Explanation: Given, f(x)=2+x2
For one-one, ‌‌f(x1)=f(x2)
⇒2+x12‌=2+x22
⇒x12‌=x2‌2
⇒x1‌=±x2
⇒‌x1‌=x2
‌‌ or ‌‌x1‌=−x2
Thus, f(x) is not one-one.
For onto
Let
∴‌f(x)‌=y‌ such that ‌
⇒x2‌=y−2
⇒x‌=±√y−2

Put y=−3, we get
x=±√−3−2=±√−5
Which is not possible as root of negative is not a real number.
Hence, x is not real.
So, f(x) is not onto.