Solve this following

Question:

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as,

$f(x)= \begin{cases}-55 x, & \text { if } x<-5 \\ 2 x^{3}-3 x^{2}-120 x, & \text { if }-5 \leq x \leq 4 \\ 2 x^{3}-3 x^{2}-36 x-336, & \text { if } x>4\end{cases}$

Let $A=\{\mathbf{x} \in \mathbf{R}: \mathrm{f}$ is increasing $\} .$ Then $A$ is equal to :

  1. $(-\infty,-5) \cup(4, \infty)$

  2. $(-5, \infty)$

  3. $(-\infty,-5) \cup(-4, \infty)$

  4. $(-5,-4) \cup(4, \infty)$


Correct Option: , 4

Solution:

$f^{\prime}(x)=\left\{\begin{array}{cc}-55 ; & x<-5 \\ 6(x-5)(x+4) ; & -54\end{array}\right.$

$\mathrm{f}(\mathrm{x})$ is increasing in

$x \in(-5,-4) \cup(4, \infty)$

 

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