CUET Preparation Today
Which of the following functions is onto ? |
$f: R→R, f(x) = x^2 $ $f: R→R, f(x) = x^2 +1$ $f: R→R, f(x) = x^2-1 $ $f: R→R, f(x) = 2x+1 $ |
$f: R→R, f(x) = 2x+1 $ |
The correct answer is Option () → $f: R→R, f(x) = 2x+1$ $\frac{y-1}{2}=x⇒f^{-1}x=\frac{x-1}{2}$ for every $y∈R$ atleast one $x∈R$ exists ⇒ it is onto function |
