Application Of Derivatives Ques 3
- Let $f(x)=\left\{\begin{array}{cc}-x^3+\frac{\left(b^3-b^2+b-1\right)}{\left(b^2+3 b+2\right)}, & 0 \leq x \leq 1 \\ 2 x-3, & 1 \leq x \leq 3\end{array}\right.$
Find all possible real values of $b$ such that $f(x)$ has the smallest value at $x=1$.
(1993, 5M)
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Answer:
Correct Answer: 3.$(b \in(-2,-1) \cup[1, \infty])$
Solution: Given, $f(x)=\left\{\begin{array}{cl}-x^3+\frac{\left(b^3-b^2+b-1\right)}{\left(b^2+3 b+2\right)} & , \text { if } 0 \leq x \leq 1 \\ 2 x-3, & \text { if } 1 \leq x \leq 3\end{array}\right.$
is smallest at $x=1$.
So, $f(x)$ is decreasing on $[0,1]$ and increasing on $[1,3]$.
Here, $f(1)=-1$ is the smallest value at $x=1$.
$\therefore \quad $ Its smallest value occur as
$ \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(-x^3\right)+\frac{\left(b^3-b^2+b-1\right)}{b^2+3 b+2} $
In order this value is not less than -1 , we must have
$ \quad \quad \frac{b^3-b^2+b-1}{b^2+3 b+2} \geq 0 $
$ \Rightarrow \quad \frac{\left(b^2+1\right)(b-1)}{(b+1)(b+2)} \geq 0 $
$ \therefore \quad b \in(-2,-1) \cup[1, \infty] $
BETA
