Consider the sets A = {(x, y) ∈ ℝ × ℝ : x² + y² = 25}, B = {(x, y) ∈ ℝ × ℝ : x² + 9y² = 144}, C = {(

Q: Wait, why do we reject the negative root here?

Actually, we don't reject it! The intersection points occur in all four quadrants because the equations are symmetric about both axes. We have four distinct points: $(\pm \sqrt{10.125}, \pm \sqrt{14.875})$. Each combination of signs gives a valid point.

Q: Why is the number of points in $C$ exactly 13? I keep miscounting.

The key is to be systematic. Start with $x=0$, then $x=\pm 1$, then $x=\pm 2$. For $x=0$, $y^2 \le 4 \implies y \in \{-2, -1, 0, 1, 2\}$ (5 points). For $x=\pm 1$, $y^2 \le 3 \implies y \in \{-1, 0, 1\}$ (3 points each, so 6 total). For $x=\pm 2$, $y^2 \le 0 \implies y=0$ (1 point each, so 2 total). Summing them up: $5 + 6 + 2 = 13$.

Q: Why is it $P(n, r)$ and not $C(n, r)$?

A one-one function (injective mapping) implies that the order of mapping matters. If element $d_1$ maps to $c_1$ and $d_2$ maps to $c_2$, that is a different function than $d_1$ mapping to $c_2$ and $d_2$ mapping to $c_1$. Since the assignment of each domain element to a unique codomain element is distinct, we use permutations.

Question

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

Options:

(A) 15120

(B) 19320

(C) 17160

(D) 18290

Sigma Percentile · Accepted Answer

### Analyzing the Setup: Set D

The first figure is a circle defined by the equation $x^2 + y^2 = 25$, which is centered at the origin with a radius of $5$. The second figure is an ellipse defined by $x^2 + 9y^2 = 144$.

By dividing the ellipse equation by $144$, we reveal its standard form:

$$
\frac{x^2}{12^2} + \frac{y^2}{4^2} = 1
$$

Set $D$ represents the intersection points of these two curves. To find these points, we substitute $x^2 = 25 - y^2$ into the ellipse equation:

$$
(25 - y^2) + 9y^2 = 144
$$

This simplifies to $8y^2 = 119$, which yields $y^2 = \frac{119}{8}$. Since $y^2$ is positive, we obtain two distinct values for $y$.

Substituting back, we find $x^2 = 25 - \frac{119}{8} = \frac{81}{8}$, which also yields two distinct values for $x$. Because both $x$ and $y$ have two possible signs, there are $2 	imes 2 = 4$ intersection points. Thus, **$n(D) = 4$**.

### The Lattice World: Counting Set C

Set $C$ is defined by the inequality $x^2 + y^2 \le 4$, where $x, y \in \mathbb{Z}$. We determine the number of integer lattice points by testing values of $x$:

If $x = 0$, then $y^2 \le 4$, implying $y \in \{-2, -1, 0, 1, 2\}$. This provides $5$ points.

If $x = \pm 1$, then $1 + y^2 \le 4$, which simplifies to $y^2 \le 3$. This allows $y \in \{-1, 0, 1\}$, providing $3$ points for each $x$, totaling $6$ points.

If $x = \pm 2$, then $4 + y^2 \le 4$, which implies $y^2 \le 0$, so $y = 0$. This provides $1$ point for each $x$, totaling $2$ points.

Summing these values, we get $5 + 6 + 2 = 13$. Therefore, **$n(C) = 13$**.

### The Final Mapping: One-One Functions

We are tasked with finding the number of one-one functions $f: D 	o C$. Since $n(D) = 4$ and $n(C) = 13$, a one-one function requires each element in $D$ to map to a unique element in $C$.

This is a permutation problem where we choose $4$ distinct elements from $13$ and arrange them. The number of such functions is given by:

$$
^{13}P_4 = \frac{13!}{(13-4)!} = 13 	imes 12 	imes 11 	imes 10
$$

Performing the calculation:

$$
13 	imes 12 = 156
$$
$$
156 	imes 11 = 1716
$$
$$
1716 	imes 10 = 17160
$$

The total number of one-one functions is **$17160$**.

Select Answer:

Visualized Solution

Visualizing Set $A$

Visualizing Set $B$

Finding Intersection Set $D$

Solving for $y$

Solving for $x$

Counting Elements in $D$

Defining Set $C$

Listing Points in $C$

Counting Elements in $C$

One-One Functions Concept

Final Calculation

Conclusion

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Lattice World

The Final Mapping

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