Application Of Derivatives Ques 61
61. Let $f(x)=5-|x-2|$ and $g(x)=|x+1|, x \in R$. If $f(x)$ attains maximum value at $\alpha$ and $g(x)$ attains minimum value of $\beta$, then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^{2}-5 x+6\right)}{x^{2}-6 x+8}$ is equal to
(a) $1 / 2$
(b) $-3 / 2$
(c) $-1 / 2$
(d) $3 / 2$
(2019 Main, 12 April II)
Show Answer
Answer:
Correct Answer: 61.(a)
Solution:
Formula:
Maxima and Minima of functions of one variable :
- Given functions are $f(x)=5-|x-2|$
and $g(x)=|x+1|$, where $x \in R$.
Clearly, maximum of $f(x)$ occurred at $x=2$, so $\alpha=2$.
and minimum of $g(x)$ occurred at $x=-1$, so $\beta=-1$.
$ \begin{aligned} & \Rightarrow \quad \alpha \beta=-2 \\ & \text { Now, } \lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^{2}-5 x+6\right)}{x^{2}-6 x+8} \\ & =\lim _{x \rightarrow 2} \frac{(x-1)(x-3)(x-2)}{(x-4)(x-2)} \quad[\because \alpha \beta=-2] \\ & =\lim _{x \rightarrow 2} \frac{(x-1)(x-3)}{(x-4)}=\frac{(2-1)(2-3)}{(2-4)}=\frac{1 \times(-1)}{(-2)}=\frac{1}{2} \end{aligned} $
BETA
