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. $-7$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is D $7$

$f\left(x\right)=x^3-3(a-2)x^2+3ax+7$
$\therefore f^{\prime }\left(x\right)=3x^2-6(a-2)x+3a$
$f^{\prime }\left(x\right)\ge 0\forall x\in (0,1]$
$f^{\prime }\left(x\right)\le 0\forall x\in [1,5)$
$\Rightarrow x=1$ is a critical point.
$f^{\prime }\left(1\right)=0$
$\Rightarrow 3-6a+12+3a=0$
​​​​​​​$\Rightarrow a=5$
​​​​​​​
​​​​​​​$\therefore \frac{f\left(x\right)-14}{(x-1{)}^2}=0$
​​​​​​​$\Rightarrow \frac{x^3-9x^2+15x-7}{(x-1{)}^2}=0$
$\Rightarrow \frac{(x-1{)}^2(x-7)}{(x-1{)}^2}=0$
​​​​​​​$\Rightarrow x=7$ is a root.