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IAL 2025 Oct Q5

A Level / Edexcel / P1

IAL 2025 Oct Paper · Question 5

题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

(a) Express 2x2−16x+502x^2-16x+50 in the form

a(x+b)2+c\begin{align*} a(x+b)^2+c \end{align*}

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

(3)

Figure 1 shows a sketch of the curve CC with equation y=2x2−16x+50y=2x^2-16x+50.

Figure 1

Given that in Figure 1

  • MM is the minimum point on CC
  • line ll passes through the origin and intersects CC at the points MM and PP

(b) find, using algebra and showing your working, the xx coordinate of PP.

(5)

The region RR is shown shaded in Figure 1.

(c) Use inequalities to fully define RR.

(2)

解答

(a)

解法一

思路

展开

先提出二次项系数 22，再对括号内完成平方。

答题过程

展开 2x2−16x+50= 2(x2−8x)+50= 2((x−4)2−16)+50= 2(x−4)2−32+50= 2(x−4)2+18.\begin{align*} 2x^2-16x+50 =&\,2(x^2-8x)+50\\ =&\,2\bigl((x-4)^2-16\bigr)+50\\ =&\,2(x-4)^2-32+50\\ =&\,2(x-4)^2+18. \end{align*}

So

a=2,b=−4,c=18.\begin{align*} a=2,\qquad b=-4,\qquad c=18. \end{align*}

(b)

解法一

思路

展开

由 (a) 可知最低点 M=(4,18)M=(4,18)。直线 ll 过原点和 MM，所以先求 ll 的方程，再与曲线联立。交点之一是 MM，另一个就是 PP。

答题过程

展开

From part (a), the minimum point is

M=(4,18).\begin{align*} M=(4,18). \end{align*}

Since ll passes through (0,0)(0,0) and M(4,18)M(4,18), its gradient is

184=92.\begin{align*} \frac{18}{4}=\frac92. \end{align*}

So

l:y=92x.\begin{align*} l:\quad y=\frac92x. \end{align*}

At intersections of ll and CC,

2x2−16x+50= 92x.\begin{align*} 2x^2-16x+50=&\,\frac92x. \end{align*}

Multiply by 22:

4x2−32x+100= 9x4x2−41x+100= 0.\begin{align*} 4x^2-32x+100=&\,9x\\ 4x^2-41x+100=&\,0. \end{align*}

Factorise:

4x2−41x+100= (x−4)(4x−25).\begin{align*} 4x^2-41x+100 =&\,(x-4)(4x-25). \end{align*}

So

(x−4)(4x−25)= 0.\begin{align*} (x-4)(4x-25)=&\,0. \end{align*}

The solution x=4x=4 is the point MM, so point PP has

x=254.\begin{align*} x=\frac{25}{4}. \end{align*}

(c)

解法一

思路

展开

区域 RR 位于直线 ll 上方、曲线 CC 下方，并且在 yy 轴和最低点 MM 之间。由于 MM 的 xx 坐标为 44，所以 0⩽x⩽40\leqslant x\leqslant4。

答题过程

展开

The lower boundary is

y=92x.\begin{align*} y=\frac92x. \end{align*}

The upper boundary is

y=2x2−16x+50.\begin{align*} y=2x^2-16x+50. \end{align*}

Also the region runs from x=0x=0 to x=4x=4.

Therefore RR is defined by

0⩽x⩽4,92x⩽y⩽2x2−16x+50.\begin{align*} 0\leqslant x\leqslant4, \qquad \frac92x\leqslant y\leqslant2x^2-16x+50. \end{align*}