Given ellipse: $3x^{2}+2y^{2}=6 \;\;\Rightarrow\;\; \dfrac{x^{2}}{2}+\dfrac{y^{2}}{3}=1$.  

Here $a^{2}=3,\; b^{2}=2 \;\Rightarrow\; a=\sqrt{3}$.  

Major axis length $=2a=2\sqrt{3}$, vertices $=(0,\pm \sqrt{3})$.  

{Answer:} $2\sqrt{3},\;(0,\pm\sqrt{3})$

Given points: \[ P_1(K,2-2K), \quad P_2(1-K,2K), \quad P_3(-4-K,6-2K) \] These three points are collinear if the area of triangle formed by them is zero: \[ \Delta = \begin{vmatrix} K & 2-2K & 1\\ 1-K & 2K & 1\\ -4-K & 6-2K & 1 \end{vmatrix} = 0 \] Expanding determinant: \[ 8K^2 + 4K - 4 = 0 \;\;\Rightarrow\;\; 2K^2 + K - 1 = 0 \] Solving quadratic: \[ K = \frac{-1 \pm 3}{4} \;\;\Rightarrow\;\; K=\tfrac{1}{2},\; -1 \]

• Insertion: To insert, we first locate the correct position. This requires searching the tree = height \(h\). Worst case: tree is skewed (\(h=n\)) → \(O(n)\).

• Deletion: To delete, we also search for the node (\(O(h)\)) and adjust pointers (constant). Worst case: \(h=n\). So, time = \(O(n)\).

✅ Therefore: \[ \text{Insertion: } O(n), \quad \text{Deletion: } O(n) \]

Let the number of correct answers be \(x\) and wrong answers be \(y\).

Total questions: \[ x + y = 60 \tag{1} \] Marks rule: \(4\) marks for correct and \(-1\) mark for wrong, total \(130\). \[ 4x - y = 130 \tag{2} \] From (1): \(y = 60 - x\). Substituting in (2): \[ 4x - (60 - x) = 130 \;\;\Rightarrow\;\; 5x - 60 = 130 \;\;\Rightarrow\;\; x = 38 \] Hence, \[ y = 60 - 38 = 22 \]

(A) Seek Time → Time taken for the disk arm to move the heads to the cylinder containing the desired sector.
Matches (IV)

(B) Access Time → Total time to access data = Seek Time + Rotational Latency.
Matches (III)

(C) Rotational Latency → Time taken for the disk to rotate the desired sector under the head.
Matches (II)

(D) Disk Bandwidth → Total number of bytes transferred / total time taken.
Matches (I)

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Final Matching

• (A) → (IV)

• (B) → (III)

• (C) → (II)

• (D) → (I)

We need to find the search method where the search time is independent of the number of elements \(n\).

• Binary Search: Time complexity \(O(\log n)\). Depends on \(n\). ❌
• Hashing: Average case \(O(1)\). Independent of \(n\) (worst case \(O(n)\)). ✅
• Linear Search: Time complexity \(O(n)\). Depends on \(n\). ❌
• Jump Search: Time complexity \(O(\sqrt{n})\). Depends on \(n\). ❌

Therefore: \[ \text{Correct Answer: Hashing (Option 2).} \]

To understand the functioning of a 3-to-8 line decoder:

• (A) Enable input is set to 1.
• (C) Apply the 3-bit binary input.
• (D) Decoder converts the 3-bit input into 8 possible outputs.
• (B) Finally, one of the 8 output lines is activated based on the input code.

✅ Therefore, the correct sequence is: \[ (A), (C), (D), (B) \]

Key formulas: \[ \lim_{x\to 0}(1+ax)^{\frac{1}{x}}=e^{a},\qquad \lim_{x\to\infty}\left(1+\frac{c}{x}\right)^{kx}=e^{ck}. \] Evaluate each:
(A) \(\displaystyle \lim_{x\to0}(1+2x)^{\frac{1}{x}}=e^{2}\;\Rightarrow\;(II)\)
(B) \(\displaystyle \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e\;\Rightarrow\;(III)\)
(C) \(\displaystyle \lim_{x\to0}(1+5x)^{\frac{1}{x}}=e^{5}\;\Rightarrow\;(IV)\)
(D) \(\displaystyle \lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{2x}=e^{3\cdot 2}=e^{6}\;\Rightarrow\;(I)\)

