A) 3
B) 9
C) 4
D) 6
Correct Answer: A
Solution :
| \[9{{x}^{3}}-9{{x}^{2}}-x+1=0\] \[\Rightarrow (x-1)\,\,(9{{x}^{2}}-1)=0\] \[\Rightarrow x=1,\,\,\pm \frac{1}{3}\] |
| \[\cos \alpha =1\] \[\Rightarrow \alpha =0\] |
| \[\cos \beta =1/3\] \[\Rightarrow \beta =co{{s}^{-1}}1/3\] |
| \[\cos \gamma =1/3\] \[\Rightarrow \gamma =\pi -{{\cos }^{-1}}1/3\] |
| \[\sum \alpha =\pi ,\]\[\sum \cos \alpha =1\] |
| \[\therefore \] center is \[(\pi ,1)\] |
| \[\therefore \] equation of the circle is \[{{(x-\pi )}^{2}}+{{(y-1)}^{2}}={{r}^{2}}\] |
| Which passes through \[(2si{{n}^{-1}}\left( \tan \frac{\pi }{4} \right),4)\equiv (\pi ,4)\] |
| \[\therefore \,\,\,0+9={{r}^{2}}\Rightarrow r=3\] |
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