GATE CS 2015 Set 2

The correct phrase is 'completely forgot' (adverb modifying verb) and 'just don't know' (adverb modifying verb phrase). 'Don't just know' changes the meaning significantly. Therefore, option (C) is the correct choice.

The word 'personnel' refers to people employed in an organization or engaged in an organized undertaking such as military service.
(A) Should be 'personal' (private).
(B) Should be 'personal' (private).
(C) Correct usage: 'All personnel' refers to the staff/employees.
(D) Should be 'personal' (individual/private).

'Apparel' is a general term for clothing or garments. Fabric, textile, and fibre are materials used to make clothing, not the clothing items themselves.

Let $W$ be the weight of one pole.
We need to find the total weight of 10 poles, which is $10W$.Statement (I): "One fourth of the weight of a pole is 5 Kg."
This can be written as $\frac{1}{4}W = 5$ Kg.
From this, we can find $W = 4 \times 5 = 20$ Kg.
Then, the total weight of 10 poles is $10W = 10 \times 20 = 200$ Kg.
So, Statement (I) alone is sufficient.Statement (II): "The total weight of these poles is 160 kg more than the total weight of two poles."
Let $N$ be the number of poles, which is 10.
Total weight of $N$ poles is $NW$.
Total weight of 2 poles is $2W$.
So, $NW = 2W + 160$.
Given $N=10$, we have $10W = 2W + 160$.
$8W = 160$
$W = \frac{160}{8} = 20$ Kg.
Then, the total weight of 10 poles is $10W = 10 \times 20 = 200$ Kg.
So, Statement (II) alone is also sufficient.Since either statement alone is sufficient to find the total weight of 10 poles, the correct option is (C).The final answer is $\boxed{C}$

The given function is $f(x) = 1 - |x|$ defined on the interval $-1 \leq x \leq 1$.To find the maximum value of $f(x)$, we need to minimize $|x|$.
The absolute value function $|x|$ is always non-negative, i.e., $|x| \geq 0$.
The minimum value of $|x|$ occurs at $x=0$, where $|0|=0$.Since $x=0$ is within the given interval $[-1, 1]$, the minimum value of $|x|$ in this interval is $0$ at $x=0$.Substituting $x=0$ into the function:
$f(0) = 1 - |0| = 1 - 0 = 1$.This is the maximum value of $f(x)$ because $f(x) = 1 - |x|$ will be largest when $|x|$ is smallest.So, the function attains its maximum value of $1$ at $x=0$.Comparing this with the given options:
(A) $0, -1$ (x value, max value)
(B) $-1,0$ (x value, max value)
(C) $0, 1$ (x value, max value)
(D) $-1, 2$ (x value, max value)The correct option is (C).The final answer is $\boxed{C}$

Let's analyze each sentence for grammar and usage:(A) "Since the report lacked needed information, it was of no use to them."

• "lacked needed information" is grammatically correct.
• "it was of no use to them" is a correct and idiomatic expression.
• The sentence is concise and grammatically sound.

(B) "The report was useless to them because there were no needed information in it."

• The phrase "no needed information" is problematic. "Information" is an uncountable noun, so it should be "no needed information" (singular agreement) or "no information needed". The use of "were" with "information" is incorrect. It should be "there was no needed information".

(C) "Since the report did not contain the needed information, it was not real useful to them"

• "not real useful" is grammatically incorrect. "Real" is an adjective, but here an adverb is needed to modify "useful". It should be "not really useful".

(D) "Since the report lacked needed information, it would not had been useful to them."

• "would not had been" is grammatically incorrect. The correct conditional perfect form is "would not have been".

