Q997: Consider the sets $A = \{(x, y) \in R \times R : x^2 + y^2 =...

COMPETENCY BASED

UNDERSTAND

4 Marks 2025 JEE Main 2025 (Online) 4th April Morning Shift MCQ SINGLE

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

(A) 15120

(B) 18290

(D) 19320

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Correct Answer: C

Explanation

$A = \{(x, y) \in R \times R : x^2 + y^2 = 25\}$, $B = \{(x, y) \in R \times R: x^2 + 9y^2 = 144\}$
$x^2 + 9y^2 - (x^2 + y^2) = 144 - 25$
Plug in $y^2 = \frac{119}{8}$ into either equation to find $x$.
$x^2 = 25 - \frac{119}{8}$
$x^2 = \frac{200 - 119}{8}$
$x^2 = \frac{81}{8}$
$x = \pm \sqrt{\frac{81}{8}}$, $y = \pm \sqrt{\frac{119}{8}}$
Now, $C = \{(x, y) \in Z \times Z: x^2 + y^2 \leq 4\}$
Valid points are $(-2, 0), (-1, -1), (-1, 0), (-1, 1), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (1, -1), (1, 0), (1, 1)$
$\therefore$ Total valid points in $C = 13$
$\Rightarrow$ There are 4 distinct real points in set $D$
$\therefore$ The number of one-one functions from $D$ to $C$
$\Rightarrow {}^{13}P_4 \Rightarrow \frac{13!}{(13-4)!} = \frac{13!}{9!} = 17160$

AI Tutor Explanation

Step-by-Step Solution

Find the intersection of sets A and B (set D):

Set A: x² + y² = 25 (circle)

Set B: x² + 9y² = 144 (ellipse)

Subtract the equation of the circle from the equation of the ellipse:

(x² + 9y²) - (x² + y²) = 144 - 25

8y² = 119

y² = 119/8

y = ±√(119/8) = ±√(238)/4 ≈ ±5.45

Since y must be real, substitute y² back into the circle equation:

x² + 119/8 = 25

x² = 25 - 119/8 = (200 - 119)/8 = 81/8

x = ±√(81/8) = ±9/(2√2) = ±9√2/4 ≈ ±3.18

Therefore, D = {(9√2/4, √(238)/4), (9√2/4, -√(238)/4), (-9√2/4, √(238)/4), (-9√2/4, -√(238)/4)}

However, since the question implies integer solutions, there must be an error in the initial approach. Let's re-examine the equations.

From A: x² = 25 - y². Substitute into B: 25 - y² + 9y² = 144 => 8y² = 119. This confirms the non-integer solutions.

Let's try to find integer solutions by inspection. If x = ±3, then x² = 9. From A, y² = 25 - 9 = 16, so y = ±4. Check if (±3, ±4) satisfy B: 9 + 9(16) = 9 + 144 = 153 ≠ 144. So, these are not solutions.

If x = ±4, then x² = 16. From A, y² = 25 - 16 = 9, so y = ±3. Check if (±4, ±3) satisfy B: 16 + 9(9) = 16 + 81 = 97 ≠ 144. So, these are not solutions.

If x = 0, then y = ±5 from A. Substituting into B: 0 + 9(25) = 225 ≠ 144. If y = 0, then x = ±12 from B. Substituting into A: 144 ≠ 25.

There seems to be an error in the problem statement, as the intersection D contains no integer solutions. However, let's assume that the intersection points are approximately (±3, ±4) and (±4, ±3) for the sake of solving the problem. Then |D| = 4.
Find the number of elements in set C:

Set C: {(x, y) ∈ Z × Z: x² + y² ≤ 4}

Possible integer values for x and y are:

x = 0: y² ≤ 4 => y = -2, -1, 0, 1, 2 (5 values)

x = ±1: y² ≤ 3 => y = -1, 0, 1 (3 values each, 6 total)

x = ±2: y² ≤ 0 => y = 0 (1 value each, 2 total)

Total elements in C: 5 + 6 + 2 = 13. So, |C| = 13.
Calculate the number of one-to-one functions from D to C:

Since |D| = 4 and |C| = 13, the number of one-to-one functions is given by:

¹³P₄ = 13! / (13-4)! = 13! / 9! = 13 × 12 × 11 × 10 = 17160

Correct Answer: 17160

AI Suggestion: Option C

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the properties of sets, equations of circles and ellipses, and the concept of one-to-one functions to solve the problem. They must interpret the given information and apply their knowledge to find the solution.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concepts of sets, equations of circles and ellipses, and one-to-one functions. It involves applying these concepts to find the intersection of sets and then calculating the number of one-to-one functions.

Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. The question requires the application of multiple concepts (sets, coordinate geometry, functions) to solve a non-standard problem. It's not a direct textbook question but requires problem-solving skills.

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