66Q1) In the interval (&ndash∞, 1), the function is defined by [\cos\pi x] which will be discontinuous whenever the argument of [.] is an integer, i.e. whenever \cos\pi x is an integer. Now cos can take only three integral values of &ndash 1, 0, 1. These values will be achieved whenever \pi x=k\dfrac{\pi}{2} for integer k. That is whenever x=k2 for some integer k. But since x<1, k must be less than 2.
For (1, ∞), the function is defined by |2x-3|[x-2]. The function |2x-3| is continuous for all x. On the other hand, [x-2] will be discontinuous whenever x-2 is an integer i.e. whenever x is an integer >1.
Hence the product |2x -3|[x-2] is discontinuous at all integers x>1.
Finally, at x=1, we have
f(1)=-1
The left hand limit = -1
The right hand limit = -1
Hence, the function is continuous at x=1.