Question

$f\left(x\right)$ $=\{\begin{array}{c}x+1;x\le 0\\ x^2;0<x<2\\ x-1;x\ge 2\end{array}$
The value of $f(-1)+f\left(0\right)$ is

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

The correct option is B $1$

$\Rightarrow$ In function $f(-1)$, the $x$ is less than $0$

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

$\Rightarrow$ $f(-1)=-1+1$

$\therefore$ $f(-1)=0$ ----- ( 1 )

$\Rightarrow$ In function $f\left(0\right)$, the $x$ is equal to $0$.

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

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

$\therefore$ $f\left(0\right)=1$ ----- ( 2 )

From ( 1 ) and ( 2 ),

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