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Let f : R → R : f(x) = x2 + 1. Find
(i) f–1 {10}
(ii) f–1 {–3}.
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