Find the interval in which function f(x) = sinx+cosx, 0 ≤ x ≤ 2π is decreasing

The correct option is (π/4, 5π/4).

To find the interval in which the function is decreasing, we follow these step-by-step mathematical calculations and reasoning:

Step 1: Understand the condition for a decreasing function
A function $f\left(x\right)$ is strictly decreasing in a given interval if its first derivative with respect to $x$ is less than zero. That is:
$f^{\prime }\left(x\right)<0$

Step 2: Differentiate the function
The given function is:
$f\left(x\right)=sinx+cosx$
Differentiating both sides with respect to $x$, we get:
$f^{\prime }\left(x\right)=\frac{d}{dx}(sinx+cosx)$
Using standard trigonometric derivative rules, $\frac{d}{dx}(sinx)=cosx$ and $\frac{d}{dx}(cosx)=-sinx$:
$f^{\prime }\left(x\right)=cosx-sinx$

Step 3: Find the critical points
To find the critical points in the domain $0\le x\le 2\pi$, we set $f^{\prime }\left(x\right)=0$:
$cosx-sinx=0$
$cosx=sinx$
Dividing both sides by $cosx$ (where $cosx\ne 0$):
$tanx=1$
Within the interval $[0,2\pi ]$, the solutions for $tanx=1$ are:
$x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$

Step 4: Determine the sign of the derivative in each subinterval
The critical points divide the domain $[0,2\pi ]$ into three subintervals:
1) $[0,\frac{\pi }{4})$
2) $(\frac{\pi }{4},\frac{5\pi }{4})$
3) $(\frac{5\pi }{4},2\pi ]$
We pick a test point from each subinterval to check the sign of $f^{\prime }\left(x\right)=cosx-sinx$:

• For the interval $[0,\frac{\pi }{4})$, let $x=0$:
  $f^{\prime }\left(0\right)=cos\left(0\right)-sin\left(0\right)=1-0=1>0$ (increasing).
• For the interval $(\frac{\pi }{4},\frac{5\pi }{4})$, let $x=\frac{\pi }{2}$:
  $f^{\prime }\left(\frac{\pi }{2}\right)=cos\left(\frac{\pi }{2}\right)-sin\left(\frac{\pi }{2}\right)=0-1=-1<0$ (decreasing).
• For the interval $(\frac{5\pi }{4},2\pi ]$, let $x=\frac{3\pi }{2}$:
  $f^{\prime }\left(\frac{3\pi }{2}\right)=cos\left(\frac{3\pi }{2}\right)-sin\left(\frac{3\pi }{2}\right)=0-(-1)=1>0$ (increasing).

Conclusion
Since $f^{\prime }\left(x\right)<0$ only in the interval $(\frac{\pi }{4},\frac{5\pi }{4})$, the function $f\left(x\right)$ is decreasing in the interval (π/4, 5π/4).