Let total students be \( n(U) = 200 \). Football players: \( n(F) = 120 \) Cricket players: \( n(C) = 50 \) Both: \( n(F \cap C) = 30 \)

Students who play one game only: \[ n(F \setminus C) + n(C \setminus F) = (n(F) - n(F \cap C)) + (n(C) - n(F \cap C)) \] \[ = (120 - 30) + (50 - 30) = 90 + 20 = 110 \]

\(\therefore\) The number of students who play one game only = 110.

Solution:

(A) Ogive is used to determine the Median. ✅ True

(B) Given:
\( P(A) = \tfrac{1}{2}, \; P(B) = \tfrac{7}{12}, \; P(\text{not A and not B}) = \tfrac{1}{4} \)

\( P(A \cup B) = 1 - \tfrac{1}{4} = \tfrac{3}{4} \)
\( P(A \cap B) = P(A) + P(B) - P(A \cup B) \) \( = \tfrac{1}{2} + \tfrac{7}{12} - \tfrac{3}{4} = \tfrac{1}{3} \)
\( P(A)\cdot P(B) = \tfrac{1}{2} \times \tfrac{7}{12} = \tfrac{7}{24} \neq \tfrac{1}{3} \)

So, A and B are not independent. ❌ False

(C) Empirical relation: \[ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \] ✅ True

(D) Two–digit even numbers from {1,2,3,4,5}:
- Units digit must be even → {2, 4} → 2 choices.
- Tens digit (if repetition allowed) → 5 choices.
\[ \text{Total} = 5 \times 2 = 10 \] ✅ True (with repetition allowed)

✔ Final Answer: (A), (C), and (D) are correct.

Solution:

Number of straight lines from \(n\) points (no three collinear) is \(\binom{n}{2}\).

Here, \(n = 15\).

\[ \binom{15}{2} = \frac{15 \times 14}{2} = 105 \]

Adjustment for collinearity:
Out of 15 points, 5 are collinear.
Lines from these 5 points = \(\binom{5}{2} = 10\).
But actually they form only 1 line.
Extra counted = \(10 - 1 = 9\).

Correct total lines:

\[ 105 - 9 = 96 \]

Final Answer: The total number of straight lines formed = 96

Final Answer: (A) → (II), (B) → (I), (C) → (III), (D) → (IV)

Correct Option: A–II, B–I, C–III, D–IV

• Number of pages = 1K = $2^10$.
• Page size = 2 KB = $2^{11}$ bytes.
• Virtual address space = Number of pages × Page size = $2^{10}×2^{11}=2^{21}$ bytes.
• Bits required =21 bits.

Step 1: For (A)

\( y = \dfrac{1}{2 - \sin 3x} \)
Since \( \sin 3x \in [-1,1] \), we get \( 2 - \sin 3x \in [1,3] \).
Hence \( y \in \left[\tfrac{1}{3}, 1\right] \). → Matches with (III).

Step 2: For (B)

\( y = \dfrac{x^2 + x + 2}{x^2 + x + 1} = 1 + \dfrac{1}{x^2 + x + 1} \)
Since denominator is always positive, \( y > 1 \).
Minimum denominator = \(\tfrac{3}{4}\) at \(x = -\tfrac{1}{2}\).
So maximum \( y = 1 + \tfrac{1}{3/4} = \tfrac{7}{3} \).
Thus, Range = \((1, \tfrac{7}{3}] \). → Matches with (I).

Step 3: For (C)

\( y = \sin x - \cos x = \sqrt{2}\sin\!\left(x - \tfrac{\pi}{4}\right) \)
Hence, Range = \([-\sqrt{2}, \sqrt{2}] \). → Matches with (IV).

Step 4: For (D)

\( y = \cot^{-1}(-x) - \tan^{-1}(x) + \sec^{-1}(x) \)
Simplifying with inverse trig identities gives Range:
\(\left[\tfrac{\pi}{2}, \pi\right) \cup \left(\pi, \tfrac{3\pi}{2}\right]\). → Matches with (II).

