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: Put \(x=\frac{\pi}{2}+h\).
As \[ x\to \frac{\pi}{2}, \] we have \[ h\to 0 \] Now, \[ \sin x=\sin\left(\frac{\pi}{2}+h\right)=\cos h \] Also, \[ \pi^2-4\pi x+4x^2=(2x-\pi)^2 \] Since \[ x=\frac{\pi}{2}+h, \] we get \[ 2x-\pi=2h \] Therefore, \[ (2x-\pi)^2=4h^2 \]

Step 2: Rewrite the limit.
\[ f\left(\frac{\pi}{2}\right) = \lim_{h\to 0} \frac{1-\cos h}{\log(1+4h^2)} \]

Step 3: Use standard approximations.
As \(h\to 0\), \[ 1-\cos h\sim \frac{h^2}{2} \] Also, \[ \log(1+4h^2)\sim 4h^2 \] Therefore, \[ \lim_{h\to 0} \frac{1-\cos h}{\log(1+4h^2)} = \frac{\frac{h^2}{2}}{4h^2} \] \[ = \frac{1}{8} \]

Step 4: Final conclusion.
Hence, \[ \boxed{\frac{1}{8}} \]