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 $f(-1)+f\left(0\right)$ is

• A. $2$
• B. $0$
• C. $-2$
• D. $1$

Answer 1

The correct option is D $1$

$\Rightarrow$ In function $f(-1)$, $x$ is lies between $-2$ and $2$.

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

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

$\Rightarrow$ In function $f\left(0\right)$, $x$ is lies between $-2$ and $2$.

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

$\Rightarrow$ $f\left(0\right)=(0{)}^2=0$ -------- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $f(-1)+f\left(0\right)=1+0=1$