Question

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

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

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is D $2$

$\Rightarrow$ In function $f(-1)$, $x$ is equal to $-1$

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

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

$\Rightarrow$ In function $f\left(0\right)$, $x$ is between $-1$ and $1$.

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

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

So, from ( 1 ) and ( 2 ),

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