Question

Show that the function f : R → R defined by f(x) = x/x² +1,∀ x ∈ R is neither one-one nor onto. Also, if g : R → R is defined as g(x) = 2x – 1, find fog(x).

Answer 1

For a function to be one to one, if we assume f(x₁​) = f(x₂​), then x_1​ = x_2​

Given, f: R→R defined by f(x) = \(\frac x{x^2 +1}\)​, ∀x∈R

Thus for one-one function, consider

f(x₁​) = f(x₂​)

⇒ \(\frac {x_1}{x_1^2 +1} = \frac{x_2}{x_2^2 +1}\)

⇒ x_1​x₂​(x_2​ − x₁​) = x_2​ − x₁​

⇒ x₂​x₁​ = 1, if x₂​ \(\ne\) x_1​ 

⇒ f is not one-one function.

Also, a function is onto if and only if for every y in the co-domain, there is x in the domain such that f(x) = y

Let f(x) = y

⇒ \(\frac x{x^2 +1} = y\)

⇒ \(x = \frac{1 \pm \sqrt{1 - 4y^2}}{2y}\)

Now, substituting this x in f(x) = y we can see that this function is not onto.

Now, fog(x) = f(g(x))

= f(2x − 1)

\(= \frac{2x - 1}{4x^2 - 4x + 2}\)

Answer 2

Given, f : 1R → R; f(x) = x/1 x²

g : 1R → 1R; g(x) = 2x – 1.

To show that f is neither one-one nor onto

(i) f is one-one : Let x₁, x₂ ∈ R (domain) and

f(x₁) = f(x₂)

Hence f is neither one-one nor not.

Now, fog(x) = f(g(x))