Mathematics - Functions Question with Solution | TestHub

$\left\{\frac{3}{2},\frac{11}{4}\right\}$ $|\mathrm{x}-1|=2[\mathrm{x}]-3\{\mathrm{x}\}$

(i) if $\mathrm{x}\ge 1$, then $\mathrm{x}-1=2[\mathrm{x}]-3\{\mathrm{x}\}$

$$\begin{array}{llccc}& [x]+\{x\}-1=2[x]-3\{x\}& 4\{x\}=1+[x]\therefore & 0\le 1+[x]<4\therefore & -1\le [x]<3\end{array}$$

$\therefore$ possible value of $[\mathrm{x}]$ are 1 and 2

if $[x]=1$, then $\{x\}=\frac{1}{2}$

if $[x]=2$, then $\{x\}=3/4$

$$\begin{array}{llc}\therefore & \mathrm{x}=3/2\therefore & \mathrm{x}=\frac{11}{4}\end{array}$$

(ii) $x<1$, then the equation becomes

$$\begin{array}{rlrl}& 1-x=2[x]-3\{x\}& & \end{array}$$

$$1-[x]-\{x\}=2[x]-3\{x\}$$

$$2\{x\}=3[x]-1$$

$\therefore 0\le 3[\mathrm{x}]-1<2$

i.e. $1\le 3[x]<3$

i.e. $\frac{1}{3}\le [x]<1$ which is not possible

$\therefore x=3/2,\frac{11}{4}$ are the only solutions