The set of all values of p for which the function f(x)=2x³+(2p−7)x²+3(2p−9)x−6 has a maxima for some value of x<0 and a minima for some value of x>0 is (∞,9/2). Option B is the right choice.
To determine the set of all values of p for which the function f(x)=2x³+(2p−7)x²+3(2p−9)x−6 has a maximum for some value of x<0 and a minimum for some value of x>0, we need to analyze the function's behavior.
First, let's find the critical points of the function by setting its derivative f'(x)=0 and solving for x:
f'(x)=6x²(p-1)+3(2p-9)=0
Factoring out 3x², we get:
3x²(p-1)+(2p-9)=0
3x²(p-1)+2(p-9)=0
3x²(p-5)=0
This equation implies three possible scenarios:
x=0: This is a critical point for all values of p.
p=1: In this case, the derivative becomes 3(2p-9)=0, which implies x=3. However, this value of x is not within the specified ranges (x<0 and x>0), so it is not relevant for the problem.
p=5: In this case, the derivative becomes 3(2p-9)=30, which means there are no critical points other than x=0.
Since the function's degree is odd, it has a vertical asymptote at x=0. Therefore, we need to analyze the function's behavior in the intervals (-∞,0) and (0,∞).
For x<0, the function is increasing if p<5 and decreasing if p>5.
For x>0, the function is decreasing if p<5 and increasing if p>5.
To have a maximum for x<0 and a minimum for x>0, we need the function to be increasing in the interval (0,∞) and decreasing in the interval (-∞,0). This implies p>5.
Therefore, the set of all values of p for which the function has the desired behavior is (5,∞). The answer is B. (∞,9/2).