JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 21)
Let $$\lambda$$ $$\ne$$ 0 be in R. If $$\alpha$$ and $$\beta$$ are the roots of the equation x2 $$-$$ x + 2$$\lambda$$ = 0, and $$\alpha$$ and $$\gamma$$ are the roots of equation 3x2 $$-$$ 10x + 27$$\lambda$$ = 0, then $${{\beta \gamma } \over \lambda }$$ is equal to ____________.
Answer
18
Explanation
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
$${{\beta \gamma } \over \lambda } = {{{2 \over 3} \times 3} \over {{1 \over 9}}} = 18$$
$$\alpha$$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
$${{\beta \gamma } \over \lambda } = {{{2 \over 3} \times 3} \over {{1 \over 9}}} = 18$$
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