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IAL 2020 Oct Q9

A Level / Edexcel / P3

IAL 2020 Oct Paper · Question 9

题目

Problem

(a) Given that

x4−x3−10x2+3x−9x2−x−12≡x2+P+Qx−4x>−3\frac{x^4-x^3-10x^2+3x-9}{x^2-x-12}\equiv x^2+P+\frac{Q}{x-4}\qquad x>-3

find the value of the constant PP and show that Q=5Q=5.

(4)

The curve CC has equation y=g(x)y=g(x), where

g(x)=x4−x3−10x2+3x−9x2−x−12−3<x<3.5, x∈Rg(x)=\frac{x^4-x^3-10x^2+3x-9}{x^2-x-12}\qquad -3<x<3.5,\ x\in\mathbb{R}

(b) Find the equation of the tangent to CC at the point where x=2x=2.

Give your answer in the form y=mx+cy=mx+c, where mm and cc are constants to be found.

(5)

Figure 4 shows a sketch of the curve CC.

The region RR, shown shaded in Figure 4, is bounded by CC, the yy-axis, the xx-axis and the line with equation x=2x=2.

(c) Find the exact area of RR, writing your answer in the form a+bln⁡2a+b\ln 2, where aa and bb are constants to be found.

(5)
题目中文翻译

(a) 已知

x4−x3−10x2+3x−9x2−x−12≡x2+P+Qx−4x>−3\frac{x^4-x^3-10x^2+3x-9}{x^2-x-12}\equiv x^2+P+\frac{Q}{x-4}\qquad x>-3

求常数 PP 的值，并证明 Q=5Q=5。

曲线 CC 的方程为 y=g(x)y=g(x)，其中

g(x)=x4−x3−10x2+3x−9x2−x−12−3<x<3.5, x∈Rg(x)=\frac{x^4-x^3-10x^2+3x-9}{x^2-x-12}\qquad -3<x<3.5,\ x\in\mathbb{R}

(b) 求曲线 CC 在 x=2x=2 处的切线方程。

将答案写成 y=mx+cy=mx+c 的形式，其中 mm 和 cc 为待求常数。

图 4 给出了曲线 CC 的草图。

图中阴影部分区域 RR 由曲线 CC、yy 轴、xx 轴以及直线 x=2x=2 围成。

(c) 求区域 RR 的精确面积，并将答案写成 a+bln⁡2a+b\ln 2 的形式，其中 aa 和 bb 为待求常数。

解答

(a)

First factorise the denominator:

x2−x−12=(x−4)(x+3)x^2-x-12=(x-4)(x+3)

We are given

x4−x3−10x2+3x−9x2−x−12≡x2+P+Qx−4\frac{x^4-x^3-10x^2+3x-9}{x^2-x-12} \equiv x^2+P+\frac{Q}{x-4}

Multiply both sides by (x−4)(x+3)(x-4)(x+3):

x4−x3−10x2+3x−9≡(x2+P)(x−4)(x+3)+Q(x+3)x^4-x^3-10x^2+3x-9 \equiv (x^2+P)(x-4)(x+3)+Q(x+3)

Since

(x−4)(x+3)=x2−x−12(x-4)(x+3)=x^2-x-12

we get

x4−x3−10x2+3x−9≡(x2+P)(x2−x−12)+Q(x+3)x^4-x^3-10x^2+3x-9 \equiv (x^2+P)(x^2-x-12)+Q(x+3)

Expand:

(x2+P)(x2−x−12)+Q(x+3)=x4−x3+(P−12)x2+(−P+Q)x+(−12P+3Q)(x^2+P)(x^2-x-12)+Q(x+3) =x^4-x^3+(P-12)x^2+(-P+Q)x+(-12P+3Q)

Compare coefficients with

x4−x3−10x2+3x−9x^4-x^3-10x^2+3x-9

From the coefficient of x2x^2,

P−12=−10P-12=-10

so

P=2P=2

From the coefficient of xx,

−P+Q=3-P+Q=3

Substitute P=2P=2:

−2+Q=3-2+Q=3

Therefore

Q=5Q=5

Hence

P=2,Q=5\boxed{P=2,\qquad Q=5}

So

g(x)=x2+2+5x−4\boxed{g(x)=x^2+2+\frac{5}{x-4}}

(b)

From part (a),

g(x)=x2+2+5x−4g(x)=x^2+2+\frac{5}{x-4}

Differentiate:

g′(x)=2x−5(x−4)2g'(x)=2x-\frac{5}{(x-4)^2}

At x=2x=2,

g′(2)=2(2)−5(2−4)2g'(2)=2(2)-\frac{5}{(2-4)^2}

So

g′(2)=4−54=114g'(2)=4-\frac54=\frac{11}{4}

This is the gradient of the tangent.

Now find the yy coordinate when x=2x=2:

g(2)=22+2+52−4=4+2−52=72\begin{aligned} g(2) &=2^2+2+\frac{5}{2-4}\\ &=4+2-\frac52\\ &=\frac72 \end{aligned}

So the tangent passes through

(2,72)\left(2,\frac72\right)

Using

y−y1=m(x−x1)y-y_1=m(x-x_1)

we get

y−72=114(x−2)y-\frac72=\frac{11}{4}(x-2)

Therefore

y=114x−112+72y=\frac{11}{4}x-\frac{11}{2}+\frac72

Hence

y=114x−2\boxed{y=\frac{11}{4}x-2}

(c)

The region RR lies between x=0x=0 and x=2x=2, under the curve CC and above the xx-axis.

Using the expression from part (a),

g(x)=x2+2+5x−4g(x)=x^2+2+\frac{5}{x-4}

So the area is

∫02(x2+2+5x−4) dx\int_0^2 \left(x^2+2+\frac{5}{x-4}\right)\,\mathrm{d}x

Integrate:

∫(x2+2+5x−4) dx=13x3+2x+5ln⁡∣x−4∣\int \left(x^2+2+\frac{5}{x-4}\right)\,\mathrm{d}x =\frac13x^3+2x+5\ln|x-4|

Therefore

Area=[13x3+2x+5ln⁡∣x−4∣]02=(83+4+5ln⁡2)−(0+0+5ln⁡4)=203+5ln⁡2−5ln⁡4\begin{aligned} \text{Area} &=\left[\frac13x^3+2x+5\ln|x-4|\right]_0^2\\ &=\left(\frac83+4+5\ln2\right)-\left(0+0+5\ln4\right)\\ &=\frac{20}{3}+5\ln2-5\ln4 \end{aligned}

Since

ln⁡4=ln⁡(22)=2ln⁡2\ln4=\ln(2^2)=2\ln2

we get

Area=203+5ln⁡2−10ln⁡2\text{Area}=\frac{20}{3}+5\ln2-10\ln2

Hence

Area=203−5ln⁡2\boxed{\text{Area}=\frac{20}{3}-5\ln2}