Question

If $f(x)=\{\begin{array}{ll}1+\cos x,& x\le 0\\ a-x,& 0<x\le 2\\ x^2-b^2,& x>2\end{array}$ is continuous everywhere, then $a^2+b^2=$

• A. 4
• B. 8
• C. 6
• D. 12

Answer 1

Answer: (B)

$\because f(x)$ is continuous everywhere
$\Rightarrow f(x)$ is continuous at $x=0$ and $x=2$
$\therefore LHL$ at $(x=0)=RHL$ at $(x=0)=f(0)$
$\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }f(x)=2$
$\underset{x\to 0^{-}}{\lim }(1+\cos x)=\underset{x\to 0^{+}}{\lim }(a-x)=2$
$\Rightarrow 1+1=a-0=2$
$\Rightarrow a=2$
and LHL at $(x=2)=RHL$ at $(x=2)=f(2)$
$\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2^{+}}{\lim }f(x)=a-2$
$\Rightarrow \underset{x\to 2^{-}}{\lim }(a-x)=\underset{x\to 2^{+}}{\lim }\left(x^2-b^2\right)=2-2$
or $2-2=2^2-b^2=0$
$\Rightarrow b^2=4$
$\therefore a^2+b^2=4+4=8$