A value of c for which the minimum value of f(x)=x2−4cx+8c is greater than the maximum value of $g(x) = -x^2 + 3cx - 2c , is
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Official Correct Answer: C. We have the functions first to get a clear picture.
The first function is f(x) = x² - 4c x + 8c. Since the coefficient of x² is positive (1), this is a parabola opening upwards, so it will have a minimum point.
The second function is g(x) = -x² + 3c x - 2c. Here, the coefficient of x² is negative (-1), so this parabola opens downward, meaning it has a maximum point.
The question is asking for the value of c such that the minimum of f(x) is greater than the maximum of g(x). So, We find the minimum of f(x) and the maximum of g(x), set up an inequality where min f(x) > max g(x), and solve for c.
Alright, let's recall how to find the vertex of a quadratic function. For any quadratic function in the form ax² + bx + c, the vertex occurs at x = -b/(2a). The y-coordinate of the vertex will give the minimum or maximum value.
Starting with f(x) = x² - 4c x + 8c. Here, a = 1, b = -4c. So, the x-coordinate of the vertex is -b/(2a) = -(-4c)/(2*1) = 4c/2 = 2c.
Now, plug this back into f(x) to find the minimum value.
f(2c) = (2c)² - 4c*(2c) + 8c
= 4c² - 8c² + 8c
= -4c² + 8c
So, the minimum value of f(x) is -4c² + 8c.
Next, let's find the maximum value of g(x) = -x² + 3c x - 2c. Here, a = -1, b = 3c. The x-coordinate of the vertex is -b/(2a) = -(3c)/(2*(-1)) = 3c/2.
Plugging this back into g(x):
g(3c/2) = -(3c/2)² + 3c*(3c/2) - 2c
= -(9c²/4) + (9c²/2) - 2c
Compute each term step by step:
First term: -(9c²/4)
Second term: 9c²/2 = 18c²/4
Third term: -2c
So, combining them:
= (-9c²/4 + 18c²/4) - 2c
= (9c²/4) - 2c
So, the maximum value of g(x) is (9c²)/4 - 2c.
Now, the problem states that the minimum of f(x) is greater than the maximum of g(x). So, we set up the inequality:
-4c² + 8c > (9c²)/4 - 2c
Let me write that down:
-4c² + 8c > (9c²)/4 - 2c
To solve this inequality, I'll bring all terms to one side. Let's subtract (9c²)/4 and add 2c to both sides:
-4c² + 8c - (9c²)/4 + 2c > 0
Combine like terms. First, let's combine the c² terms:
-4c² - (9c²)/4. To combine these, I'll convert -4c² to quarters: -16c²/4 - 9c²/4 = (-25c²)/4.
Next, the c terms: 8c + 2c = 10c.
So, the inequality becomes:
(-25c²)/4 + 10c > 0
Let me write that as:
(-25/4)c² + 10c > 0
To make it easier, multiply both sides by 4 to eliminate the denominator:
-25c² + 40c > 0
Let me rearrange this:
-25c² + 40c > 0
Factoring out a -5c:
-5c(5c - 8) > 0
Hmm, okay. So, the inequality is -5c(5c - 8) > 0.
Let me think about how to solve this. When dealing with inequalities involving quadratics, it's helpful to find the critical points and test intervals.
First, find the roots:
-5c(5c - 8) = 0
So, c = 0 or 5c - 8 = 0 => c = 8/5 = 1.6
So, the critical points are c = 0 and c = 1.6.
These divide the number line into three intervals:
1. c < 0
2. 0 < c < 1.6
3. c > 1.6
Now, let's test each interval.
1. c < 0: Let's pick c = -1.
Plug into -5c(5c - 8):
-5*(-1)(5*(-1) - 8) = 5*( -5 -8 ) = 5*(-13) = -65 < 0
So, the expression is negative here.
2. 0 < c < 1.6: Let's pick c = 1.
-5*1*(5*1 -8) = -5*(5 -8) = -5*(-3) = 15 > 0
So, positive in this interval.
3. c > 1.6: Let's pick c = 2.
-5*2*(5*2 -8) = -10*(10 -8) = -10*2 = -20 < 0
Negative here.
So, the inequality -25c² + 40c > 0 holds when 0 < c < 1.6.
But wait, the original inequality was -4c² + 8c > (9c²)/4 - 2c, which we transformed into -25c² + 40c > 0, which is equivalent to 25c² - 40c < 0.
Wait, actually, when I multiplied both sides by 4, the inequality sign remains the same because 4 is positive. So, the inequality is correct as -25c² + 40c > 0.
So