If the function $f$ is given by $\mathrm{f}(x)=x^3-3(\mathrm{a}-2) x^2+3 \mathrm{a} x+7$, for some a $\in…
If the function $f$ is given by $\mathrm{f}(x)=x^3-3(\mathrm{a}-2) x^2+3 \mathrm{a} x+7$, for some a $\in \mathbb{R}$, is increasing in $(0,1]$ and decreasing in $[1,5)$, then a root of the equation $\frac{\mathrm{f}(x)-14}{(x-1)^2}=0(x \neq 1)$ is
$-7$
$6$
$7$
$5$
Solution
$f(x)=x^3-3(a-2) x^2+3 a x+7$
As $\mathrm{f}(x)$ is increasing in $(0,1]$ and decreasing in $[1,5)$, we get that $\mathrm{f}(x)$ has critical point at $x=1$ $\Rightarrow \mathrm{f}^{\prime}(1)=0$
$\mathrm{f}^{\prime}(x)=3 x^2-6(\mathrm{a}-2) x+3 \mathrm{a}$
$\begin{aligned}
& \therefore \quad 3(1)^2-6(a-2)+3 a=0 \\
& \therefore \quad \mathrm{a}=5 \\
& \therefore \quad \frac{\mathrm{f}(x)-14}{(x-1)^2}=\frac{x^3-9 x^2+15 x-7}{(x-1)^2} \\
& =\frac{(x-1)^2(x-7)}{(x-1)^2} \\
& =x-7 \\
&
\end{aligned}$
$\therefore \quad$ The required root is 7 .