If $f\left( x \right) = {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} – 1$ , $x \in R$ , then the equation $f(x) = 0$ has

Solve $|x\,-\,2| + |x\,-\,1| = x\,-\,3$

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

Suppose $\quad f : R \rightarrow(0, \infty)$ be a differentiable function such that $5 f ( x + y )= f ( x ) \cdot f ( y ), \forall x , y \in R$. If $f(3)=320$, then $\sum \limits_{n=0}^5 f(n)$ is equal to :

Let $f\,:\,R \to R$ be a function such that $f\left( x \right) = {x^3} + {x^2}f'\left( 1 \right) + xf''\left( 2 \right) + f'''\left( 3 \right)$, $x \in R$. Then $f(2)$ equals