# If $|b| \ge 1$ and $x=-|a|b$, then which one of the following is necessarily true? [closed]

If $|b| \ge 1$ and $x=-|a|b$, then which one of the following is necessarily true?

(1)$a-xb \lt 0$

(2)$a-xb \ge 0$

(3)$a-xb \gt 0$

(4)$a-xb \le 0$

$|b|\ge 1$, means if $b$ is positive then, $b \ge 1$, else $b \le -1$

-

## closed as too localized by Did, LVK, Steve D, William, VelvetThunderSep 9 '12 at 15:26

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What do you know? What did you try? Can you really answer none of these items? (Subsidiary question: when will you start conforming to the recommandations about the proper ways of asking questions on this site? And start answering comments, by the way?) –  Did Aug 25 '12 at 8:20
Why is this tagged functional equations? –  t.b. Aug 25 '12 at 8:24
Why is this tagged linear-algebra? –  Did Aug 25 '12 at 8:49
The new tag is ludicrous. Where do you see some linear algebra in the question? By the way, what do you view as belonging to linear algebra? If you have no clue, why do you see proper to edit tags? –  Did Aug 25 '12 at 9:38
Pretty sure a homework tag should be up there somewhere, that's for sure... –  user641 Aug 26 '12 at 15:55

$a-xb=a+|a|b\cdot b=a+|a|b^2$

If $a≤0,|a|=-a=>a-xb=(-a)(b^2-1)$ which is clearly $≥0$ as $b^2≥1$ as $|b|≥1$ assuming $b$ is real.

If $a>0, |a|=a=>a-xb=a(b^2+1)$ which is >0

So. $a-xb≥0$

Alternatively, $a-xb=a+|a|b\cdot b=a+|a|b^2≥a+|a|$ as $b^2≥1$ as $|b|≥1$ assuming $b$ is real.

which is 0 if $a≤0$,

and if $a>0$, $a-xb≥2a>0$

So. $a-xb≥0$

-