I tried doing it by area concept. Since the function varies from 1/2 to 1 in 0 to 1 it's area should also vary from 1/2 to 1 in 0 to 1 and similarly for 1 to 2 as well. When I add the answer should be from 1/2 to 3/2 but it is not in the option and the answer given is C. Where I am going wrong.
You're not going wrong - it's just that they're being tricky with the answers, and not providing the best possible range. The interval $[\frac12,\frac32]$ is contained in $[0,2)$, so that inequality is true. We just don't have the full range of possibilities inside that.
You are right and the answer given is wrong , c can not be even a possible range to have a subset of it as the actual answer. Don't worry and good luck.
