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 solve the problem, we need to find the quadratic equation whose roots are given by the limits $l = \lim_{\theta \to 0} \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta}$ and $m = \lim_{\theta \to 0} \frac{2 \tan \theta}{\theta (1 - \tan^2 \theta)}$.

1. Evaluate the Limit for $l$:
Notice that $3 \sin \theta - 4 \sin^3 \theta = \sin(3\theta)$.
Thus, $l = \lim_{\theta \to 0} \frac{\sin(3\theta)}{\theta} = \lim_{\theta \to 0} \frac{\sin(3\theta)}{3\theta} \cdot 3 = 1 \cdot 3 = 3$.

2. Evaluate the Limit for $m$:
We have $m = \lim_{\theta \to 0} \frac{2 \tan \theta}{\theta (1 - \tan^2 \theta)}$.
Rewrite using $\frac{2 \tan \theta}{1 - \tan^2 \theta} = \tan(2\theta)$, so $m = \lim_{\theta \to 0} \frac{\tan(2\theta)}{\theta} = \lim_{\theta \to 0} \frac{\tan(2\theta)}{2\theta} \cdot 2 = 1 \cdot 2 = 2$.

3. Form the Quadratic Equation:
The roots of the quadratic equation are $l = 3$ and $m = 2$.
The equation is $(x - 3)(x - 2) = 0$, which expands to:
$x^2 - 5x + 6 = 0$.

Final Answer:
The quadratic equation is $x^2 - 5x + 6 = 0$.