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P1 2020 Oct Q9

A Level / Edexcel / P1

IAL 2020 Oct Paper · Question 9

题目

Problem

A curve has equation y=f(x)y=f(x).

Given that

f′(x)=27x2−21x3−5x2x,x>0,\begin{align*} f'(x)=27x^2-\frac{21x^3-5x}{2\sqrt{x}},\qquad x>0, \end{align*}

and that the curve passes through the point (9,10)(9,10),

find f(x)f(x), giving each term in its simplest form.

(6)

解答

解法一

思路

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先把含根号的分式拆开,并写成指数形式。这里要特别小心负号:整个分式前面有一个减号。积分后用点 (9,10)(9,10) 求常数。

答题过程

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Since x=x1/2\sqrt{x}=x^{1/2},

21x3−5x2x= 21x32x1/2−5x2x1/2= 212x5/2−52x1/2.\begin{align*} \frac{21x^3-5x}{2\sqrt{x}} =&\,\frac{21x^3}{2x^{1/2}}-\frac{5x}{2x^{1/2}}\\ =&\,\frac{21}{2}x^{5/2}-\frac52x^{1/2}. \end{align*}

So

f′(x)= 27x2−212x5/2+52x1/2.\begin{align*} f'(x) =&\,27x^2-\frac{21}{2}x^{5/2}+\frac52x^{1/2}. \end{align*}

Integrating,

f(x)= 9x3−3x7/2+53x3/2+c.\begin{align*} f(x) =&\,9x^3-3x^{7/2}+\frac53x^{3/2}+c. \end{align*}

Use f(9)=10f(9)=10:

10= 9(9)3−3(9)7/2+53(9)3/2+c= 6561−6561+45+c= 45+c.\begin{align*} 10 =&\,9(9)^3-3(9)^{7/2}+\frac53(9)^{3/2}+c\\ =&\,6561-6561+45+c\\ =&\,45+c. \end{align*}

Hence

c=−35.\begin{align*} c=-35. \end{align*}

Therefore

f(x)=9x3−3x7/2+53x3/2−35.\begin{align*} f(x)=9x^3-3x^{7/2}+\frac53x^{3/2}-35. \end{align*}