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 \(\frac{120}{\pi} \sum_{\theta \in \mathrm{s}} \theta\) is equal to

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Determinants

Let S be the set of all values of

θ \in [- π, π]

for which the system of linear equations

x + y + 3 z = 0

- x + (tan θ) y + 7 z = 0

x + y + (tan θ) z = 0

has non-trivial solution. Then

\frac{120}{π} \sum_{θ \in s} θ

is equal to

Answer: (C)

For a non-trivial solution, the determinant must be zero:

1 - 1 1 1 tan θ 1 37 tan θ = 0

Expanding:

1 (tan^{2} θ - 7) - 1 (- tan θ - 7) + 3 (- 1 - tan θ) = 0

tan^{2} θ - 7 + tan θ + 7 - 3 - 3 tan θ = 0

tan^{2} θ + tan θ (1 - 3) - 3 = 0

This simplifies to:

(tan θ - 3) (tan θ + 1) = 0

tan θ = 3

tan θ = - 1

.

In

[- π, π]

θ = - \frac{2 π}{3}, \frac{π}{3}, - \frac{π}{4}, \frac{3 π}{4}

Sum:

\sum θ = - \frac{2 π}{3} + \frac{π}{3} - \frac{π}{4} + \frac{3 π}{4} = - \frac{π}{3} + \frac{π}{2} = \frac{π}{6}

Therefore:

\frac{120}{π} \cdot \frac{π}{6} = 20

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option