The number of points f(x) = cases [ x], & 0 x 1 |2x - 3| [x - 2], & 1 < x 2 cases is…

MATHEMATICS · Single correct · AP · Calculus AB · Limits and Continuity · Limits and Continuity

Problem

The number of points

f(x) = \begin{cases} [\cos \pi x], & 0 \leq x \leq 1 \\ |2x - 3| [x - 2], & 1 < x \leq 2 \end{cases}

is discontinuous at is (

[\cdot]

denotes the greatest integral function) (a) two (b) three (c) four (d) zero