 2).
• S gets Power, Q gets Telecom (consistent with constraint 3).
• P gets Defense, T gets Finance. Both P and T are assigned (consistent with constraint 4).
• All constraints are satisfied. So (B) is correct.

(C) P-Home, Q-Power, T-Defense, S-Telecom, U-Finance

• U gets Finance. This violates our deduction that U is unassigned. So (C) is incorrect.

(D) Q-Home, U-Power, T-Defense, R-Telecom, P-Finance

• U gets Power. This violates our deduction that U is unassigned. So (D) is incorrect.

The final answer is $\boxed{B}$

Let's trace the execution of the C program:

• An array s1 of 7 characters is declared and initialized with the string "1234".
• In memory, s1 will be {'1', '2', '3', '4', '\0', ?, ?}. The \0 is the null terminator automatically added after "1234". The last two elements are uninitialized.
• The indices are s1[0] = '1', s1[1] = '2', s1[2] = '3', s1[3] = '4', s1[4] = '\0'.

2. char *p;

• A character pointer p is declared.

3. p = s1 + 2;

• The pointer p is made to point to the memory location of s1[2]. So, p now points to the character '3'.

4. *p = '0';

• The character at the memory location pointed to by p is changed to '0'. Since p points to s1[2], s1[2] is updated from '3' to '0'.
• The s1 array now looks like: {'1', '2', '0', '4', '\0', ?, ?}.

5. printf("%s", s1);

• This statement prints the string starting from the address of s1 until a null terminator (\0) is encountered.
• It will print s1[0], s1[1], s1[2], s1[3], and then stop because s1[4] is \0.
• The characters printed will be '1', '2', '0', '4'.

Therefore, the output of the program will be "1204".The final answer is $\boxed{C}$

Given: $S = \{1,2,3,4,5,6\}$, so $|S|=6$. $U$ is the power set of $S$. $T'$ is the complement of $T$ with respect to $S$, so $|T'| = |S| - |T| = 6 - |T|$. $T \setminus R$ means elements in $T$ but not in $R$.Let's evaluate each option:(A) $\forall X \in U (|X| = |X'|)$

• This statement claims that for every set $X$ in the power set $U$, its cardinality is equal to the cardinality of its complement.
• If $|X| = |X'|$, then $|X| = 6 - |X|$, which implies $2|X| = 6$, so $|X| = 3$.
• This statement would only be true if every set $X \in U$ had exactly 3 elements. This is false. For example, if $X = \{1\} \in U$, then $|X|=1$ and $|X'|=5$. Clearly, $1 \neq 5$.
• Therefore, statement (A) is false.

(B) $\exists X \in U \exists Y \in U (|X| = 5, |Y| = 5 \text{ and } X \cap Y = \emptyset)$

• This statement claims there exist two sets $X, Y$ in $U$ such that both have 5 elements and they are disjoint.
• If $|X|=5$ and $|Y|=5$, and $X \cap Y = \emptyset$, then the union of $X$ and $Y$ would have $|X \cup Y| = |X| + |Y| = 5 + 5 = 10$ elements.
• However, $X$ and $Y$ are subsets of $S$, so $X \cup Y$ must also be a subset of $S$. This means $|X \cup Y| \le |S| = 6$.
• Since $10 \le 6$ is false, it is impossible for two disjoint sets to each have 5 elements from a universal set of 6 elements.
• Therefore, statement (B) is false.

(C) $\forall X \in U \forall Y \in U (|X| = 2, |Y| = 3 \text{ and } X \setminus Y = \emptyset)$

• This statement claims that for any sets $X, Y$ in $U$ such that $|X|=2$ and $|Y|=3$, it must be that $X \setminus Y = \emptyset$.
• $X \setminus Y = \emptyset$ means that all elements of $X$ are also in $Y$, i.e., $X \subseteq Y$.
• So the statement claims: for all $X, Y$ with $|X|=2$ and $|Y|=3$, it must be that $X \subseteq Y$.
• This is false. Consider $S = \{1,2,3,4,5,6\}$. Let $X = \{1,2\}$ and $Y = \{3,4,5\}$. Here, $|X|=2$ and $|Y|=3$. However, $X$ is not a subset of $Y$, so $X \setminus Y = \{1,2\} \neq \emptyset$.
• Therefore, statement (C) is false.

