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) = x2 and g(x) = sin πx/2. Show that g-1 exists but f-1 does not exist. Also, find g-1.

We have f : A A where A = {x ϵ R | –1 ≤ x ≤ 1} defined by f(x) = x2.

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 x1, x2ϵ A (domain) such that f(x1) = f(x2)

x12 = x22

x12 – x22 = 0

(x1 – x2)(x1 + x2) = 0

x1 – x2 = 0 or x1 + x2 = 0

x1 = ±x2

So, we have f(x1) = f(x2) x1 = ±x2.

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) = 12 = 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 x1, x2ϵ A (domain) such that g(x1) = g(x2)

(in the given range)

x1 = x2

So, we have g(x1) = g(x2) x1 = x2.

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

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