Question

Let $[ t ]$ denote the greatest integer $\leq t$ and $\{ t \}$ denote the fractional part of $t$. Then integral value of $\alpha$ for which the left hand limit of the function $f(x)=[1+x]+\frac{\alpha^{2[x]+\{x \mid}+[x]-1}{2\lfloor x]+\{x\}}$ at $x=0$ is equal to $\alpha-\frac{4}{3}$ is___

Answer 1

$f ( x )=[1+ x ]+\frac{\alpha^{2[ x ]+\{x\}}+[ x ]-1}{2[ x ]+\{ x \}}$
$\displaystyle\lim _{ x \rightarrow 0^{-}} f ( x )=\alpha-\frac{4}{3}$
$\Rightarrow 0+\frac{\alpha^{-1}-2}{-1}=\alpha-\frac{4}{3}$
$\Rightarrow 2-\frac{1}{\alpha}=\alpha-\frac{4}{3}$
$\Rightarrow \alpha+\frac{1}{\alpha}=\frac{10}{3}$
$\Rightarrow \alpha=3 ; \alpha \in I$