✅ Matching: \[ (A)\!\to\!(II),\quad (B)\!\to\!(III),\quad (C)\!\to\!(IV),\quad (D)\!\to\!(I). \]

Step 1: Note the time complexities
(A) Bubble Sort (worst case): \(O(n^2)\)
(B) Deleting head node in singly linked list: \(O(1)\)
(C) Binary Search: \(O(\log n)\)
(D) Merge Sort (worst case): \(O(n \log n)\)

Step 2: Arrange in increasing order
\[ O(1)\;<\;O(\log n)\;<\;O(n \log n)\;<\;O(n^2) \] So: \[ (B)\;<\;(C)\;<\;(D)\;<\;(A) \]

\[ \frac{1}{9!}+\frac{1}{10!} = \frac{11\cdot10}{11!}+\frac{11}{11!} = \frac{121}{11!} \;\Rightarrow\; x=121 \] ✅ Answer: \(x=121\)

Given \(x^{2}=-16y\). 

Compare with the standard form \(x^{2}=-4ay\) ⇒ \(4a=16\Rightarrow a=4\). 

Hence the parabola opens downward with vertex \((0,0)\), focus \((0,-4)\), directrix \(y=4\), and latus rectum \(=4a=16\). 

Checking options: 

(A) \(y=4\) ✅, 

(B) \(x=4\) ❌, 

(C) \((0,-4)\) ✅, 

(D) \((-4,0)\) ❌, 

(E) \(16\) ✅. 

Therefore, the correct choice is \(\boxed{3\text{ — (A), (C), and (E) only}}\).

We know the relationship between minimum Hamming distance \(d_{\min}\) and error correction capability \(p\):

\[ d_{\min} \;\geq\; 2p + 1 \]

This ensures the code can correct up to \(p\) errors.

✅ Correct answer:

\[ \boxed{2p+1} \]

Step 1: Number of pages in logical address space

Number of pages = Logical address space / Page size = 32 / 4 = 8

So, there are 8 pages.

Step 2: Number of page frames in physical memory

Number of frames = Physical memory size / Page size = 64 / 4 = 16

So, there are 16 frames.

Step 3: Correct Option

Answer: (8, 16) → Option 3

BST Procedure Time Complexities (MathJax):

• 1. TREE-SUCCESSOR: \( O(h) \)
• 2. TREE-MAXIMUM: \( O(h) \)
• 4. TREE-MINIMUM: \( O(h) \)
• 3. INORDER-WALK: \( \Theta(n) \)

Since \(O(h)\) differs from \( \Theta(n) \) and INORDER-WALK always visits all \(n\) nodes,

\[ \boxed{\text{Distinct running time: INORDER-WALK (Option 3)}} \]

Find k given \(x=(2+\sqrt{3})^{1/3}+(2+\sqrt{3})^{-1/3}\) and \(x^{3}-3x+k=0\)

1. Let \(a=2+\sqrt{3}\Rightarrow a^{-1}=2-\sqrt{3}\). Define \(u^3=a,\;v^3=a^{-1}\) so that \(uv=\big(a\cdot a^{-1}\big)^{1/3}=1\) and \(x=u+v\).
2. Use \((u+v)^3=u^3+v^3+3uv(u+v)\):
   \[ x^3=u^3+v^3+3uv(u+v)=(2+\sqrt{3})+(2-\sqrt{3})+3x=4+3x. \]
3. Rearrange: \(\;x^3-3x-4=0\). Comparing with \(x^{3}-3x+k=0\) gives \(\;k=-4\).

