Q

# Need explanation for: - Complex numbers and quadratic equations - JEE Main-7

If $\alpha$ and $\beta$ are the roots of the quadratic equation,

$x^{2}+xsin\theta -2sin\theta =0,\theta \epsilon (0,\frac{\pi}{2})$, then

$\frac{\alpha ^{12}+\beta ^{12}}{(\alpha ^{-12}+\beta ^{-12})\cdot (\alpha -\beta )^{24}}$ is equal to :

• Option 1)

$\frac{2^{12}}{(sin\theta-4)^{12}}$

• Option 2)

$\frac{2^{12}}{(sin\theta+8)^{12}}$

• Option 3)

$\frac{2^{12}}{(sin\theta-8)^{6}}$

• Option 4)

$\frac{2^{6}}{(sin\theta+8)^{12}}$

Views

$\alpha$ and $\beta$ are the roots of the equation

$x^{2}+xsin\theta -2sin\theta =0$

$\alpha +\beta =-sin\theta$

$\alpha \beta =-2sin\theta$

Now,

$\frac{\alpha ^{12}+\beta ^{12}}{(\alpha ^{-12}+\beta ^{-12})\cdot (\alpha -\beta )^{24}}=\frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}}$

$=\frac{(\alpha \beta )^{12}}{[(\alpha +\beta )^{2}-4\alpha \beta ]^{12}}$

$=[\frac{-2sin\theta}{sin^{2}\theta+8sin\theta}]^{12}$

$=\frac{2^{12}}{(sin\theta+8)^{12}}$

correct option (2)

Option 1)

$\frac{2^{12}}{(sin\theta-4)^{12}}$

Option 2)

$\frac{2^{12}}{(sin\theta+8)^{12}}$

Option 3)

$\frac{2^{12}}{(sin\theta-8)^{6}}$

Option 4)

$\frac{2^{6}}{(sin\theta+8)^{12}}$

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