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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & f(x)=x^3-3(a-2) x^2+3 a x+7 \ & f^{\prime}(x)=3 x^2-6(a-2) x+3 a\end{aligned}$ $f^{\prime}(x)$ changes its behaving at $x=1$ $\begin{aligned} & f^{\prime}(1)=0 \ & \Rightarrow 3 	imes 1^2-6(a-2) 	imes 1+3 a=0 \ & \Rightarrow a=b \Rightarrow f(x)=x^3-9 x^2+15 x+7 \ & \Rightarrow \frac{f(x)-14}{(x-1)^2}=0 \Rightarrow \frac{\left(x^3-9 x^2+15 x+7ight)-14}{(x-1)^2}=0 \ & \Rightarrow x^3-9 x^2+15 x-7=0 \ & \Rightarrow(x-1)(x-1)(x-7) \ & \Rightarrow x=1,1,7\end{aligned}$

If the function $f(x)=x^3-3(a-2) x^2+3 a x+7$, for some $a \in R$ is increasing in $(0,1]$ and decreasing in…

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