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IAL 2022 Jan Q3

A Level / Edexcel / P1

IAL 2022 Jan Paper · Question 3

题目

Problem

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

Solutions relying on calculator technology are not acceptable.

(i) Given that

f(x)=(x+2)2+(3x−58)2\begin{align*} f(x)=(x+\sqrt2)^2+(3x-5\sqrt8)^2 \end{align*}

express f(x)f(x) in the form

ax2+bx2+c\begin{align*} ax^2+bx\sqrt2+c \end{align*}

where aa, bb and cc are integers to be found.

(3)

(ii) Solve

3(4y−33)=5y+3\begin{align*} \sqrt3(4y-3\sqrt3)=5y+\sqrt3 \end{align*}

giving your answer in the form

p+q3\begin{align*} p+q\sqrt3 \end{align*}

where pp and qq are simplified fractions.

(4)

解答

(i)

解法一

思路

展开

关键是先化简 8=22\sqrt8=2\sqrt2，否则中间的 x8x\sqrt8 项很容易算错。展开后把 x2x^2 项、x2x\sqrt2 项和常数项分别合并。

答题过程

展开

Since 8=22\sqrt8=2\sqrt2,

f(x)= (x+2)2+(3x−58)2= (x+2)2+(3x−102)2.\begin{align*} f(x) =&\,(x+\sqrt2)^2+(3x-5\sqrt8)^2\\ =&\,(x+\sqrt2)^2+(3x-10\sqrt2)^2. \end{align*}

Now expand each bracket:

(x+2)2= x2+2x2+2,(3x−102)2= 9x2−60x2+200.\begin{align*} (x+\sqrt2)^2 =&\,x^2+2x\sqrt2+2,\\ (3x-10\sqrt2)^2 =&\,9x^2-60x\sqrt2+200. \end{align*}

Therefore

f(x)= x2+2x2+2+9x2−60x2+200= 10x2−58x2+202.\begin{align*} f(x) =&\,x^2+2x\sqrt2+2 +9x^2-60x\sqrt2+200\\ =&\,10x^2-58x\sqrt2+202. \end{align*}

(ii)

解法一

思路

展开

先展开左边，把含 yy 的项移到同一边，再把 yy 单独留下。得到含根号的分母后，用共轭式有理化。

答题过程

展开 3(4y−33)= 5y+34y3−9= 5y+34y3−5y= 9+3y(43−5)= 9+3.\begin{align*} \sqrt3(4y-3\sqrt3)=&\,5y+\sqrt3\\ 4y\sqrt3-9=&\,5y+\sqrt3\\ 4y\sqrt3-5y=&\,9+\sqrt3\\ y(4\sqrt3-5)=&\,9+\sqrt3. \end{align*}

Hence

y= 9+343−5= 9+343−5⋅43+543+5= (9+3)(43+5)(43)2−52= 363+45+12+5348−25= 57+41323= 5723+41233.\begin{align*} y =&\,\frac{9+\sqrt3}{4\sqrt3-5}\\ =&\,\frac{9+\sqrt3}{4\sqrt3-5} \cdot\frac{4\sqrt3+5}{4\sqrt3+5}\\ =&\,\frac{(9+\sqrt3)(4\sqrt3+5)} {(4\sqrt3)^2-5^2}\\ =&\,\frac{36\sqrt3+45+12+5\sqrt3}{48-25}\\ =&\,\frac{57+41\sqrt3}{23}\\ =&\,\frac{57}{23}+\frac{41}{23}\sqrt3. \end{align*}

解法二

思路

展开

题目要求答案写成 p+q3p+q\sqrt3，所以可以直接设 y=p+q3y=p+q\sqrt3。代入后把有理部分和 3\sqrt3 部分分别比较，就会得到两个关于 p,qp,q 的一次方程。

答题过程

展开

Let

y=p+q3.\begin{align*} y=p+q\sqrt3. \end{align*}

Substitute this into the equation:

3(4y−33)= 5y+33{4(p+q3)−33}= 5(p+q3)+33(4p+4q3−33)= 5p+5q3+34p3+12q−9= 5p+(5q+1)3.\begin{align*} \sqrt3(4y-3\sqrt3)=&\,5y+\sqrt3\\ \sqrt3\{4(p+q\sqrt3)-3\sqrt3\} =&\,5(p+q\sqrt3)+\sqrt3\\ \sqrt3(4p+4q\sqrt3-3\sqrt3) =&\,5p+5q\sqrt3+\sqrt3\\ 4p\sqrt3+12q-9 =&\,5p+(5q+1)\sqrt3. \end{align*}

Compare the rational parts and the 3\sqrt3 parts:

12q−9= 5p,4p= 5q+1.\begin{align*} 12q-9=&\,5p,\\ 4p=&\,5q+1. \end{align*}

From 4p=5q+14p=5q+1,

p=5q+14.\begin{align*} p=\frac{5q+1}{4}. \end{align*}

Substitute into 12q−9=5p12q-9=5p:

12q−9= 5(5q+14)48q−36= 25q+523q= 41q= 4123.\begin{align*} 12q-9=&\,5\left(\frac{5q+1}{4}\right)\\ 48q-36=&\,25q+5\\ 23q=&\,41\\ q=&\,\frac{41}{23}. \end{align*}

