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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 $b \le -1$

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

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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
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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
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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
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Pretty sure a homework tag should be up there somewhere, that's for sure... –  user641 Aug 26 '12 at 15:55
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1 Answer

up vote 1 down vote accepted

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

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. $a-xb≥0$

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