Question

If the function f given by $f\left(x\right)=x^3-3(a-2)x^2+3ax+7$, for some $a\in R$ is increasing in $(0,1]$ and decreasing in $[1,5)$, then a root of the equation, $\frac{f\left(x\right)-14}{(x-1{)}^2}=0(x\ne 1)$ is

• A. $6$
• B. $5$
• C. $7$
• D. $-7$

Answer 1

The correct option is C $7$

$f^{\prime }\left(x\right)=3x^2-6(a-2)x+3a$
$f^{\prime }\left(x\right)\ge 0\forall x\in (0,1]$
$f^{\prime }\left(x\right)\le 0\forall x\in [1,5)$
$\Rightarrow f^{\prime }\left(x\right)=0$ at $x=1\Rightarrow a=5$
$f\left(x\right)-14=(x-1{)}^2(x-7)$
$\frac{f\left(x\right)-14}{(x-1{)}^2}=x-7$

$\therefore$ answer is $7$