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: An upward parabola has its minimum at the vertex; a downward parabola has its maximum at the vertex. Find both vertex values as functions of \(c\) and impose min \(>\) max.

Step 1: \(f(x) = x^2 - 4cx + 8c\) opens upward, so its minimum is at the vertex \(x = \dfrac{4c}{2} = 2c\). Plugging in: \[ f_{\min} = (2c)^2 - 4c(2c) + 8c = 4c^2 - 8c^2 + 8c = -4c^2 + 8c. \]

Step 2: \(g(x) = -x^2 + 3cx - 2c\) opens downward, so its maximum is at \(x = \dfrac{3c}{2}\). Plugging in: \[ g_{\max} = -\frac{9c^2}{4} + \frac{9c^2}{2} - 2c = \frac{9c^2}{4} - 2c. \]

Step 3: Impose \(f_{\min} > g_{\max}\): \[ -4c^2 + 8c > \frac{9c^2}{4} - 2c. \] Multiply by 4: \(-16c^2 + 32c > 9c^2 - 8c\), i.e. \(25c^2 - 40c < 0\), so \(5c(5c - 8) < 0\).

Step 4: The product is negative between the roots \(c = 0\) and \(c = \dfrac{8}{5} = 1.6\). So \(0 < c < 1.6\). Among the options only \(\dfrac{1}{2}\) lies here.

Answer: \(\boxed{c = \dfrac{1}{2}}\)

Answer 2

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 \[ \boxed{\dfrac{1}{2}}. \]

Answer 3

Concept:

• For an upward-opening parabola $y=Ax^2+Bx+C$ with $A>0$, the line $y=m$ touches the graph at exactly one point exactly when $m$ equals the minimum value; this happens precisely when $Ax^2+Bx+(C-m)=0$ has a repeated root, that is, when its discriminant equals zero.
• The same idea, with the inequality direction reversed, pins down the maximum value of a downward-opening parabola.
• This discriminant test finds the extreme value directly, without needing to recall or apply the vertex formula $x=-B/(2A)$.

Step 1: Find the minimum value of f(x) using the discriminant test.
$f(x)=x^2-4cx+8c$. Setting $f(x)=m$ gives $x^2-4cx+(8c-m)=0$. This quadratic in $x$ has a repeated root exactly at the minimum, so its discriminant is zero:
$(-4c)^2-4(8c-m)=0 \Rightarrow 16c^2-32c+4m=0 \Rightarrow m=8c-4c^2$.
So $f_{\min}=8c-4c^2$.

Step 2: Find the maximum value of g(x) using the same test.
$g(x)=-x^2+3cx-2c$. Setting $g(x)=M$ gives $-x^2+3cx-(2c+M)=0$, or equivalently $x^2-3cx+(2c+M)=0$. Its discriminant is zero at the maximum:
$(-3c)^2-4(2c+M)=0 \Rightarrow 9c^2-8c-4M=0 \Rightarrow M=\dfrac{9c^2-8c}{4}$.
So $g_{\max}=\dfrac{9c^2}{4}-2c$.

Step 3: Impose the condition $f_{\min}>g_{\max}$.
$8c-4c^2>\dfrac{9c^2}{4}-2c$. Multiply through by $4$: $32c-16c^2>9c^2-8c$, which rearranges to $25c^2-40c<0$, i.e. $5c(5c-8)<0$.

Step 4: Solve the inequality and match with the options.
$5c(5c-8)<0$ holds between the roots $c=0$ and $c=\dfrac{8}{5}=1.6$, so $0<c<1.6$. Among the given options, only $c=\dfrac12$ lies in this range.

Final Answer: $c=\dfrac12$