JEE Mains · Maths · STD 12 - 1. relation and function
Consider the sets \(\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} \times \mathbb{R}: \mathrm{x}^2+\mathrm{y}^2=25\right\}\), \(\mathrm{B}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} \times \mathbb{R}: \mathrm{x}^2+9 \mathrm{y}^2=144\right\}, \mathrm{C}=\{(\mathrm{x}, \mathrm{y})\) \(\left.\in \mathbb{Z} \times \mathbb{Z}: x^2+y^2 \leq 4\right\}\), 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
Answer & Solution
Correct Answer
(C) 17160
Step-by-step Solution
Detailed explanation
\(\mathrm{A}: \mathrm{x}^2+\mathrm{y}^2=25...(i)\) B : \(\frac{x^2}{144}+\frac{y^2}{16}=1...(ii)\) C: \(x^2+y^2 \leq 4...(iii)\) Solve (1) & (2) \(\begin{aligned} & x^2+9\left(25-x^2\right)=144 -8 x^2=144-225=-81 \\ & x= \pm \frac{9}{2 \sqrt{2}} \end{aligned}\)…
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