Per unit cost comparison (lowest → highest):

• (E) Magnetic tape → Cheapest (backup/archive storage)
• (B) Magnetic disk → Hard disk, inexpensive
• (C) Optical disk → CD/DVD/Blu-ray, costlier than HDD
• (A) DRAM → Main memory, more expensive
• (D) SRAM → Cache memory, fastest and most expensive

Final Order (increasing per unit cost):

E < B < C < A < D

Shortcut (Formula):

For a parabola \(y^2=4ax\), an equilateral triangle inscribed with one vertex at the parabola’s vertex has side

\[ s = 8a\sqrt{3}. \]

Here \(y^2=8x \Rightarrow 4a=8 \Rightarrow a=2\). Hence

\[ s = 8\cdot 2\sqrt{3}=16\sqrt{3}. \]

Final Answer: \(16\sqrt{3}\)

Let the common ratio be \(r\).

\[ x_1 = a, \; x_2 = ar, \; x_3 = ar^2 \] \[ y_1 = b, \; y_2 = br, \; y_3 = br^2 \]

So the points are \((a,b), \; (ar,br), \; (ar^2,br^2)\).

Slopes:

Between first two points: \[ m_{12} = \frac{br - b}{ar - a} = \frac{b(r-1)}{a(r-1)} = \frac{b}{a} \] Between second and third points: \[ m_{23} = \frac{br^2 - br}{ar^2 - ar} = \frac{br(r-1)}{ar(r-1)} = \frac{b}{a} \]

Since \(m_{12} = m_{23}\), the points are collinear.

Final Answer: The points \((x_1,y_1), (x_2,y_2), (x_3,y_3)\) are collinear.

Solution:

Final Answer: (A) → (IV), (B) → (I), (C) → (II), (D) → (III)

Final Matching (List–I → List–II)

Answer: (A) → (IV), (B) → (I), (C) → (II), (D) → (III)

Solutions (with steps)

(A) \(x^2-4x+4y+4y^2=12\) \[ (x-2)^2-4+4\!\left[(y+\tfrac12)^2-\tfrac14\right]=12 \;\Rightarrow\; (x-2)^2+4(y+\tfrac12)^2=17 \] \[ \frac{(x-2)^2}{17}+\frac{(y+\tfrac12)^2}{17/4}=1 \] Ellipse with \(a^2=17,\; b^2=\tfrac{17}{4}\Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\tfrac14}=\frac{\sqrt{3}}{2}. \)

(B) \(5x^2+9y^2=45 \Rightarrow \frac{x^2}{9}+\frac{y^2}{5}=1\). Here \(a=\sqrt{9}=3,\; b=\sqrt{5}\). Latus rectum length (ellipse) \(= \dfrac{2b^2}{a}=\dfrac{2\cdot5}{3}=\dfrac{10}{3}.\)

(C) Tangent: \(x+y=a \Rightarrow y=-x+a\). Touches \(y=x-x^2\): \[ x-x^2=-x+a \Rightarrow x^2-2x+a=0 \] For tangency, discriminant \(=0\): \(4-4a=0 \Rightarrow a=1.\)

(D) \(3x^2-y^2=4 \Rightarrow \dfrac{x^2}{4/3}-\dfrac{y^2}{4}=1\). Hyperbola with \(a^2=\tfrac{4}{3},\, b^2=4\). \[ e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{4}{4/3}}=\sqrt{1+3}=2. \]

Curves: \(y^2=4x\) (right-opening) and \(x^2=4y\) (upward). Intersection: From \(x=\dfrac{y^2}{4}\) in \(x^2=4y\) ⇒ \(\dfrac{y^4}{16}=4y \Rightarrow y(y^3-64)=0\Rightarrow y=0,4\). Thus points are \((0,0)\) and \((4,4)\) in the first quadrant.

For \(0\le y\le 4\): right curve is \(x=2\sqrt{y}\) (from \(x^2=4y\)), left curve is \(x=\dfrac{y^2}{4}\) (from \(y^2=4x\)).

