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IAL 2023 Jan Q5

A Level / Edexcel / P4

IAL 2023 Jan Paper · Question 5

题目

Problem

Figure 2 shows a sketch of the curve with equation

y2=2x2+15x+10yy^2=2x^2+15x+10y

(a) Find

dydx\frac{dy}{dx}

in terms of xx and yy.

(4)

The curve is not defined for values of xx in the interval (p,q)(p,q), as shown in Figure 2.

(b) Using your answer to part (a) or otherwise, find the value of pp and the value of qq.

(Solutions relying entirely on calculator technology are not acceptable.)

(3)
题目中文翻译

图 2 给出了曲线的示意图,其方程为

y2=2x2+15x+10yy^2=2x^2+15x+10y

(a) 用 x,yx,y 表示

dydx\frac{dy}{dx}

如图 2 所示,当 xx 取区间 (p,q)(p,q) 内的值时,曲线无定义。

(b) 利用 (a) 的结果或用其他方法,求 ppqq 的值。

(完全依赖计算器技术的解法不被接受。)

解答

(a)

解法一

思路

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对曲线方程两边关于 xx 隐式求导。所有含 yy 的项都要乘上 dydx\frac{\mathrm{d}y}{\mathrm{d}x},然后把导数项移到同一边并提取公因式。

答题过程

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Differentiating

y2=2x2+15x+10yy^2=2x^2+15x+10y

implicitly with respect to xx gives

2ydydx=4x+15+10dydx.2y\frac{\mathrm{d}y}{\mathrm{d}x} =4x+15+10\frac{\mathrm{d}y}{\mathrm{d}x}.

Therefore,

(2y10)dydx=4x+15,dydx=4x+152y10.\begin{align*} (2y-10)\frac{\mathrm{d}y}{\mathrm{d}x} =&\,4x+15,\\ \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\boxed{\frac{4x+15}{2y-10}}. \end{align*}

(b)

解法一

思路

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由图形可知,无定义区间的两个端点是曲线左右两支的竖直切线位置。在这些位置,dydx\frac{\mathrm{d}y}{\mathrm{d}x} 无定义,因此令上一问导数的分母为零,先求 yy,再代回曲线方程求两个端点的 xx 坐标。

答题过程

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At the two endpoints of the interval, the curve has vertical tangents. Therefore, the denominator of

dydx=4x+152y10\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{4x+15}{2y-10}

is zero. Hence

2y10=0,2y-10=0,

so

y=5.y=5.

Substituting y=5y=5 into the equation of the curve,

25=2x2+15x+50,2x2+15x+25=0,(2x+5)(x+5)=0.\begin{align*} 25=&\,2x^2+15x+50,\\ 2x^2+15x+25=&\,0,\\ (2x+5)(x+5)=&\,0. \end{align*}

Thus

x=5orx=52.x=-5 \qquad\text{or}\qquad x=-\frac52.

Since p<qp<q,

p=5,q=52.\boxed{p=-5,\qquad q=-\frac52}.

解法二

思路

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这是官方评分资料接受的另一条路线。把曲线方程整理成关于 yy 的二次方程;对于给定的 xx,曲线上存在实点的条件是该二次方程的判别式不小于零。因此,判别式小于零的 xx 范围正是曲线无定义的区间。

答题过程

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Rearranging the equation as a quadratic in yy gives

y210y(2x2+15x)=0.y^2-10y-(2x^2+15x)=0.

For real values of yy, its discriminant must satisfy

Δ=(10)24(1)[(2x2+15x)]=100+8x2+60x=4(2x2+15x+25)=4(2x+5)(x+5).\begin{align*} \Delta =&\,(-10)^2 -4(1)\big[-(2x^2+15x)\big]\\ =&\,100+8x^2+60x\\ =&\,4(2x^2+15x+25)\\ =&\,4(2x+5)(x+5). \end{align*}

The curve is not defined when Δ<0\Delta<0, so

(2x+5)(x+5)<0.(2x+5)(x+5)<0.

The two critical values are x=5x=-5 and x=52x=-\frac52. Since the product is negative between these values,

5<x<52.-5<x<-\frac52.

Therefore,

p=5,q=52.\boxed{p=-5,\qquad q=-\frac52}.