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 determine the values of \(p\) for which the function \(f(x)\) has both a maximum at some \(x < 0\) and a minimum at some \(x > 0\), we analyze the first derivative of \(f(x)\). First, compute the first derivative:

\( f'(x) = \frac{d}{dx} (2x^3 + (2p-7)x^2 + 3(2p-9)x - 6) = 6x^2 + 2(2p-7)x + 3(2p-9) \)
For \(f(x)\) to have both a maximum and a minimum, the equation \(f'(x) = 0\) must have two distinct real roots—one positive and one negative.

1. Discriminant Condition: The discriminant \(D\) of the quadratic equation \(6x^2 + 2(2p-7)x + 3(2p-9) = 0\) must be positive for two distinct real roots:

\( D = [2(2p-7)]^2 - 4 \cdot 6 \cdot 3(2p-9) = 4(2p-7)^2 - 72(2p-9) > 0 \)

Simplifying:

\( 4(4p^2 - 28p + 49) - 144p + 648 > 0 \Rightarrow 16p^2 - 112p + 196 - 144p + 648 > 0 \Rightarrow 16p^2 - 256p + 844 > 0 \Rightarrow 4p^2 - 64p + 211 > 0 \)

Solving the inequality \(4p^2 - 64p + 211 > 0\), we find the critical points and determine the intervals where the inequality holds.

2. Product of Roots Condition: For the roots to have opposite signs, the product of the roots must be negative:

\( \text{Product of roots} = \frac{3(2p - 9)}{6} = \frac{6p - 27}{6} = p - \frac{9}{2} < 0 \Rightarrow p < \frac{9}{2} \)

Combining both conditions, the set of all values of \(p\) that satisfy both is:

\( p \in \left( -\infty, \frac{9}{2} \right) \)

Thus, the correct answer is option (4).

Answer 2

The correct answer is (D) : $\left(-\infty, \frac{9}{2}\right)$
$f(x)=2x^3+(2p-7)x^2+3(2p-9)x-6$
$f^{\prime }(x)=6x^2+2(2p-7)x+3(2p-9)$
$f^{\prime }(0)<0$
$\therefore 3(2p-9)<0$
$p<\frac{9}{2}$
$p\in \left(-\infty ,\frac{9}{2}\right)$