JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Sets Relations and Functions

For the left-hand limit as $x\to 0^{-}$, we have $[x]=-1$ and $\{x\}=1+x$ (since $x$ is slightly less than 0). Substituting into the function:

${\lim }_{x\to 0^{-}}f(x)=[1+x]+\frac{{\alpha }^{2(-1)+(1+x)}+(-1)-1}{2(-1)+(1+x)}$

As $x\to 0^{-}$, $[1+x]=0$, and the fraction becomes:

$\frac{{\alpha }^{-1}-2}{-1}=2-\frac{1}{\alpha }$

Setting this equal to $\alpha -\frac{4}{3}$:

$2-\frac{1}{\alpha }=\alpha -\frac{4}{3}$

$\alpha +\frac{1}{\alpha }=\frac{10}{3}$

Multiplying by $\alpha$: ${\alpha }^2-\frac{10}{3}\alpha +1=0$

Solving: $\alpha =3$ or $\alpha =\frac{1}{3}$. Since we need an integral value, $\alpha =3$.