Quod Erat Demonstrandum

2009/10/18

致 5E 同學

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 1:40 下午

在測驗中,我隨便擬了一道極顯淺的題:

設直線 L 及曲線 C 的方程分別是 x + 4y = 03x + (y - 1)^3 = 4。若 L (的圖像) 切 C (的圖像) 於 P,求 P 的坐標。另設 C 上一點 Q,其 y-坐標為 1,求 CQ 處的法線(normal)之方程。

測驗時你們問:如何解三次方程?嗯,我知出事了,大家應該走錯方向。改卷時知道,學生純粹把 L 的方程代入 C 的方程,隨即解一個三次方程,得 y = 5, -1。然而,這題只有 y = -1 才正確。有人代入後,不明所以地進行求導,幸運地,他們也得到 (y - 1)^2 = 4,但解出了 y = 3, -1,也是沒有 reject y = 3

5E 同學,marking scheme 可到 download page 下載。

3 則迴響 »

  1. 我唔合格, 仲要低過mean…

    迴響 由 FTL — 2009/10/19 @ 11:56 下午 | 回覆

  2. Here is my way. (I didn’t read your marking, this may coincide with what you did.)

    Since x + 4y = 0 is the tangent to 3x + (y – 1)^3 = 4. We have 3(-4y) + (y – 1)^3 = 4 has a repeated root.
    Hence we have: -12 + 3(y – 1)^2 = 0. Solving yields: y = 3 or -1.
    However, y = 3 does not satisfy the original equation, therefore the repeated root is -1.
    The rest is trivial.

    迴響 由 koopakoo — 2009/10/20 @ 1:46 上午 | 回覆

    • Thank you Koopa, it is a trick appears in the topic of polynomials in pure mathematics syllabus, I did not used this method in the F.5 quiz as expected. Thank you again!

      迴響 由 johnmayhk — 2009/10/20 @ 12:47 下午 | 回覆


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