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IAL 2025 May A Q10

A Level / Edexcel / P1

IAL 2025 May A Paper · Question 10

题目

Problem

The curve CC has equation y=f(x)y=f(x).

Given that

  • f′(x)=kx(x−3)5f'(x)=\dfrac{k\sqrt{x}(x-3)}{5}, where kk is a constant
  • the point PP with xx coordinate 44 lies on CC
  • the equation of the normal to CC at PP is y=−54x+2y=-\dfrac54x+2

(a) find an equation of the tangent to CC at PP, giving your answer in the form y=mx+cy=mx+c, where mm and cc are constants.

(3)

(b) find the value of kk.

(2)

(c) find f(x)f(x), writing the answer in simplest form.

(5)

解答

(a)

解法一

思路

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法线斜率是 −54-\frac54，所以切线斜率是它的负倒数 45\frac45。

点 PP 的 xx 坐标是 44，而 PP 在给出的法线上，所以先代入法线方程求 PP 的 yy 坐标。

答题过程

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The normal has gradient

−54.\begin{align*} -\frac54. \end{align*}

So the tangent has gradient

45.\begin{align*} \frac45. \end{align*}

Since PP has x=4x=4 and lies on the normal,

y= −54(4)+2= −5+2= −3.\begin{align*} y=&\,-\frac54(4)+2\\ =&\,-5+2\\ =&\,-3. \end{align*}

So

P=(4,−3).\begin{align*} P=(4,-3). \end{align*}

The tangent is

y+3= 45(x−4)y= 45x−165−3y= 45x−315.\begin{align*} y+3=&\,\frac45(x-4)\\ y=&\,\frac45x-\frac{16}{5}-3\\ y=&\,\frac45x-\frac{31}{5}. \end{align*}

(b)

解法一

思路

展开

切线斜率就是 f′(4)f'(4)。把 x=4x=4 代入 f′(x)f'(x)，令它等于 45\frac45，即可求 kk。

答题过程

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Since the tangent gradient is 45\frac45,

f′(4)= 45.\begin{align*} f'(4)=&\,\frac45. \end{align*}

Using

f′(x)=kx(x−3)5,\begin{align*} f'(x)=\frac{k\sqrt{x}(x-3)}{5}, \end{align*}

we get

k4(4−3)5= 452k5= 452k= 4k= 2.\begin{align*} \frac{k\sqrt4(4-3)}{5}=&\,\frac45\\ \frac{2k}{5}=&\,\frac45\\ 2k=&\,4\\ k=&\,2. \end{align*}

(c)

解法一

思路

展开

把 k=2k=2 代入导函数后，先展开成幂函数再积分。积分后用点 P=(4,−3)P=(4,-3) 求常数。

答题过程

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Using k=2k=2,

f′(x)= 2x(x−3)5= 25x1/2(x−3)= 25x3/2−65x1/2.\begin{align*} f'(x)=&\,\frac{2\sqrt{x}(x-3)}{5}\\ =&\,\frac25x^{1/2}(x-3)\\ =&\,\frac25x^{3/2}-\frac65x^{1/2}. \end{align*}

Integrating,

f(x)= 25⋅x5/25/2−65⋅x3/23/2+c= 425x5/2−45x3/2+c.\begin{align*} f(x) =&\,\frac25\cdot\frac{x^{5/2}}{5/2} -\frac65\cdot\frac{x^{3/2}}{3/2}+c\\ =&\,\frac4{25}x^{5/2}-\frac45x^{3/2}+c. \end{align*}

Use P=(4,−3)P=(4,-3):

−3= 425(4)5/2−45(4)3/2+c.\begin{align*} -3=&\,\frac4{25}(4)^{5/2}-\frac45(4)^{3/2}+c. \end{align*}

Now

(4)1/2= 2,(4)3/2= 8,(4)5/2= 32.\begin{align*} (4)^{1/2}=&\,2,\\ (4)^{3/2}=&\,8,\\ (4)^{5/2}=&\,32. \end{align*}

So

−3= 425(32)−45(8)+c= 12825−325+c= 12825−16025+c= −3225+c.\begin{align*} -3=&\,\frac4{25}(32)-\frac45(8)+c\\ =&\,\frac{128}{25}-\frac{32}{5}+c\\ =&\,\frac{128}{25}-\frac{160}{25}+c\\ =&\,-\frac{32}{25}+c. \end{align*}

Hence

c= −3+3225= −4325.\begin{align*} c=&\,-3+\frac{32}{25}\\ =&\,-\frac{43}{25}. \end{align*}

Therefore

f(x)=425x5/2−45x3/2−4325.\begin{align*} f(x)=\frac4{25}x^{5/2}-\frac45x^{3/2}-\frac{43}{25}. \end{align*}