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Let $\mathrm{[t]}$ denote the greatest integer $\mathrm{\leq t}$ and $\mathrm{{t}}$ denote the fractional part of $\mathrm{t}$ . The integral value of $\mathrm{\alpha }$ for which the left hand limit of the function $\mathrm{f(x)=[1+x]+\frac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}}$ at $\mathrm{x=0}$ is equal to $\mathrm{\alpha -\frac{4}{3}}$ , is _______
Option: 1
3

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

$\mathrm{\lim _{x \rightarrow 0^{-}}[x]=-1, \lim _{x \rightarrow 0^{-}}\{x\}=1}$

$\mathrm{So\: \lim _{x \rightarrow 0^{-}} f(x)}$

$\mathrm{=\lim _{x \rightarrow-}\left(1+[x]+\frac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}\right)=\alpha-\frac{4}{3}}$

$\mathrm{\Rightarrow \quad 1-1+\frac{\alpha^{2(-1)+1}+(-1)-1}{2(-1)+1}=\alpha-\frac{4}{3} }\\$

$\mathrm{\Rightarrow \quad \frac{\frac{1}{\alpha}-2}{-1}=\alpha-\frac{4}{3} \Rightarrow-\frac{1}{2}+2=\alpha-\frac{4}{3} }$

$\mathrm{\Rightarrow \alpha+\frac{1}{\alpha}=\frac{10}{3} \Rightarrow 3 \alpha^{2}-10 \alpha+3=0}\\$

$\mathrm{\Rightarrow \alpha=3\: O R\: \alpha=1 / 3}\\$

$\mathrm{Since \: \alpha= integer},\\$

$\mathrm{ \alpha=3}$

Hence answer is $\mathrm{ 3}$

Posted by

Sumit Saini

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