(D) $\forall X \in U \forall Y \in U (X \setminus Y = Y' \setminus X')$

• Let's analyze the left side: $X \setminus Y$ is the set of elements in $X$ that are not in $Y$. This can be written as $X \cap Y'$.
• Let's analyze the right side: $Y' \setminus X'$ is the set of elements in $Y'$ that are not in $X'$. This can be written as $Y' \cap (X')'$.
• By the double complement rule, $(X')' = X$.
• So, $Y' \setminus X' = Y' \cap X$.
• Since set intersection is commutative, $Y' \cap X = X \cap Y'$.
• Thus, $X \setminus Y = X \cap Y'$ and $Y' \setminus X' = X \cap Y'$. Both sides are equal.
• This is a fundamental set identity (also known as the definition of set difference in terms of intersection and complement, applied to both sides).
• Therefore, statement (D) is true.

The final answer is $\boxed{D}$

A functional dependency X → Y is called trivial if $Y \subseteq X$.Checking the options:
(A) $\{S, T\} \not\subseteq \{P, R\}$ (Non-trivial)
(B) $\{R, T\} \not\subseteq \{P, R\}$ (Non-trivial)
(C) $\{S\} \subseteq \{P, S\}$ (Trivial)
(D) $\{Q\} \not\subseteq \{P, S, U\}$ (Non-trivial)Since trivial dependencies are always valid and included in the closure $F^+$, option (C) is the correct answer.

The Ready state contains processes that are prepared to execute when given the opportunity. The number of processes in the Ready state is determined by the number of jobs submitted to the system and the available memory, not by the number of CPUs.The number of processes in the Running state is limited by the number of CPUs ($n$), but the Ready queue can hold any number of processes (independent of $n$).

Therefore, the easiest to implement is SLR and the most powerful is canonical LR.

The binary operator # is defined as $X \# Y = X' + Y'$.
Using De Morgan's laws, we can write this as $X' + Y' = (X \cdot Y)'$. This is the NAND operation.Statement (S2): Commutativity
We need to check if the operator # is commutative, i.e., if $Q \# R = R \# Q$.LHS = $Q \# R = Q' + R'$.
RHS = $R \# Q = R' + Q'$.Since Boolean OR (+) is a commutative operation, $Q' + R' = R' + Q'$.
Therefore, the operator # is commutative.
Statement (S2) is true.Statement (S1): Associativity
We need to check if the operator # is associative, i.e., if $(P \# Q) \# R = P \# (Q \# R)$.LHS = $(P \# Q) \# R$
$= (P' + Q') \# R$
$= (P' + Q')' + R'$
Using De Morgan's law on $(P' + Q')'$, we get $(P'') \cdot (Q'') = PQ$.
So, LHS = $PQ + R'$.RHS = $P \# (Q \# R)$
$= P \# (Q' + R')$
$= P' + (Q' + R')'$
Using De Morgan's law on $(Q' + R')'$, we get $(Q'') \cdot (R'') = QR$.
So, RHS = $P' + QR$.Now we check if $PQ + R' = P' + QR$ for all P, Q, R.
Let's test with a counter-example. Let P=1, Q=1, R=0.
LHS = $1 \cdot 1 + 0' = 1 + 1 = 1$.
RHS = $1' + 1 \cdot 0 = 0 + 0 = 0$.
Since LHS $\neq$ RHS, the equality does not hold in general. The operator # is not associative.
Statement (S1) is false.Since only statement (S2) is true, the correct option is (B).

UFP = $(30 \times 4) + (60 \times 5) + (23 \times 4) + (8 \times 10) + (2 \times 7)$
UFP = $120 + 300 + 92 + 80 + 14 = 606$. There are 14 factors in total.

• 4 factors are not applicable (value = 0).
• 4 factors have value 3.
• Remaining factors ($14 - 4 - 4 = 6$) have value 4.

