The interval in which the $f(x)=\sin x-\cos x, 0 \leq x \leq 2 \pi$ is strictly decreasing is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) - $\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$
$f(x)=\sin x-\cos x$
$f'(x)=\cos x+\sin x=0$
so $x=\frac{3π}{4},\frac{7π}{4}$
$f'(x)<0$ for $x∈(\frac{3π}{4},\frac{7π}{4})$
so f is decreasing in $(\frac{3π}{4},\frac{7π}{4})$