Let $\begin{aligned} f(x) & = \begin{cases}|x|-3 & , ~x<1 \\ |x-2|+a & , ~x \geq 1\end{cases} \\ g(x) & = \begin{cases}2-|x| & , ~x<1 \\ Sgn(x)-b & , ~x \geq 1\end{cases} \end{aligned}$

If $h(x)=f(x)+g(x)$ is discontinuous at exactly one point, then which of the following are correct?

(a) $a=-3, b=0$

(b) $a=-3, b=-1$

(c) $a=2, b=1$

(d) $a=0, b=3$

(b), (c)

We have,

$f(x)= \begin{cases}-x-3 &, ~x<0 \\ x-3 & , ~0 \leq x<1 \\ -x+2+a &, ~1 \leq x<2 \\ x-2+a &, ~x \geq 2\end{cases}$

and, $g(x)= \begin{cases}2+x & , ~x<0 \\ 2-x & , ~0 \leq x<1 \\ 1-b & , ~1 \leq x<2 \\ 1-b & , ~x \geq 2\end{cases}$

∴   $h(x)=f(x)+g(x)= \begin{cases}-1 & , ~x<0 \\ -1  & , ~0 \leq x<1 \\ a-b+3-x & , ~\leq x<2 \\ a-b-1+x & , ~x \geq 2\end{cases}$

We observe that h(x) is continuous for all values of x, except at x = 1. At x = 1, it will be discontinuous, if $\lim\limits_{x \rightarrow 1^{-}} h(x) \neq \lim\limits_{x \rightarrow 1^{+}} h(x)$ i.e. if $-1 \neq a-b+2$ or, if $a-b \neq-3$

Clearly, values in options (b) and (c) satisfy this relation.