Let [t] denote the greatest integer $\le$ t and {t} denote the fractional part of t. The integral value of $\alpha$ for which the left hand limit of the function

$f(x) = [1 + x] + {{{\alpha ^{2[x] + {\{x\}}}} + [x] - 1} \over {2[x] + \{ x\} }}$ at x = 0 is equal to $\alpha - {4 \over 3}$, is _____________.

<p>$f(x) = [1 + x] + {{{a^{2[x] + \{ x\} }} + [x] - 1} \over {2[x] + \{ x\} }}$</p> <p>$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \alpha - {4 \over 3}$</p> <p>$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} 1 + [x] + {{{\alpha ^{x + [x]}} + [x] - 1} \over {x + [x]}} = \alpha - {4 \over 3}$$</p> <p>$$ \Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} 1 - 1 + {{{\alpha ^{ - h - 1}} - 1 - 1} \over { - h - 1}} = \alpha - {4 \over 3}$$</p> <p>$\therefore$ ${{{\alpha ^{ - 1}} - 2} \over { - 1}} = \alpha - {4 \over 3}$</p> <p>$\Rightarrow 3{\alpha ^2} - 10\alpha + 3 = 0$</p> <p>$\therefore$ $\alpha = 3$ or ${1 \over 3}$</p> <p>$\because$ $\alpha$ in integer, hence $\alpha$ = 3</p>