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    				| 1. | If f : R → R, f(x) = 2x + 1 and g : R → R, g(x) = x3, then (gof)-1(27) is equal to :(a) 2(b) 1(c) -1(d) 0 | 
| Answer» Answer is (b) Let (gof)-1(27) = x ∴ (gof)(x) = 27 g{f(x)} = 27 g{2x + 1} = 27 (2x + 1)3 = 27 2x + 1 = 271/3 2x + 1 = 33 x 1/3 2x + 1 = 3 ∴ 2x = 2 x = 1 | |