TDI = $(4 \times 0) + (4 \times 3) + (6 \times 4) = 0 + 12 + 24 = 36$. FP = $\text{UFP} \times (0.65 + 0.01 \times \text{TDI})$
FP = $606 \times (0.65 + 0.01 \times 36) = 606 \times (0.65 + 0.36) = 606 \times 1.01 = 612.06$.The computed value is approximately 612.

When a webpage is saved locally, relative URLs (like "2.html") resolve relative to the local file path (e.g., file:///C:/Users/.../2.html). To make them resolve to the original web server location, the <base> tag must be used in the HTML head.The <base> tag specifies the base URL for all relative URLs in a document. By adding <base href="http://www.yourname.com/">, a relative link like "2.html" will be interpreted as "http://www.yourname.com/2.html", which correctly points to the server.Option (B) represents this tag (despite the non-standard colon syntax used in the question options, it is the only semantically correct choice for setting a base URL).

Statement I: TCP is a full-duplex protocol, meaning data can be sent and received simultaneously on a single connection. This is correct.
Statement II: TCP does have an option for Selective Acknowledgement (SACK), which allows the receiver to inform the sender about all segments that have been received successfully, so the sender only needs to retransmit the lost segments. The statement says it has 'no option', so it is incorrect.
Statement III: TCP is a byte-stream oriented protocol, not a message-stream oriented protocol. UDP is message-oriented. So this statement is incorrect.
Therefore, only statement I is correct.

The sum of cubes of the first $n$ natural numbers is given by:$$\sum_{i=0}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4} = \frac{n^4 + 2n^3 + n^2}{4}$$This expression is of the order $n^4$, so:

• It is $\Theta(n^4)$ (Statement I is correct).
• It is NOT $\Theta(n^5)$ (Statement II is incorrect).
• It is $O(n^5)$ because $n^4 \leq c \cdot n^5$ for large $n$ (Statement III is correct).
• It is $\Omega(n^3)$ because $n^4 \geq c \cdot n^3$ for large $n$ (Statement IV is correct).

Thus, I, III, and IV are correct, but II is not. This matches option (C).

In any binary tree, let:

• $n_0$ be the number of leaf nodes (nodes with 0 children).
• $n_1$ be the number of nodes with exactly 1 child.
• $n_2$ be the number of nodes with exactly 2 children.
• $N$ be the total number of nodes.
• $E$ be the total number of edges.

We know that $N = n_0 + n_1 + n_2$.
Also, the number of edges $E = N - 1$.
From the sum of degrees (or counting edges from parents), $E = 0 \cdot n_0 + 1 \cdot n_1 + 2 \cdot n_2 = n_1 + 2n_2$.
Equating the two expressions for $E$:
$N - 1 = n_1 + 2n_2$
$(n_0 + n_1 + n_2) - 1 = n_1 + 2n_2$
$n_0 - 1 = n_2$
$n_2 = n_0 - 1$Given $n_0 = 200$ leaf nodes:
$n_2 = 200 - 1 = 199$.
So, the number of nodes with exactly two children is 199.

The load factor $\alpha$ of a hash table is defined as the ratio of the number of elements stored in the table to the total number of slots (or buckets) in the table.Given:
Number of elements = 2000
Number of slots = 25Formula for load factor:
$\alpha = \frac{\text{Number of elements}}{\text{Number of slots}}$Substitute the given values:
$\alpha = \frac{2000}{25}$Calculate the value:
$\alpha = 80$Thus, the load factor for the hash table is 80.

