Question

Given $f\left(x\right)=x^2+a$, if $x\le 0$
$=2\sqrt{1+x^2}+b$, if $x>0$
and $f(-1)=2$, if $f$ is continuous at $x=0$, then $(a,b)\equiv$____

• A. $(-1,1)$
• B. $(0,1)$
• C. $(1,0)$
• D. $(1,-1)$

Answer 1

Answer: (D)

$(1,-1)$

Given $f(-1)=2$ then from the definition of $f\left(x\right)$ we have, $1+a=2$ or, $a=1$.

Since the function is continuous at $x=0$ then,

$\underset{x\to 0-}{lim}f\left(x\right)=\underset{x\to 0+}{lim}f\left(x\right)$

or, $a=2+b$

or, $1=2+b$ [ Using the value of $a$]

or, $b=-1$.

Then $(a,b)=(1,-1)$.