Question

The number of solution(s) of $\left\{x\right\}+\left\{x^2\right\}+\left\{x^3\right\}=3$ is
(where {.} denotes fractional part function)

• A. $2$
• B. $0$
• C. $1$
• D. $3$

Answer 1

The correct option is B $0$

We know that, fractional part function lies in $[0,1)$.

$0\le \left\{x\right\}<1$
$0\le \left\{x^2\right\}<1$
$0\le \left\{x^3\right\}<1$

Adding the three inequalities, we get
$0\le \left\{x\right\}+\left\{x^2\right\}+\left\{x^3\right\}<3$
Hence, $\left\{x\right\}+\left\{x^2\right\}+\left\{x^3\right\}=3$ has no solution.