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 Question:
The function is declared continuous at $x = 0$. By definition of continuity, the mathematical limit of the function as $x \to 0$ must exactly equal the defined function value at that point, $f(0) = -1$. We must evaluate the limit and equate it.

Step 2: Detailed Explanation:
Let's evaluate the limit $L = \lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos bx}$.
Direct substitution of $x = 0$ yields $\frac{1 - 1}{1 - 1} = \frac{0}{0}$.
We can use L'Hôpital's Rule or Maclaurin series expansion. Series expansion is much faster and cleaner here.
The Taylor series expansion for cosine near 0 is $\cos(\theta) \approx 1 - \frac{\theta^2}{2}$.
Substitute expansions into the numerator:
$\cos(ax) - \cos(bx) \approx \left(1 - \frac{a^2 x^2}{2}\right) - \left(1 - \frac{b^2 x^2}{2}\right)$
Numerator $\approx \frac{b^2 - a^2}{2} x^2$
Substitute expansions into the denominator:
$\cos(cx) - \cos(bx) \approx \left(1 - \frac{c^2 x^2}{2}\right) - \left(1 - \frac{b^2 x^2}{2}\right)$
Denominator $\approx \frac{b^2 - c^2}{2} x^2$
Now, evaluate the limit with these expanded approximations:
$L = \lim_{x \to 0} \frac{\frac{b^2 - a^2}{2} x^2}{\frac{b^2 - c^2}{2} x^2}$
The $\frac{x^2}{2}$ terms cancel out perfectly:
$L = \frac{b^2 - a^2}{b^2 - c^2}$
Since the function is continuous, this limit must equal $f(0) = -1$:
$\frac{b^2 - a^2}{b^2 - c^2} = -1$
Cross-multiply and solve:
$b^2 - a^2 = -1 \cdot (b^2 - c^2)$
$b^2 - a^2 = -b^2 + c^2$
Bring the $b^2$ terms to one side:
$2b^2 = a^2 + c^2$
This algebraic relationship ($2 \times \text{Middle Term} = \text{First Term} + \text{Last Term}$) is the fundamental defining condition for an Arithmetic Progression (A.P.).
Therefore, the sequence of terms $a^2, b^2, c^2$ forms an Arithmetic Progression.

Step 3: Final Answer:
The terms $a^2, b^2, c^2$ are in Arithmetic progression, matching option (b).