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IAL 2025 Oct Q8

A Level / Edexcel / P1

IAL 2025 Oct Paper · Question 8

题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

A curve has equation y=f(x)y=f(x), x>0x>0.

Given that

  • f′(x)=2x+8x2+kf'(x)=2x+\dfrac{8}{x^2}+k, where kk is a constant
  • the equation of the tangent to the curve at x=2x=\sqrt2 is y=5x−32y=5x-3\sqrt2

(a) find the exact value of kk.

(2)

(b) Find an equation of the normal to the curve at x=2x=\sqrt2.

(2)

(c) Find f(x)f(x), writing your answer in simplest form.

(4)

解答

(a)

解法一

思路

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切线方程 y=5x−32y=5x-3\sqrt2 的斜率是 55。所以在 x=2x=\sqrt2 时，f′(x)=5f'(x)=5，代入即可求 kk。

答题过程

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Since the tangent has gradient 55,

f′(2)=5.\begin{align*} f'(\sqrt2)=5. \end{align*}

So

22+8(2)2+k= 522+4+k= 5k= 1−22.\begin{align*} 2\sqrt2+\frac{8}{(\sqrt2)^2}+k=&\,5\\ 2\sqrt2+4+k=&\,5\\ k=&\,1-2\sqrt2. \end{align*}

(b)

解法一

思路

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法线斜率是切线斜率的负倒数，所以斜率为 −15-\frac15。先用切线方程求出切点坐标，再写法线方程。

答题过程

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At x=2x=\sqrt2, the yy coordinate is

y= 52−32= 22.\begin{align*} y=&\,5\sqrt2-3\sqrt2\\ =&\,2\sqrt2. \end{align*}

So the point is

(2,22).\begin{align*} (\sqrt2,2\sqrt2). \end{align*}

The gradient of the normal is

−15.\begin{align*} -\frac15. \end{align*}

Thus

y−22= −15(x−2)y= −15x+152+22y= −15x+1125.\begin{align*} y-2\sqrt2=&\,-\frac15(x-\sqrt2)\\ y=&\,-\frac15x+\frac15\sqrt2+2\sqrt2\\ y=&\,-\frac15x+\frac{11\sqrt2}{5}. \end{align*}

(c)

解法一

思路

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把 (a) 的 kk 代入后积分，再用切点 (2,22)(\sqrt2,2\sqrt2) 求积分常数。

答题过程

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Using

k=1−22,\begin{align*} k=1-2\sqrt2, \end{align*}

we have

f′(x)=2x+8x−2+1−22.\begin{align*} f'(x)=2x+8x^{-2}+1-2\sqrt2. \end{align*}

Integrating,

f(x)= x2−8x−1+(1−22)x+c= x2−8x+(1−22)x+c.\begin{align*} f(x) =&\,x^2-8x^{-1}+(1-2\sqrt2)x+c\\ =&\,x^2-\frac8x+(1-2\sqrt2)x+c. \end{align*}

Use the point (2,22)(\sqrt2,2\sqrt2):

22= (2)2−82+(1−22)2+c= 2−42+2−4+c= −2−32+c.\begin{align*} 2\sqrt2 =&\,(\sqrt2)^2-\frac8{\sqrt2} +(1-2\sqrt2)\sqrt2+c\\ =&\,2-4\sqrt2+\sqrt2-4+c\\ =&\,-2-3\sqrt2+c. \end{align*}

So

c=2+52.\begin{align*} c=2+5\sqrt2. \end{align*}

Therefore

f(x)=x2−8x+(1−22)x+2+52.\begin{align*} f(x)=x^2-\frac8x+(1-2\sqrt2)x+2+5\sqrt2. \end{align*}