Question

Let f: R→R be defined as f(x) = 2x – 1 and g: R – {1}→R be defined as g(x) = `(x - 1/2)/(x - 1)`. Then the composition function f (g(x)) is ______.

Options

• Both one-one and onto

• Onto but not one-one

• Neither one-one nor onto

• One-one but not onto

Answer 1

Let f: R→R be defined as f(x) = 2x – 1 and g: R – {1}→R be defined as g(x) = `(x - 1/2)/(x - 1)`. Then the composition function f (g(x)) is one-one but not onto.

Explanation:

Given: f(x) = 2x – 1 and g: R – {1}→R

g(x) = `(x - 1/2)/(x - 1)`

f(g(x)) = 2g(x) – 1 = `(2(x - 1/2))/(x - 1) - 1`

= `(2x - 1 - x + 1)/(x - 1)`

= `x/(x - 1)`

= `(x - 1 + 1)/(x - 1)`

= `(x - 1)/(x - 1) + 1/(x - 1)`

y = f(g(x)) = `1 + 1/(x - 1)`

Differentiate both sides w.r.t.x

`d/(dx)f(g(x)) = 0 - 1/(x - 1)^2`

⇒ `d/(dx)f(g(x)) = -1/(x - 1)^2`

which is always negative.

So, f(g(x)) is a strictly decreasing function.

Hence, f(g(x)) is one-one function.

To check for onto function:

Let y = `1 + 1/(x - 1)`

⇒ y – 1 = `1/(x - 1)`

⇒ x – 1 = `1/(y - 1)`

⇒ x = `1 + 1/(y - 1), y ≠ 1`

∴ x is defined for all y ∈ R – {1} ⇒ y = 1 has no pre-image.

Hence, f(g(x)) is into function.