If $\left\{\begin{matrix}=0&at&x=0\\=\frac{1}{2}-x+[x]&if&0<x<\frac{1}{2}\\=\frac{1}{2}&if&x=\frac{1}{2}\\=\frac{2}{3}-x&if&\frac{1}{2}<x<1\\=1&if&x=1\end{matrix}\right.$ Then f (x) is
Answer & explanation
Correct answer: option 4
$\underset{x→0^+}{\lim}f(x)=\underset{x→0}{\lim}\frac{1}{2}-x+[x]=\frac{1}{2}-0+[0+]=\frac{1}{2}$
f(0) = 0
∴ f(x) is discontinuous at x = 0
Since $\underset{x→{\frac{1}{2}}^+}{\lim}f(x)=\underset{x→{\frac{1}{2}}^-}{\lim}(\frac{2}{3}-x)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$
∴ f(x) is not continuous at x = 1/2
$\underset{x→1^-}{\lim}f(x)\underset{x→1}{\lim}(\frac{2}{3}-x)=-\frac{1}{3}≠f(1)$
∴ f(x) is not continuous at x = 1.