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 $x \geq 2$, L.H.S. is always non negative and R.H.S. is always - ve. Hence for $x \geq 2$ no solution.
If $1 \leq x< 2$ then $(x-2)=(x-1)-1=x-2$,
which is an identity $\Rightarrow$(D)
For $0 \leq x<1$, 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.