Area \(=\displaystyle \int_{0}^{4}\!\left(2\sqrt{y}-\frac{y^2}{4}\right)\,dy = \left[\frac{4}{3}y^{3/2}-\frac{y^3}{12}\right]_{0}^{4} = \frac{32}{3}-\frac{16}{3}=\frac{16}{3}.\)

Final Answer: \(\displaystyle \frac{16}{3}\) square units.

Consider regions for \(x\) around 1 and 2.

1) \(x\le 1:\quad |x-1|+|x-2|=(1-x)+(2-x)=3-2x \ge 4 \Rightarrow x\le -\tfrac12.\)

2) \(1\le x\le 2:\quad |x-1|+|x-2|=(x-1)+(2-x)=1\) (not \(\ge4\)). No solutions.

3) \(x\ge 2:\quad |x-1|+|x-2|=(x-1)+(x-2)=2x-3 \ge 4 \Rightarrow x\ge \tfrac{7}{2}.\)

Final Answer: \(x \in (-\infty,\,-\tfrac12] \,\cup\, [\tfrac{7}{2},\,\infty)\).

Correct Option: 1

1. Kamal is to the left of Amrita ⇒ K is somewhere left of A.
2. Dipak is to the right of Amrita ⇒ A is left of D.
3. Ankit is between Amrita and Dipak ⇒ order among these three must be A – N – D (from left to right).
4. Now place Kamal (K) to the left of Amrita as required. With four seats, the only arrangement that satisfies all:
   K  –  A  –  N  –  D

Students at the corners: Kamal (left corner) and Dipak (right corner).

A completely skewed BST (all nodes on one side) has height \(n\). Searching may traverse every node in the worst case.

Worst-case time complexity: \(O(n)\)

Given parametric equations:

To find the last node of a singly linked list, we must keep moving until the link (next pointer) becomes NULL. This ensures we stop exactly at the last node.

// Structure of node
struct Node {
    int data;
    struct Node* link;
};

// Function to get last node
struct Node* getLastNode(struct Node* head) {
    struct Node* temp = head;

    // Traverse until last node
    while (temp -> link != NULL) {
        temp = temp -> link;
    }

    return temp;  // Now temp points to last node
}
  

Explanation:

• Initialize a pointer temp with head.
• Keep moving forward using temp = temp → link while temp → link != NULL.
• When loop ends, temp is pointing to the last node.

Correct Code Logic: while (temp → link != NULL) temp = temp → link;

We are asked to evaluate

\(\displaystyle I = \int_a^b x f(x)\, dx \quad \text{given } f(a+b-x)=f(x).\)

Step 1: Put substitution \(t=a+b-x\). Then \(dx=-dt\).

When \(x=a \Rightarrow t=b\), when \(x=b \Rightarrow t=a\).

So, \[ I = \int_a^b x f(x)\, dx = \int_b^a (a+b-t) f(t)(-dt) = \int_a^b (a+b-t) f(t)\, dt. \]

Step 2: Add both forms of \(I\):

\[ 2I = \int_a^b [x f(x) + (a+b-x) f(x)] dx = \int_a^b (a+b) f(x)\, dx. \]

Step 3: Simplify:

\[ I = \frac{a+b}{2} \int_a^b f(x)\, dx. \]

Final Answer: \(\displaystyle \frac{a+b}{2}\int_a^b f(x)\, dx\) → matches Option 1.

Given \(x^2+\dfrac{1}{x^2}=2\). Let \(t=x+\dfrac{1}{x}\). Then \(x^2+\dfrac{1}{x^2}=t^2-2\), so \(t^2-2=2 \Rightarrow t^2=4 \Rightarrow t=\pm2\).

From \(x+\dfrac{1}{x}=2 \Rightarrow x=1\) and from \(x+\dfrac{1}{x}=-2 \Rightarrow x=-1\).

Hence \(x^{256}+\dfrac{1}{x^{256}}= \begin{cases} 1+1=2,& x=1\\ (-1)^{256}+(-1)^{-256}=1+1=2,& x=-1 \end{cases}\).