Based on this analysis, sentence (A) is the most suitable with respect to grammar and usage.The final answer is $\boxed{A}$

Let $PS = l$. Let $PQ = q$ and $PR = r$.
Given that PS is the angle bisector of $\angle QPR$ and $\angle QPS = 60^\circ$.
Since PS is the angle bisector, $\angle RPS = \angle QPS = 60^\circ$.
Therefore, the total angle $\angle QPR = \angle QPS + \angle RPS = 60^\circ + 60^\circ = 120^\circ$.The area of $\triangle PQR$ can be expressed as the sum of the areas of $\triangle PQS$ and $\triangle PRS$.
Area($\triangle PQR$) = $\frac{1}{2} \times PQ \times PR \times \sin(\angle QPR) = \frac{1}{2} qr \sin(120^\circ)$
Area($\triangle PQS$) = $\frac{1}{2} \times PQ \times PS \times \sin(\angle QPS) = \frac{1}{2} ql \sin(60^\circ)$
Area($\triangle PRS$) = $\frac{1}{2} \times PR \times PS \times \sin(\angle RPS) = \frac{1}{2} rl \sin(60^\circ)$So, $\frac{1}{2} qr \sin(120^\circ) = \frac{1}{2} ql \sin(60^\circ) + \frac{1}{2} rl \sin(60^\circ)$
Multiply by 2:
$qr \sin(120^\circ) = ql \sin(60^\circ) + rl \sin(60^\circ)$
$qr \sin(120^\circ) = l(q+r) \sin(60^\circ)$We know that $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ)$.
Substitute this into the equation:
$qr \sin(60^\circ) = l(q+r) \sin(60^\circ)$Since $\sin(60^\circ) \neq 0$, we can divide both sides by $\sin(60^\circ)$:
$qr = l(q+r)$Solving for $l$ (the length of PS):
$l = \frac{qr}{q+r}$This matches option (B).The final answer is $\boxed{\text{B}}$

We need to evaluate $fg (h (2,5,7,3), 4,6,8)$.
First, let's calculate $h(2,5,7,3)$.
Here, $p=2, q=5, r=7, s=3$.
Calculate $p \times q = 2 \times 5 = 10$.
Calculate $r \times s = 7 \times 3 = 21$.Since $r \times s (21) > p \times q (10)$, we use the second condition for $h(p,q,r,s)$:
$h(2,5,7,3) = \text{remainder of } (r \times s) / (p \times q) = 21 \pmod{10}$.
$21 = 2 \times 10 + 1$, so the remainder is $1$.
Thus, $h(2,5,7,3) = 1$.Now we need to evaluate $fg(1, 4, 6, 8)$.
Based on the definition $fgh(p, q, r, s) = f(p, q, r, s) \times g(p, q, r, s) \times h(p, q, r, s)$, it implies that $fg(A, B, C, D)$ would mean $f(A,B,C,D) \times g(A,B,C,D)$.
So, we need to find $f(1, 4, 6, 8) \times g(1, 4, 6, 8)$.$f(1, 4, 6, 8) = \max(1, 4, 6, 8) = 8$.
$g(1, 4, 6, 8) = \min(1, 4, 6, 8) = 1$.Finally, $fg(1, 4, 6, 8) = f(1, 4, 6, 8) \times g(1, 4, 6, 8) = 8 \times 1 = 8$.The value is 8.

