Question

Let $f\left(x\right)=\frac{x^2+3}{[x+1]},0\le x\le 2$, where $[.]$ is the greatest integer function. Then the sum of the least value and the greatest value of $f\left(x\right)$ is

Answer 1

$f\left(x\right)=\{\begin{array}{cc}{x}^2+3,& 0\le x<1\\ \frac{x^2+3}{2},& 1\le x<2\\ \frac{x^2+3}{3},& x=2\end{array}$

From graph $\therefore f_{max}=4,f_{min}=2$
$f_{max}+f_{min}=4+2=6.$