Quod Erat Demonstrandum

2011/04/08

致中五修 M2 的同學

Filed under: NSS,Pure Mathematics — johnmayhk @ 3:35 下午

這是今年 HKALE Pure Mathematics (II) Q.8 的題目,你應可取滿分的。

(a)

(i) Prove that \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin x}dx=1.

(ii) Evaluate \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{1+\sin x}dx=1. (5 marks)

(b)

Let f:[0,\pi]\rightarrow \mathbb{R} be a continuous function such that f(\pi-x)=f(x) for all x\in [0,\pi].

Using integration by substitution, prove that \int_{0}^{\pi}f(x)dx=2\int_{0}^{\frac{\pi}{2}}f(x)dx. (3 marks)

(c)

Let g:[0,\pi]\rightarrow \mathbb{R} be a continuous function such that g(\pi-x)=-g(x) for all x\in [0,\pi].

Using the substitution u=\pi-x, prove that \int_{0}^{\pi}g(x)\ln(1+e^{\cos x})dx=\frac{1}{2}\int_{0}^{\pi}g(x)\cos xdx. (3 marks)

(d)

Evaluate \int_{0}^{\pi}\frac{\cos x\ln(1+e^{\cos x})}{(1+\sin x)^2}dx. (4 marks)

11 則迴響 »

  1. John Sir (c)好似d問題,in(g(x)cos(x))的上限好像應該是兀。

    迴響 由 Jaychan — 2011/04/08 @ 5:49 下午 | 回覆

  2. 唔怪得之熟口熟面,原來係今年份卷!
    岩岩考完既我覺得,今年份PII算淺,除左無端端喺PII翻抄86年既jensen’s inequality,同埋好煩既function graph

    迴響 由 eric — 2011/04/08 @ 9:12 下午 | 回覆

  3. 唉,啲pure math無掂一年真係生疏哂,第一part係咪先上下乘1-sinx???

    迴響 由 6901goldenbrother — 2011/04/09 @ 12:37 上午 | 回覆

  4. 是否t-formula較快?

    迴響 由 eric — 2011/04/09 @ 4:19 下午 | 回覆

    • I think u are right lo, t-formula may be a better option

      迴響 由 Matthew Cheung — 2011/04/09 @ 10:33 下午 | 回覆

  5. if directly 第一part係咪先上下乘1-sinx, we need to handle tan(pi/2)- sec(pi/2).
    But Assume an integral,
    int.(1/(1+sinx) = int.(1/1+sinx) + int.(1/1+sinx)
    Using 先上下乘1-sinx, show that int.(1/(1+sinx) =2
    And using substituition, to show int.(1/1+sinx) = int.(1/1+sinx)

    Therefore, int.(1/(1+sinx) =2=2int.(1/1+sinx)

    At last, int.(1/1+sinx) =1

    迴響 由 Matthew Cheung — 2011/04/09 @ 10:15 下午 | 回覆

  6. computer mistake

    int.1/(1+sinx) from 0 to pi

    int.1/(1+sinx) from pi/2 to pi
    +
    int.1/(1+sin from 0 to pi/2

    迴響 由 Matthew Cheung — 2011/04/09 @ 10:17 下午 | 回覆

  7. I think the question wants us to make use of
    sin (pi/2-x)=cos x;
    and 1+cos x=2(cos x/2)^2.
    You will obtain sth in form of (sec x/2)^2
    which can be easily integrated.

    迴響 由 Ivan — 2011/04/10 @ 8:48 下午 | 回覆

  8. Beside the best solution that have mentioned above,we can also using even function to prove it.
    Using the fact that int.(1/(1 sinx))=int.(1/1 cosx))
    As well as 1/(1 cosx) is a even function, the anwser can be found by consider the function,1/(1 cosx),for the interval that. -pi/2<=x<=pi/2.

    迴響 由 Jaychan — 2011/04/11 @ 7:05 上午 | 回覆

  9. it’s not appropriate to multiply (1-sinx) / (1-sinx) because the function will then undefined at the upper limit pi/2 (denominator zero)

    迴響 由 Tony — 2011/05/10 @ 2:59 下午 | 回覆


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