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.