CalcGospel 國際數學圖譜
返回

IAL 2025 Oct A Q3

A Level / Edexcel / P3

IAL 2025 Oct A Paper · Question 3

题目

Problem

g(x)=x4+x3−7x2+8x−48x2+x−12x>3,x∈Rg(x)=\frac{x^4+x^3-7x^2+8x-48}{x^2+x-12} \qquad x>3,\quad x\in\mathbb{R}

(a) Given that

x4+x3−7x2+8x−48x2+x−12≡x2+A+Bx−3\frac{x^4+x^3-7x^2+8x-48}{x^2+x-12} \equiv x^2+A+\frac{B}{x-3}

find the values of the constants AA and BB.

(4)

(b) Hence, or otherwise, find the equation of the tangent to the curve with equation

y=g(x)y = g(x)

at the point where x=4x = 4. Give your answer in the form y=mx+cy = mx + c, where mm and cc are constants to be determined.

(Solutions relying on calculator technology are not acceptable.)

(5)
题目中文翻译 g(x)=x4+x3−7x2+8x−48x2+x−12x>3,x∈Rg(x)=\frac{x^4+x^3-7x^2+8x-48}{x^2+x-12} \qquad x>3,\quad x\in\mathbb{R}

(a) 已知

x4+x3−7x2+8x−48x2+x−12≡x2+A+Bx−3\frac{x^4+x^3-7x^2+8x-48}{x^2+x-12} \equiv x^2+A+\frac{B}{x-3}

求常数 AA 和 BB 的值。

(b) 由此或用其他方法，求曲线

y=g(x)y = g(x)

在 x=4x = 4 处的切线方程。答案写成 y=mx+cy = mx + c，其中 mm 和 cc 为待定常数。

解答

(a)

解法一

思路

展开

先用分子多项式除以分母，得到二次商式与一次余式。再把分母因式分解；余式恰好含有因子 x+4x+4，可以约去，从而读出 AA 和 BB。

答题过程

展开

Dividing the numerator by x2+x−12x^2+x-12, first subtract x2(x2+x−12)x^2(x^2+x-12):

 x4+x3−7x2+8x−48 −(x4+x3−12x2)= 5x2+8x−48.\begin{align*} &\,x^4+x^3-7x^2+8x-48\\ &\,-(x^4+x^3-12x^2)\\ =&\,5x^2+8x-48. \end{align*}

Then subtract 5(x2+x−12)5(x^2+x-12):

 5x2+8x−48 −(5x2+5x−60)= 3x+12.\begin{align*} &\,5x^2+8x-48\\ &\,-(5x^2+5x-60)\\ =&\,3x+12. \end{align*}

Hence

g(x)= x2+5+3x+12x2+x−12= x2+5+3(x+4)(x+4)(x−3)= x2+5+3x−3.\begin{align*} g(x) =&\,x^2+5+\frac{3x+12}{x^2+x-12}\\ =&\,x^2+5 +\frac{3(x+4)}{(x+4)(x-3)}\\ =&\,x^2+5+\frac{3}{x-3}. \end{align*}

Comparing with the given form,

A=5,B=3.\boxed{A=5,\qquad B=3}.

解法二

思路

展开

官方评分资料也接受先把恒等式两边乘以分母，再比较多项式系数。比较 x2x^2 与 xx 的系数便能依次求出 AA、BB，常数项可作为核对。

答题过程

展开

Since

x2+x−12=(x+4)(x−3),x^2+x-12=(x+4)(x-3),

multiplying the identity by x2+x−12x^2+x-12 gives

 x4+x3−7x2+8x−48≡ (x2+A)(x2+x−12)+B(x+4).\begin{align*} &\,x^4+x^3-7x^2+8x-48\\ \equiv&\,(x^2+A)(x^2+x-12)+B(x+4). \end{align*}

Expanding the right-hand side,

 (x2+A)(x2+x−12)+B(x+4)≡ x4+x3+(A−12)x2 +(A+B)x−12A+4B.\begin{align*} &\,(x^2+A)(x^2+x-12)+B(x+4)\\ \equiv&\,x^4+x^3+(A-12)x^2\\ &\,+(A+B)x-12A+4B. \end{align*}

Comparing the coefficients of x2x^2,

A−12=−7,A-12=-7,

so A=5A=5. Comparing the coefficients of xx,

A+B=8,A+B=8,

so B=3B=3. The constant term also checks:

−12(5)+4(3)=−48.-12(5)+4(3)=-48.

Therefore,

A=5,B=3.\boxed{A=5,\qquad B=3}.

(b)

解法一

思路

展开

承接 (a) 的简单形式求导，在 x=4x=4 处求出切线斜率，同时代入函数求切点纵坐标。最后使用点斜式并整理成题目指定的 y=mx+cy=mx+c。

答题过程

展开

From part (a),

g(x)=x2+5+3x−3.g(x)=x^2+5+\frac{3}{x-3}.

Therefore,

g′(x)=2x−3(x−3)2.g'(x)=2x-\frac{3}{(x-3)^2}.

At x=4x=4, the gradient is

g′(4)=2(4)−3(4−3)2=5.g'(4)=2(4)-\frac{3}{(4-3)^2}=5.

The corresponding point on the curve is

g(4)= 42+5+34−3= 24.\begin{align*} g(4) =&\,4^2+5+\frac{3}{4-3}\\ =&\,24. \end{align*}

Using the point-slope form,

y−24= 5(x−4)y= 5x+4.\begin{align*} y-24=&\,5(x-4)\\ y=&\,5x+4. \end{align*}

Hence the equation of the tangent is

y=5x+4.\boxed{y=5x+4}.