Let the given matrix be $A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{pmatrix}$.
We are given that $\lambda = 1$ is an eigenvalue.
To find the eigenvectors corresponding to $\lambda = 1$, we need to solve the equation $(A - \lambda I)v = 0$, where $I$ is the identity matrix and $v$ is the eigenvector.Substitute $\lambda = 1$:
$(A - I)v = 0$$A - I = \begin{pmatrix} 1-1 & -1 & 2 \\ 0 & 1-1 & 0 \\ 1 & 2 & 1-1 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 2 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{pmatrix}$Let the eigenvector be $v = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$.
The system of equations becomes:
1) $0x - 1y + 2z = 0 \implies -y + 2z = 0 \implies y = 2z$
2) $0x + 0y + 0z = 0$ (This equation is trivial and provides no new information)
3) $1x + 2y + 0z = 0 \implies x + 2y = 0$Now, substitute $y = 2z$ from equation (1) into equation (3):
$x + 2(2z) = 0$
$x + 4z = 0$
$x = -4z$So, the eigenvector $v$ can be written as:
$v = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -4z \\ 2z \\ z \end{pmatrix}$We can factor out $z$:
$v = z \begin{pmatrix} -4 \\ 2 \\ 1 \end{pmatrix}$Since an eigenvector must be non-zero, $z \neq 0$. Let $z = \alpha$, where $\alpha$ is any non-zero real number.
Thus, the set of eigenvectors corresponding to the eigenvalue 1 is $\{\alpha(-4,2,1)|\alpha \neq 0, \alpha \in \mathbb{R}\}$.This matches option (B).

We need to evaluate the limit: $\lim_{x\to\infty} (1 + x^2)e^{-x}$First, rewrite the expression to identify its form as $x \to \infty$:
$(1 + x^2)e^{-x} = \frac{1 + x^2}{e^x}$As $x \to \infty$, the numerator $(1 + x^2) \to \infty$ and the denominator $e^x \to \infty$.
This is an indeterminate form of type $\frac{\infty}{\infty}$, so we can apply L'Hôpital's Rule.First application of L'Hôpital's Rule:
Take the derivative of the numerator and the denominator:
$\frac{d}{dx}(1 + x^2) = 2x$
$\frac{d}{dx}(e^x) = e^x$So, the limit becomes:
$\lim_{x\to\infty} \frac{2x}{e^x}$This is still an indeterminate form of type $\frac{\infty}{\infty}$. We apply L'Hôpital's Rule again.Second application of L'Hôpital's Rule:
Take the derivative of the new numerator and denominator:
$\frac{d}{dx}(2x) = 2$
$\frac{d}{dx}(e^x) = e^x$So, the limit becomes:
$\lim_{x\to\infty} \frac{2}{e^x}$Now, as $x \to \infty$, $e^x \to \infty$. Therefore, $\frac{2}{e^x} \to 0$.Thus, $\lim_{x\to\infty} (1 + x^2)e^{-x} = 0$.The correct option is (A).

We need to form a 4-digit number $d_1 d_2 d_3 d_4$ such that $d_1 \le d_2 \le d_3 \le d_4$ using digits from the set $\{1, 2, 3\}$.
This is a problem of combinations with repetition (stars and bars), as the order is fixed (non-decreasing) and determined solely by the count of each digit chosen.The number of ways to choose $r=4$ digits from $n=3$ types with repetition is given by:$$\binom{n+r-1}{r} = \binom{3+4-1}{4} = \binom{6}{4}$$$$\binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2} = 15$$Alternatively, we can list the combinations:

• All same (e.g., 1111): 3 ways
• 3 same, 1 different (e.g., 1112): $3 \times 2 = 6$ ways
• 2 same, 2 same (e.g., 1122): 3 ways
• 2 same, 2 different (e.g., 1123): 3 ways

Total = $3 + 6 + 3 + 3 = 15$.

Let $H$ be the proposition "The result is head" and $T$ be the proposition "The person is telling the truth".
The statement given by the person is $S: H \iff T$.Case 1: The person is Type 1 (Truth-teller).
Then $T$ is true. Since they tell the truth, the statement $S$ must be true.
So, $H \iff \text{True}$ is true, which implies $H$ is true.Case 2: The person is Type 2 (Liar).
Then $T$ is false. Since they lie, the statement $S$ must be false.
So, $H \iff \text{False}$ is false. The biconditional $P \iff Q$ is false if and only if $P$ and $Q$ have different truth values.
Since $T$ is false, $H$ must be true for the statement to be false.In both cases, $H$ is true. Therefore, the result is head.

Let's insert the elements sequentially into an empty BST:The levels are:
Level 1: 71
Level 2: 65, 84
Level 3: 69, 83
Level 4: 67The lowest level (deepest) is Level 4, which contains the element 67.