Question

If $f(x) = \frac{\cos ax - \cos bx}{\cos cx - \cos bx}$ for $x \ne 0$ and $f(0) = -1$ is continuous at $x = 0$, then $a^2, b^2, c^2$ are in ______.

• A. Geometric progression
• B. Arithmetic progression
• C. Harmonic progression
• D. Arithmetico-Geometric progression

Hint:

Memorize this incredibly useful standard limit result: $\lim_{x\to0} \frac{\cos Ax - \cos Bx}{x^2} = \frac{B^2 - A^2}{2}$. It bypasses double L'Hôpital applications and solves massive problems in seconds!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For continuity, $\lim_{x \to 0} f(x) = f(0) = -1$. We use L'Hôpital's rule or expansion.

Step 2: Formula Application:
Using the expansion $\cos \theta \approx 1 - \frac{\theta^2}{2}$: $f(x) \approx \frac{(1 - a^2x^2/2) - (1 - b^2x^2/2)}{(1 - c^2x^2/2) - (1 - b^2x^2/2)} = \frac{b^2 - a^2}{b^2 - c^2}$.

Step 3: Explanation:
Given the limit is $-1$: $\frac{b^2 - a^2}{b^2 - c^2} = -1 \implies b^2 - a^2 = -(b^2 - c^2)$ $b^2 - a^2 = -b^2 + c^2$ $2b^2 = a^2 + c^2$. This is the standard condition for $a^2, b^2, c^2$ to be in Arithmetic Progression (A.P.).

Step 4: Final Answer:
$a^2, b^2, c^2$ are in Arithmetic Progression.