Question:
Let \( f(x) = x^2 + 6x + 1 \) and let \( R \) denote the set of points \( (x, y) \) in the coordinate plane such that \( f(x) + f(y) \) so and \( f(x) - f(y) \leq 0 \). Which of the following is closest to the area of \( R \)?
Let \( f(x) = x^2 + 6x + 1 \) and let \( R \) denote the set of points \( (x, y) \) in the coordinate plane such that \( f(x) + f(y) \) so and \( f(x) - f(y) \leq 0 \). Which of the following is closest to the area of \( R \)?
Show Hint
In inequalities involving quadratic functions, you can complete the square to make the analysis easier and find the area of the region.
Updated On: Apr 1, 2025
- 22
- 23
- 24
- 25
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The Correct Option is D
Solution and Explanation
The area of the region \( R \) can be found by solving the inequality \( f(x) + f(y) \leq 0 \) and \( f(x) - f(y) \leq 0 \). The closest value to the area is \( 25 \).
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