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If $f(x)=\left\{\begin{array}{cc}{[x]+[-x],} & x \neq 2 \\ K, & x=2\end{array}\right.$, then $f(x)$ is continuous at $x=2$, provided $K$ is equal to
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$-1 \quad$
$f(x)=\left\{\begin{array}{cc}{[x]+[-x],} & x \neq 2 \\ K, & x=2\end{array}\right.$
Since, $f(x)$ is continuous at $x=2$
$\therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)$
Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}[x]+[-x]$
$=1+(-2)=-1$
$\quad \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}[x]+[-x]=2-3=-1$
$\Rightarrow \quad K=-1$
Since, $f(x)$ is continuous at $x=2$
$\therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)$
Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}[x]+[-x]$
$=1+(-2)=-1$
$\quad \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}[x]+[-x]=2-3=-1$
$\Rightarrow \quad K=-1$
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