Question

If the function f is continuous at \(x = π\), where \(f(x) = \frac{1-cos[7(x-π)]}{5(x-π)^2}\), for \(x\neq π\), then \(f(π) =\)

• A. \(\frac{49}{4}\)
• B. \(\frac{4}{49}\)
• C. \(\frac{49}{10}\)
• D. \(\frac{10}{49}\)

Hint:

Continuity needs f(pi) to equal the limit; use 1 - cos t over t squared tends to one half.

Answer 1

The Correct Option is C

Step 1: Set up the limit
Let \(h = x-\pi\). As \(x\to\pi\), \(h\to0\). The limit is \(\lim_{h\to0}\frac{1-\cos7h}{5h^2}\).

Step 2: Use half-angle
\(1-\cos7h = 2\sin^2\frac{7h}{2}\), so the limit is \(\frac25\lim\frac{\sin^2(7h/2)}{h^2} = \frac25\cdot\frac{49}{4}\).

Step 3: Evaluate
\[ \frac{2}{5}\cdot\frac{49}{4} = \frac{49}{10} \]

Step 4: Continuity
For continuity \(f(\pi)\) equals this limit, so \(f(\pi)=\frac{49}{10}\). Option (C).

Final Answer:
f(pi) is 49/10. \[ \boxed{\text{(C)}\ \frac{49}{10}} \]