Question

If the function

$ f(x) = \begin{cases} \frac{\cos ax - \cos 9x}{x^2}, & \text{if } x \neq 0 \\ 16, & \text{if } x = 0 \end{cases} $

is continuous at $ x = 0 $, then $ a = ? $

• A. \(\pm 8\)
• B. \(\pm 7\)
• C. \(\pm 6\)
• D. \(\pm 5\)

Hint:

 L’Hˆopital’s Rule is a powerful tool for evaluating limits of indeterminate forms. Remember to check the conditions for applying the rule before using it.

Answer 1

The Correct Option is B

For continuity at \(x = 0\), the limit as \(x \to 0\) of \(f(x)\) must equal \(f(0)\). We'll use L'Hôpital's Rule since the limit is in the indeterminate form $\frac{0}{0}$: \[ f(0) = 16 \] \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\cos ax - \cos 9x}{x^2} \] Apply L'Hôpital's Rule (differentiate numerator and denominator): \[ \lim_{x \to 0} \frac{\cos ax - \cos 9x}{x^2} = \lim_{x \to 0} \frac{-a\sin ax + 9\sin 9x}{2x} \] This is still in the indeterminate form $\frac{0}{0}$, so apply L'Hôpital's Rule again: \[ \lim_{x \to 0} \frac{-a\sin ax + 9\sin 9x}{2x} = \lim_{x \to 0} \frac{-a^2\cos ax + 81\cos 9x}{2} \] Now, evaluate the limit: \[ \lim_{x \to 0} \frac{-a^2\cos ax + 81\cos 9x}{2} = \frac{-a^2 + 81}{2} \] Set this equal to \(f(0) = 16\): \[ \frac{-a^2 + 81}{2} = 16 \] \[ -a^2 + 81 = 32 \] \[ a^2 = 49 \] \[ a = \pm 7 \]