Let the arithmetic sequence be $a_1, a_2, a_3, a_4, a_5$ with common difference $d$. So, $P=a_1, R=a_1+d, S=a_1+2d, T=a_1+3d, U=a_1+4d$.Statement I: $2P, 2R, 2S, 2T, 2U$
The terms are $2a_1, 2(a_1+d), 2(a_1+2d), 2(a_1+3d), 2(a_1+4d)$.
This can be written as $2a_1, 2a_1+2d, 2a_1+4d, 2a_1+6d, 2a_1+8d$.
This is an arithmetic sequence with the first term $2a_1$ and a common difference of $2d$.
So, Statement I is TRUE.Statement II: $P-3, R-3, S-3, T-3, U-3$
The terms are $a_1-3, (a_1+d)-3, (a_1+2d)-3, (a_1+3d)-3, (a_1+4d)-3$.
This can be written as $(a_1-3), (a_1-3)+d, (a_1-3)+2d, (a_1-3)+3d, (a_1-3)+4d$.
This is an arithmetic sequence with the first term $a_1-3$ and a common difference of $d$.
So, Statement II is TRUE.Statement III: $P^2, R^2, S^2, T^2, U^2$
The terms are $a_1^2, (a_1+d)^2, (a_1+2d)^2, (a_1+3d)^2, (a_1+4d)^2$.
Let's check the differences between consecutive terms:
Difference between the first two terms: $(a_1+d)^2 - a_1^2 = a_1^2 + 2a_1d + d^2 - a_1^2 = 2a_1d + d^2$.
Difference between the second and third terms: $(a_1+2d)^2 - (a_1+d)^2 = (a_1^2 + 4a_1d + 4d^2) - (a_1^2 + 2a_1d + d^2) = 2a_1d + 3d^2$.
For the sequence to be an arithmetic sequence, these differences must be equal:
$2a_1d + d^2 = 2a_1d + 3d^2$
$d^2 = 3d^2$
$2d^2 = 0$
This implies $d=0$. If $d=0$, then all terms $P, R, S, T, U$ are identical, and their squares would also form an arithmetic sequence (with common difference 0). However, for a general arithmetic sequence where $d \neq 0$, the squares do not form an arithmetic sequence. For example, if $P=1, R=2, S=3$ (common difference $d=1$), then $P^2=1, R^2=4, S^2=9$. The differences are $4-1=3$ and $9-4=5$, which are not equal. Thus, Statement III is generally FALSE.Based on the analysis, only statements I and II are true.The final answer is $\boxed{\text{B}}$

Let the position of O be at the origin $(0,0)$. $P = (0, -2)$ $N = (1 \cdot \cos(45^\circ), -1 \cdot \sin(45^\circ)) = \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \approx (0.707, -0.707)$ $M = (N_x, N_y + 4) = \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} + 4\right) \approx (0.707, 3.293)$ Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$\Delta x = M_x - P_x = 0.707 - 0 = 0.707$
$\Delta y = M_y - P_y = 3.293 - (-2) = 5.293$
$d = \sqrt{(0.707)^2 + (5.293)^2} = \sqrt{0.5 + 28.016} = \sqrt{28.516} \approx 5.34$The distance is approximately 5.34 km.

Let the propositions be:
$C$: Candidate is known to be corrupt
$E$: Candidate will be elected
$K$: Candidate is kindThe given statements are:
$S1: C \implies \neg E$
$S2: K \implies E$From $S2$, the contrapositive is $\neg E \implies \neg K$.
Using hypothetical syllogism on $S1$ ($C \implies \neg E$) and the contrapositive of $S2$ ($\neg E \implies \neg K$), we get:
$C \implies \neg K$ (If a candidate is known to be corrupt, he is not kind).The contrapositive of this derived statement ($C \implies \neg K$) is:
$K \implies \neg C$ (If a candidate is kind, he is not known to be corrupt).This matches option (C).

The set is $S = \{ 0, 1, 2, \dots, 10 \}$.
The number of elements in the set is $n = 10 - 0 + 1 = 11$.The cardinality of the power set of a set with $n$ elements is $2^n$.
Here, $n = 11$, so the cardinality is $2^{11}$.$2^{10} = 1024$
$2^{11} = 2 \times 1024 = 2048$.