Final Answer: \(2\).

The system of equations is:

2x + 2y + z = 1
4x + ky + 2z = 2
kx + 4y + z = 1

The coefficient matrix is

\(A = \begin{bmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{bmatrix}\).

The determinant is

\(\Delta = \begin{vmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{vmatrix}\).

Expanding:

\(\Delta = 2\begin{vmatrix} k & 2 \\ 4 & 1 \end{vmatrix} - 2\begin{vmatrix} 4 & 2 \\ k & 1 \end{vmatrix} + 1\begin{vmatrix} 4 & k \\ k & 4 \end{vmatrix}\).

\(\Delta = 2(k-8) - 2(4-2k) + (16-k^2)\).

\(\Delta = -k^2 + 6k - 8 = -(k-2)(k-4)\).

• If \(k \neq 2,4\), then \(\Delta \neq 0\) and the system has a unique solution.
• If \(k=4\): equations (1) and (2) are dependent, equation (3) reduces to the same relation, hence infinitely many solutions.
• If \(k=2\): substituting gives \(y=0\) and \(2x+z=1\), equation (3) is the same, hence infinitely many solutions.

Correct Statements: (A), (C), (D)

K-map (3 variables: A rows, BC columns in Gray order 00,01,11,10)

Groupings and terms:

• Cells (A=1, BC=11) & (A=1, BC=10) → pair → AB
• Cells (A=1, BC=00) & (A=1, BC=10) → pair → AC'
• Cells (A=0, BC=11) & (A=1, BC=11) → pair → BC

Simplified function: \(F = AB \;+\; AC' \;+\; BC\)

Correct option: 1) BC + AB + AC'

Solution:

(A) \(A + AB = A(1 + B) = A \cdot 1 = A\)   (Absorption law) ✅

(B) \((A + B)' = A'B'\)   (De Morgan’s law: complement of sum is product of complements) ✅

(C) \((A')' = A\)   (Double complement law) ✅

(D) \((AB)' = A' + B'\)   (De Morgan’s law: complement of product is sum of complements) ✅

Final Answer: All are true — (A), (B), (C), (D).

The function is

\( f(x) = [x]^n , \quad n \geq 2 \)

The GIF \([x]\) is discontinuous at all integers. Raising it to the integer power \(n \geq 2\) does not remove this discontinuity, because the jump still exists at each integer value of \(x\).

For non-integer \(x\), the function is constant over intervals \((m, m+1)\) where \(m \in \mathbb{Z}\), so it is continuous within each open interval between integers.

Final Answer: The function is discontinuous at all integers.

A circuit that adds two 1-bit numbers and produces outputs for Sum and Carry but does not handle carry input is called a Half Adder.

Boolean Expressions:

• Sum = \( A \oplus B \)
• Carry = \( A \cdot B \)

Final Answer: The circuit is a Half Adder.

Logical address is given as (2, 212).

Final Answer: The physical address corresponding to logical address (2, 212) is 2712.

Ways to Implement a Priority Queue

The correct answer is:

(A) Arrays, (C) Heap Data Structure, and (D) Linked List

Explanation:

• Arrays:
  • Unsorted Array: Insertion is O(1), Deletion is O(n).
  • Sorted Array: Insertion is O(n), Deletion is O(1).
• Heap Data Structure:
  • Binary heaps are commonly used for efficient implementation.
  • Insertion and Deletion take O(log n) time.
• Linked List:
  • Can be implemented as sorted or unsorted.
  • Efficiency depends on the choice of implementation.

Why not Fibonacci Tree?

Fibonacci trees are not used directly for priority queues. However, Fibonacci heaps (a separate data structure) can implement priority queues efficiently.

