Application of Derivatives 2 Question 5
5. 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, $\frac{f(x)-14}{(x-1)^{2}}=0(x \neq 1)$ is
(a) -7
(b) 6
(c) 7
(d) 5
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Solution:
- Given that 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 $[1,5)$.
$$ \begin{aligned} & f^{\prime}(1)=0 \quad[\because \text { tangent at } x=1 \text { will be } \\ & \Rightarrow\left(3 x^{2}-6(a-2) x+3 a\right) _{x=1}=0 \\ & \Rightarrow \quad 3-6(a-2)+3 a=0 \\ & \Rightarrow \quad 3-6 a+12+3 a=0 \\ & \Rightarrow \quad 15-3 a=0 \\ & \Rightarrow \quad a=5 \\ & \text { So, } \quad f(x)=x^{3}-9 x^{2}+15 x+7 \\ & \Rightarrow f(x)-14=x^{3}-9 x^{2}+15 x-7 \\ & \Rightarrow f(x)-14=(x-1)\left(x^{2}-8 x+7\right)=(x-1)(x-1)(x-7) \\ & \Rightarrow \quad \frac{f(x)-14}{(x-1)^{2}}=(x-7) \end{aligned} $$
Now, $\quad \frac{f(x)-14}{(x-1)^{2}}=0,(x \neq 1)$
$$ \begin{array}{rrr} \Rightarrow & x-7=0 \\ \Rightarrow & x=7 \end{array} $$
