If $f(x) = \sin x -\cos x, x ∈ [0,2\pi]$ then
(A) f(x) is increasing in $\left(0,\frac{3\pi}{4}\right)$
(B) f(x) is decreasing in $\left(0,\frac{3\pi}{4}\right)$
(C) f(x) is decreasing in $\left(\frac{3\pi}{4},\frac{7\pi}{4}\right)$
(D) f(x) is decreasing in $\left(\frac{7\pi}{4},2\pi \right)$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (C) only
Let \( f(x) = \sin x - \cos x \)
We compute the derivative:
\( f'(x) = \cos x + \sin x \)
To find critical points, set \( f'(x) = 0 \):
\( \cos x + \sin x = 0 \Rightarrow \tan x = -1 \)
This occurs at \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \) in the interval \( [0, 2\pi] \)
Now analyze the sign of \( f'(x) \):
- For \( x \in (0, \frac{3\pi}{4}) \), take \( x = \frac{\pi}{4} \):
- For \( x \in (\frac{3\pi}{4}, \frac{7\pi}{4}) \), take \( x = \pi \):
- For \( x \in (\frac{7\pi}{4}, 2\pi) \), take \( x = \frac{15\pi}{8} \):
\( f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} > 0 \)
⇒ Increasing
\( f'(\pi) = \cos(\pi) + \sin(\pi) = -1 + 0 = -1 < 0 \)
⇒ Decreasing
\( f'(x) > 0 \) (as both sine and cosine are positive in this range)
⇒ Increasing
Correct options: (A) and (C)