Quod Erat Demonstrandum

2014/02/10

無聊改卷後

Filed under: NSS — johnmayhk @ 5:55 下午

二次方程某基本題:

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

某同學的解:

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

得到正確答案,可惜方法不對

johnmayhk-sosad

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 | 回覆

  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 | 回覆

    • 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 | 回覆

  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 | 回覆


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