Answer & explanation
Correct answer: option 1
$f:R→R$ $f(x)=x^2$
preimage of negative real numbers don't exist have to option is (NOT Surjective)
$f(1)=f(-1)=1$ (Not injective)
Correct answer: option 1
$f:R→R$ $f(x)=x^2$
preimage of negative real numbers don't exist have to option is (NOT Surjective)
$f(1)=f(-1)=1$ (Not injective)