Question

Let f(x) = [x] + $\sqrt{\left\{x\right\}}$, where [x] denotes the greatest integer part of x and {x} denotes the fractional part of $x$. Then $f^{-1}\left(x\right)$is

• A. $\left[x\right]+\sqrt{\left\{x\right\}}$
• B. ${\left[x\right]}^2+\left\{x\right\}$
• C. $\left[x\right]+{\left\{x\right\}}^2$
• D. $\left\{x\right\}$

Answer 1

The correct option is C

$\left[x\right]+{\left\{x\right\}}^2$

Let y = f(x) and [x] = I $y=I+\sqrt{x-I}$

$\Rightarrow \sqrt{x-I}=(y-I)$

$\Rightarrow x-I=(y-I{)}^2$

$\Rightarrow x=(y-I{)}^2+I$

$\Rightarrow x={\left\{y\right\}}^2+\left[y\right]$

$\therefore f^{-1}\left(x\right)=\left[x\right]+{\left\{x\right\}}^2$

Alternatively

$y=\left[x\right]+\sqrt{\left\{x\right\}}$

$\Rightarrow \left[y\right]+\left\{y\right\}=\left[x\right]+\sqrt{\left\{x\right\}}$

$\Rightarrow \left\{y\right\}=\sqrt{\left\{x\right\}}.....(\because [y]=[x\left]\right)$

$\Rightarrow \left\{x\right\}={\left\{y\right\}}^2$

$\Rightarrow \left\{x\right\}+\left[x\right]=\left[y\right]+{\left\{y\right\}}^2$

$\Rightarrow x=\left[y\right]+{\left\{y\right\}}^2\therefore f^{-1}\left(x\right)=\left[x\right]+{\left\{x\right\}}^2$