Then

p= 5⋅4123+14= 20523+23234= 22823⋅14= 5723.\begin{align*} p =&\,\frac{5\cdot\frac{41}{23}+1}{4}\\ =&\,\frac{\frac{205}{23}+\frac{23}{23}}{4}\\ =&\,\frac{228}{23}\cdot\frac14\\ =&\,\frac{57}{23}. \end{align*}

Therefore

y=5723+41233.\begin{align*} y=\frac{57}{23}+\frac{41}{23}\sqrt3. \end{align*}

解法三

思路

展开

通过方程两边平方消去根号，化为一元二次方程求解的方法（Otherwise）。 我们也可以先展开原方程得到 4y3−9=5y+34y\sqrt3 - 9 = 5y + \sqrt3。 将等式两边同时平方以消去其中的 3\sqrt3 项：(4y3−9)2=(5y+3)2(4y\sqrt3 - 9)^2 = (5y + \sqrt3)^2。 展开后合并同类项，得到一个关于 yy 的一元二次方程：23y2−823y+78=023y^2 - 82\sqrt3 y + 78 = 0。 用求根公式解出 y=413±5723y = \frac{41\sqrt3 \pm 57}{23}。为了排除平方带来的增根，将解代入原方程进行检验，最终确定符合要求的唯一解为 y=5723+41233y = \frac{57}{23} + \frac{41}{23}\sqrt3。

答题过程

展开

Expand the equation:

4y3−9=5y+3\begin{align*} 4y\sqrt3 - 9 = 5y + \sqrt3 \end{align*}

Square both sides of the equation:

(4y3−9)2=  (5y+3)248y2−72y3+81=  25y2+10y3+323y2−823y+78=  0.\begin{align*} (4y\sqrt3 - 9)^2 =&\,\, (5y + \sqrt3)^2\\[3mm] 48y^2 - 72y\sqrt3 + 81 =&\,\, 25y^2 + 10y\sqrt3 + 3\\[3mm] 23y^2 - 82\sqrt3 y + 78 =&\,\, 0. \end{align*}

Solve for yy using the quadratic formula:

y=  823±(−823)2−4(23)(78)2(23)=  823±20172−717646=  823±1299646=  823±11446=  413±5723.\begin{align*} y =&\,\, \frac{82\sqrt3 \pm \sqrt{(-82\sqrt3)^2 - 4(23)(78)}}{2(23)}\\[3mm] =&\,\, \frac{82\sqrt3 \pm \sqrt{20172 - 7176}}{46}\\[3mm] =&\,\, \frac{82\sqrt3 \pm \sqrt{12996}}{46}\\[3mm] =&\,\, \frac{82\sqrt3 \pm 114}{46}\\[3mm] =&\,\, \frac{41\sqrt3 \pm 57}{23}. \end{align*}

Verify both solutions in the original equation:

If y=413−5723y = \frac{41\sqrt3 - 57}{23}:

LHS=  4(413−5723)3−9=  492−228323−9=  285−228323,RHS=  5(413−5723)+3=  2053−28523+23323=  2283−28523.\begin{align*} \text{LHS} =&\,\, 4\left(\frac{41\sqrt3 - 57}{23}\right)\sqrt3 - 9\\[3mm] =&\,\, \frac{492 - 228\sqrt3}{23} - 9\\[3mm] =&\,\, \frac{285 - 228\sqrt3}{23},\\[3mm] \text{RHS} =&\,\, 5\left(\frac{41\sqrt3 - 57}{23}\right) + \sqrt3\\[3mm] =&\,\, \frac{205\sqrt3 - 285}{23} + \frac{23\sqrt3}{23}\\[3mm] =&\,\, \frac{228\sqrt3 - 285}{23}. \end{align*}

Since LHS≠RHS\text{LHS} \neq \text{RHS}, y=413−5723y = \frac{41\sqrt3 - 57}{23} is an extraneous solution.

If y=413+5723y = \frac{41\sqrt3 + 57}{23}:

LHS=  4(413+5723)3−9=  492+228323−9=  285+228323,RHS=  5(413+5723)+3=  2053+28523+23323=  2283+28523.\begin{align*} \text{LHS} =&\,\, 4\left(\frac{41\sqrt3 + 57}{23}\right)\sqrt3 - 9\\[3mm] =&\,\, \frac{492 + 228\sqrt3}{23} - 9\\[3mm] =&\,\, \frac{285 + 228\sqrt3}{23},\\[3mm] \text{RHS} =&\,\, 5\left(\frac{41\sqrt3 + 57}{23}\right) + \sqrt3\\[3mm] =&\,\, \frac{205\sqrt3 + 285}{23} + \frac{23\sqrt3}{23}\\[3mm] =&\,\, \frac{228\sqrt3 + 285}{23}. \end{align*}

\begin{align*} Since LHS=RHS\text{LHS} = \text{RHS}, the correct solution is: \end{align*}

y=5723+41233.y = \frac{57}{23} + \frac{41}{23}\sqrt3.