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IAL 2019 May Q3

A Level / Edexcel / P1

IAL 2019 May Paper · Question 3

题目

Problem

Figure 1 shows the plan of a garden. The marked angles are right angles.

Figure 1

The six edges are straight lines.

The lengths shown in the diagram are given in metres.

Given that the perimeter of the garden is greater than 2929 m,

(a) show that x>1.5x>1.5 m.

(3)

Given also that the area of the garden is less than 72 m272\text{ m}^2,

(b) form and solve a quadratic inequality in xx.

(5)

(c) Hence state the range of possible values of xx.

(1)

解答

(a)

解法一

思路

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把六条边相加得到周长。图中相对的竖直长度合起来是两个 5x5x，相对的水平长度合起来是两个 (6x−2)(6x-2)。

答题过程

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The perimeter is

2(5x)+2(6x−2)= 10x+12x−4= 22x−4.\begin{align*} 2(5x)+2(6x-2) =&\,10x+12x-4\\ =&\,22x-4. \end{align*}

Since the perimeter is greater than 2929,

22x−4>2922x>33x>3322x>1.5.\begin{align*} 22x-4&>29\\ 22x&>33\\ x&>\frac{33}{22}\\ x&>1.5. \end{align*}

(b)

解法一

思路

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把图形分成两个矩形：一个是 2x2x by (2x−1)(2x-1)，另一个是 3x3x by (6x−2)(6x-2)。面积小于 7272，所以得到一个二次不等式。

答题过程

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The area is

2x(2x−1)+3x(6x−2)= 4x2−2x+18x2−6x= 22x2−8x.\begin{align*} 2x(2x-1)+3x(6x-2) =&\,4x^2-2x+18x^2-6x\\ =&\,22x^2-8x. \end{align*}

Since the area is less than 7272,

22x2−8x<7222x2−8x−72<011x2−4x−36<0.\begin{align*} 22x^2-8x&<72\\ 22x^2-8x-72&<0\\ 11x^2-4x-36&<0. \end{align*}

Factorise:

11x2−4x−36= (11x+18)(x−2).\begin{align*} 11x^2-4x-36 =&\,(11x+18)(x-2). \end{align*}

So

(11x+18)(x−2)<0.\begin{align*} (11x+18)(x-2)&<0. \end{align*}

The critical values are

x=−1811,x=2.\begin{align*} x=-\frac{18}{11},\qquad x=2. \end{align*}

Since the quadratic is negative between its roots,

−1811<x<2.\begin{align*} -\frac{18}{11}<x<2. \end{align*}

(c)

解法一

思路

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同时满足上一问的面积限制和 (a) 的周长限制。把两个范围取交集即可。

答题过程

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From part (a),

x>1.5.\begin{align*} x>1.5. \end{align*}

From part (b),

−1811<x<2.\begin{align*} -\frac{18}{11}<x<2. \end{align*}

Therefore the possible range is

1.5<x<2.\begin{align*} 1.5<x<2. \end{align*}