Q41 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

Which one of the following functions is a bijection?

  • A
    $f: \mathbb{R}-\mathbb{Z} \rightarrow[0,1]$ defined by $\mathrm{f}(\mathrm{x})=\sqrt{x-[x]}$. (Here $[x]$ represents the greatest integer function)
  • B
    $f: \mathbb{R} \rightarrow(-\infty, 2)$ defined by $\mathrm{f}(\mathrm{x})=4 x-x^2-3$
  • C
    $f:(5, \infty) \rightarrow \mathbb{R}-\{0\}$ defined by $f(x)=\frac{1}{\sqrt{x-5}}$
  • Correct answer
    D
    $f:[0,4] \rightarrow[0,4]$ defined by $f(x)=\sqrt{16-x^2}$

Solution

(a) $f(x)=\sqrt{x-[x]}=\sqrt{\{x\}}$ $\because\{1,2\}=\{2,2\}$ $\therefore f(x)$ is many one. (b) $f(x)=4 x-x^2-3$ Clearly it is a quadratic expression with real roots hence many one function. (c) $f(x)=\frac{1}{\sqrt{x-5}}$ Clearly Range have only positive real numbers. Hence it is into. $\therefore f(x)=\sqrt{16-x^2}$ is bijective.

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

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