### Application Of Derivatives Question 3

#### Question 3 - 25 January - Shift 2

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

(1) $(\frac{9}{2}, \infty)$

(2) $(0, \frac{9}{2})$

(3) $(-\infty, \frac{9}{2})$

(4) $(-\frac{9}{2}, \frac{9}{2})$

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: Maxima and minima of a function

$ \begin{aligned} & 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 \\ & \quad p<\frac{9}{2} \\ & \quad p \in(-\infty, \frac{9}{2}) \end{aligned} $