Q. 184.9( 9 Votes )

# Let f : R – → R be a function defined as. Show that f : R – → range(f) is one-one and onto. Hence, find f^{-1}.**[CBSE 2017]**

**[CBSE 2017]**

Answer :

We have f : R – → R and

We need to prove f : R – → range(f) is invertible.

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

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

⇒ (4x_{1})(3x_{2} + 4) = (3x_{1} + 4)(4x_{2})

⇒ 12x_{1}x_{2} + 16x_{1} = 12x_{1}x_{2} + 16x_{2}

⇒ 16x_{1} = 16x_{2}

∴ 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 range(f) (co-domain) such that f(x) = y

⇒ 4x = 3xy + 4y

⇒ 4x – 3xy = 4y

⇒ x(4 – 3y) = 4y

Clearly, for every y ϵ range(f), there exists x ϵ A (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

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