Question

The quadratic equation whose roots are \[ l = \lim_{\theta \to 0} \left( \frac{3\sin\theta - 4\sin^3\theta}{\theta} \right) \] \[ m = \lim_{\theta \to 0} \left( \frac{2\tan\theta}{\theta(1-\tan^2\theta)} \right) \] is:

• A. \( x^2+5x+6=0 \)
• B. \( x^2-5x+6=0 \)
• C. \( x^2-5x-6=0 \)
• D. \( x^2+5x-6=0 \)

Hint:

For limit evaluations of trigonometric functions, use approximations: \[ \sin\theta \approx \theta, \tan\theta \approx \theta, \cos\theta \approx 1 \] for small values of \( \theta \).

Answer 1

The Correct Option is C

Evaluating the first limit: \[ l = \lim_{\theta \to 0} \frac{3\sin\theta - 4\sin^3\theta}{\theta} \] Approximating \( \sin\theta \approx \theta \) for small \( \theta \): \[ = \lim_{\theta \to 0} \frac{3\theta - 4\theta^3}{\theta} = 3 \] Evaluating the second limit: \[ m = \lim_{\theta \to 0} \frac{2\tan\theta}{\theta(1-\tan^2\theta)} \] Approximating \( \tan\theta \approx \theta \): \[ = \lim_{\theta \to 0} \frac{2\theta}{\theta(1-\theta^2)} \] \[ = \lim_{\theta \to 0} \frac{2}{1-\theta^2} = 2 \] Thus, the quadratic equation with roots \( l \) and \( m \) is: \[ (x - 3)(x - 2) = 0 \] \[ x^2 - 5x + 6 = 0 \] Thus, the correct answer is: \[ x^2 - 5x - 6 = 0 \]