GATE CS 2014 Set 3

• Sticky: adhesive.
• Well-connected: having influential connections.
• Friendly: kind and pleasant.

• Q1 (2 marks): 21 students answered correctly. Marks = $21 \times 2 = 42$
• Q2 (3 marks): 15 students answered correctly. Marks = $15 \times 3 = 45$
• Q3 (2 marks): 23 students answered correctly. Marks = $23 \times 2 = 46$

• Statement (i): "Our understanding of the universe changes based on the positions of stars"
• Analysis: The passage states that the position of a specific star behind the sun provided the "clinching evidence" for a particular paradigm shift. It does not state or imply a general rule that our understanding of the universe changes based on the positions of stars. This is an overgeneralization. Thus, (i) cannot be inferred.
• Statement (ii): "Paradigm shifts usually occur at the beginning of centuries"
• Analysis: The passage mentions that several hypotheses were proposed "by the beginning of the $20^{\text{th}}$ century." This is a specific historical observation about one event and does not support the general claim that paradigm shifts usually occur at the beginning of centuries. Thus, (ii) cannot be inferred.
• Statement (iii): "Stars are important objects in the universe"
• Analysis: While this statement is a generally accepted fact in astronomy, it is not an inference that can be drawn directly from the logical structure of the passage. The passage uses a star's position merely as a tool for experimental measurement in a specific instance. Thus, (iii) cannot be inferred.
• Statement (iv): "Experimental evidence was important in confirming this paradigm shift"
• Analysis: The passage explicitly states that "the clinching evidence was provided by experimental measurements of the position of a star...". The word "clinching" means decisive or confirming. Therefore, experimental evidence was indeed crucial in confirming this paradigm shift. Thus, (iv) is a valid inference.

• $G_1$ be the GDP in Rupees at the start of the period (2012).
• $G_2$ be the GDP in Rupees at the end of the period (2013).
• $E_1$ be the exchange rate at the start of the period: $E_1 = 50 \text{ Rs./USD}$.
• $E_2$ be the exchange rate at the end of the period: $E_2 = 60 \text{ Rs./USD}$.

Step 1: Express the final GDP in Rupees

Step 2: Calculate the GDP in USD for both periods

• Initial GDP in USD ($U_1$):

• Final GDP in USD ($U_2$):

Step 3: Calculate the percentage change in USD GDP

Step 1: Extract values from the graph

• For the year 2011:

• For the year 2012:

Step 2: Use the given condition

• $M_{2011} = 1.0 \times F$
• $M_{2012} = 1.5 \times F$

Step 3: Calculate the required ratio

• L: P implies Q is TRUE (since they are equivalent).
• M: Q implies P is TRUE (since they are equivalent).
• N: P is equivalent to Q is TRUE.

• 1: $< 4$ (False)
• 3: $< 4$ (False)
• 5: $\ge 4$ (True) and $\neq 15$ (True)
• 15: $= 15$ (False, since $L \neq G$)

• (B) Consider eigenvalues $\{10, -2\}$. The trace is $8 > 0$, but one eigenvalue is negative.
• (C) Consider eigenvalues $\{-1, -1\}$. The determinant is $1 > 0$, but both eigenvalues are negative.
• (D) Consider eigenvalues $\{-1, -1, 4\}$. The trace is $2$ and the determinant is $4$. Their product is $8 > 0$, but two eigenvalues are negative.

• For $x \in [0, \pi]$, $\sin x \ge 0 \implies |\sin x| = \sin x$.
• For $x \in [\pi, 2\pi]$, $\sin x \le 0 \implies |\sin x| = -\sin x$.

• $P$ varies ($0, 1$), $R$ varies ($0, 1$).
• $Q$ is constant $0$, $S$ is constant $0$.
• This group yields the term $\bar{Q}\bar{S}$.

• $P$ varies ($0, 1$), $R$ varies ($0, 1$).
• $Q$ is constant $1$, $S$ is constant $1$.
• This group yields the term $QS$.

• If $x$ is 1, $y = a$.
• If $x$ is 0 (else), $y = b$.

• Visit T.
• Visit R.
• Visit U's subtree: U has child W. Treating W as left child, in-order is W → U.
• Visit V.
• Sequence for R's subtree: $T \rightarrow R \rightarrow W \rightarrow U \rightarrow V$.

• The left grid is a $3 \times 3$ grid, which has $3 \times 3 = 9$ vertices.
• There is a bridge consisting of $1$ vertex connecting the two main components.
• The right component consists of a $3 \times 3$ grid with the center vertex missing ($8$ vertices) and one additional vertex connected to the right side.
• Total vertices $V = 9 \text{ (left)} + 1 \text{ (bridge)} + 8 \text{ (right grid)} + 1 \text{ (far right)} = 19$ vertices.

• Length 0: $\epsilon$ is in $L$ (take all exponents as $0$).
• Length 1: $a$ ($a^1$), $b$ ($b^1$) are in $L$.
• Length 2:
• $aa$ ($a^2$)
• $ab$ ($a^1 b^1$)
• $ba$ ($(ba)^1$)
• $bb$ ($b^2$)

• Length 3:
• $aaa$ ($a^3$)
• $aab$ ($a^2 b^1$)
• $aba$ ($a^1 (ba)^1$)
• $abb$ ($a^1 b^2$)
• $baa$ ($(ba)^1 a^1$)
• $bab$: To form $bab$, the first $b$ must come from $b^*$ or $(ba)^*$. If from $b^*$, we need $ab$ from $(ba)^* a^*$, which is impossible. If from $(ba)^*$, it must be $ba$, leaving $b$ to be matched by $a^*$, which is impossible. Thus, $bab \notin L$.
• $bba$ ($b^2 a^1$)
• $bbb$ ($b^3$)