Question

The solution of $\frac{d y}{d x}=\cos (x-y)$ is

• A. $y +\cot \frac{x-y}{2}+ c$
• B. $x +\cot \frac{x-y}{2}+ c$
• C. $x+\tan \frac{x-y}{2}+c$
• D. $y-\cot \frac{x-y}{2}+c$

Answer 1

Answer: (C)

since $f$ is continuous at $x=\pi, f(\pi)=\underset{x \rightarrow \pi}{lt}\left[\frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}\right]$
$=\underset{x \rightarrow \pi}{lt}\left[\frac{1(-\sin x)}{2 \sqrt{2+\cos x}} / 2(\pi-x)(-1)\right]($ by L.H. Rule $)$
$=\underset{x \rightarrow \pi}{lt}\left[\frac{\sin (x)}{4 \sqrt{2+\cos x}}\left(\frac{1}{\pi-x}\right)\right]=\frac{1}{4} \cdot \frac{1}{\sqrt{2-1}} \cdot \underset{x \rightarrow \pi}{lt}\left(\frac{\sin (\pi-x)}{(\pi-x)}\right)($ by L.H. Rule $)$
$=\frac{1}{4} .(1)=\frac{1}{4}$