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IAL 2025 Jan Q6

A Level / Edexcel / P1

IAL 2025 Jan Paper · Question 6

题目

Problem

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

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

Given that

  • the point P(4,−5)P(4,-5) lies on CC
  • f′(x)=2x2+ax+b4xf'(x)=\dfrac{2x^2+ax+b}{4\sqrt{x}}, where aa and bb are constants
  • the gradient of the tangent to CC at PP is 77

(a) show that

4a+b=24\begin{align*} 4a+b=24 \end{align*}
(2)

Given also that a+b=−9a+b=-9,

(b) find, in simplest form, f(x)f(x).

(7)

Curve CC is transformed to the curve with equation y=f(x−3)y=f(x-3).

Given that point PP is transformed to the point QQ,

(c) state the coordinates of QQ.

(1)

解答

(a)

解法一

思路

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点 PP 的 xx 坐标是 44，并且该点处切线斜率是 77，所以 f′(4)=7f'(4)=7。把 x=4x=4 代入即可推出目标式。

答题过程

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Since the gradient at PP is 77,

f′(4)=7.\begin{align*} f'(4)=7. \end{align*}

So

2(4)2+a(4)+b44= 732+4a+b8= 732+4a+b= 564a+b= 24.\begin{align*} \frac{2(4)^2+a(4)+b}{4\sqrt4}=&\,7\\ \frac{32+4a+b}{8}=&\,7\\ 32+4a+b=&\,56\\ 4a+b=&\,24. \end{align*}

This is the required result.

(b)

解法一

思路

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先联立 4a+b=244a+b=24 和 a+b=−9a+b=-9，求出 a,ba,b。再把 f′(x)f'(x) 拆成幂函数形式积分。最后用点 P(4,−5)P(4,-5) 求积分常数。

答题过程

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Solve

4a+b= 24,a+b= −9.\begin{align*} 4a+b=&\,24,\\ a+b=&\,-9. \end{align*}

Subtract the second equation from the first:

3a= 33a= 11.\begin{align*} 3a=&\,33\\ a=&\,11. \end{align*}

Then

11+b= −9b= −20.\begin{align*} 11+b=&\,-9\\ b=&\,-20. \end{align*}

Therefore

f′(x)= 2x2+11x−204x= 2x24x1/2+11x4x1/2−204x1/2= 12x3/2+114x1/2−5x−1/2.\begin{align*} f'(x) =&\,\frac{2x^2+11x-20}{4\sqrt{x}}\\ =&\,\frac{2x^2}{4x^{1/2}} +\frac{11x}{4x^{1/2}} -\frac{20}{4x^{1/2}}\\ =&\,\frac12x^{3/2} +\frac{11}{4}x^{1/2} -5x^{-1/2}. \end{align*}

Integrate:

f(x)= 12⋅x5/25/2+114⋅x3/23/2−5⋅x1/21/2+c= 15x5/2+116x3/2−10x1/2+c.\begin{align*} f(x) =&\,\frac12\cdot\frac{x^{5/2}}{5/2} +\frac{11}{4}\cdot\frac{x^{3/2}}{3/2} -5\cdot\frac{x^{1/2}}{1/2}+c\\ =&\,\frac15x^{5/2} +\frac{11}{6}x^{3/2} -10x^{1/2}+c. \end{align*}

Use P(4,−5)P(4,-5):

−5= 15(4)5/2+116(4)3/2−10(4)1/2+c= 15(32)+116(8)−20+c= 325+443−20+c= 96+220−30015+c= 1615+c.\begin{align*} -5 =&\,\frac15(4)^{5/2} +\frac{11}{6}(4)^{3/2} -10(4)^{1/2}+c\\ =&\,\frac15(32)+\frac{11}{6}(8)-20+c\\ =&\,\frac{32}{5}+\frac{44}{3}-20+c\\ =&\,\frac{96+220-300}{15}+c\\ =&\,\frac{16}{15}+c. \end{align*}

So

c= −5−1615= −9115.\begin{align*} c=&\,-5-\frac{16}{15}\\ =&\,-\frac{91}{15}. \end{align*}

Therefore

f(x)=15x5/2+116x3/2−10x1/2−9115.\begin{align*} f(x)=\frac15x^{5/2} +\frac{11}{6}x^{3/2} -10x^{1/2} -\frac{91}{15}. \end{align*}

(c)

解法一

思路

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y=f(x−3)y=f(x-3) 是把 y=f(x)y=f(x) 向右平移 33 个单位，所以点的 xx 坐标加 33，yy 坐标不变。

答题过程

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The transformation from y=f(x)y=f(x) to y=f(x−3)y=f(x-3) is a translation 33 units to the right.

So

P(4,−5)↦Q(7,−5).\begin{align*} P(4,-5)\mapsto Q(7,-5). \end{align*}