A) \[\left( 1,\frac{11}{18} \right)\]
B) \[\left( 0,\frac{11}{8} \right)\]
C) \[\left( 2,\frac{11}{18} \right)\]
D) None of these
Correct Answer: C
Solution :
| [c] it is given that \[\alpha ,\beta \] are the roots of the equation \[{{x}^{2}}-6{{p}_{1}}x+2=0.\] |
| \[\therefore \alpha +\beta =6{{p}_{1}},\alpha \beta =2\] ? (i) |
| \[\beta ,\gamma \] are the roots of the equation \[{{x}^{2}}-6{{p}_{2}}x+3=0.\] |
| \[\therefore \beta +\gamma =6{{p}_{2}},\beta \gamma =3\] ?. (ii) |
| \[\gamma ,\alpha \] are the roots of the equation \[{{x}^{2}}-6{{p}_{3}}x+6=0.\] |
| \[\therefore \,\,\,\,\gamma +\alpha =6{{p}_{3}},\,\,\,\gamma \alpha =6\] ?. (iii) |
| From Eqs. (i), (ii) and (iii), we get |
| \[\Rightarrow \alpha \beta \gamma =6\] \[[\therefore \alpha ,\beta ,\gamma >0]\] |
| Now, \[\alpha \beta =2\] and \[\alpha \beta \gamma =6\] |
| \[\Rightarrow \gamma =3\] |
| \[\beta \gamma =3\] and \[\alpha \beta \gamma =6\] |
| \[\alpha =3\] \[\alpha =6\alpha \beta \gamma =6\] |
| \[\Rightarrow \beta =1\] |
| \[\therefore \,\,\,\,\,\,\alpha +\beta =6{{p}_{1}}\Rightarrow 3=6{{p}_{1}}\] |
| \[\Rightarrow {{p}_{1}}=\frac{1}{2}\] |
| \[\beta +\gamma =6{{p}_{2}}\Rightarrow 4=6{{p}_{2}}\] |
| \[\Rightarrow \,\,\,\,{{p}_{2}}=\frac{2}{3}\] |
| and \[\gamma +\alpha =6{{p}_{3}}\Rightarrow 5=6{{p}_{3}}\] |
| \[\Rightarrow {{p}_{3}}=\frac{5}{6}\] |
| The coordinates of the centroid of triangle are |
| \[\left( \frac{\alpha +\beta +\gamma }{3},\frac{1}{3}\left( \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma } \right) \right)\] or |
| \[\left( \frac{6}{3},\frac{1}{3}\left( \frac{1}{2}+1+\frac{1}{3} \right) \right)or\left( 2,\frac{11}{18} \right)\] |
You need to login to perform this action.
You will be redirected in
3 sec
