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 a quadratic, its extreme value (minimum if the leading coefficient is positive, maximum if negative) can be found using the vertex formula, or by setting the quadratic equal to that value and requiring a repeated root (discriminant zero). Once you have both extreme values in terms of c, combine them into a single inequality.

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.