Question

Let $[t]$ denote the greatest integer $\le t$ 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

Answer: 3

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