QJEEM22Ma27f3: JEE Main 2022 Mathematics - Class 11 JEEM22Ma27f3 Solution: If the sum of the squares of the reciprocals of the roots and…

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

Given, two roots are $\alpha$ and $\beta$ .

$\therefore$ Sum of roots $= \alpha + \beta = {-\lambda \over 3}$

And product of roots $= \alpha \beta = {-1 \over 3}$

Given that,

Sum of square of reciprocal of roots $\alpha$ and $\beta$ is 15.

$\therefore$ ${1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 15$

$\Rightarrow {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} = 15$

$\Rightarrow {{{{(\alpha + \beta )}^2} - 2\alpha \beta } \over {{{(\alpha \beta )}^2}}} = 15$

$\Rightarrow {{{{{\lambda ^2}} \over 9} + 2 \times {1 \over 3}} \over {{1 \over 9}}} = 15$

$\Rightarrow {{{{{\lambda ^2} + 6} \over 9}} \over {{1 \over 9}}} = 15$

$\Rightarrow {\lambda ^2} + 6 = 15$

$\Rightarrow {\lambda ^2} = 9$

Now, $6{({\alpha ^3} + {\beta ^3})^2}$

$= 6{\{ (\alpha + \beta )({\alpha ^2} + {\beta ^2} - \alpha \beta )\} ^2}$

$= 6{(\alpha + \beta )^2}{\left[ {{{(\alpha + \beta )}^2} - 2\alpha \beta - \alpha \beta } \right]^2}$

$= 6{\left( {{-\lambda \over 3}} \right)^2}{\left[ {{{\left( {{-\lambda \over 3}} \right)}^2} - 3\,.\,{-1 \over 3}} \right]^2}$

$= 6 \times {{{\lambda ^2}} \over 9} \times \left[ {{{{\lambda ^2}} \over 9} + 1} \right]$

$= 6 \times {{9} \over 9} \times {\left[ {{{9} \over 9} + 1} \right]^2}$

$= 6 \times {\left( {{2}} \right)^2}$

$= {{6 \times 4}} = 24$