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IAL 2022 Jan Q6

A Level / Edexcel / P1

IAL 2022 Jan Paper · Question 6

题目

Problem

The curve CC has equation y=f(x)y=f(x) where x>0x>0.

Given that

  • f′(x)=(x+3)2xxf'(x)=\dfrac{(x+3)^2}{x\sqrt{x}}
  • the point P(4,20)P(4,20) lies on CC

(a)

(i) find the value of the gradient at PP,

(ii) Hence find the equation of the tangent to CC at PP, giving your answer in the form ax+by+c=0ax+by+c=0 where aa, bb and cc are integers to be found.

(4)

(b) Find f(x)f(x), simplifying your answer.

(7)

解答

(a)(i)

解法一

思路

展开

点 PP 的 xx 坐标是 44，所以梯度就是 f′(4)f'(4)。

答题过程

展开 f′(4)= (4+3)244= 494⋅2= 498.\begin{align*} f'(4) =&\,\frac{(4+3)^2}{4\sqrt4}\\ =&\,\frac{49}{4\cdot2}\\ =&\,\frac{49}{8}. \end{align*}

So the gradient at PP is

498.\begin{align*} \frac{49}{8}. \end{align*}

(a)(ii)

解法一

思路

展开

切线经过 P(4,20)P(4,20)，斜率是上一小题的 498\frac{49}{8}。先写点斜式，再整理成整数系数的一边为零形式。

答题过程

展开

The tangent has gradient 498\frac{49}{8} and passes through (4,20)(4,20), so

y−20= 498(x−4).\begin{align*} y-20=&\,\frac{49}{8}(x-4). \end{align*}

Multiply by 88:

8y−160= 49x−19649x−8y−36= 0.\begin{align*} 8y-160=&\,49x-196\\ 49x-8y-36=&\,0. \end{align*}

Therefore the tangent is

49x−8y−36=0.\begin{align*} 49x-8y-36=0. \end{align*}

(b)

解法一

思路

展开

已知的是导函数，所以要求 f(x)f(x) 需要积分。积分前先把分式拆成幂函数：

(x+3)2xx\begin{align*} \frac{(x+3)^2}{x\sqrt{x}} \end{align*}

分母是 xx=x3/2x\sqrt{x}=x^{3/2}，分子展开后有三项，拆开会比较容易积分。最后用 P(4,20)P(4,20) 求常数。

答题过程

展开

First simplify f′(x)f'(x):

f′(x)= (x+3)2xx= x2+6x+9x3/2= x1/2+6x−1/2+9x−3/2.\begin{align*} f'(x) =&\,\frac{(x+3)^2}{x\sqrt{x}}\\ =&\,\frac{x^2+6x+9}{x^{3/2}}\\ =&\,x^{1/2}+6x^{-1/2}+9x^{-3/2}. \end{align*}

Integrate:

f(x)= ∫(x1/2+6x−1/2+9x−3/2) dx= x3/23/2+6⋅x1/21/2+9⋅x−1/2−1/2+c= 23x3/2+12x1/2−18x−1/2+c.\begin{align*} f(x) =&\,\int\left(x^{1/2}+6x^{-1/2}+9x^{-3/2}\right)\,dx\\ =&\,\frac{x^{3/2}}{3/2} +6\cdot\frac{x^{1/2}}{1/2} +9\cdot\frac{x^{-1/2}}{-1/2} +c\\ =&\,\frac23x^{3/2}+12x^{1/2}-18x^{-1/2}+c. \end{align*}

Since P(4,20)P(4,20) lies on CC,

20= 23(4)3/2+12(4)1/2−18(4)−1/2+c= 23(8)+12(2)−18(12)+c= 163+24−9+c= 163+15+c= 613+c.\begin{align*} 20 =&\,\frac23(4)^{3/2}+12(4)^{1/2}-18(4)^{-1/2}+c\\ =&\,\frac23(8)+12(2)-18\left(\frac12\right)+c\\ =&\,\frac{16}{3}+24-9+c\\ =&\,\frac{16}{3}+15+c\\ =&\,\frac{61}{3}+c. \end{align*}

So

c=−13.\begin{align*} c=-\frac13. \end{align*}

Therefore

f(x)=23x3/2+12x1/2−18x−1/2−13.\begin{align*} f(x)=\frac23x^{3/2}+12x^{1/2}-18x^{-1/2}-\frac13. \end{align*}