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

Approach (plug each option and test directly): The whole problem is one inequality, $f_{\min}(c) > g_{\max}(c)$. Get clean formulas for both, then just substitute each candidate $c$.

Step 1: Completing the square (or vertex formula): for the upward parabola $f$, $f_{\min} = 8c - 4c^2$; for the downward parabola $g$, $g_{\max} = \dfrac{9c^2}{4} - 2c$.

Step 2: Test $c = \dfrac{1}{2}$: \[ f_{\min} = 8(0.5) - 4(0.25) = 4 - 1 = 3, \qquad g_{\max} = \frac{9(0.25)}{4} - 2(0.5) = 0.5625 - 1 = -0.4375. \] Here $3 > -0.4375$ holds. This option works.

Step 3 (rule out the rest quickly): For any $c \le 0$ (options $-\tfrac12$ and $-2$), $f_{\min} = 8c - 4c^2 \le 0$ while $g_{\max} = \dfrac{9c^2}{4} - 2c \ge 0$, so the inequality fails. For $c = 2$: $f_{\min} = 16 - 16 = 0$ but $g_{\max} = 9 - 4 = 5$, and $0 > 5$ is false.

Answer: Only $c = \dfrac{1}{2}$ satisfies the condition.