Let $f:(1,3) \rightarrow \mathrm{R}$ be a function defined by
$f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},$ where $[\mathrm{x}]$ denotes the greatest
integer $\leq \mathrm{x} .$ Then the range of $f$ is
$\left(\frac{3}{5}, \frac{4}{5}\right)$
$\left(\frac{2}{5}, \frac{3}{5}\right] \cup\left(\frac{3}{4}, \frac{4}{5}\right)$
$\left(\frac{2}{5}, \frac{4}{5}\right]$
$\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]$
The graph of $y = f(x)$ is shown then number of solutions of the equation $f(f(x)) =2$ is
$f(x,\;y) = \frac{1}{{x + y}}$ is a homogeneous function of degree
The domain of $f(x) = [\sin x] \cos \left( {\frac{\pi }{{[x - 1]}}} \right)$ is (where $[.]$ denotes $G.I.F.$)
Given the function $f(x) = \frac{{{a^x} + {a^{ - x}}}}{2},\;(a > 2)$. Then $f(x + y) + f(x - y) = $
Consider a function $f : N \rightarrow R$, satisfying $f(1)+2 f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x) ; x \geq 2$ with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to