# Quod Erat Demonstrandum

## 2008/04/08

### PM Q

Filed under: Pure Mathematics — johnmayhk @ 11:37 下午

Haven’t seen the 2008 AL Pure Math questions yet, just know a little bit, the following is a 3-mark question in paper I.

Let $p(x)$ be a polynomial such that $p(x) = p(x - 1) + x^{100}, \forall x \in \mathbb{R}$, determine the degree and the leading coefficient of $p(x)$.

Suppose $p(x) = ax^n +$ terms with powers less than $n$.

Note that, for any natural number $r$, $ax^r - a(x - 1)^r \equiv rax^{r - 1} +$ terms with powers less than $r - 1$. Thus,

$p(x) - p(x - 1) \equiv nax^{n - 1} +$ terms with powers less than $n - 1$. From given,

$nax^{n - 1} +$ terms with powers less than $n - 1 \equiv x^{100}$, hence

$n - 1 = 100$ and $na = 1$, hence we have $n = 101$ (the degree) and $a = \frac{1}{101}$ (the leading coefficent).

## 11 則迴響 »

1. That’s a neat problem.

迴響 由 koopa — 2008/04/09 @ 1:47 上午 | 回應

2. line 3, $p(x)=ax^n+$ terms with power less than “$n$“.

聽我的學生說，今年的Pure Math好像有點難度，不過我未看過卷，所以都未肯定。

迴響 由 Kam — 2008/04/09 @ 10:41 上午 | 回應

3. Thank you Kam sir for the debugging, I’ll make that amendment. I’d just have a glance at the questions. No complex no. long question in paper 1, short questions in both paper 1 and 2 are quite similar to questions in previous years. The degree of difficulty? Not yet commented.

迴響 由 johnmayhk — 2008/04/09 @ 12:07 下午 | 回應

4. John Sir客氣了。

每年公開試之後，我都希望可以親手試做一下試卷。除了感受一下學生的體會外，也試一試自己有沒有退步，然而通常都有些不懂的。

迴響 由 Kam — 2008/04/09 @ 7:28 下午 | 回應

5. 我就覺得paper II難…但係d人話易 =]

迴響 由 Ho Ki — 2008/04/09 @ 10:29 下午 | 回應

6. Alternative solution to paper I 11a.

(i) By rational root theorem, it suffices to check that +/-1, +/-2, +/-5, +/-10 are not roots of x^2 – 10, which is obvious.
(ii) Since x^2 – 10 is irreducible over Q, we have [Q(sqrt{10}) : Q] = 2, and that 1, sqrt{10} forms a basis of Q(sqrt{10}} as a Q-vector space. Therefore m + nsqrt{10} = 0 with m, n in Q implies m = n = 0 since 1 and sqrt{10} are linearly independent.

However, I wonder such a solution is permitted in the exam, any comments?

Also, I admit that this solution is not really shorter, but imagine a question like: suppose a + bt + ct^2 = 0, where t = 2^{1/3}, and a, b, c in Q. Prove that a = b = c = 0. The same trick works, since 1, t, t^2 forms a basis for [Q(t): Q] = 3.

迴響 由 koopa — 2008/04/11 @ 8:19 上午 | 回應

7. To koopa,

(i) is fine,
(ii)? Urm… if I were a marker, I’ll give full marks. But…

“Vector" vanishes in the syllabus of pure math, not to mention the concept of vector spaces and something much more general in abstract algebra. (Just curious, is abstract algebra taught in year 1 now?)

Could I use the case that using differentiation in HKCEE math paper as a kind of analogy?

迴響 由 johnmayhk — 2008/04/11 @ 5:59 下午 | 回應

8. Is it true that for HKCEE or HKAL exams, one can only use techniques within the syllabus to receive full credit?

As another example, if one uses knowledge learned in A.maths, and apply those techniques to questions in a general math paper in HKCEE, he/she will receive no credit?

迴響 由 koopa — 2008/04/12 @ 12:26 上午 | 回應

9. Also, I am not sure whether algebra is taught in year 1. (I guess one can take it if he/she wishes)

迴響 由 koopa — 2008/04/12 @ 12:31 上午 | 回應

10. I guess algebra is usually not taught in year 1 (for hk)

To koopa, what if the marker does not know fields at all? It’s possible. Moreover, when a marker has to mark many papers, I don’t think he will read a solution over that carefully. So if the solution is too different from the marking scheme, they may just mark it as wrong.

迴響 由 Soarer — 2008/04/12 @ 1:38 上午 | 回應

11. […] Filed under: HKALE, Pure Mathematics — johnmayhk @ 4:26 pm 網友 Koopa 提供了兩個方法證明 是無理數[這（竟）是 2008-AL-Pure Mathematics (I) […]

通告 由 數學平常談：證某類無理數的小法@濟濟一堂 « Quod Erat Demonstrandum — 2008/04/12 @ 4:27 下午 | 回應