Question

Consider the sets $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 = 25\}$, $B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + 9y^2 = 144\}$, $C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \leq 4\}$, and $D = A \cap B$. The total number of one-one functions from the set $D$ to the set $C$ is:

• A. 15120
• B. 19320
• C. 17160
• D. 18290

Hint:

The number of one-one functions from a set with $m$ elements to a set with $n$ elements is given by $n \times (n-1) \times (n-2) \times \ldots \times (n-m+1)$.

Answer 1

The Correct Option is C

1. Identify the sets $A$ and $B$: - $A: x^2 + y^2 = 25$ - $B: \frac{x^2}{144} + \frac{y^2}{16} = 1$ 
2. Solve for the intersection $D = A \cap B$: 
- Substitute $x^2 + y^2 = 25$ into $x^2 + 9y^2 = 144$: \[ x^2 + 9(25 - x^2) = 144 \] \[ x^2 + 225 - 9x^2 = 144 \] \[ -8x^2 = 144 - 225 \] \[ -8x^2 = -81 \] \[ x^2 = \frac{81}{8} \] \[ x = \pm \frac{9}{2\sqrt{2}} \] - Substitute $x$ back into $x^2 + y^2 = 25$: \[ y^2 = 25 - \frac{81}{8} \] \[ y = \pm \frac{\sqrt{119}}{2\sqrt{2}} \] 
3. Determine the elements of set $D$: \[ D = \left\{\left(\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}}\right), \left(\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}}\right), \left(-\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}}\right), \left(-\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}}\right)\right\} \] - Number of elements in set $D = 4$. 
4. Identify the set $C$: \[ C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \leq 4\} \] - Possible pairs $(x, y)$: \[ \{(0, 2), (2, 0), (0, -2), (-2, 0), (1, 1), (-1, -1), (1, -1), (-1, 1), (1, 0), (0, 1), (-1, 0), (0, -1), (0, 0)\} \] - Number of elements in set $C = 13$. 
5. Calculate the total number of one-one functions from set $D$ to set $C$: \[ \text{Total number of one-one functions} = 13 \times 12 \times 11 \times 10 = 17160 \] Therefore, the correct answer is (3) 17160.

Answer 2

We need the number of one–one (injective) functions from \(D=A\cap B\) to \(C\), where

\[ A=\{(x,y)\in\mathbb{R}^2:x^2+y^2=25\},\quad B=\{(x,y)\in\mathbb{R}^2:x^2+9y^2=144\},\quad C=\{(x,y)\in\mathbb{Z}^2:x^2+y^2\le 4\}. \]

Concept Used:

Find \(|D|\) by solving the system for intersection points. Count \(|C|\) by enumerating integer lattice points with \(x^2+y^2\le 4\). The number of injective functions from a set of size \(m\) to a set of size \(n\) is the permutation \( ^nP_m = \dfrac{n!}{(n-m)!} \).

Step-by-Step Solution:

Step 1: Compute \(D=A\cap B\) by solving the system:

\[ \begin{cases} x^2+y^2=25,\\ x^2+9y^2=144. \end{cases} \] \[ \text{Subtract: }(x^2+9y^2)-(x^2+y^2)=144-25 \;\Rightarrow\; 8y^2=119 \;\Rightarrow\; y^2=\frac{119}{8}. \] \[ x^2=25-y^2=25-\frac{119}{8}=\frac{81}{8}. \]

Thus the intersection points are \( (\pm \sqrt{81/8},\,\pm \sqrt{119/8}) \). All sign choices are allowed, giving

\[ |D|=4. \]

Step 2: Count \(|C|\): integer pairs with \(x^2+y^2\le 4\).

\[ \begin{aligned} x^2+y^2=0 &: (0,0)\quad \Rightarrow 1,\\ x^2+y^2=1 &: (\pm1,0),(0,\pm1)\quad \Rightarrow 4,\\ x^2+y^2=2 &: (\pm1,\pm1)\quad \Rightarrow 4,\\ x^2+y^2=4 &: (\pm2,0),(0,\pm2)\quad \Rightarrow 4. \end{aligned} \] \[ |C|=1+4+4+4=13. \]

Step 3: Number of one–one functions from \(D\) to \(C\):

\[ {}^{13}P_{4}=\frac{13!}{(13-4)!}=13\cdot 12\cdot 11\cdot 10=17160. \]

Final Computation & Result

The total number of one–one functions from \(D\) to \(C\) is 17160.