Answer : Given: f(x) = x2 + 1 To find: (i) f-1{10}

We know that, if f: X → Y such that y ∈ Y. Then f-1(y) = {x ∈ X: f(x) = y}. In other words, f-1(y) is the set of pre – images of y

Let f-1{10} = x. Then, f(x) = 10 …(i) and it is given that f(x) = x2 + 1 …(ii)

So, from (i) and (ii), we get x2 + 1 = 10

⇒ x2 = 10 – 1

⇒ x2 = 9

⇒ x = √9

⇒ x = ± 3

∴ f-1{10} = {-3, 3}

To find: (ii) f-1{-3}

Let f-1{-3} = x. Then, f(x) = -3 …(iii) and it is given that f(x) = x2 + 1 …(iv)

So, from (iii) and (iv), we get x2 + 1 = -3

⇒ x2 = -3 – 1

⇒ x2 = -4

Clearly, this equation is not solvable in R

∴ f-1{-3} = ɸ