A) \[f\]is continuous at \[x=\frac{1}{2}\]
B) \[f\]is continuous at \[x=0\]
C) \[f\]is differentiable in \[\left( -1,0 \right)\]
D) \[f\]is differentiable in \[\left( 0,1 \right)\]
Correct Answer: B
Solution :
| \[f(x)=\cos \pi (\left| x \right|+[x])\] |
| \[f\left( 0 \right)=1,f\left( 0-0 \right)=\cos \pi \left( -1 \right)=-1\] |
| \[f\left( \frac{1}{2} \right)=\cos \pi \left( \frac{1}{2} \right)=0,\] |
| \[f\left( \frac{1}{2}-0 \right)=\cos \pi \left( \frac{1}{2} \right)=0\] |
| \[f\left( \frac{1}{2}+0 \right)=\cos \frac{\pi }{2}=0\] |
| for \[x\in \left( 0,1 \right),f\left( x \right)=\cos \pi x\] |
| Which is differentiable for \[x\in \left( -1,0 \right),\]\[f\left( x \right)=\cos \pi \left( -\pi -1 \right)=-\cos \pi x\] |
| Which is differentiable |
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