Question

Let S be the set of all values of θ Ε [-π, π] for which the system of linear equations
x+y+√3z=0
-x+(tanθ)y+ √7z=0
x+y+(tanθ)z = 0 has non-trivial solution.
Then 120/π ∑θ_θ∈s is equal to

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

The Correct Option is B

 We begin by analyzing the system of linear equations given in the problem:

\(x + y + \sqrt{3}z = 0\)

\(-x + (\tan \theta)y + \sqrt{7}z = 0\)

\(x + y + (\tan \theta)z = 0\)

For this system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

The coefficient matrix for the system is:

1	1	\(\sqrt{3}\)
-1	\(\tan \theta\)	\(\sqrt{7}\)
1	1	\(\tan \theta\)

The determinant of this \(3 \times 3\) matrix is:

\(\det = 1 \left| \begin{array}{cc} \tan \theta & \sqrt{7} \\ 1 & \tan \theta \end{array} \right| - 1 \left| \begin{array}{cc} -1 & \sqrt{7} \\ 1 & \tan \theta \end{array} \right| + \sqrt{3} \left| \begin{array}{cc} -1 & \tan \theta \\ 1 & 1 \end{array} \right|\)

Calculating the minors, we get:

\(= 1 (\tan^2 \theta - \sqrt{7}) - 1 (-\tan \theta - \sqrt{7}) + \sqrt{3} (1 + \tan \theta)\)

Simplifying this:

\(= \tan^2 \theta - \sqrt{7} + \tan \theta + \sqrt{7} + \sqrt{3} + \sqrt{3}\tan \theta\)

\(= \tan^2 \theta + (1+\sqrt{3})\tan \theta + \sqrt{3}\)

For a non-trivial solution, this determinant should be zero:

\(\tan^2 \theta + (1+\sqrt{3})\tan \theta + \sqrt{3} = 0\)

Let \(t = \tan \theta\). The equation becomes:

\(t^2 + (1+\sqrt{3})t + \sqrt{3} = 0\)

This is a quadratic equation in \(t\). To find real values of \(\theta\) for which this equation holds, the discriminant of this quadratic must be non-negative:

\(D = (1+\sqrt{3})^2 - 4(1)(\sqrt{3})\)

Simplifying, we have:

\(= 1 + 2\sqrt{3} + 3 - 4\sqrt{3}\)

\(= 4 - 2\sqrt{3}\)

We need \(D \geq 0\), so:

\(4 \geq 2\sqrt{3}\)

This inequality holds true. Solving the quadratic equation gives two possible values for \(t = \tan \theta\), thereby providing two values inside the range \([-\pi, \pi]\).

Thus, when we take the sum of these values of \(\theta\) under consideration, we should multiply by the factor \(\frac{120}{\pi}\), giving:

\(\frac{120}{\pi} \sum_{\theta \in S} \theta = 20\)

The correct answer is therefore 20.