Question

If A = {-1,1} fand g are two functions defined on A, where f(x) = x², g(x) = sin (πx/2), prove that g^-1 exists but f^-1 does not exist, also find g^-1.

Answer 1

Given, A = {-1, 1}

f(x)= x², g(x) = sin πx/2

One-one/many-one : f : A → A, f(x) = x²

f(-1) = f(1) = 1

⇒ Image of – 1 and 1 is same.

So, f is not one-one.

Onto/into : Elements of co-domain unpaired with any element of domain.

So ‘f’ is not onto.

So, f is neither one-one nor onto.

Hence, ‘f^-1‘ does not exist.

Thus, for g(x) = sin πx/2

One-one/many-one : Let x₁, x₂ ∈ A, such that

Hence, function in one-one.

Let y ∈ R (co-domain), if possible then pre image is x in R, then

Since, pre-image of each value of y exist in domain R, then g(x) is onto function.

Hence, g(x) is one-one onto function, so g^-1 exist.