A) \[(1/2,\,1/2)\]
B) \[(0,-1)\]
C) \[(0,2)\]
D) \[(1,0)\]
Correct Answer: A
Solution :
For \[x=-\frac{\pi }{2},\,f\left( -\frac{\pi }{2} \right)=0\] \[LHL=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f\left( -\frac{\pi }{2}-h \right)\] \[\underset{h\to 0}{\mathop{\lim }}\,\,\,2\cos \,\left( -\frac{\pi }{2}-h \right)=0\] \[RHL=\underset{x\to \frac{{{\pi }^{+}}}{2}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f\left( -\frac{\pi }{2}+h \right)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,a\,\sin \,\left( -\frac{\pi }{2}+h \right)+b\] \[=-a+b\] Since, function is continuous. \[\therefore \] \[RHL=LHL\] \[\Rightarrow \] \[-a+b=0\] \[\Rightarrow \] \[a=b\] From the given options only ie, \[\left( \frac{1}{2},\frac{1}{2} \right)\] satisfies this condition.You need to login to perform this action.
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