A) \[A=0,\,B=1\]
B) \[A=1,\,B=1\]
C) \[A=-1,\,B=1\]
D) \[A=-1,\,B=0\]
Correct Answer: C
Solution :
For continuity at all \[x\in R,\] we must have \[f\left( -\frac{\pi }{2} \right)=\underset{x\to {{\left( \frac{\pi }{2} \right)}^{-}}}{\mathop{\lim }}\,(-2\sin x)=\underset{x\to {{\left( -\frac{\pi }{2} \right)}^{+}}}{\mathop{\lim }}\,(A\sin x+B)\] \[\Rightarrow \,\,2=-A+B\] ?(i) and \[f\left( \frac{\pi }{2} \right)=\underset{x\to {{\left( \frac{\pi }{2} \right)}^{-}}}{\mathop{\lim }}\,(A\sin x+B)=\underset{x\to {{\left( \frac{\pi }{2} \right)}^{+}}}{\mathop{\lim }}\,(cosx)\] \[\Rightarrow \,\,0=A+B\] ??(ii) From (i) and (ii), \[A=-1\] and \[B=1\]You need to login to perform this action.
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