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Prove that the Function F : N → N, Defined by F(X) = X2 + X + 1 is One-one but Not Onto. Find Inverse of F : N → S, Where S is Range of F. - Mathematics

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Sum

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)`

Concept: Increasing and Decreasing Functions
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