Find the minimum integral value of alpha for | vedclass

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(A) To find the $x-$intercepts,we set $f(x) = 0$.$||x - 2| - \alpha| - 5 = 0$$||x - 2| - \alpha| = 5$This implies $|x - 2| - \alpha = 5$ or $|x - 2| -

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) To find the $x-$intercepts,we set $f(x) = 0$.$||x - 2| - \alpha| - 5 = 0$$||x - 2| - \alpha| = 5$This implies $|x - 2| - \alpha = 5$ or $|x - 2| - \alpha = -5$.$|x - 2| = \alpha + 5$ or $|x - 2| = \alpha - 5$.For the equation $|x - 2| = k$ to have two distinct solutions,we must have $k > 0$.If $\alpha - 5 > 0$,then $|x - 2| = \alpha - 5$ gives two solutions,and $|x - 2| = \alpha + 5$ also gives two solutions (since $\alpha + 5 > \alpha - 5 > 0$).Thus,we need $\alpha - 5 > 0$,which means $\alpha > 5$.Since we are looking for the minimum integral value of $\alpha$,and $\alpha > 5$,the smallest integer $\alpha$ is $6$.

Find the minimum integral value of $\alpha$ for which the graph of $f(x) = ||x - 2| - \alpha| - 5$ has exactly four $x-$intercepts.

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