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P1 2024 Oct Q3

A Level / Edexcel / P1

IAL 2024 Oct Paper · Question 3

题目

Problem

In this question you must show all stages of your working.

f(x)=(x+5)2x,x>0\begin{align*} f(x)=\frac{(x+5)^2}{\sqrt{x}},\qquad x>0 \end{align*}

(a) Find ∫f(x) dx\displaystyle \int f(x)\,\mathrm{d}x.

(4)

(b) (i) Show that when f′(x)=0f'(x)=0,

3x2+10x−25=0\begin{align*} 3x^2+10x-25=0 \end{align*}

(ii) Hence state the value of xx for which

f′(x)=0\begin{align*} f'(x)=0 \end{align*}
(5)
题目中文翻译

本题必须写出所有解题步骤。

已知

f(x)=(x+5)2x,x>0.\begin{align*} f(x)=\frac{(x+5)^2}{\sqrt{x}},\qquad x>0. \end{align*}

(a) 求 ∫f(x) dx\displaystyle \int f(x)\,\mathrm{d}x。

(b)

(i) 证明当 f′(x)=0f'(x)=0 时,

3x2+10x−25=0.\begin{align*} 3x^2+10x-25=0. \end{align*}

(ii) 由此写出使 f′(x)=0f'(x)=0 的 xx 值。

解答

(a)

解法一

思路

展开

先展开 (x+5)2(x+5)^2,再除以 x\sqrt{x},把每一项都写成 xx 的幂。这样就可以逐项积分。

答题过程

展开 f(x)= x2+10x+25x1/2= x3/2+10x1/2+25x−1/2.\begin{align*} f(x) =&\,\frac{x^2+10x+25}{x^{1/2}}\\ =&\,x^{3/2}+10x^{1/2}+25x^{-1/2}. \end{align*}

Therefore

∫f(x) dx= ∫(x3/2+10x1/2+25x−1/2) dx= x5/25/2+10⋅x3/23/2+25⋅x1/21/2+c= 25x5/2+203x3/2+50x1/2+c.\begin{align*} \int f(x)\,\mathrm{d}x =&\,\int\big(x^{3/2}+10x^{1/2}+25x^{-1/2}\big)\,\mathrm{d}x\\ =&\,\frac{x^{5/2}}{5/2} +10\cdot\frac{x^{3/2}}{3/2} +25\cdot\frac{x^{1/2}}{1/2}+c\\ =&\,\frac25x^{5/2}+\frac{20}{3}x^{3/2}+50x^{1/2}+c. \end{align*}

(b)(i)

解法一

思路

展开

沿用 (a) 中已经化好的幂形式,逐项求导。令导数等于 00 后,为了去掉负指数和分数,可以乘以 2x3/22x^{3/2}。

因为题目给了 x>0x>0,所以 x3/2x^{3/2} 不为 00,这样乘过去不会改变方程的解。

答题过程

展开

From part (a),

f(x)=x3/2+10x1/2+25x−1/2.\begin{align*} f(x)=x^{3/2}+10x^{1/2}+25x^{-1/2}. \end{align*}

Differentiate:

f′(x)= 32x1/2+10⋅12x−1/2+25⋅(−12)x−3/2= 32x1/2+5x−1/2−252x−3/2.\begin{align*} f'(x) =&\,\frac32x^{1/2} +10\cdot\frac12x^{-1/2} +25\cdot\left(-\frac12\right)x^{-3/2}\\ =&\,\frac32x^{1/2}+5x^{-1/2}-\frac{25}{2}x^{-3/2}. \end{align*}

When f′(x)=0f'(x)=0,

32x1/2+5x−1/2−252x−3/2=0.\begin{align*} \frac32x^{1/2}+5x^{-1/2}-\frac{25}{2}x^{-3/2}=0. \end{align*}

Multiply by 2x3/22x^{3/2}:

3x2+10x−25=0.\begin{align*} 3x^2+10x-25=0. \end{align*}

This is the required result.

解法二

思路

展开

也可以直接从 f(x)=(x+5)2x−1/2f(x)=(x+5)^2x^{-1/2} 使用乘积法则求导。把所得导数提出公因式后,再令其等于 00,即可得到题目要求的二次方程。

答题过程

展开

Write

f(x)=(x+5)2x−1/2.\begin{align*} f(x)=(x+5)^2x^{-1/2}. \end{align*}

Using the product rule,

f′(x)= 2(x+5)x−1/2−12(x+5)2x−3/2= 12x−3/2(4x(x+5)−(x+5)2)= 12x−3/2(x+5)(3x−5).\begin{align*} f'(x) =&\,2(x+5)x^{-1/2} -\frac12(x+5)^2x^{-3/2}\\ =&\,\frac12x^{-3/2} \big(4x(x+5)-(x+5)^2\big)\\ =&\,\frac12x^{-3/2}(x+5)(3x-5). \end{align*}

Since x>0x>0, the factor 12x−3/2\frac12x^{-3/2} is non-zero. Hence, when f′(x)=0f'(x)=0,

(x+5)(3x−5)= 03x2+10x−25= 0.\begin{align*} (x+5)(3x-5)=&\,0\\ 3x^2+10x-25=&\,0. \end{align*}

This is the required result.

(b)(ii)

解法一

思路

展开

由 (b)(i) 解二次方程即可。最后一定要用 x>0x>0 排除负根。

答题过程

展开

From part (b)(i),

3x2+10x−25=0.\begin{align*} 3x^2+10x-25=0. \end{align*}

Factorise:

3x2+10x−25= (3x−5)(x+5).\begin{align*} 3x^2+10x-25 =&\,(3x-5)(x+5). \end{align*}

So

(3x−5)(x+5)= 0x= 53orx=−5.\begin{align*} (3x-5)(x+5)=&\,0\\ x=&\,\frac53\quad\text{or}\quad x=-5. \end{align*}

Since x>0x>0, reject x=−5x=-5.

Therefore

x=53.\begin{align*} x=\frac53. \end{align*}