Question

Let \(f(x)\) be a function continuous at \(x=0\), defined for \(x\neq 0\) as \(f(x)=\frac{1-\cos (1-\cos (6x))}{x^{2}\sin ^{2}(3x)}\). Then the value of \(f(0)\) is...

• A. \(9\)
• B. \(18\)
• C. \(36\)
• D. \(162\)

Hint:

For continuity, f(0) equals the limit of the given expression as x tends to 0. Use the small-angle expansion of cosine.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Since \(f\) is continuous at \(x=0\), \(f(0)=\lim_{x\rightarrow 0}f(x)\).
We use \(1-\cos\theta\approx \dfrac{\theta^{2}}{2}\) and \(\sin\theta\approx\theta\) for small \(\theta\).

Step 2: Simplify the Inner Cosine
\(1-\cos 6x\approx \dfrac{(6x)^{2}}{2}=18x^{2}\).

Step 3: Simplify the Numerator
Numerator \(=1-\cos(18x^{2})\approx \dfrac{(18x^{2})^{2}}{2}=162x^{4}\).

Step 4: Calculation
Denominator \(=x^{2}\sin^{2}3x\approx x^{2}\cdot 9x^{2}=9x^{4}\).
So \(f(0)=\dfrac{162x^{4}}{9x^{4}}=18\).

Final Answer:
The value of f(0) is 18. \[ \boxed{\text{(B) }18} \]