If the function $f(x)=\frac{1-\sin 2 x+\cos 2 x}{1+\sin 2 x+\cos 2 x}$ if $\quad x \neq \frac{\pi}{2}$…

If the function $f(x)=\frac{1-\sin 2 x+\cos 2 x}{1+\sin 2 x+\cos 2 x}$ if $\quad x \neq \frac{\pi}{2}$ $=\mathrm{k} \quad$ if $\quad x=\frac{\pi}{2}$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{k}=$
  1. 2
  2. 1
  3. 0
  4. -1

Solution

$\begin{aligned} \lim _{x \rightarrow \frac{\pi}{2}} f(x)=& \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin 2 x+\cos 2 x}{1+\sin 2 x+\cos 2 x} \\ =& \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1+\cos 2 x)-\sin 2 x}{(1+\cos 2 x)+\sin 2 x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^{2} x-2 \sin x \cos x}{x+2 \sin x \cos x} \\ =& \lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos x(\cos x-\sin x)}{2 \cos x(\cos x+\sin x)}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x-\sin x}{\cos x+\sin x}=\frac{0-1}{0+1}=-1 \end{aligned}$ Since $f(x)$ is continuous at $x=\frac{\pi}{2}$, we get $k=-1$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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