Question

The function defined by \[\mathrm{f}(x)=\frac{2x+3}{3x+4},x\neq-\frac{4}{3}\] is

Options

• only one one

• only onto

• onto for \[y\neq\frac{2}{3}\] and one-one

• neither one-one nor onto

Answer 1

onto for \[y\neq\frac{2}{3}\] and one-one

Explanation:

Assume f (x₁) = f (x₂)

\[\therefore\quad\frac{2x_1+3}{3x_1+4}=\frac{2x_2+3}{3x_2+4}\]

\[\therefore\quad(2x_1+3)(3x_2+4)=(2x_2+3)(3x_1+4)\]

\[\therefore\quad8x_1+9x_2=8x_2+9x_1\]

\[\therefore\quad x_1=x_2\]

Function is one-one.

Let y = f (x)

\[y=\frac{2x+3}{3x+4}\]

∴ y(3x + 4) = 2x + 3
∴ 3xy + 4y = 2x + 3
∴ 3xy − 2x = 3 − 4y

\[\therefore\quad x=\frac{3-4y}{3y-2}\]

∴ For every y in the co-domain, there exists x in the domain such that

f(x) = y except at y = 2/3

⇒ Function is onto for y ≠ 2/3