If (f:\mathbb{R}\to\mathbb{R}) is given by (f(x)=\begin{cases}x+2,&x\geq0\x^2+2,&x<0\end{cases}), which statement about (f) being one-one is correct?

Correct answer: It is not one-one because (f(-1)=f(1))

Step 1: For a piecewise function, values from different pieces must also be compared. Step 2: (f(-1)=(-1)^2+2=3) and (f(1)=1+2=3). Step 3: Equal outputs for different inputs break one-one behaviour.