Let f: R→ R be defined as f(x)=3x+4. Choose the correct answer.

The correct answer is Option (1) → f is injective and surjective

for $f(x_1)=f(x_2)$

$3x_1+4=3x_2+4$

$⇒x_1=x_2$ strictly

⇒ it is injective

$y=3x+4⇒x=\frac{y-4}{3}$

so for any y in R

atleast 1 value of x exist in R

⇒ it is surjective

⇒ f is injective and surjective