Question

If $f(x)=\begin{cases}1+\cos x, & x \leq 0 \\ a-x, & 0< x \leq 2 \\ x^2-b^2, & x>2\end{cases}$ 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 $(x=0)=f(0)$
$ \displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=2$
$ \displaystyle\lim _{x \rightarrow 0^{-}}(1+\cos x)=\displaystyle\lim _{x \rightarrow 0^{+}}(a-x)=2$
$ \Rightarrow 1+1=a-0=2$
$\Rightarrow a=2$
and LHL at $(x=2)= RHL$ at $(x=2)=f(2)$
$ \displaystyle\lim _{x \rightarrow 2^{-}} f(x)=\displaystyle\lim _{x \rightarrow 2^{+}} f(x)=a-2$
$ \Rightarrow \displaystyle\lim _{x \rightarrow 2^{-}}(a-x)=\displaystyle\lim _{x \rightarrow 2^{+}}\left(x^2-b^2\right)=2-2$
or $2-2=2^2-b^2=0$
$ \Rightarrow b^2=4 $
$\therefore a^2+b^2=4+4=8$