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IAL 2021 Jan Q7

A Level / Edexcel / P1

IAL 2021 Jan Paper · Question 7

题目

Problem

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

Solutions relying on calculator technology are not acceptable.

f(x)=2x−3x−5,x>0.\begin{align*} f(x)=2x-3\sqrt{x}-5,\qquad x>0. \end{align*}

(a) Solve the equation

f(x)=9.\begin{align*} f(x)=9. \end{align*}
(4)

(b) Solve the equation

f′′(x)=6.\begin{align*} f''(x)=6. \end{align*}
(5)

解答

(a)

解法一

思路

展开

设 u=xu=\sqrt{x}，则 x=u2x=u^2。这样原方程会变成关于 uu 的二次方程。因为 u=x≥0u=\sqrt{x}\ge0，负根要舍去。

答题过程

展开 2x−3x−5= 92x−3x−14= 0.\begin{align*} 2x-3\sqrt{x}-5=&\,9\\ 2x-3\sqrt{x}-14=&\,0. \end{align*}

Let

u=x.\begin{align*} u=\sqrt{x}. \end{align*}

Then x=u2x=u^2, so

2u2−3u−14= 0(2u−7)(u+2)= 0.\begin{align*} 2u^2-3u-14=&\,0\\ (2u-7)(u+2)=&\,0. \end{align*}

Hence

u=72oru=−2.\begin{align*} u=\frac72\quad\text{or}\quad u=-2. \end{align*}

Since u=x≥0u=\sqrt{x}\ge0,

u=72.\begin{align*} u=\frac72. \end{align*}

Therefore

x=u2=(72)2=494.\begin{align*} x=u^2=\left(\frac72\right)^2=\frac{49}{4}. \end{align*}

解法二

思路

展开

通过移项并平方消去根号，转化为二次方程求解的方法。 我们不需要设辅助变量。先将原方程移项整理为 2x−14=3x2x - 14 = 3\sqrt{x}。 然后将等式两边同时平方以消去根号：

(2x−14)2=(3x)2  ⟹  4x2−56x+196=9x\begin{align*} (2x - 14)^2 = (3\sqrt{x})^2 \implies 4x^2 - 56x + 196 = 9x \end{align*}

化简得到一元二次方程 4x2−65x+196=04x^2 - 65x + 196 = 0。 通过因式分解求出 x=4x = 4 或 x=494x = \frac{49}{4}。由于平方是同解变形，有可能引入增根，我们将这两个值代回原方程检验，确定 x=4x=4 为增根舍去，最终的正解为 x=494x = \frac{49}{4}。

答题过程

展开

Rearrange the equation to isolate the surd term on one side:

2x−14=3x\begin{align*} 2x - 14 = 3\sqrt{x} \end{align*}

Square both sides of the equation:

(2x−14)2=  (3x)24x2−56x+196=  9x4x2−65x+196=  0.\begin{align*} (2x - 14)^2 =&\,\, (3\sqrt{x})^2\\[3mm] 4x^2 - 56x + 196 =&\,\, 9x\\[3mm] 4x^2 - 65x + 196 =&\,\, 0. \end{align*}

Factorise the quadratic equation:

(x−4)(4x−49)=0(x - 4)(4x - 49) = 0

So

x=4orx=494.x = 4 \quad\text{or}\quad x = \frac{49}{4}.

We must verify these solutions in the original equation 2x−3x−5=92x - 3\sqrt{x} - 5 = 9.

For x=4x = 4:

LHS=2(4)−34−5=8−6−5=−3≠9.\begin{align*} \text{LHS} = 2(4) - 3\sqrt{4} - 5 = 8 - 6 - 5 = -3 \neq 9. \end{align*}

Thus, x=4x = 4 is an extraneous solution.

For x=494x = \frac{49}{4}:

LHS=2(494)−3494−5=492−212−5=14−5=9.\begin{align*} \text{LHS} = 2\left(\frac{49}{4}\right) - 3\sqrt{\frac{49}{4}} - 5 = \frac{49}{2} - \frac{21}{2} - 5 = 14 - 5 = 9. \end{align*}

This solution is valid. Therefore:

x=494.\begin{align*} x = \frac{49}{4}. \end{align*}

(b)

解法一

思路

展开

把 x\sqrt{x} 写成 x1/2x^{1/2}，连续求两次导数。最后解幂方程时要小心负指数。

答题过程

展开 f(x)= 2x−3x1/2−5,f′(x)= 2−32x−1/2,f′′(x)= −32(−12)x−3/2= 34x−3/2.\begin{align*} f(x)=&\,2x-3x^{1/2}-5,\\ f'(x)=&\,2-\frac32x^{-1/2},\\ f''(x)=&\,-\frac32\left(-\frac12\right)x^{-3/2}\\ =&\,\frac34x^{-3/2}. \end{align*}

Set f′′(x)=6f''(x)=6:

34x−3/2= 6x−3/2= 81x3/2= 8x3/2= 18.\begin{align*} \frac34x^{-3/2}=&\,6\\ x^{-3/2}=&\,8\\ \frac1{x^{3/2}}=&\,8\\ x^{3/2}=&\,\frac18. \end{align*}

Raise both sides to the power 23\frac23:

x= (18)2/3= (12)2= 14.\begin{align*} x =&\,\left(\frac18\right)^{2/3}\\ =&\,\left(\frac12\right)^2\\ =&\,\frac14. \end{align*}