Question

$f\left(x\right)$ $=\{\begin{array}{c}{x}^2-1;x\le -2\\ x^2;-2<x<2\\ x^2+1;x\ge 2\end{array}$
The value of $f\left(2\right)+f(-4)$ is

• A. $20$
• B. $21$
• C. $15$
• D. $0$

Answer 1

The correct option is A $20$

$\Rightarrow$ In function $f\left(2\right)$, $x$ is equal to $2$.

$\Rightarrow$ So, $f\left(x\right)=x^2+1$

$\Rightarrow$ $f\left(2\right)=(2{)}^2+1$

$\therefore$ $f\left(2\right)=4+1=5$ ------- ( 1 )

$\Rightarrow$ In function $f(-4)$, $x$ is less tha $-2$.

$\Rightarrow$ So, $f\left(x\right)=x^2-1$

$\Rightarrow$ $f(-4)=(-4{)}^2-1$

$\therefore$ $f(-4)=16-1=15$ ------ ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $f\left(2\right)+f(-4)=5+15=20$