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1. Let x be a real number such that x^2 + 7x + 12 <= 0. Find the largest possible value of x^2 + 5x + 6.

2. For 0 <= x <= 1, let f(x) = max{ x^2, 2x(1 - x), (1 - x)^2}.Find the minimum value of f(x) for 0 <= x <= 1. Note: For real numbers a, b, c,...,

max {a, b, c, . . . } denotes the maximum (or largest) number among a, b, c, . . .

May 25, 2024

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Here's the solution for 1.

We can approach this problem by analyzing the given inequality and maximizing the expression for x^2 + 5x + 6 within the constraints.

Factoring the Inequality: The inequality x^2 + 7x + 12 <= 0 can be factored as (x + 3)(x + 4) <= 0. This means either both factors are less than or equal to zero, or one factor is positive and the other is negative (the product becomes zero).

Identifying the Range of x: From the factored inequality, we know:

x ≤ -3 (when x + 3 is less than or equal to zero)

x ≤ -4 (when x + 4 is less than or equal to zero)

However, the second condition (x ≤ -4) is redundant as -3 is already less than or equal to -4. Therefore, the valid range for x is x ≤ -3.

Maximizing x^2 + 5x + 6: We want to maximize the value of x^2 + 5x + 6 within the constraint x ≤ -3. We can rewrite the expression as a completed square:

x^2 + 5x + 6 = (x^2 + 5x + 25/4) + 11/4 = (x + 5/2)^2 + 11/4

Since the square of any real number (x + 5/2) is non-negative (including zero), adding a constant positive value (11/4) will always result in a positive or zero value.

Largest Possible Value: The largest possible value for x^2 + 5x + 6 occurs when (x + 5/2)^2 is zero. This happens when x = -5/2. However, this value of x (-5/2) violates the constraint x ≤ -3.

Therefore, the largest possible value for x^2 + 5x + 6 within the allowed range of x is achieved when x is as close to -3 as possible (without exceeding it). This occurs when x = -3.

Evaluating the Maximum Value: Substituting x = -3 in the expression:

x^2 + 5x + 6 = (-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0

Answer: The largest possible value of x^2 + 5x + 6 is 0.

May 25, 2024