Question

The function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}x+2,& 1\le x\le 2\end{array}\\ \begin{array}{cc}4,& x=2\end{array}\\ \begin{array}{cc}3x-2,& x>2\end{array}\end{array}$ is continuous at

• A. $x=2$ only
• B. $x\le 1$
• C. $x\ge 1$
• D. None of these

Answer 1

The correct option is C $x\ge 1$

At $x=2,f\left(x\right)=4$

$\Rightarrow \underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(3x-2)=4$

$\Rightarrow \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(x+2)=4$

Therefore, the function is contniuous for $x\ge 1$.