Which of the following is NOT true for the function $f(x)=x^2-4x, x\in [0, 4]$ ?

$f(x)= 3$ has two positive solutions

The correct answer is Option (3) → $f(x)= 3$ has two positive solutions

$f(x)=x^2-4x$

and, $f(x)=3$

$∴x^2-4x=3$

$⇒x^2-4x-3=0$

$⇒x=\frac{4±\sqrt{16+12}}{2}=\frac{4±\sqrt{28}}{2}$

$⇒x=\frac{4-\sqrt{28}}{2}<0$ and $x=\frac{4+\sqrt{28}}{2}>0$

$∴f(x)=3$ has one positive and one negative solutions.