Let $S$ be the set of positive integers.
$R$ is a relation on $S$ such that $aRb$ if and only if $a$ and $b$ are distinct and have a common divisor other than 1.Let's check the properties of $R$: Consider $a=2$, $b=4$, $c=6$.
$aRb$: $2$ and $4$ are distinct, and have a common divisor $2 > 1$. So, $2R4$ is true.
$bRc$: $4$ and $6$ are distinct, and have a common divisor $2 > 1$. So, $4R6$ is true.
Now, let's check $aRc$: $2$ and $6$ are distinct, and have a common divisor $2 > 1$. So, $2R6$ is true.
This example suggests transitivity might hold. However, we need to be careful with the 'distinct' condition. Let's try another example:
Consider $a=2$, $b=4$, $c=8$.
$aRb$: $2$ and $4$ are distinct, common divisor $2 > 1$. True.
$bRc$: $4$ and $8$ are distinct, common divisor $4 > 1$. True.
$aRc$: $2$ and $8$ are distinct, common divisor $2 > 1$. True. Let's try to find a counterexample for transitivity.
Suppose $aRb$ and $bRc$.
This means $a \neq b$, $b \neq c$.
There exists $d_1 > 1$ such that $d_1 | a$ and $d_1 | b$.
There exists $d_2 > 1$ such that $d_2 | b$ and $d_2 | c$. We need to check if $aRc$ is true, i.e., $a \neq c$ and there exists $d_3 > 1$ such that $d_3 | a$ and $d_3 | c$. Consider $a=2$, $b=6$, $c=3$.
$aRb$: $2$ and $6$ are distinct, common divisor $2 > 1$. True.
$bRc$: $6$ and $3$ are distinct, common divisor $3 > 1$. True.
Now, check $aRc$: $2$ and $3$ are distinct. Do they have a common divisor greater than 1? No, their only common divisor is 1. So, $2R3$ is false.
Therefore, $R$ is not transitive.Combining these findings:

• $R$ is not reflexive.
• $R$ is symmetric.
• $R$ is not transitive.

This matches option (D).

To find the number of divisors of 2100, we first find the prime factorization of 2100.$2100 = 21 \times 100$
$21 = 3 \times 7$
$100 = 10^2 = (2 \times 5)^2 = 2^2 \times 5^2$So, $2100 = (3 \times 7) \times (2^2 \times 5^2) = 2^2 \times 3^1 \times 5^2 \times 7^1$If a number $N$ has a prime factorization $N = p_1^{e_1} \times p_2^{e_2} \times \dots \times p_k^{e_k}$, then the number of divisors of $N$ is given by the product of one more than each exponent:
Number of divisors $= (e_1+1)(e_2+1)\dots(e_k+1)$For $2100 = 2^2 \times 3^1 \times 5^2 \times 7^1$:
$e_1 = 2$ (for prime 2)
$e_2 = 1$ (for prime 3)
$e_3 = 2$ (for prime 5)
$e_4 = 1$ (for prime 7)Number of divisors $= (2+1)(1+1)(2+1)(1+1)$
$= (3)(2)(3)(2)$
$= 6 \times 6$
$= 36$Thus, the number of divisors of 2100 is 36.

Let the given matrix be $A = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$.
To find the eigenvalues, we need to solve the characteristic equation $\det(A - \lambda I) = 0$, where $I$ is the identity matrix and $\lambda$ represents the eigenvalues.$A - \lambda I = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4-\lambda & 5 \\ 2 & 1-\lambda \end{bmatrix}$Now, calculate the determinant:
$\det(A - \lambda I) = (4-\lambda)(1-\lambda) - (5)(2) = 0$
$4 - 4\lambda - \lambda + \lambda^2 - 10 = 0$
$\lambda^2 - 5\lambda - 6 = 0$This is a quadratic equation. We can solve it by factoring or using the quadratic formula.
Factoring the quadratic equation:
We need two numbers that multiply to -6 and add to -5. These numbers are -6 and 1.
$(\lambda - 6)(\lambda + 1) = 0$This gives us two eigenvalues:
$\lambda_1 = 6$
$\lambda_2 = -1$The eigenvalues are $6$ and $-1$.
The question asks for the larger of the two eigenvalues.
Comparing $6$ and $-1$, the larger eigenvalue is $6$.Alternatively, using the trace and determinant:
For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the characteristic equation is $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$.
Here, $a=4, b=5, c=2, d=1$.
Trace $(A) = a+d = 4+1 = 5$.
Determinant $(A) = ad-bc = (4)(1) - (5)(2) = 4 - 10 = -6$.
So, the characteristic equation is $\lambda^2 - 5\lambda - 6 = 0$, which is the same as obtained above.
Solving $\lambda^2 - 5\lambda - 6 = 0$ yields $\lambda = 6$ and $\lambda = -1$.
The larger eigenvalue is $6$.

