Let f(x)=5-|x-2| and g(x)=|x+1|

Question:

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 at $\beta$,

then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^{2}-5 x+6\right)}{x^{2}-6 x+8}$ is equal to :

  1. (1) $1 / 2$

  2. (2) $-3 / 2$

  3. (3) $-1 / 2$

  4. (4) $3 / 2$


Correct Option: 1,

Solution:

$f(x)=5-|x-2|$

Graph of $y=f(x)$

By the graph $f(x)$ is maximum at $x=2 \therefore \alpha=2$

$g(x)=|x+1|$

Graph of $y=g(x)$

By the graph $g(x)$ is minimum at $x=-1$

$\therefore \quad \beta=-1$

Now, $\lim _{x \rightarrow 2} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}$

$=\lim _{x \rightarrow 2} \frac{(x-1)(x-3)}{x-4}=\frac{1}{2}$

 

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