Let f(x) = 5 -- |x -- 2| and g(x) = |x + 1|, x [ ] R.

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
$$\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$$ is equal to :

Accepted Answer

From f(x) = 5 -, x - 2 |

maximum value of f(x) is at x = 2

From g(x) =, x + 1 |

minimum value of g(x) is at x = -1

$$ \therefore $$ $$\alpha \beta $$ = - 2

$$ \Rightarrow $$  $$\mathop {\lim }\limits_{x \to 2} {{(x - 1)(x - 2)(x - 3)} \over {(x - 2)(x - 4)}}$$

$$ \Rightarrow $$ $${{(2 - 1)(2 - 3)} \over {(2 - 4)}} = {1 \over 2}$$

Solutions · 1