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    				| 1. | If f : R → R,f(x) = cos (x + 2), is f-1 exists. | 
| Answer» Given function f : R → R, f(x) = cos (x + 2). Putting x = 2π f(2π) = cos (2π+ 2) = cos (2) Putting x = 0 f(0) = cos (0 + 2) = cos 2 Here, only one image is obtained for 0 and 2π. So, ‘f’ is not one-one. Thus, ‘f’ is not one-one onto. Hence, f-1 : R → R does not exist. | |