QPaperGen

Correct Answer: only (S2) is true

Explanation: To evaluate the relation \(R\) on the set \(A = \{0, 1, 2, 3, 4, 5\}\), we first need to understand the conditions for an element \((x, y)\) to be in \(R\). Specifically, \((x, y) \in R\) if and only if \(\max\{x, y\} \in \{3, 4\}\).


\[\] Number of elements in R (\(S_1\)):

For \(\max\{x, y\} = 3\), the possible pairs are \((0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3)\). There are 7 pairs.\[\]
For \(\max\{x, y\} = 4\), the possible pairs are \((0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)\). There are 9 pairs.

\[\] Combining these, the set \(R\) consists of a total of \(7 + 9 = 16\) elements. Therefore, statement \((S_1)\) is false.

\[\] Properties of R (\(S_2\)):

\[\] Reflexivity: A relation is reflexive if \((x, x) \in R\) for all \(x \in A\). For a pair \((x, x)\) to be in \(R\), \(\max\{x, x\} = x\) must be in \(\{3, 4\}\). This is not true for all elements in \(A\) (e.g., \((0, 0)\) is not in \(R\)). So, \(R\) is not reflexive.
\[\] Symmetry: A relation is symmetric if whenever \((x, y) \in R\), then \((y, x) \in R\). Since \(\max\{x, y\} = \max\{y, x\}\), if \((x, y) \in R\), then \((y, x)\) is also in \(R\). So, \(R\) is symmetric.
\[\] Transitivity: A relation is transitive if whenever \((x, y) \in R\) and \((y, z) \in R\), then \((x, z) \in R\). For example, \((0, 3) \in R\) and \((3, 1) \in R\), but \((0, 1) \notin R\) because \(\max\{0, 1\} = 1 \notin \{3, 4\}\). So, \(R\) is not transitive.

\[\] Statement \((S_2)\) claims that \(R\) is symmetric but neither reflexive nor transitive, which is correct. Therefore, statement \((S_2)\) is true.

Correct Answer: \(B \subset A\)

Explanation: For set \(A\):
\[
|x - 1| \leq 4 \;\; \Rightarrow \;\; -3 \leq x \leq 5
\]
\[
|y - 5| \leq 6 \;\; \Rightarrow \;\; -1 \leq y \leq 11
\]
So \(A\) represents a rectangle with vertices \((-3,-1), (5,-1), (5,11), (-3,11)\).

For set \(B\):
\[
16(x - 2)^2 + 9(y - 6)^2 \leq 144
\]
\[
\Rightarrow \;\; \frac{(x-2)^2}{9} + \frac{(y-6)^2}{16} \leq 1
\]
This is an ellipse centered at \((2,6)\) with semi-axes \(3\) (along \(x\)-axis) and \(4\) (along \(y\)-axis).

Clearly, the ellipse \(B\) lies entirely inside the rectangle \(A\).

\[
\therefore \quad B \subset A
\]

Correct Answer: 17

Explanation: We are given:
\[
x R y \;\;\Leftrightarrow\;\; 2x - y \in \{0,1\}.
\]

Thus,
\[
y = 2x \quad \text{or} \quad y = 2x - 1.
\]

For \(A = \{-3,-2,-1,0,1,2,3\}\), we list all possible pairs:

\[
R = \{(-1,-2),(0,0),(1,2),(-1,-3),(0,-1),(1,1),(2,3)\}.
\]

Hence,
\[
l = |R| = 7.
\]

\[ \textbf{Reflexivity:}\]
For reflexivity, we need \((a,a) \in R\) for all \(a \in A\).

But the rule requires \(2a - a = a \in \{0,1\}\), so only \((0,0)\) and \((1,1)\) are present.

The remaining \(5\) pairs \((a,a)\) for \(a \in \{-3,-2,-1,2,3\}\) are missing.
Thus, \(m = 5\) must be added.

\[ \textbf{Symmerty:}\]
A relation is symmetric if \((x,y) \in R \Rightarrow (y,x) \in R\).

Currently, the following pairs are missing:
\[
(-1,-2) \Rightarrow (-2,-1), \quad
(1,2) \Rightarrow (2,1), \quad
(-1,-3) \Rightarrow (-3,-1), \quad
(0,-1) \Rightarrow (-1,0), \quad
(2,3) \Rightarrow (3,2).
\]

So \(n = 5\) pairs must be added.

Final Count
\[
l + m + n = 7 + 5 + 5 = 17.
\]

\[
\therefore \quad \text{Correct Answer is (A) 17.}
\]

Correct Answer: 17160

Explanation: We are given:
\[
A = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 25\}, \quad
B = \{(x, y) \in \mathbb{R}^2 : x^2 + 9y^2 = 144\}.
\]

Subtracting the two equations:
\[
x^2 + 9y^2 - (x^2 + y^2) = 144 - 25 \quad \Rightarrow \quad 8y^2 = 119
\]
\[
\Rightarrow \quad y^2 = \tfrac{119}{8}.
\]

Substitute into \(x^2 + y^2 = 25\):
\[
x^2 = 25 - \tfrac{119}{8} = \tfrac{200 - 119}{8} = \tfrac{81}{8}.
\]

Hence,
\[
x = \pm \sqrt{\tfrac{81}{8}}, \quad y = \pm \sqrt{\tfrac{119}{8}}.
\]

So, the set \(D = A \cap B\) has **4 distinct real points**.

