题目

Problem

(i) The curve $C_1$ has equation

y=3\ln(x^2-5)-4x^2+15\qquad x>\sqrt{5}

Show that $C_1$ has a stationary point at $x=\dfrac{\sqrt{p}}{2}$ where $p$ is a constant to be found.

(4)

(ii) A different curve $C_2$ has equation

y=4x-12\sin^2x

(a) Show that, for this curve,

\frac{dy}{dx}=A+B\sin 2x

where $A$ and $B$ are constants to be found.

(b) Hence, state the maximum gradient of this curve.

(4)

题目中文翻译

(i) 曲线 $C_1$ 的方程为

y=3\ln(x^2-5)-4x^2+15\qquad x>\sqrt{5}

证明 $C_1$ 在 $x=\dfrac{\sqrt{p}}{2}$ 处有一个驻点，其中 $p$ 是需要求出的常数。

(ii) 另一条曲线 $C_2$ 的方程为

y=4x-12\sin^2x

(a) 证明对这条曲线，

\frac{dy}{dx}=A+B\sin 2x

其中 $A$ 与 $B$ 是需要求出的常数。

(b) 由此写出这条曲线的最大梯度。

解答

For

y=3\ln(x^2-5)-4x^2+15

求导得

\frac{dy}{dx} =3\cdot\frac{2x}{x^2-5}-8x

\frac{dy}{dx} =\frac{6x}{x^2-5}-8x

At a stationary point,

\frac{dy}{dx}=0

Hence

\frac{6x}{x^2-5}-8x=0

Since $x>\sqrt5$ , we have $x\ne0$ . Divide by $x$ :

\frac{6}{x^2-5}-8=0

\frac{6}{x^2-5}=8

and therefore

x^2-5=\frac34

Thus

x^2=\frac{23}{4}

Since $x>\sqrt5$ , take the positive root:

x=\frac{\sqrt{23}}{2}

This is of the form

x=\frac{\sqrt{p}}{2}

where

\boxed{p=23}

For

y=4x-12\sin^2x

求导得

\frac{dy}{dx}=4-12\cdot 2\sin x\cos x

\frac{dy}{dx}=4-24\sin x\cos x

Using

\sin 2x=2\sin x\cos x

we get

\frac{dy}{dx}=4-12\sin 2x

Therefore

\boxed{A=4,\qquad B=-12}

Since

-1\le \sin 2x\le 1

the expression

4-12\sin 2x

is greatest when

\sin 2x=-1

Therefore the maximum gradient is

4-12(-1)=16

\boxed{16}