Q. 205.0( 1 Vote )

# Let A = {x ϵ R | –1 ≤ x ≤ 1} and let f : A → A, g : A → A be two functions defined by f(x) = x^{2} and g(x) = sin πx/2. Show that g^{-1} exists but f^{-1} does not exist. Also, find g^{-1}.

Answer :

We have f : A → A where A = {x ϵ R | –1 ≤ x ≤ 1} defined by f(x) = x^{2}.

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

First, we will check if f is one-one.

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

⇒ x_{1}^{2} = x_{2}^{2}

⇒ x_{1}^{2} – x_{2}^{2} = 0

⇒ (x_{1} – x_{2})(x_{1} + x_{2}) = 0

⇒ x_{1} – x_{2} = 0 or x_{1} + x_{2} = 0

∴ x_{1} = ±x_{2}

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

This means that two different elements of the domain are mapped to the same element by the function f.

For example, consider f(–1) and f(1).

We have f(–1) = (–1)^{2} = 1 and f(1) = 1^{2} = 1 = f(–1)

Thus, f is not one-one and hence f^{-1} doesn’t exist.

Now, let us consider g : A → A defined by g(x) = sin

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

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

(in the given range)

∴ x_{1} = x_{2}

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

Thus, function g is one-one.

Let y ϵ A (co-domain) such that g(x) = y

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

Thus, the function g has an inverse.

We have g(x) = y ⇒ x = g^{-1}(y)

But, we found g(x) = y ⇒

Hence,

Thus, g(x) is invertible and

Rate this question :

Let f :R → R be given by f (x) = tan x. Then f^{–1} (1) is

| Let * be a binary operation on Q defined by

Show that * is commutative as well as associative. Also find its identity elements, if it exists.

Mathematics - Board PapersFind the identity element in the set Q + of all positive rational numbers for the operation * defined by for all a, b ∈ Q +.

Mathematics - Board PapersLet f : A → B and g : B → C be the bijective functions. Then (g o f)^{–1} is