JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 13)
Let S be the set of all values of $$\theta \in[-\pi, \pi]$$ for which the system of linear equations
$$x+y+\sqrt{3} z=0$$
$$-x+(\tan \theta) y+\sqrt{7} z=0$$
$$x+y+(\tan \theta) z=0$$
has non-trivial solution. Then $$\frac{120}{\pi} \sum_\limits{\theta \in \mathrm{s}} \theta$$ is equal to :
40
30
10
20
Explanation
Since, the given system has a non trivial solution,
$$ \text { So, } \Delta=0 $$
$$ \Rightarrow \Delta=\left|\begin{array}{ccc} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{array}\right|=0 $$
$$ \begin{aligned} & \Rightarrow 1\left(\tan ^2 \theta-\sqrt{7}\right)-1(-\tan \theta-\sqrt{7})+\sqrt{3}(-1-\tan \theta)=0 \\\\ & \Rightarrow \tan ^2 \theta-\sqrt{7}+\tan \theta+\sqrt{7}-\sqrt{3}-\sqrt{3} \tan \theta=0 \\\\ & \Rightarrow \tan \theta(\tan \theta-\sqrt{3})+1(\tan \theta-\sqrt{3})=0 \\\\ & \Rightarrow \tan \theta=\sqrt{3} \text { or } \tan \theta=-1 \\\\ & \therefore \theta=\left\{\frac{\pi}{3}, \frac{-2 \pi}{3}, \frac{-\pi}{4}, \frac{3 \pi}{4}\right\} \\\\ & \text { So, } \frac{120}{\pi} \sum \theta=\frac{120}{\pi}\left\{\frac{4 \pi-8 \pi-3 \pi+9 \pi}{12}\right\} \\\\ & =\frac{120}{\pi}\left[\frac{2 \pi}{12}\right]=20 \end{aligned} $$
$$ \text { So, } \Delta=0 $$
$$ \Rightarrow \Delta=\left|\begin{array}{ccc} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{array}\right|=0 $$
$$ \begin{aligned} & \Rightarrow 1\left(\tan ^2 \theta-\sqrt{7}\right)-1(-\tan \theta-\sqrt{7})+\sqrt{3}(-1-\tan \theta)=0 \\\\ & \Rightarrow \tan ^2 \theta-\sqrt{7}+\tan \theta+\sqrt{7}-\sqrt{3}-\sqrt{3} \tan \theta=0 \\\\ & \Rightarrow \tan \theta(\tan \theta-\sqrt{3})+1(\tan \theta-\sqrt{3})=0 \\\\ & \Rightarrow \tan \theta=\sqrt{3} \text { or } \tan \theta=-1 \\\\ & \therefore \theta=\left\{\frac{\pi}{3}, \frac{-2 \pi}{3}, \frac{-\pi}{4}, \frac{3 \pi}{4}\right\} \\\\ & \text { So, } \frac{120}{\pi} \sum \theta=\frac{120}{\pi}\left\{\frac{4 \pi-8 \pi-3 \pi+9 \pi}{12}\right\} \\\\ & =\frac{120}{\pi}\left[\frac{2 \pi}{12}\right]=20 \end{aligned} $$
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