Answer:
Option D
Explanation:
Given,
$f(x)= \begin{cases}\sin x & if x\leq0\\x^{2}+a^{2}, &if 0<x<1\\bx+2, & if 1\leq x \leq 2\\0&,ifx>2\end{cases}$
is continuous on IR,
$\therefore$ $\lim_{x \rightarrow {0^{-}}}f(x)=\lim_{x \rightarrow {0^{+}}}f(x) $ and
$\lim_{x \rightarrow {2^{-}}} f(x)=\lim_{x \rightarrow {2^{+}}}f(x)$
$\Rightarrow$ $\lim_{x \rightarrow {0^{-}}}\sin x=\lim_{x \rightarrow {0^{-}}}x^{2}+a^{2} $
and $\lim_{x \rightarrow {2^{-}}}bx+2=\lim_{x \rightarrow {2^{+}}}0 $
$\Rightarrow$ $0=0+a^{2}$ and $2b+2=0$
$\Rightarrow$ a=0 and b=-1
Now, $a+b+ab=0+(-1)+0=-1$
