Question

If the function f is continuous at \(x = 1\), where \(f(x) = \frac{1+cos(πx)}{π(1-x)^2}\), for \(x\neq 1\), then the value of \(f(1)\) is....

• A. \(\frac{π}{2}\)
• B. \(\frac{π}{4}\)
• C. \(\frac{π}{6}\)
• D. \(\frac{π}{9}\)

Hint:

Put t = 1 - x and use 1 - cos u = 2 sin^2(u/2).

Answer 1

The Correct Option is A

Step 1: L Hopital:
At $x=1$ both numerator and denominator are $0$. Differentiate twice.

Step 2: First pass:
Numerator derivative $-\pi\sin\pi x$, denominator derivative $-2\pi(1-x)$. Both are still 0.

Step 3: Second pass:
Numerator $-\pi^2\cos\pi x\to\pi^2$ at $x=1$. Denominator $2\pi$. Ratio $=\dfrac{\pi^2}{2\pi}=\dfrac\pi2$.

Final Answer:
Two applications of L'Hopital also give pi/2. \[ \boxed{A} \]