A) \[f(x)\]is continuous at \[x=\frac{\pi }{2}\]
B) (b \[f(x)\]is continuous at\[x=0\]
C) \[f(x)\] is discontinuous at \[x=0\]
D) \[f(x)\] is continuous over the whole real number
Correct Answer: D
Solution :
Continuity at \[x=0\] LHL At \[x=0,\,\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,(1)=1\] RHL At \[x=0,\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)\] \[=\underset{x\to 0}{\mathop{\lim }}\,(1+\sin x)=1\] \[f(0)=1+\sin 0=1\] \[=L=RHL=f(0),\]so \[f(x)\]is continuous at \[x=0.\] Continuity at \[x=\frac{\pi }{2}\] LHL \[\text{At}\,x=\frac{\pi }{2},\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f(x)\] \[=\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,(1+\sin x)=1+1=2\] RHL At \[x=\frac{\pi }{2},\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f(x)=2+{{\left( \frac{\pi }{2}-\frac{\pi }{2} \right)}^{2}}=2\] \[r\left( \frac{\pi }{2} \right)=2+{{\left( \frac{\pi }{2}-\frac{\pi }{2} \right)}^{2}}=2\] \[\therefore \] \[\text{LHL = RHL}=f\left( \frac{\pi }{2} \right)\] So,\[f(x)\] is continuous at \[x=\frac{\pi }{2}.\] Hence, \[f(x)\]is continuous over the whole real number.
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