A) \[2{{p}^{3}}=pq-r\]
B) \[2{{p}^{3}}=pq+r\]
C) \[{{p}^{3}}=pq-r\]
D) None of these
Correct Answer: A
Solution :
| [a] Given \[\alpha ,\beta ,\gamma \] are the zeroes of the polynomial \[f(x)={{x}^{3}}-3p{{x}^{2}}+qx-r.\]Then Sum of zeroes |
| \[=-\frac{\text{Coefficient of}\,{{x}^{2}}}{\text{Coefficient of}\,{{x}^{3}}}=-\frac{(-3p)}{1}=3p\] |
| \[\Rightarrow \,\,\,\,\,\,\,\alpha \beta \gamma =3p\Rightarrow 2\beta +\beta =3p\] |
| [\[2\beta =\alpha +\gamma \] (Given)] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,3\beta =3p\Rightarrow \beta =p\] ...(1) |
| Since, \[\beta \] is a zero of polynomial \[f(x)\]. |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f(\beta )=0\] |
| \[\Rightarrow \,\,\,\,\,\,{{\beta }^{3}}-3p{{\beta }^{2}}+q\beta -r=0\] |
| \[\Rightarrow \,\,\,{{p}^{3}}-3{{p}^{3}}+qp-r=0\] [using (1)] |
| \[\Rightarrow \,\,\,-2{{p}^{3}}+pq-r=0\] |
| \[\Rightarrow \,\,\,2{{p}^{3}}=pq-r\] |
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