If the function \( f(x)=\sin x+\cos x,0\le x\le 2\pi \) is increasing in the interval \( (\alpha,\beta)\cup(v,\delta) \), then \( \alpha+\beta+v+\delta \) is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{7\pi}{2} \)
\( \displaystyle f'(x)=\cos x-\sin x \)
\( f'(x)=0 \) where \( \tan x=1 \), i.e. at \( x=\frac{\pi}{4} \) and \( x=\frac{5\pi}{4} \) in [0, 2π]. These points split [0, 2π] into three parts:
\( \displaystyle 0\lt x\lt\frac{\pi}{4}:\ \cos x\gt\sin x,\ f'(x)\gt 0 \ (\text{increasing}) \)
\( \displaystyle \frac{\pi}{4}\lt x\lt\frac{5\pi}{4}:\ \cos x\lt\sin x,\ f'(x)\lt 0 \ (\text{decreasing}) \)
\( \displaystyle \frac{5\pi}{4}\lt x\lt 2\pi:\ \cos x\gt\sin x,\ f'(x)\gt 0 \ (\text{increasing}) \)
So f is increasing in \( \left(0,\frac{\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right) \), giving \( \alpha=0,\ \beta=\frac{\pi}{4},\ v=\frac{5\pi}{4},\ \delta=2\pi \):
\( \displaystyle \alpha+\beta+v+\delta=0+\frac{\pi}{4}+\frac{5\pi}{4}+2\pi \)
\( \displaystyle =\frac{7\pi}{2} \)