A) \[a=2n+(3/2);b\in R;n\in I\]
B) \[a=4n+2;b\in R;n\in I\]
C) \[a=4n+(3/2);b\in {{R}^{+1}};n\in I\]
D) \[a=4n+1;b\in {{R}^{+}};n\in I\]
Correct Answer: A
Solution :
[a] \[f(-1)=b(1-1)+1=1;\underset{h\to 0}{\mathop{\lim }}\,f(-1+h)=1\] \[\underset{h\to 0}{\mathop{\lim }}\,f(-1-h)=\underset{h\to 0}{\mathop{\lim }}\,\sin (\pi (-1-h)+\pi a)\] \[=\sin (-\pi +\pi a)=-sin\pi a\] For continuous \[\sin \pi a=-1=\sin \left( 2n\pi +\frac{3\pi }{2} \right)\] \[\Rightarrow \pi a=2n\pi +\frac{3\pi }{2}\Rightarrow a=2n+\frac{3}{2}\] Hence, \[a=2n+\frac{3}{2},n\in I\] and \[b\in R\].
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