If x_1, x_2, x_3 are the real roots of the eq | vedclass

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(A) Given the cubic equation $x^3 - x^2 	an 	heta + x 	an^2 	heta + 	an 	heta = 0$. By Vieta's formulas,the sum of roots,sum of roots taken two

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) Given the cubic equation $x^3 - x^2 	an 	heta + x 	an^2 	heta + 	an 	heta = 0$. By Vieta's formulas,the sum of roots,sum of roots taken two at a time,and product of roots are: $x_1 + x_2 + x_3 = 	an 	heta$ $x_1 x_2 + x_2 x_3 + x_3 x_1 = 	an^2 	heta$ $x_1 x_2 x_3 = -	an 	heta$ We know the identity $	an^{-1} x_1 + 	an^{-1} x_2 + 	an^{-1} x_3 = 	an^{-1} \left( \frac{(x_1 + x_2 + x_3) - x_1 x_2 x_3}{1 - (x_1 x_2 + x_2 x_3 + x_3 x_1)} ight)$. Substituting the values: $= 	an^{-1} \left( \frac{	an 	heta - (-	an 	heta)}{1 - 	an^2 	heta} ight) = 	an^{-1} \left( \frac{2 	an 	heta}{1 - 	an^2 	heta} ight)$. Using the double angle formula $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,we get: $= 	an^{-1} (	an 2	heta) = 2	heta$. At $	heta = \frac{\pi}{12}$,the value is $2 	imes \frac{\pi}{12} = \frac{\pi}{6}$. Thus,option $A$ is correct.

If $x_1, x_2, x_3$ are the real roots of the equation $x^3-x^2 \tan \theta+x \tan ^2 \theta+\tan \theta=0$ and $0 < \theta < \frac{\pi}{4}$,then the value of $\tan ^{-1} x_1+\tan ^{-1} x_2+\tan ^{-1} x_3$ at $\theta=\frac{\pi}{12}$ is

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