Question

The equation $x^2-4 x+[x]+3=x[x]$, where $[x]$ denotes the greatest integer function, has :

• A. no solution
• B. exactly two solutions in $(-\infty, \infty)$
• C. a unique solution in $(-\infty, 1)$
• D. a unique solution in $(-\infty, \infty)$

Answer 1

The Correct Option is D

$x^2-4x+[x]+3=x[x]$
$\Rightarrow x^2-4x+3=x[x]-[x]$
$\Rightarrow (x-1)(x-3)=[x].(x-1)$
$\Rightarrow x=1\text{or}x-3=[x]$
$\Rightarrow x-[x]=3$
$\Rightarrow \{x\}=3\text{(Not Possible)}$
Only one solution $x=1$ in $(-\infty ,\infty )$

Answer 2

Rewrite the equation: \[ x^2 - 4x + [x] + 3 = x[x]. \] For \( [x] = n \), \( x = n + f \), where \( f \in [0, 1) \). Substituting: \[ (n+f)^2 - 4(n+f) + n + 3 = (n+f)n. \] Simplify and solve for \( f \) for each integer \( n \): \[ f^2 + (2n - 4)f + n^2 - 4n + 3 = n^2 + nf. \] This quadratic equation in \( f \) yields a unique solution for certain \( n \). Final Answer: (4)