Quadratic Equations
Quadratic Equations
star_batch_jee_advanced_2025
Grade None
Question:
The greatest value of the function $f(x) = \frac{1}{2bx^2 - x^3 - 3b^2}$ on the interval $[-2, 1]$ depending on the parameter $b$ is(are):
-\frac{1}{3b^2} if b \in [0, 2]
\frac{1}{4b - 4 - 3b^2} if b \in [0, 4]
\frac{1}{8b - 16 - 3b^2} if b \leq 2
-\frac{1}{3b^2} if b \geq 2
Step-by-Step Solution
Key Concept: Since $f$ is the reciprocal of a cubic, maximizing $f$ requires finding the minimum value of the denominator across the interval, considering both endpoints and critical points.
To find the maximum of $f(x) = \frac{1}{2bx^2 - x^3 - 3b^2}$, we need to minimize the denominator $g(x) = 2bx^2 - x^3 - 3b^2$ on $[-2, 1]$. Taking $g'(x) = 4bx - 3x^2 = x(4b - 3x) = 0$ gives critical points at $x = 0$ and $x = \frac{4b}{3}$. We evaluate $g$ at critical points and endpoints: $g(-2) = 8b + 8 - 3b^2$, $g(0) = -3b^2$, $g(1) = 2b - 1 - 3b^2$, and $g(\frac{4b}{3}) = \frac{16b^2}{9} - \frac{64b^3}{27} - 3b^2$ (when in interval). For $b \leq 2$, the minimum occurs at $x = -2$ giving $f_{max} = \frac{1}{8b - 16 - 3b^2}$. For $b \geq 2$, the minimum occurs at $x = 0$ giving $f_{max} = -\frac{1}{3b^2}$.
Correct Answer: 3,4