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Question
Prove that the function f : N → N, defined by f(x) = x2 + x + 1 is one-one but not onto. Find the inverse of f: N → S, where S is range of f.
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Solution
The given function is
f : N → N
f(x) = x2 + x + 1
Let x1, x2 6N
So let f (x1) = f (x2)
`x_1^2 + x_1 + 1 = x_2^2 + x_2 + 1`
`x_1^2 - x_2^2 + x_1 - x_2 = 0`
(x1 - x2) (x1 + x2 + 1) = 0
∵ x2 = x1
or x2 = - x1 - 1
∵ x1 ∈ N
∴ x1 - 1 ∈ N
So x2 ≠ -x1 - 1
∵ f (x2) = f (x1) only for x1 = x2
So f(x) is one -one function.
∵ f (x) = x2 + x + 1
`"f" ("x") = ("x" + 1/2)^2 + 3/4`
Which is an increasing function.
f(1) = 3
∵ Range of f(x) will be {3, 7, .....} Which is a subset of N.
So it is an into function. i.e., f(x) is not an onto function.
let y = x2 + x + 1
x2 + x + 1 - y = 0
`"x" = (-1± sqrt((1 - 4 )(1 - "y")))/(2)`
`"x" = (-1 ± sqrt(4"y" -3))/(2)`
So two possibilities are there for `f^-1 ("x")`
`"f"^-1 ("x") = (-1 + sqrt(4"x" -3))/(2), (-1 - sqrt(4"x" -3))/(2)` and we know `"f"^-1 (3)` = 1 because `"f"(1) = 3`
so `"f"^-1 ("x") = (-1 + sqrt(4"x" - 3))/(2)`
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