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P3 2020 Jan Q4

A Level / Edexcel / P3

IAL 2020 Jan Paper · Question 4

题目

Problem

(i)

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

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

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

(6)

(ii)

g(x)=xsin⁡4x0≤x<π4g(x)=x\sqrt{\sin 4x}\qquad 0\le x<\frac{\pi}{4}

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

Show that the xx coordinate of MM satisfies the equation

tan⁡4x+kx=0\tan 4x+kx=0

where kk is a constant to be found.

(5)
题目中文翻译

(i)

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

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

(b) 由此求 f(x)f(x) 为增函数时 xx 的取值范围。

(ii)

g(x)=xsin⁡4x0≤x<π4g(x)=x\sqrt{\sin 4x}\qquad 0\le x<\frac{\pi}{4}

曲线 y=g(x)y=g(x) 在点 MM 处取得最大值。

证明点 MM 的 xx 坐标满足方程

tan⁡4x+kx=0\tan 4x+kx=0

其中 kk 为待求常数。

解答

(i)(a)

解法一

思路

展开

用商法则求导,再从分子提取公因式。

答题过程

展开

We have

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

Use the quotient rule:

f′(x)=(x−3)ddx(2x+5)2−(2x+5)2ddx(x−3)(x−3)2f'(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

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

and

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

So

f′(x)=4(2x+5)(x−3)−(2x+5)2(x−3)2=(2x+5)[4(x−3)−(2x+5)](x−3)2=(2x+5)(4x−12−2x−5)(x−3)2=(2x+5)(2x−17)(x−3)2\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

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

解法二

思路

展开

先将分式拆成多项式与一个简单分式,逐项求导后再通分。

答题过程

展开

Since 2x+5=2(x−3)+112x+5=2(x-3)+11, expanding the square and dividing by x−3x-3 gives

f(x)=4(x−3)+44+121x−3.f(x)=4(x-3)+44+\frac{121}{x-3}.

Differentiating and putting the result over a common denominator,

f′(x)= 4−121(x−3)2= [2(x−3)]2−112(x−3)2= (2x+5)(2x−17)(x−3)2.\begin{align*} f'(x)=&\,4-\frac{121}{(x-3)^2}\\ =&\,\frac{[2(x-3)]^2-11^2}{(x-3)^2}\\ =&\,\frac{(2x+5)(2x-17)}{(x-3)^2}. \end{align*}

(i)(b)

For f(x)f(x) to be increasing,

f′(x)>0f'(x)>0

From part (a),

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

Since

(x−3)2>0(x-3)^2>0

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

(2x+5)(2x−17)(2x+5)(2x-17)

So we solve

(2x+5)(2x−17)>0(2x+5)(2x-17)>0

The critical values are

x=−52,x=172x=-\frac52,\qquad x=\frac{17}{2}

The product is positive outside these two values. Therefore

x<−52orx>172\boxed{x<-\frac52\quad \text{or}\quad x>\frac{17}{2}}

(ii)

解法一

思路

展开

对乘积和根式分别用乘积法则与链式法则。极大值点在区间内部,且该点的余弦因子非零。

答题过程

展开

Let xMx_M be the xx-coordinate of MM. Since g(0)=0g(0)=0 and g(x)>0g(x)>0 for 0<x<π/40<x<\pi/4, the maximum occurs at an interior point.

Using the product rule and chain rule,

g′(x)=sin⁡4x+2xcos⁡4xsin⁡4x(0<x<π/4).g'(x)=\sqrt{\sin 4x}+\frac{2x\cos 4x}{\sqrt{\sin 4x}} \qquad (0<x<\pi/4).

At MM, set g′(xM)=0g'(x_M)=0 and multiply by sin⁡4xM\sqrt{\sin 4x_M}:

sin⁡4xM+2xMcos⁡4xM=0.\sin 4x_M+2x_M\cos 4x_M=0.

If cos⁡4xM=0\cos 4x_M=0, then xM=π/8x_M=\pi/8 and the left-hand side is 11, so this is impossible. Dividing by cos⁡4xM\cos 4x_M gives

tan⁡4xM+2xM=0,\tan 4x_M+2x_M=0,

so k=2\boxed{k=2}.

解法二

思路

展开

函数在给定区间非负,最大值的位置在平方后不变。先平方,可以避免对根式直接求导。

答题过程

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On the given interval g(x)≥0g(x)\geq0. Its maximum occurs at the same point as the maximum of

H(x)=[g(x)]2=x2sin⁡4x.H(x)=[g(x)]^2=x^2\sin 4x.

The maximum is interior because H(0)=0H(0)=0 and H(x)>0H(x)>0 for 0<x<π/40<x<\pi/4. At its xx-coordinate xMx_M,

0=H′(xM)= 2xMsin⁡4xM +4xM2cos⁡4xM.\begin{align*} 0=H'(x_M)=&\,2x_M\sin 4x_M\\ &\,+4x_M^2\cos 4x_M. \end{align*}

As xM>0x_M>0, divide by 2xM2x_M to get sin⁡4xM+2xMcos⁡4xM=0\sin 4x_M+2x_M\cos 4x_M=0. The cosine factor cannot be zero, since then the sine would be 11. Therefore

tan⁡4xM+2xM=0,k=2.\tan 4x_M+2x_M=0, \qquad \boxed{k=2}.