Question

A value of $c$ for which the minimum value of $f(x) = x^2 - 4cx + 8c$ is greater than the maximum value of $g(x) = -x^2 + 3cx - 2c$, is:

• A. $\dfrac{1}{2}$
• B. $-\dfrac{1}{2}$
• C. $-2$
• D. $2$

Hint:

For quadratic functions, compare minimum and maximum values by: \begin{itemize} \item Finding vertex values using $x = -\frac{b}{2a}$, \item Substituting back to get extremum values, \item Then solving the resulting inequality in the parameter (here, $c$). \end{itemize}

Answer 1

The Correct Option is A

Step 1: Minimum value of $f(x)$. \[ f(x) = x^2 - 4cx + 8c \] is an upward-opening parabola (coefficient of $x^2$ is positive). Its vertex is at \[ x_{\min} = -\frac{b}{2a} = -\frac{-4c}{2\cdot 1} = 2c. \] So the minimum value is \[ f_{\min} = f(2c) = (2c)^2 - 4c(2c) + 8c = 4c^2 - 8c^2 + 8c = -4c^2 + 8c. \]
Step 2: Maximum value of $g(x)$. \[ g(x) = -x^2 + 3cx - 2c \] is a downward-opening parabola (coefficient of $x^2$ is negative). Its vertex is at \[ x_{\max} = -\frac{b}{2a} = -\frac{3c}{2(-1)} = \frac{3c}{2}. \] So the maximum value is \[ g_{\max} = g\!\left(\frac{3c}{2}\right) = -\left(\frac{3c}{2}\right)^2 + 3c\left(\frac{3c}{2}\right) - 2c = -\frac{9c^2}{4} + \frac{9c^2}{2} - 2c. \] Combine the $c^2$ terms: \[ -\frac{9c^2}{4} + \frac{9c^2}{2} = -\frac{9c^2}{4} + \frac{18c^2}{4} = \frac{9c^2}{4}. \] Thus, \[ g_{\max} = \frac{9c^2}{4} - 2c. \]
Step 3: Impose the condition $f_{\min} > g_{\max}$. \[ -4c^2 + 8c > \frac{9c^2}{4} - 2c. \] Multiply both sides by 4 to clear the denominator: \[ 4(-4c^2 + 8c) > 9c^2 - 8c \] \[ -16c^2 + 32c > 9c^2 - 8c. \] Bring all terms to one side: \[ 0 > 9c^2 + 16c^2 - 8c - 32c \Rightarrow 0 > 25c^2 - 40c. \] So \[ 25c^2 - 40c < 0. \] Factor: \[ 5c^2 - 8c < 0 \Rightarrow c(5c - 8) < 0. \]
Step 4: Solve the inequality in $c$. Critical points: $c = 0$, $c = \frac{8}{5} = 1.6$. Since the parabola $5c^2 - 8c$ opens upward, it is negative between the roots: \[ 0 < c < \frac{8}{5}. \]
Step 5: Check options. We need $c$ in the interval $(0, 1.6)$: - (A) $\dfrac{1}{2} = 0.5$ lies in $(0, 1.6)$ ✓ - (B) $-\dfrac{1}{2}$ is not in the interval ✗ - (C) $-2$ is not in the interval ✗ - (D) $2$ is not in the interval ✗ Therefore, the correct choice is \[ \boxed{\dfrac{1}{2}}. \]