QJEEM20Ma588b: JEE Main 2020 Mathematics - Class 11 JEEM20Ma588b Solution: Let be in R. If and are the roots of the equation, x 2 - x + 2…

$\alpha$ and $\beta$ are the roots of the
equation x² - x + 2 $\lambda$ = 0 .....(1)

$\therefore$ $\alpha + \beta = 1,\,\alpha \beta = 2\lambda$

$\alpha$ and $\gamma$ are the roots of
the equation, $3{x^2} - 10x + 27\lambda = 0$ ......(2)

$\therefore$ $\alpha + \gamma$ = ${{10} \over 3}$ , $\alpha \gamma$ = ${{27\lambda } \over 3}$ = 9 $\lambda$

Multiplying equation (1) by 3 and subtracting form equation (2) we get

-7x + 21 $\lambda$ = 0

$\Rightarrow$ x = 3 $\lambda$

$\therefore$ $\alpha$ = 3 $\lambda$

As $\alpha$ is root of equation (1) so

$\alpha$ ² - $\alpha$ + 2 $\lambda$ = 0

$\Rightarrow$ 9 $\lambda$ ² - 3 $\lambda$ + 2 $\lambda$ = 0

$\Rightarrow$ $\lambda$ = ${1 \over 9}$

$\Rightarrow$ $\alpha$ = 3 $\times$ ${1 \over 9}$ = ${1 \over 3}$

Also $\alpha \beta = 2\lambda$ = ${2 \over 9}$

$\Rightarrow$ $\beta$ = ${2 \over 3}$

Also $\alpha \gamma$ = 9 $\lambda$ = 9 $\times$ ${1 \over 9}$ = 1

$\Rightarrow$ $\gamma$ = 3

$\therefore$ ${{\beta \gamma } \over \lambda }$ = ${{{2 \over 3}.3} \over {{1 \over 9}}}$ = 18