JEE MAIN - Mathematics (2022 - 27th June Evening Shift - No. 18)
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 _____________.
Explanation
$$f(x) = [1 + x] + {{{a^{2[x] + \{ x\} }} + [x] - 1} \over {2[x] + \{ x\} }}$$
$$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \alpha - {4 \over 3}$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} 1 + [x] + {{{\alpha ^{x + [x]}} + [x] - 1} \over {x + [x]}} = \alpha - {4 \over 3}$$
$$ \Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} 1 - 1 + {{{\alpha ^{ - h - 1}} - 1 - 1} \over { - h - 1}} = \alpha - {4 \over 3}$$
$$\therefore$$ $${{{\alpha ^{ - 1}} - 2} \over { - 1}} = \alpha - {4 \over 3}$$
$$ \Rightarrow 3{\alpha ^2} - 10\alpha + 3 = 0$$
$$\therefore$$ $$\alpha = 3$$ or $${1 \over 3}$$
$$\because$$ $$\alpha$$ in integer, hence $$\alpha$$ = 3
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