Answer :

We have f : N → N and f(x) = 9x^{2} + 6x – 5.

We need to prove f : N → S is invertible.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x_{1}, x_{2}ϵ N (domain) such that f(x_{1}) = f(x_{2})

⇒ 9x_{1}^{2} + 6x_{1} – 5 = 9x_{2}^{2} + 6x_{2} – 5

⇒ 9x_{1}^{2} + 6x_{1} = 9x_{2}^{2} + 6x_{2}

⇒ 9x_{1}^{2} – 9x_{2}^{2} + 6x_{1} – 6x_{2} = 0

⇒ 9(x_{1}^{2} – x_{2}^{2}) + 6(x_{1} – x_{2}) = 0

⇒ 9(x_{1} – x_{2})(x_{1} + x_{2}) + 6(x_{1} – x_{2}) = 0

⇒ (x_{1} – x_{2})[9(x_{1} + x_{2}) + 6] = 0

⇒ x_{1} – x_{2} = 0 (as x_{1}, x_{2}ϵ R^{+})

∴ x_{1} = x_{2}

So, we have f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2}.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ S (co-domain) such that f(x) = y

⇒ 9x^{2} + 6x – 5 = y

Adding 6 to both sides, we get

9x^{2} + 6x – 5 + 6 = y + 6

⇒ 9x^{2} + 6x + 1 = y + 6

⇒ (3x + 1)^{2} = y + 6

Clearly, for every y ϵ S, there exists x ϵ N (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^{-1}(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

Hence, we have

Thus, f^{-1}(43) = 2 and f^{-1}(163) = 4.

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