Question

The quadratic equation whose roots are 

\(l = \lim_{\theta\to0} \frac{3sin\theta - 4sin^3\theta}{\theta}\)

m = \(\lim_{\theta\to0} \frac{2tan\theta}{\theta(1-tan^2\theta)}\) is

• A. x² - 5x + 6 = 0
• B. x² - 5x + 5 = 0
• C. x² - 5x - 6 = 0
• D. x² + 5x - 6 = 0

Answer 1

The Correct Option is A

To determine the quadratic equation given the problem, we must first find the roots of the quadratic equation, i.e., \(l\) and \(m\). These are derived using the provided limits:

1. First, calculate \(l = \lim_{\theta \to 0} \frac{3\sin \theta - 4\sin^3 \theta}{\theta}\).

   Using the limit property \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\), we can express the limit as follows:

   \(l = \lim_{\theta \to 0} \frac{\sin \theta (3 - 4\sin^2 \theta)}{\theta}\)

   This simplifies to:

   \(l = \lim_{\theta \to 0} \left(\frac{\sin \theta}{\theta} \cdot (3 - 4\sin^2 \theta)\right)\)

   Applying the limit, we have:

   \(l = 1 \cdot (3 - 4 \cdot 0^2) = 3\)

2. Next, calculate \(m = \lim_{\theta \to 0} \frac{2\tan \theta}{\theta(1 - \tan^2 \theta)}\).

   Given that \(\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1\):

   \(m = \lim_{\theta \to 0} \frac{\tan \theta (2)}{\theta(1 - \tan^2 \theta)} = \lim_{\theta \to 0} \frac{2 \tan \theta}{\theta} \cdot \frac{1}{1 - \tan^2 \theta}\)

   Simplifying using limits:

   \(m = 2 \cdot 1 \cdot \frac{1}{1 - 0^2} = 2\)

Hence, the roots of the quadratic equation are \(l = 3\) and \(m = 2\).

Since the roots of a quadratic equation \(ax^2 + bx + c = 0\) are known to relate as follows:

• Sum of roots \(l + m = -\frac{b}{a}\)
• Product of roots \(l \cdot m = \frac{c}{a}\)

For \(l = 3\) and \(m = 2\),

• Sum of roots: \(3 + 2 = 5\)
• Product of roots: \(3 \cdot 2 = 6\)

Thus, the quadratic equation is:

x^2 - 5x + 6 = 0

Hence, the correct option is x² - 5x + 6 = 0.