Question

Let the function \(f(x)=2 x^3+(2 p-7) x^2+3(2 p-9) x-6\) have a maxima for some value of \(x<0\) and a minima for some value of \(x>0\) Then, the set of all values of \(p\) is

• A. $\left(\frac{9}{2}, \infty\right)$
• B. $\left(0, \frac{9}{2}\right)$
• C. $\left(-\frac{9}{2}, \frac{9}{2}\right)$
• D. $\left(-\infty, \frac{9}{2}\right)$

Answer 1

The Correct Option is D

To find the solution, we must first understand the behavior of the function \(f(x) = 2x^3 + (2p-7)x^2 + 3(2p-9)x - 6\) in terms of its maxima and minima. The critical points of a function are found by differentiating the function and setting the derivative equal to zero.

First, we find the first derivative of the function, \(f(x)\):

\(f'(x) = \frac{d}{dx}(2x^3 + (2p-7)x^2 + 3(2p-9)x - 6)\)

\(= 6x^2 + 2(2p-7)x + 3(2p-9)\)

\(= 6x^2 + (4p-14)x + 6p - 27\)

Set \(f'(x) = 0\) to find critical points:

\(6x^2 + (4p-14)x + 6p - 27 = 0\)

The quadratic equation \(6x^2 + (4p-14)x + 6p - 27 = 0\) has roots given by:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Where:

• \(a = 6\)
• \(b = 4p - 14\)
• \(c = 6p - 27\)

To have a maxima at \(x < 0\) and minima at \(x > 0\), the product of the roots must be negative. Using the properties of quadratic equations, the product of the roots (\(\frac{c}{a}\)) must be negative:

\(\frac{6p - 27}{6} < 0\)

Solving this inequality gives:

\(6p - 27 < 0\)

\(\Rightarrow 6p < 27\)

\(\Rightarrow p < \frac{27}{6}\)

\(\Rightarrow p < \frac{9}{2}\)

This inequality implies that the values of \(p\) for which the function has a maxima at \(x < 0\) and a minima at \(x > 0\) is \((-∞, \frac{9}{2})\).

Therefore, the correct answer is \((-∞, \frac{9}{2})\), which corresponds to the option:

\((-∞, \frac{9}{2})\)