# Quod Erat Demonstrandum

## 2014/02/10

### 無聊改卷後

Filed under: NSS — johnmayhk @ 5:55 下午

Suppose $2x^2-4x-a+1\ge 0$ for any real value of $x$, find the range of values of $a$.

The inequality holds for any real value of $x$, so we may put $x=1$, yield

$2-4-a+1\ge 0$

$\Rightarrow a\le -1$

## 4 則迴響 »

1. The point is: Find the range of a such that the inequality holds for any real x. Not: For any real x, the inequality is true so find a.

迴響 由 Simon YAU YAU — 2014/02/16 @ 11:09 上午 | 回應

2. The point is: Find the range of a such that the inequality holds for any real x. Not: For any real x, the inequality is true so find “"the corresponding value of"" a.

迴響 由 Simon YAU YAU — 2014/02/16 @ 12:51 下午 | 回應

• Mr YAU your statement “For any real x, the inequality is true so find “”the corresponding value of”” a” is not “Not” (i.e. it is correct).
When we put a real value x in the inequality, we can have the corresponding values of ‘a’ (as there are many values of ‘a’ corresponding to that x, so it becomes a range of values of ‘a’). If we can really put EVERY real value x in the inequality, combine all corresponding ranges of ‘a’, it is the answer.

迴響 由 Wong Wing Cheung — 2014/02/20 @ 7:11 下午 | 回應

3. From $2x^2-4x-a+1\geq0$, we have $a\leq2(x-1) ^2-1$.
Consider $b=2(x-1) ^2-1\geq-1$ (minimum is attained when $x=1$)
Then, the range of is $a\leq-1$, the intersection of $(-\infty, b]$ with $b\geq-1$.

迴響 由 CHIN — 2014/02/24 @ 1:26 上午 | 回應