Quod Erat Demonstrandum

2012/10/08

錯在哪裡系列之求角度

Filed under: mathematics,NSS — johnmayhk @ 12:01 下午

Question

For real numbers x, define f(x+\frac{1}{x})=\frac{x^2}{1+x^2+x^4}.

Solve f(\sin\theta)=-\frac{4}{3}

for 0^o\le \theta \le 360^o.

Solution

f(x+\frac{1}{x})

=\frac{x^2}{1+x^2+x^4}

=\frac{1}{(1/x)^2+1+x^2}

=\frac{1}{(x+1/x)^2-1}

Hence

f(u)=\frac{1}{u^2-1}

Now

f(\sin\theta)=\frac{1}{\sin^2\theta-1}=-\frac{4}{3}

\sin^2\theta=\frac{1}{4}

\sin\theta=\pm \frac{1}{2}

\therefore \theta=30^o,150^o,210^o,330^o

但其實有些東西出了問題,請同學找找看。

6 則迴響 »

  1. x cannot be zero, so the subject change has problem.
    Is it correct, Ng sir?

    迴響 由 Little Maths — 2012/10/11 @ 12:47 下午 | 回覆

    • Thank you for trying!

      As you mentioned, the change is a problem.

      Precisely, the absolute value of x+\frac{1}{x} is not less than 2 for any non-zero real numbers x. (Why?)

      Hence the substitution of x+\frac{1}{x}=\sin\theta is invalid!

      迴響 由 johnmayhk — 2012/10/11 @ 4:20 下午 | 回覆

      • (sqrt(x)-1/sqrt(x))^2 > or =0
        x-2+1/x > or = 0
        x+1/x not less than2

        迴響 由 jeffrey yan — 2012/10/12 @ 6:27 下午

  2. My try:
    for real x, \frac{x^2}{1+x^2+x^4} cannot be negative. So no definition of f is given when f is negative.
    Therefore, the equation cannot be solved.

    迴響 由 mm100100 — 2012/10/11 @ 7:21 下午 | 回覆

  3. Good jeffrey,

    But be careful, you need to consider the case for the value of x being less than zero.

    As I mentioned, it should be the ABSOLUTE value of (x+1/x).

    迴響 由 johnmayhk — 2012/10/13 @ 1:07 下午 | 回覆


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