---

Now, for set \(C\):
\[
C = \{(x, y) \in \mathbb{Z}^2 : x^2 + y^2 \leq 4\}.
\]

Valid integer points are:
\[
(-2,0), \; (-1,-1), \; (-1,0), \; (-1,1), \; (0,-2), \; (0,-1), \; (0,0), \; (0,1), \; (0,2), \; (1,-1), \; (1,0), \; (1,1), \; (2,0).
\]

Thus, \(|C| = 13\).

---

We need the number of one-one functions from \(D\) (with \(|D|=4\)) to \(C\) (with \(|C|=13\)):
\[
= {}^{13}P_{4} = \frac{13!}{(13-4)!} = \frac{13!}{9!}.
\]

\[
= 13 \times 12 \times 11 \times 10 = 17160.
\]

\[
\therefore \quad \text{The total number of one-one functions is } 17160.
\]

Correct Answer: 12

Explanation: We are given \(A = \{-2,-1,0,1,2,3\}\) and
\[
xRy \iff y = \max\{x,1\}.
\]



\(\textbf{Step 1: Find \(R\)}\)

\[
\begin{aligned}
x = -2 &\implies y = \max(-2,1) = 1 \quad \Rightarrow (-2,1) \in R, \\
x = -1 &\implies y = 1 \quad \Rightarrow (-1,1) \in R, \\
x = 0 &\implies y = 1 \quad \Rightarrow (0,1) \in R, \\
x = 1 &\implies y = 1 \quad \Rightarrow (1,1) \in R, \\
x = 2 &\implies y = 2 \quad \Rightarrow (2,2) \in R, \\
x = 3 &\implies y = 3 \quad \Rightarrow (3,3) \in R.
\end{aligned}
\]

Thus,
\[
R = \{(-2,1), (-1,1), (0,1), (1,1), (2,2), (3,3)\}, \quad l = 6.
\]


\(\textbf{Step 2: Make \(R\) reflexive:}\)

A relation is reflexive if \((a,a) \in R\) for all \(a \in A\).

Currently, we already have \((1,1), (2,2), (3,3)\).

Missing pairs: \((-2,-2), (-1,-1), (0,0)\).

Thus, \[m = 3\].



\(\textbf{Step 3: Make \(R\) symmetric:}\)

A relation is symmetric if \((x,y) \in R \implies (y,x) \in R\).

Check:
\[\begin{aligned}
(-2,1) \implies (1,-2) \qquad & \text{missing.}\\
(-1,1) \implies (1,-1)\qquad & \text{missing.}\\
(0,1) \implies (1,0)\qquad & \text{missing.}\\
(1,1), (2,2), (3,3) \qquad & \text{are already symmetric.}\\
\end{aligned}
\]
So, we need to add \(3\) pairs. Hence \[n=3\]

\(
\textbf{Step 4: Final answer}\)

\[
l+m+n = 6+3+3 = 12.
\]

\[
\therefore \quad \text{The required value is } \boxed{12}.
\]

Correct Answer: 18

Explanation: We are given \(A = \{-3,-2,-1,0,1,2,3\}\) and
\[
xRy \iff 0 \leq x^2 + 2y \leq 4.
\]

\(\textbf{Step 1: Find valid pairs:}\)

Check for each \(y \in A\):
\[\]
- For \(y = -3\): condition \(\;0 \leq x^2 - 6 \leq 4 \;\Rightarrow\; 6 \leq x^2 \leq 10\).
Thus \(x = \pm 3\).
Pairs: \((3,-3), (-3,-3)\).
\[\]
- For \(y = -2\): \(0 \leq x^2 - 4 \leq 4 \;\Rightarrow\; 4 \leq x^2 \leq 8\).
Thus \(x = \pm 2\).
Pairs: \((-2,-2), (2,-2)\).
\[\]
- For \(y = -1\): \(0 \leq x^2 - 2 \leq 4 \;\Rightarrow\; 2 \leq x^2 \leq 6\).
Thus \(x = \pm 2\).
Pairs: \((-2,-1), (2,-1)\).
\[\]
- For \(y = 0\): \(0 \leq x^2 \leq 4\).
Thus \(x = -2,-1,0,1,2\).
Pairs: \((-2,0), (-1,0), (0,0), (1,0), (2,0)\).
\[\]
- For \(y = 1\): \(0 \leq x^2 + 2 \leq 4 \;\Rightarrow\; -2 \leq x^2 \leq 2\).
Thus \(x^2 \leq 2 \;\Rightarrow\; x = -1,0,1\).
Pairs: \((-1,1), (0,1), (1,1)\).
\[\]
- For \(y = 2\): \(0 \leq x^2 + 4 \leq 4 \;\Rightarrow\; x^2 = 0\).
Thus \(x=0\).
Pair: \((0,2)\).

\[\]
\(\textbf{Step 2: Count elements of \(R\)}\)

So the relation is
\(
R = \{(3,-3), (-3,-3), (-2,-2), (2,-2), (-2,-1), (2,-1), (-2,0), (-1,0), (0,0), (1,0), (2,0), (-1,1), (0,1), (1,1), (0,2)\}.
\)

Hence \(l = 15\).

\[\]

\(\textbf{Step 3: Make \(R\) reflexive:}\)

A relation is reflexive if \((a,a) \in R\) for all \(a \in A\).

Currently in \(R\):
\((-3,-3), (-2,-2), (0,0), (1,1)\) are present.

Missing: \((-1,-1), (2,2), (3,3)\).

So we must add \(m = 3\) elements.

\[\]

\(\textbf{Step 4: Final Answer}\)

\[
l+m = 15+3 = 18.
\]

\[
\therefore \quad \boxed{18}
\]

Correct Answer: 5