1.

Let f : R → R be defined as f (x) = x2 + 1. Then find f –1(– 5)(a) {0} (b)  ϕ   (c) {5} (d) {– 5, 5} 

Answer»

Answer: (b) =  ϕ  

f (x) = x2 + 1 

Let y = f (x) = x2 + 1 

⇒ y – 1 = x

⇒ x = ± \(\sqrt{y-1}\) 

⇒ f –1(x) = ± \(\sqrt{x-1}\)

∴  f –1(– 5) = ±  \(\sqrt{-5-1}\)  ± \(\sqrt{-6}\) ∉  R 

∴  f –1(– 5) is the null set.



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