To find an element that is neither maximum nor minimum in an unordered list of $n$ distinct elements (assuming $n \ge 3$, as for $n < 3$ such an element does not exist):Alternatively, both the maximum and minimum can be found in approximately $3n/2$ comparisons by comparing elements in pairs.Once the maximum ($M$) and minimum ($m$) elements are identified, any other element in the list (if $n \ge 3$) is guaranteed to be neither the maximum nor the minimum. We can then iterate through the list to find the first element $x_i$ such that $x_i \ne M$ and $x_i \ne m$. This iteration takes at most $n$ comparisons.Combining these steps, the total number of comparisons will be proportional to $n$. Therefore, the complexity is $\Theta(n)$.For example, if the list is $[5, 1, 10, 2, 8]$:

• Find max (10) and min (1). This takes $O(n)$ comparisons.
• Iterate: 5 is not 10 and not 1. So 5 is an element that is neither max nor min. This takes $O(n)$ comparisons in the worst case to find such an element after max/min are known.

Thus, the overall complexity is $\Theta(n)$.

The given count sequence is $(0,0,1,1,2,2,3,3,0,0,...)$.Let's list the unique states in this sequence:

• The number 0 appears twice.
• The number 1 appears twice.
• The number 2 appears twice.
• The number 3 appears twice.

This means the counter cycles through 8 distinct states before repeating:To represent 8 distinct states, we need $N$ flip-flops such that $2^N \ge 8$.
$2^N \ge 8 \implies N \ge \log_2 8 \implies N \ge 3$.Therefore, a minimum of 3 JK flip-flops are required to construct a synchronous counter with this sequence.

Given values:

• Cache miss time ($T_{miss}$) = 50 nanoseconds
• Cache hit time ($T_{hit}$) = 5 nanoseconds
• Cache hit rate ($H$) = 80% = 0.8

The cache miss rate ($M$) is $1 - H$.

• Cache miss rate ($M$) = $1 - 0.8 = 0.2$

The formula for average memory access time ($T_{avg}$) is:
$T_{avg} = (H \times T_{hit}) + (M \times T_{miss})$Substitute the given values into the formula:
$T_{avg} = (0.8 \times 5 \text{ ns}) + (0.2 \times 50 \text{ ns})$
$T_{avg} = 4 \text{ ns} + 10 \text{ ns}$
$T_{avg} = 14 \text{ ns}$Thus, the average read access time is 14 nanoseconds.

To find the length of the TLB tag, we first determine the components of the virtual address:

• Total entries = 128
• Associativity = 4-way set associative
• Number of sets = $\frac{128}{4} = 32 = 2^5$
• Therefore, the TLB Index requires 5 bits.

5. TLB Tag: VPN bits - TLB Index bits = $27 - 5 = 22$ bits.Thus, the minimum length of the TLB tag is 22 bits.

Let's evaluate each statement:

• Statement I: Turing decidable languages (also known as Recursive languages) are closed under complementation. If a language $L$ is decidable, there exists a Turing machine that halts on all inputs. We can construct a machine for $\bar{L}$ by simply swapping the accept and reject states of the machine for $L$. Thus, Statement I is true.
• Statement II: The class NP is a subset of the class of decidable languages ($NP \subseteq PSPACE \subseteq EXPTIME \subseteq Decidable$). Therefore, every language in NP is decidable. There is no language in NP that is undecidable. Thus, Statement II is false.
• Statement III: As established above, $NP \subseteq Decidable$. Therefore, if $L \in NP$, then $L$ is Turing decidable. Thus, Statement III is true.

Statements I and III are true, which corresponds to option (D).

The function foo is a recursive function that processes a string character by character.