Question

If \[ f(x)=\frac{1-\sin x}{\log(1+\pi^2-4\pi x+4x^2)} \] is continuous at \(x=\frac{\pi}{2}\), then \(f\left(\frac{\pi}{2}\right)=\)

• A. \(\dfrac{1}{4}\)
• B. \(\dfrac{1}{8}\)
• C. \(\dfrac{1}{16}\)
• D. \(\dfrac{1}{32}\)

Hint:

Use standard limits: \[ 1-\cos x\sim \frac{x^2}{2} \] and \[ \log(1+x)\sim x \] as \(x\to 0\).

Answer 1

The Correct Option is B

Step 1: Set up the limit for continuity at $ x=\pi/2 $.
We need $ f(\pi/2)=\lim_{x\to\pi/2}f(x) $. Direct substitution gives $ 0/0 $.
Step 2: Recognize $ \pi^2-4\pi x+4x^2=(2x-\pi)^2 $.
This is a perfect square! So the denominator is $ \log(1+(2x-\pi)^2) $.
Step 3: Let $ t=x-\pi/2 $, $ t\to 0 $.
Then $ 2x-\pi=2t $, denominator $ =\log(1+4t^2) $, and $ 1-\sin x=1-\cos t $ (since $ \sin(\pi/2+t)=\cos t $).
Step 4: Apply Taylor expansions for small $ t $.
$ 1-\cos t \approx \frac{t^2}{2} $ and $ \log(1+4t^2)\approx 4t^2 $.
Step 5: Evaluate the limit.
\[ \lim_{t\to 0}\frac{t^2/2}{4t^2} = \frac{1}{8} \]
Step 6: Assign $ f(\pi/2) $ for continuity.
\[ f(\pi/2)=\frac{1}{8} \] \[ \boxed{\dfrac{1}{8}} \]