Abhishek Singh
Last Activity: 3 Years ago
Clearly
Now Draw the graph of |Cos(x)| and check the injectivity by Horizontal line test and see it is injective for option (C)
12) f(x) = |x|, Clearly it becomes onto for optiond (D) by graph
13) For explicitly writing the function, check that
if x is integer f(x) = [x] + [-x] = x+ (-x) = 0
if x is non integer f(x) = [x] + [-x] = [x] + (-[x] -1) = -1
Its range is {0,-1} which is not in options