Q54 / 89 · KVPY 2017 (5 Nov SB/SX) · Mathematics

Let $\mathrm{R}$ be the set of real numbers and $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined by $f(x)=\frac{\{x\}}{1+[x]^{2}}$, where $[x]$ is the greatest integer less than or equal to $\mathrm{x}$, and $\{x\}=x-[x] .$ Which of the following statements are true ? I. The range of $f$ is a closed interval II. $\mathrm{f}$ is continuous on $\mathrm{R}$. III. $\mathrm{f}$ is one-one on $\mathrm{R}$.

  • A
    I only
  • B
    II only
  • C
    III only
  • Correct answer
    D
    None of I, II and III

Solution

$f(x)=\frac{\{x\}}{1+[x]^{2}}$ $f(x)=\left\{\begin{array}{cc}\frac{x+1}{2} ; & -1 \leq x < 0 \\ x ; & 0 \leq x < 1 \\ \frac{x-1}{2} ; & 1 \leq x < 2 \\ \frac{x-2}{5} ; & 2 \leq x < 3 \\ & \vdots \\ & \text { So }-\text { on }\end{array}\right.$ Now check accordingly

Hence, the correct answer is option: 'D'.

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