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If the sum of the squares of the reciprocals of the roots \mathrm{\alpha } and \mathrm{\beta } of the equation \mathrm{3x^{2}+\lambda x-1= 0} is 15, then \mathrm{6\left ( \alpha ^{3}+\beta ^{3} \right )^{2}} is equal to:

Option: 1

18


Option: 2

24


Option: 3

36


Option: 4

96


Answers (1)

best_answer

\mathrm{3x^{2}+\lambda x-1= 0}

\mathrm{\frac{1}{\alpha ^{2}}+\frac{1}{\beta ^{2}}= 15}
\mathrm{\alpha ^{2}+\beta ^{2}= \frac{15}{9}= \frac{5}{3}}

\mathrm{\therefore \; \left ( \alpha +\beta \right )^{2}-2\, \alpha\, \beta= \frac{5}{3} }
         \mathrm{\frac{\lambda ^{2}}{9}= 1 \quad \Rightarrow \quad \lambda ^{2}= 9 }

\mathrm{\therefore \; \; 6\left ( \alpha ^{3}+\beta ^{3} \right )^{2} = 6\left [ \left ( \alpha +\beta \right )\left ( \alpha ^{2}+\beta ^{2}-\alpha \, \beta \right ) \right ]^{2}}
                                   \mathrm{= 6\left [ \left ( \frac{-\lambda }{3} \right )\left ( \frac{5}{3}+\frac{1}{3} \right ) \right ]^{2}}
                                    \mathrm{= 6\times 4\times \frac{\lambda ^{2}}{9}}\
                                     \mathrm{= 24}\

The correct answer is option (B)

Posted by

seema garhwal

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