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Question
A function f: R→ R defined by f(x) = `(3x) /5 + 2`, x ∈ R. Show that f is one-one and onto. Hence find f−1.
Solution
f: R → R defined by f(x) = `(3x)/5 + 2`
First we have to prove that f is one-one function for that we have to prove if f(x1) = f(x2) then x1 = x2
Here f(x) = `(3x)/5 + 2`
Let f(x1) = f(x2)
∴ `(3x_1)/5 + 2 = (3x_2)/5 + 2`
∴ `(3x_1)/5 = (3x_2)/5`
∴ x1 = x2
∴ f is a one-one function.
Now, we have to prove that f is an onto function. Let y ∈ R be such that
y = f(x)
∴ y = `(3x)/5 + 2`
∴ y – 2 =`(3x)/5`
∴ `x = (5(y-2))/3 ∈ R`
∴ for any y ∈ co-domain R, there exist an element x = `(5(y-2))/3` ∈ domain R such that f(x) = y
∴ f is an onto function.
∴ f is one-one onto function.
∴ f-1 exists
∴ f-1(y) = `(5(y-2))/3`
∴ f -1(x) = `(5(x-2))/3`
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