Let f:[0,3]→R be defined by f(x)=min{x–[x],1+[x]–x} where [x] is the greatest integer less than or equal to x. Let P denote the set containing all x∈[0,3] where f is discontinuous, and Q denote the set containing all x∈(0,3) where f is not differentiable. Then the sum of number of elements in P and Q is equal to

Let f:[0,3]→R be defined by f(x)=min{x–[x],1+[x]–x} where [x] is

1.	Let f:[0,3]→R be defined by f(x)=min{x–[x],1+[x]–x} where [x] is the greatest integer less than or equal to x. Let P denote the set containing all x∈[0,3] where f is discontinuous, and Q denote the set containing all x∈(0,3) where f is not differentiable. Then the sum of number of elements in P and Q is equal to
Answer» Let $f : [0, 3] \to R$ be defined by $f (x) = min {x - [x], 1 + [x] - x}$ where $[x]$ is the greatest integer less than or equal to $x$ . Let $P$ denote the set containing all $x \in [0, 3]$ where $f$ is discontinuous, and $Q$ denote the set containing all $x \in (0, 3)$ where $f$ is not differentiable. Then the sum of number of elements in $P$ and $Q$ is equal to