Discussion Forum

1)

If   $f(x)= \begin{cases}\sin x & if x\leq0\\x^{2}+a^{2}, &if 0<x<1\\bx+2, & if 1\leq x \leq 2\\0&,ifx>2\end{cases}$ is continuous

On IR, then a+b+ab=

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, $a+b+ab=0+(-1)+0=-1$