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

Answer: (D)

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