(a) 求 $x^4+16$ 除以 $x^2+4$ 所得的商和余数。

(b) 由此证明

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

先除最高次项，再除常数项，余式为常数。

Subtracting $x^2(x^2+4)$ from $x^4+16$ leaves $-4x^2+16$.

Subtracting $-4(x^2+4)$ then leaves $32$. Thus

The quotient is $\boxed{x^2-4}$ and the remainder is $\boxed{32}$.

承接(a)拆成多项式与反正切型分式。代入上限时多项式两项相消。

By (a),

The last term integrates to $16\tan^{-1}(x/2)$, since $\mathrm{d}[\tan^{-1}(x/2)]/\mathrm{d}x=2/(x^2+4)$. Hence

At $x=2\sqrt3$, the polynomial terms give $8\sqrt3-8\sqrt3=0$, while $\tan^{-1}\sqrt3=\pi/3$.

At $x=2$, the polynomial terms give $8/3-8=-16/3$, while $\tan^{-1}1=\pi/4$.

Therefore

as required.