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Question

If $$\tan\theta_1, \tan \theta_2, \tan\theta_3$$ are the real roots of the $$x^3-(a+1)x^2+(b-a)x-b=0$$, where $$\theta_1+\theta_2+\theta_3\in (0, \pi)$$, then $$\theta_1+\theta_2+\theta_3$$ is equal to


A
π2
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B
π4
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C
3π4
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D
π
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Solution

The correct option is B $$\dfrac {\pi}{4}$$
$$ \tan \theta_{1}+\tan \theta_{2}+\tan \theta_{3}=(a+1) $$
$$ \Sigma \tan \theta_{1} \tan \theta_{2}=(b-a) $$
$$ \tan \theta_{1} \tan \theta_{2} \tan \theta_{3}=b $$
using identity
$$ \quad \tan \left(\theta_{1}+\theta_{2}+\theta_{3}\right)=\dfrac{\Sigma \tan \theta_{1}-\Pi \tan \theta_{1}}{1-\Sigma \tan \theta_{1} \tan \theta_{2}} $$
                                     $$ =\dfrac{a+1-b}{1-(b-a)}\\=1 $$
$$ \Rightarrow \theta_{1}+\theta_{2}+\theta_{3}=\dfrac{\pi}{4} $$

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