Let the function $f(x)=\tan ^{-1}(\sin x+\cos x)$ be defined on $[0,2 \pi]$. Then, f(x) is
Answer & explanation
Correct answer: option 1
We have,
$f(x) =\tan ^{-1}(\sin x+\cos x)$
$\Rightarrow f'(x) =\frac{1}{1+(\sin x+\cos x)^2} \times(\cos x-\sin x)$
Now,
$f'(x)>0 \Rightarrow \cos x-\sin x>0 \Rightarrow x \in[0, \pi / 4) \cup(5 \pi / 4,2 \pi]$
and, $f'(x)<0 \Rightarrow \cos x-\sin x<0 \Rightarrow x \in(\pi / 4,5 \pi / 4)$
Hence, f(x) is increasing on $[0, \pi / 4) \cup(5 \pi / 4,2 \pi]$ and decreasing on $(\pi / 4,5 \pi / 4)$.