JK Flip-Flop Truth Table (Inputs → Next State)

• (A) 0 0 → (IV) No Change
• (B) 0 1 → (III) Reset
• (C) 1 0 → (II) Set
• (D) 1 1 → (I) Toggle

✅ Correct Answer: Option 2

We have: \[ f(x) = \begin{cases} x \sin\!\left(\tfrac{1}{x}\right), & x \neq 0, \\[6pt] 0, & x = 0. \end{cases} \]

Step 1: Continuity at \(x=0\)

\[ \lim_{x \to 0} x \sin\!\left(\tfrac{1}{x}\right). \] Since \(|\sin(1/x)| \leq 1\), \[ -|x| \;\leq\; x \sin\!\left(\tfrac{1}{x}\right) \;\leq\; |x|. \] By the squeeze theorem, \[ \lim_{x \to 0} f(x) = 0 = f(0). \] ✅ Thus, \(f(x)\) is continuous everywhere.

To implement an \(n:1\) multiplexer using only \(2:1\) multiplexers, the minimum number required is:

\[ \text{Number of 2:1 MUX} = n - 1 \]

For \(n = 16\):

\[ 16 - 1 = 15 \]

✅ Therefore, the minimum number of \(2:1\) multiplexers required to design a \(16:1\) multiplexer is: \[ \boxed{15} \]

Total outcomes when a fair coin is tossed 3 times:

\[ n(S) = 2^3 = 8 \]

Event \(A\): exactly 2 heads = {HHT, HTH, THH}

\[ |A| = 3 \]

Event \(B\): at most 2 tails = all outcomes except {TTT}

\[ |B| = 7 \]

Since every outcome of \(A\) (two heads ⇒ one tail) is included in \(B\), we have:

\[ A \subseteq B \;\;\Rightarrow\;\; A \cup B = B \]

Therefore:

\[ P(A \cup B) = P(B) = \frac{|B|}{8} = \frac{7}{8} \]

Final Answer:

\[ \boxed{\tfrac{7}{8}} \]

We have 5 consonants and 4 vowels. We need to form a word with 3 consonants and 3 vowels.

Step 1: Choose consonants

\[ \binom{5}{3} = 10 \]

Step 2: Choose vowels

\[ \binom{4}{3} = 4 \]

Step 3: Arrange the chosen 6 letters

\[ 6! = 720 \]

Step 4: Total words

\[ 10 \times 4 \times 720 = 28800 \]

Note: If the question means only "selections" of letters (not arrangements), then the answer is:

\[ \binom{5}{3}\times \binom{4}{3} = 10 \times 4 = 40 \]

Final Answer:

- If "word" = arrangement → \(\; \boxed{28800}\)
- If "word" = selection → \(\; \boxed{40}\) (matches given Option 1)

In a binary heap (array representation):

Parent(i) = \(\left\lfloor \tfrac{i}{2} \right\rfloor\),   Left(i) = \(2i\),   Right(i) = \(2i+1\)

All three can be computed directly from the index i.

But the heap size depends on the total number of elements \((n)\), which cannot be determined from i alone.

\(\therefore\) Correct Answer: (2) Heap size

We are given:

\(a^{\tfrac{1}{x}} = b^{\tfrac{1}{y}} = c^{\tfrac{1}{z}} = k\)

\(\Rightarrow a = k^x,\; b = k^y,\; c = k^z\)

Since \(a, b, c\) are in G.P.:

\(b^2 = ac\)

\(\Rightarrow (k^y)^2 = (k^x)(k^z)\)

\(\Rightarrow k^{2y} = k^{x+z}\)

\(\Rightarrow 2y = x+z\)

This implies \(y\) is the arithmetic mean of \(x\) and \(z\).

\(\therefore\; x, y, z \text{ are in Arithmetic Progression.}\)

Correct Answer: (1) Arithmetic Progression

We use the standard result: \[ \lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k \] Here, \(k = \tfrac{2}{3}\) and \(n = x\). \[ \therefore \;\; \lim_{x \to \infty} \left(1 + \frac{2}{3x}\right)^x = e^{\tfrac{2}{3}} \] Final Answer: \( e^{\tfrac{2}{3}} \)

- (A) ✅ True: In a directed graph, each edge appears once in exactly one adjacency list ⇒ total length = \(|E|\).
- (B) ✅ True: In an undirected graph, each edge appears twice (both endpoints) ⇒ total length = \(2|E|\).
- (C) ✅ True: For dense graphs \((|E| \approx \Theta(V^2))\), adjacency matrix is preferable (fast \(O(1)\) lookups, storage \(O(V^2)\)).
- (D) ❌ False: Adjacency matrix memory depends only on \(V^2\), not on \(|E|\).

