question_answer

\[y={{\{x(x-3)\}}^{2}}\] increases for all values of \[x\] lying in the interval

A)  \[\left( 0,\,\frac{3}{2} \right)\cup \,(3,\,\infty )\]          

B) \[0<x<\infty \]

C)  \[-\infty <x<0\]                     

D)  \[(1,\,3)\,\cup \,(4,\,\infty )\]

Correct Answer: A

Solution :

 Since, \[y={{\{x(x-3)\}}^{2}}\] \[\therefore \]    \[\frac{dy}{dx}=2x(x-3)\,(2x-3)\] For increasing function. \[\frac{dy}{dx}>0\] \[\Rightarrow \]            \[2x(x-3)\,(2x-3)\,>0\] \[\Rightarrow \]            \[x(x-3)\,\left( x-\frac{3}{2} \right)>0\] \[\therefore \]    \[x\in \,\left( 0,\,\frac{3}{2} \right)\,\cup \,(3,\,\infty )\]