QJEEM21Ma0b6e: JEE Main 2021 Mathematics - Class 11 JEEM21Ma0b6e Solution: Let 0 be in R. If and are the roots of the equation x 2 x + 2 = …

3 $\alpha$ ² $-$ 10 $\alpha$ + 27 $\lambda$ = 0 ..... (1)

$\alpha$ ² $-$ $\alpha$ + 2 $\lambda$ = 0 ...... (2)

(1) $-$ 3(2) gives

$-$ 7 $\alpha$ + 21 $\lambda$ = 0 $\Rightarrow$ $\alpha$ = 3 $\lambda$

Put $\alpha$ = 3 $\lambda$ in equation (1) we get

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

9 $\lambda$ ² = $\lambda$ $\Rightarrow$ $\lambda$ = ${1 \over 9}$ as $\lambda$ $\ne$ 0

Now, $\alpha$ = 3 $\lambda$ $\Rightarrow$ $\lambda$ = ${1 \over 3}$

$\alpha$ + $\beta$ = 1 $\Rightarrow$ $\beta$ = 2/3

$\alpha$ + $\gamma$ = ${10 \over 3}$ $\Rightarrow$ $\gamma$ = 3

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