If $f : R \to R$ be defined by $f(x) = \begin{cases} 2x : x > 3 \\ x^2 : 1 < x \le 3 \\ 3x : x \le 1 \end{cases}$ Then, $f(-1) + f(2) + f(4)$ is

$9$

The correct answer is Option (1) → $9$ ##

Given that, $f(x) = \begin{cases} 2x : x > 3 \\ x^2 : 1 < x \le 3 \\ 3x : x \le 1 \end{cases}$

$f(-1) + f(2) + f(4) = 3(-1) + (2)^2 + 2 \times 4$

$= -3 + 4 + 8 = 9$