A positive square denominator still has a hole · source reference

Complete question

(a) Use graphs to solve $\sqrt x\ge2x$. [2] (b) Solve $(x-1)(x-3)/(x-2)^2<0$.

[modified 2011/RI/Prelim/I/1]

Printed working

Part (a)

$\sqrt{x}\ge2x$. From the graph, $0\le x\le\frac14$.

Part (b): Method 1

$$\frac{(x-1)(x-3)}{(x-2)^2}<0 \quad\Longrightarrow\quad(x-2)^2(x-1)(x-3)<0,\qquad x\ne2.$$

The printed quartic sketch crosses the axis at 1 and 3 and touches it at 2. The two intervals below the axis exclude the point 2. Thus $1<x<2$ or $2<x<3$.

Part (b): Method 2

Since $(x-2)^2>0$ for all real $x\ne2$,

$$(x-1)(x-3)<0,\qquad x\ne2 \quad\Longrightarrow\quad1<x<3,\qquad x\ne2.$$

Teacher margin notes

Write down the values that $x$ cannot take. In this case, $x\ne2$.

Source diagrams

Provenance

Question page references: 1. Solution page references: 2. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.