A) x = 0
B) x = 1, 2
C) x = 0, 2, 4
D) None of these
Correct Answer: B
Solution :
[b] Clearly f(x) is not continuous at x = 0 Now \[f(1)=cos\,\,3\] \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\cos 3\cos \left\{ \frac{\pi }{2}(-h) \right\}=\cos 3\] \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\cos 3\cos \left\{ \frac{\pi }{2}(h) \right\}=\cos 3\] Further \[f(2)=cos1.cos\,\,\frac{\pi }{2}=0\] \[\underset{x\to {{2}^{-}}}{\mathop{\lim \,}}\,f(x)=\underset{h\to 0}{\mathop{lim}}\,\,\,cos1.cos\left\{ \frac{\pi }{2}(1-h) \right\}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\cos 1.0=0\] \[=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\cos 1.\cos \left\{ \frac{\pi }{2}(1+h) \right\}=0\] \[\therefore f(x)\] is continuous at \[x=1\] and \[x=2\] both
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