Question

If $[x]$ and $\{x\}$ denotes the greatest integer function less than or equal to $x$ and fractional part function respectively, then the number of real $x$, satisfying the equation $(x-2)[x]=\{x\}-1$, is

• A. 0
• B. 1
• C. 2
• D. infinite

Answer 1

Answer: (D)

For $x\ge 2$, L.H.S. is always non negative and R.H.S. is always - ve. Hence for $x\ge 2$ no solution.
If $1\le x<2$ then $(x-2)=(x-1)-1=x-2$,
which is an identity $\Rightarrow$(D)
For $0\le x<1$, LHS is ' 0 ' and RHS is $(-)$ ve $\Rightarrow$ Nosolution.
For $x<0$, LHS is $(+)$ ve, RHS is $(-)$ ve $\Rightarrow$ Nosolution. $(x-2)[x]=x-[x]-1$
$\Rightarrow (x-1)[x]=x-1$
$\Rightarrow (x-1)([x]-1)=0$ Now interpret.