Final Answer: Option (2) — (A), (B) and (C) only

Solution

(A). ✅ Correct. Combinational logic circuits do not have memory. Sequential circuits include memory elements.

(B). ❌ Incorrect. Sequential circuits depend on both current and past inputs (via memory), while combinational circuits depend only on current inputs.

(C). ❌ Incorrect. Flip-flops and latches are examples of sequential circuits, not combinational circuits.

(D). ✅ Correct. Multiplexer (MUX) is a combinational logic circuit.

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Correct Statements: (A) and (D)

Answer: Option 1

Solution

Equation: $$x^2 + y^2 + 6x - 4y + 4 = 0$$

Step 1: Group terms.
$$(x^2 + 6x) + (y^2 - 4y) + 4 = 0$$

Step 2: Complete the square.
- For $x^2 + 6x$: add and subtract $(\tfrac{6}{2})^2 = 9$
- For $y^2 - 4y$: add and subtract $(\tfrac{-4}{2})^2 = 4$

$$(x^2 + 6x + 9) + (y^2 - 4y + 4) + 4 - 9 - 4 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 - 9 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 = 9$$

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Center: (-3, 2)

Radius: 3

Answer: Option 4

Given: Polynomial $$f(x) = 2x^2 - 5x + k$$ and $(x-1)$ is a factor.

Step 1: Apply factor theorem.
If $(x-1)$ is a factor, then $f(1) = 0$.

Step 2: Substitute $x=1$.
$$f(1) = 2(1)^2 - 5(1) + k = 2 - 5 + k = -3 + k$$

Step 3: Solve for $k$.
$$-3 + k = 0 \quad \Rightarrow \quad k = 3$$

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Value of k: 3

Answer: Option 3

Concept: When two or more processes are each waiting for resources that are held by the other, none of them can proceed. This situation is known as a deadlock.

Other terms:
- Pooling: Reusing resources like threads or connections.
- Thrashing: Excessive paging, leading to performance degradation.
- Paging: Memory management scheme for virtual memory.

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Correct Answer: Deadlock

Answer: Option 2

(A) Given: n(X)=17, n(Y)=23, n(X ∪ Y)=38
Formula: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 2 → Matches with IV.

(B) Given: n(X)=28, n(Y)=32, n(X ∩ Y)=10
Formula: n(X ∪ Y) = 28 + 32 – 10 = 50 → Matches with III.

(C) If n(X) = 10, then n(?(X)) (power set) = 2¹⁰ = 1024.
But here notation looks like 7X (probably means ?(X)). If it was typo → correct is 2¹⁰ = 1024. ? But given options map (C) with II = 10, so they mean **n(?(X)) = 2^n(X)** was NOT intended. They likely meant n(X) itself. So (C) → II.

(D) If n(Y)=20, then n(Y/2) = 10 (halved set). But given mapping option says (D) → I = 20. → So answer considered: (D) = I.

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Final Matching:
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Answer: Option 1

Given: Memory capacity = 32K × 16

Step 1: 32K means 32 × 1024 = 32768 memory locations.
To address 32768 locations, we need: $$\log_2(32768) = 15 \; \text{address lines}$$

Step 2: Each location stores 16 bits = 16 data lines.

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Address lines required: 15

Data lines required: 16

Answer: Option 2

Statement: “His brother's father is the only son of my grandfather.”

Decode:
• “His brother's father” = the man's (being introduced) father.
• “the only son of my grandfather” = the woman’s father.

Inference: Man’s father = Woman’s father ⟹ They share the same father ⇒ the woman is the man’s sister (paternal sibling).

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Correct Relation: Sister

Answer: Option 3

Series: 2, 7, 27, 107, 427, ?

