Verified Solution • 2025–26 Session moderate

QJEEM22Maa9d2: JEE Main 2022 Mathematics - Class 12 JEEM22Maa9d2 Solution: Let [t] denote the greatest integer t and {t} denote the…

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

f(x)=[1+x]+α2[x]+{x}+[x]−12[x]+{x}f(x) = [1 + x] + {{{\alpha ^{2[x] + {\{x\}}}} + [x] - 1} \over {2[x] + \{ x\} }} at x = 0 is equal to α−43\alpha - {4 \over 3} , is _____________.

Step-by-Step Verified Solution

f(x)=[1+x]+a2[x]+{x}+[x]−12[x]+{x}f(x) = [1 + x] + {{{a^{2[x] + \{ x\} }} + [x] - 1} \over {2[x] + \{ x\} }}

lim⁡x→0−f(x)=α−43\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \alpha - {4 \over 3}

⇒lim⁡x→0−1+[x]+αx+[x]+[x]−1x+[x]=α−43\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} 1 + [x] + {{{\alpha ^{x + [x]}} + [x] - 1} \over {x + [x]}} = \alpha - {4 \over 3}

⇒lim⁡h→0−1−1+α−h−1−1−1−h−1=α−43\Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} 1 - 1 + {{{\alpha ^{ - h - 1}} - 1 - 1} \over { - h - 1}} = \alpha - {4 \over 3}

∴\therefore α−1−2−1=α−43{{{\alpha ^{ - 1}} - 2} \over { - 1}} = \alpha - {4 \over 3}

⇒3α2−10α+3=0\Rightarrow 3{\alpha ^2} - 10\alpha + 3 = 0

∴\therefore α=3\alpha = 3 or 13{1 \over 3}

∵\because α\alpha in integer, hence α\alpha = 3

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