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IAL 2024 Jan Q4

A Level / Edexcel / P2

IAL 2024 Jan Paper · Question 4

题目

Problem

(a) Sketch the curve with equation

y=a−x+4y=a^{-x}+4

where aa is a constant and a>1a>1

On your sketch show

  • the coordinates of the point of intersection of the curve with the yy-axis
  • the equation of the asymptote to the curve.
(3)
x−4−1.513.568.5y136.2804.5774.1464.0374.009\begin{array}{c|cccccc} x&-4&-1.5&1&3.5&6&8.5\\ \hline y&13&6.280&4.577&4.146&4.037&4.009 \end{array}

The table above shows corresponding values of xx and yy for

y=3−12x+4y=3^{-\frac12x}+4

The values of yy are given to four significant figures, as appropriate.

Using the trapezium rule with all the values of yy in the table,

(b) find an approximate value for

∫−48.5(3−12x+4) dx\int_{-4}^{8.5}\left(3^{-\frac12x}+4\right)\,\mathrm{d}x

giving your answer to two significant figures.

(3)

(c) Using the answer to part (b), find an approximate value for

(i)

∫−48.53−12x dx\int_{-4}^{8.5}3^{-\frac12x}\,\mathrm{d}x

(ii)

∫−48.5(3−12x+4) dx+∫−8.54(312x+4) dx\int_{-4}^{8.5}\left(3^{-\frac12x}+4\right)\,\mathrm{d}x +\int_{-8.5}^{4}\left(3^{\frac12x}+4\right)\,\mathrm{d}x
(3)

解答

(a)

解法一

思路

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因为 a>1a>1，所以 a−xa^{-x} 是递减指数曲线。令 x=0x=0 可得 yy-截距；当 xx 很大时，a−xa^{-x} 趋近于 00，所以曲线趋近于 y=4y=4。

答题过程

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When x=0x=0,

y=a0+4=1+4=5.\begin{align*} y=a^0+4=1+4=5. \end{align*}

So the curve intersects the yy-axis at

(0,5).\begin{align*} (0,5). \end{align*}

As x→∞x\to \infty,

a−x→0,\begin{align*} a^{-x}\to 0, \end{align*}

so

y→4.\begin{align*} y\to 4. \end{align*}

Therefore, the asymptote is

y=4.\begin{align*} y=4. \end{align*}

The sketch should be a decreasing exponential curve passing through (0,5)(0,5) and approaching the horizontal asymptote y=4y=4.

(b)

解法一

思路

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表格中的 xx 每次增加 2.52.5，所以梯形法的宽度是 h=2.5h=2.5。使用所有 yy 值时，首尾各一次，中间值乘 22。

答题过程

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The strip width is

h=2.5.\begin{align*} h=2.5. \end{align*}

Using the trapezium rule,

∫−48.5(3−12x+4) dx≈2.52{13+4.009 +2(6.280+4.577 +4.146+4.037)}= 68.86125.\begin{align*} \int_{-4}^{8.5}\left(3^{-\frac12x}+4\right)\,\mathrm{d}x &\approx \frac{2.5}{2} \{13+4.009\\ &\,\hspace{2pt}+2(6.280+4.577\\ &\,\hspace{4pt}+4.146+4.037)\}\\ =&\,68.86125. \end{align*}

To two significant figures,

69.\begin{align*} 69. \end{align*}

(c)(i)

解法一

思路

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第 (b) 题算的是 3−12x+43^{-\frac12x}+4 的积分。现在只要 3−12x3^{-\frac12x} 的积分，所以要减去常数 44 在区间 [−4,8.5][-4,8.5] 上的面积。

答题过程

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Using the answer to part (b),

∫−48.53−12x dx≈69−∫−48.54 dx= 69−4(8.5−(−4))= 69−50= 19.\begin{align*} \int_{-4}^{8.5}3^{-\frac12x}\,\mathrm{d}x &\approx 69-\int_{-4}^{8.5}4\,\mathrm{d}x\\ =&\,69-4(8.5-(-4))\\ =&\,69-50\\ =&\,19. \end{align*}

(c)(ii)

解法一

思路

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第二个积分可通过代换看成和第 (b) 题同样大小的面积，因此两个积分相加就是第 (b) 题答案的两倍。

答题过程

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Using symmetry under the substitution u=−xu=-x,

∫−8.54(312x+4) dx=∫−48.5(3−12u+4) du.\int_{-8.5}^{4}\left(3^{\frac12x}+4\right)\,\mathrm{d}x =\int_{-4}^{8.5}\left(3^{-\frac12u}+4\right)\,du.

Therefore,

∫−48.5(3−12x+4) dx+∫−8.54(312x+4) dx≈2(69)= 138.\begin{align*} &\int_{-4}^{8.5}\left(3^{-\frac12x}+4\right)\,\mathrm{d}x +\int_{-8.5}^{4}\left(3^{\frac12x}+4\right)\,\mathrm{d}x\\ &\approx 2(69)\\ =&\,138. \end{align*}