Banker's Algorithm Solution

Step 1: Need Matrix

The Need Matrix is calculated as:

Need = Max - Allocation

Step 2: Initial Available Resources

The Available resources at the start are:

• R1 = 4
• R2 = 4
• R3 = 5

Step 3: Evaluate Each Sequence

Sequence (A): P2 → P4 → P1 → P3

• P2: Need = [1, 0, 1], Available = [4, 4, 5]. Allocate resources. New Available = [7, 5, 7].
• P4: Need = [0, 1, 0], Available = [7, 5, 7]. Allocate resources. New Available = [10, 7, 9].
• P1: Need = [4, 5, 2], Available = [10, 7, 9]. Allocate resources. New Available = [14, 8, 11].
• P3: Need = [6, 0, 0], Available = [14, 8, 11]. Allocate resources. New Available = [17, 12, 14].

Result: Sequence (A) is valid.

Sequence (B): P2 → P1 → P3 → P4

• P2: Need = [1, 0, 1], Available = [4, 4, 5]. Allocate resources. New Available = [7, 5, 7].
• P1: Need = [4, 5, 2], Available = [7, 5, 7]. Allocate resources. New Available = [11, 6, 9].
• P3: Need = [6, 0, 0], Available = [11, 6, 9]. Allocate resources. New Available = [14, 10, 12].
• P4: Need = [0, 1, 0], Available = [14, 10, 12]. Allocate resources. New Available = [17, 12, 14].

Result: Sequence (B) is valid.

Sequence (C): P4 → P2 → P3 → P1

• P4: Need = [0, 1, 0], Available = [4, 4, 5]. Allocate resources. New Available = [7, 6, 7].
• P2: Need = [1, 0, 1], Available = [7, 6, 7]. Allocate resources. New Available = [10, 7, 9].
• P3: Need = [6, 0, 0], Available = [10, 7, 9]. Allocate resources. New Available = [14, 8, 11].
• P1: Need = [4, 5, 2], Available = [14, 8, 11]. Allocate resources. New Available = [17, 12, 14].

Result: Sequence (C) is valid.

Sequence (D): P1 → P4 → P2 → P3

• P1: Need = [4, 5, 2], Available = [4, 4, 5]. Cannot proceed as Need > Available.

Result: Sequence (D) is invalid.

Final Answer:

Safe Sequences: (A), (B), (C)

Invalid Sequence: (D)

Solution

Given: 4/5 of the students are boys, and the rest are girls. 2/5 of boys and 1/4 of girls are absent.

Step-by-Step Solution

• Total students = N
• Boys: (4/5) × N = 4N/5
• Girls: (1/5) × N = N/5
• Boys absent: (2/5) × (4N/5) = 8N/25
• Girls absent: (1/4) × (N/5) = N/20
• Boys present: (4N/5 - 8N/25 = 12N/25)
• Girls present: (N/5 - N/20 = 3N/20)
• Total present = (12N/25 + 3N/20)

Final Calculation

LCM of 25 and 20 is 100:
(12N/25 = 48N/100), (3N/20 = 15N/100)
Total present = (48N/100 + 15N/100 = 63N/100)

Answer: 63/100

Given: 4 calls to fork().

Formula: Total processes created = 2^n

Calculation:

• Total processes = 2^4 = 16
• Child processes = 16 - 1 = 15

Answer: 15 child processes

Matching List-I with List-II:

Final Answer:

(A) → (IV), (B) → (III), (C) → (I), (D) → (II)

Given:

• Amit cycled 15 km south.
• Then turned right (west) and cycled 7 km.
• Turned right again (north) and cycled 10 km.
• Finally, turned left (west) and cycled 5 km.

Calculation:

Vertical displacement: \( 15 - 10 = 5 \, \text{km (south)} \)
Horizontal displacement: \( 7 + 5 = 12 \, \text{km (west)} \)
Using Pythagoras theorem:
Distance = √(5² + 12²) = √(25 + 144) = √169 = 13 \, \text{km}

Answer:

Amit needs to cycle 13 km to return home.

Given Series: 7, 14, 28, ...

Pattern: Each term is multiplied by 2.

Formula for the nth term: Tn = 7 × 2n-1

Calculation:

T₁₀ = 7 × 2^10-1 = 7 × 2⁹ = 7 × 512 = 3584

Answer:

The 10th term is 3584.