CUET Preparation Today
Let \(f:[0,\infty)\rightarrow\mathbb{R}\) defined by \(f(x)=\frac{x}{x+3}\), then f is |
Onto but not one one One one but not onto invertible Many one into |
One one but not onto |
$f(x_1)=f(x_2)⇒\frac{x_1}{x_1+3}=\frac{x_2}{x_2+3}⇒x_1x_2+3x_1=x_1x_2+3x_2$ $⇒x_1=x_2$ one-one nature $y=\frac{x}{x+3}⇒yx+3y=x$ so $x=\frac{3y}{y-1}$ so for $y=1$ x doesn't exist |