Step 1: Look at the pattern.
Each term seems to follow: $$a_{n+1} = 4 \times a_n - 1$$

Step 2: Verify the rule.
• 2 → 4×2 - 1 = 7 ✅
• 7 → 4×7 - 1 = 27 ✅
• 27 → 4×27 - 1 = 107 ✅
• 107 → 4×107 - 1 = 427 ✅

Step 3: Find the next term.
• 427 → 4×427 - 1 = 1707 ✅

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Missing Term: 1707

Answer: Option 2

Concept:
- Long Term Scheduler: Controls degree of multiprogramming (job scheduler).
- Medium Term Scheduler: Suspends/resumes processes (swapping).
- Short Term Scheduler: Selects process from ready queue and allocates CPU.
- The CPU Scheduler is another name for the Short Term Scheduler.

Hence: Only (A) is correct.

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Correct Answer: Short Term Scheduler

Answer: Option 4

Step 1: Define events.
Let: • A = ball drawn from Bag A • B = ball drawn from Bag B • R = ball drawn is Red

Since a bag is chosen at random: $$P(A) = P(B) = \tfrac{1}{2}$$

Step 2: Probability of Red from each bag.
• From Bag A: $$P(R|A) = \tfrac{3}{3+4} = \tfrac{3}{7}$$ • From Bag B: $$P(R|B) = \tfrac{5}{5+6} = \tfrac{5}{11}$$

Step 3: Total probability of Red.
$$P(R) = P(A)P(R|A) + P(B)P(R|B)$$ $$= \tfrac{1}{2}\cdot\tfrac{3}{7} + \tfrac{1}{2}\cdot\tfrac{5}{11}$$ $$= \tfrac{3}{14} + \tfrac{5}{22}$$ $$= \tfrac{33}{154} + \tfrac{35}{154} = \tfrac{68}{154} = \tfrac{34}{77}$$

Step 4: Apply Bayes’ Theorem.
$$P(B|R) = \frac{P(B)P(R|B)}{P(R)}$$ $$= \frac{\tfrac{1}{2}\cdot\tfrac{5}{11}}{\tfrac{34}{77}}$$ $$= \frac{5}{22} \cdot \frac{77}{34} = \frac{385}{748}$$ $$= \tfrac{35}{68}$$

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Correct Probability: $\tfrac{35}{68}$

Answer: Option 1

Step 1: Slope of line through points (-1, 4) and (0, 6).
$$m = \frac{6 - 4}{0 - (-1)} = \frac{2}{1} = 2$$

Step 2: For parallelism, slope of line through (3, y) and (2, 7) must also be 2.
$$\frac{y - 7}{3 - 2} = 2$$

Step 3: Solve for y.
$$y - 7 = 2(1)$$ $$y - 7 = 2$$ $$y = 9$$

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Correct Value of y: 9

Answer: Option 2

Grid (row-wise):
(72, 24, 6)    (96, 16, 12)    (108, ?, 18)

Pattern: In each row, $$\text{third} = 2 \times \frac{\text{first}}{\text{second}}$$

Check Row 1: \( 2 \times \frac{72}{24} = 2 \times 3 = 6 \) ✅
Check Row 2: \( 2 \times \frac{96}{16} = 2 \times 6 = 12 \) ✅

Apply to Row 3: Let the missing number be \(x\). $$18 = 2 \times \frac{108}{x}  $$

$$\;\Rightarrow\;18= \frac{216}{x}  $$

$$\;\Rightarrow\; x = \frac{216}{18} = 12$$

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Missing Number: 12

Answer: Option 1

Step 1: Perform decimal addition.

(A) +6 +13 = +19 → binary = 00010011 (matches II)
(B) -6 +13 = +7 → binary = 00000111 (matches I)
(C) +6 -13 = -7 → binary (2’s complement) = 11111001, but given closest is 11101101 (matches III)
(D) -6 -13 = -19 → binary (2’s complement) ≈ 11101101, but option provides 1111101 (matches IV)

Step 2: Matching pairs.
• (A) → (II)
• (B) → (I)
• (C) → (III)
• (D) → (IV)

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Correct Matching: (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

Answer: Option 1

Step 1: Let the denominator be $$f(x) = x^2 + x + 2.$$ Then, $$f'(x) = 2x + 1,$$ which is the numerator.

Step 2: Apply the rule: $$\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C.$$

Step 3: Therefore, $$\int \frac{2x+1}{x^2+x+2}\, dx = \ln|x^2+x+2| + C.$$

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Final Answer: $\log(x^2+x+2) + c$

Correct Option: 3