Which of the following functions is onto ?
Answer & explanation
Correct answer: option 4
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
Which of the following functions is onto ?
Correct answer: option 4
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