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: Understanding the Concept:
For \(f\) to be continuous at \(x=1\), \(f(1)\) must equal \(\lim_{x\to1}f(x)\).

Step 2: Substitute:
Let \(t=1-x\), so \(t\to0\) and \(x=1-t\). Then \(\cos\pi x=\cos(\pi-\pi t)=-\cos\pi t\).
\[ f=\frac{1-\cos\pi t}{\pi t^2} \]

Step 3: Use the identity:
\(1-\cos\pi t=2\sin^2\dfrac{\pi t}2\). So
\[ f=\frac{2\sin^2(\pi t/2)}{\pi t^2}=\frac{2}{\pi}\cdot\frac{\pi^2}{4}\left(\frac{\sin(\pi t/2)}{\pi t/2}\right)^2 \]

Step 4: Take the limit:
As \(t\to0\) the bracket tends to 1. So the limit is \(\dfrac{2}{\pi}\cdot\dfrac{\pi^2}4=\dfrac\pi2\).

Step 5: Choose:
\(f(1)=\dfrac\pi2\), option (A).

Final Answer:
The limit and f(1) equal pi/2. \[ \boxed{\frac{\pi}{2}} \]