题目

Problem

(i)

f(x)=\frac{(2x+5)^2}{x-3}\qquad x\ne 3

(a) Find $f'(x)$ in the form $\dfrac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are fully factorised quadratic expressions.

(b) Hence find the range of values of $x$ for which $f(x)$ is increasing.

(6)

(ii)

g(x)=x\sin 4x\qquad 0\le x<\frac{\pi}{4}

The curve with equation $y=g(x)$ has a maximum at the point $M$ .

Show that the $x$ coordinate of $M$ satisfies the equation

\tan 4x+kx=0

where $k$ is a constant to be found.

(5)

题目中文翻译

(i)

f(x)=\frac{(2x+5)^2}{x-3}\qquad x\ne 3

(a) 求 $f'(x)$ ，并写成 $\dfrac{P(x)}{Q(x)}$ 的形式，其中 $P(x)$ 和 $Q(x)$ 都是完全因式分解后的二次式。

(b) 由此求 $f(x)$ 为增函数时 $x$ 的取值范围。

(ii)

g(x)=x\sin 4x\qquad 0\le x<\frac{\pi}{4}

曲线 $y=g(x)$ 在点 $M$ 处取得最大值。

证明点 $M$ 的 $x$ 坐标满足方程

\tan 4x+kx=0

其中 $k$ 为待求常数。

解答

We have

f(x)=\frac{(2x+5)^2}{x-3}

Use the quotient rule:

f'(x)=\frac{(x-3)\dfrac{\mathrm{d}}{\mathrm{d}x}(2x+5)^2-(2x+5)^2\dfrac{\mathrm{d}}{\mathrm{d}x}(x-3)}{(x-3)^2}

Now

\frac{\mathrm{d}}{\mathrm{d}x}(2x+5)^2=4(2x+5)

and

\frac{\mathrm{d}}{\mathrm{d}x}(x-3)=1

\begin{aligned} f'(x) &=\frac{4(2x+5)(x-3)-(2x+5)^2}{(x-3)^2}\\ &=\frac{(2x+5)\left[4(x-3)-(2x+5)\right]}{(x-3)^2}\\ &=\frac{(2x+5)(4x-12-2x-5)}{(x-3)^2}\\ &=\frac{(2x+5)(2x-17)}{(x-3)^2} \end{aligned}

Therefore

\boxed{f'(x)=\frac{(2x+5)(2x-17)}{(x-3)^2}}

For $f(x)$ to be increasing,

f'(x)>0

From part (a),

f'(x)=\frac{(2x+5)(2x-17)}{(x-3)^2}

Since

(x-3)^2>0

for $x\ne 3$ , the sign of $f'(x)$ depends on

(2x+5)(2x-17)

So we solve

(2x+5)(2x-17)>0

The critical values are

x=-\frac52,\qquad x=\frac{17}{2}

The product is positive outside these two values. Therefore

\boxed{x<-\frac52\quad \text{or}\quad x>\frac{17}{2}}

We have

g(x)=x\sin4x

Differentiate using the product rule:

g'(x)=\sin4x+4x\cos4x

At the maximum point $M$ ,

g'(x)=0

\sin4x+4x\cos4x=0

For the maximum point in

0\leq x<\frac{\pi}{4}

we may divide by $\cos4x$ at the stationary point. This gives

\tan4x+4x=0

Comparing this with

\tan4x+kx=0

we get

\boxed{k=4}

Therefore the $x$ coordinate of $M$ satisfies

\boxed{\tan4x+4x=0}