Question

For \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x)=1+x^2\), which statement is correct?

Options

• It is many-one but not onto

• It is one-one but not onto

• It is onto but not many-one

• It is one-one and onto

Answer 1

From \[f(x_1)=f(x_2)\], we obtain \(x_1=\pm x_2\), so distinct inputs such as \(1\) and \(-1\) have the same image. Also, negative codomain elements have no pre-image, so the function is not onto.