Given $$f\left( x \right) = b\left( {{{\left[ x \right]}^2} + \left[ x \right]} \right) + 1$$     for $$x \geqslant - 1 = \sin \left( {\pi \left( {x + a} \right)} \right)$$     for $$x < - 1$$   where $$\left[ x \right]$$ denotes the integral part of $$x,$$ then for what values of $$a,\,b$$  the function is continuous at $$x = - 1\,?$$

Solution :
$$\eqalign{ & f\left( { - 1} \right) = b\left( {1 - 1} \right) + 1 = 1; \cr & \mathop {\lim }\limits_{h \to 0} f\left( { - 1 + h} \right) = 1 \cr & \mathop {\lim }\limits_{h \to 0} f\left( { - 1 - h} \right) = \mathop {\lim }\limits_{h \to 0} \sin \left( {\pi \left( { - 1 - h} \right) + \pi a} \right) \cr & = \sin \left( { - \pi + \pi a} \right) \cr & = - \sin \,\pi a \cr & {\text{For continuous }}\sin \,\pi a = - 1 = \sin \left( {2n\pi + \frac{{3\pi }}{2}} \right) \cr & \Rightarrow \pi a = 2n\pi + \frac{{3\pi }}{2} \cr & \Rightarrow a = 2n + \frac{3}{2} \cr & {\text{Hence, }}a = 2n + \frac{3}{2},\,n\, \in \,I{\text{ and }}b\, \in \,R \cr} $$