Question:
If the function $f$ given by $f(x)=x^{3}-3(a-2) x^{2}+$ $3 a x+7$, for some $a \in R$ is increasing in $(0,1]$ and decreasing in $[1,5)$, then a root of the equation,
Correct Option: , 3
Solution:
$f^{\prime}(x)=3 x^{2}-6(a-2) x+3 a$
$\mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \forall \mathrm{x} \in(0,1]$
$\mathrm{f}^{\prime}(\mathrm{x}) \leq 0 \forall \mathrm{x} \in[1,5)$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=0$ at $\mathrm{x}=1 \Rightarrow \mathrm{a}=5$
$f(x)-14=(x-1)^{2}(x-7)$
$\frac{f(x)-14}{(x-1)^{2}}=x-7$