1. **State the problem:** We want to find the intervals on which the function $$f(x) = \cos^2(4x) + 3 \cos(4x)$$ is increasing or decreasing for $$0 < x < \frac{\pi}{2}$$. 2. **Find the derivative:** To determine increasing/decreasing intervals, compute $$f'(x)$$. Using the chain rule and power rule: $$f(x) = (\cos(4x))^2 + 3 \cos(4x)$$ Let $$u = \cos(4x)$$, then $$f(x) = u^2 + 3u$$. Derivative: $$f'(x) = 2u \cdot u' + 3 u' = u'(2u + 3)$$ Since $$u = \cos(4x)$$, then $$u' = -4 \sin(4x)$$. So, $$f'(x) = -4 \sin(4x) (2 \cos(4x) + 3)$$. 3. **Find critical points:** Set $$f'(x) = 0$$. This happens when either - $$\sin(4x) = 0$$ or - $$2 \cos(4x) + 3 = 0$$. Solve each: - $$\sin(4x) = 0 \implies 4x = k\pi \implies x = \frac{k\pi}{4}$$. Within $$0 < x < \frac{\pi}{2}$$, possible $$k$$ are 1 and 2: $$x = \frac{\pi}{4}, \frac{\pi}{2}$$. - $$2 \cos(4x) + 3 = 0 \implies \cos(4x) = -\frac{3}{2}$$. But $$\cos(\theta)$$ ranges between -1 and 1, so no solution here. 4. **Determine sign of $$f'(x)$$ on intervals:** Intervals to test are: - $$\left(0, \frac{\pi}{4}\right)$$ - $$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$$ Pick test points: - For $$x = \frac{\pi}{8}$$ in $$\left(0, \frac{\pi}{4}\right)$$: Calculate $$f'(\frac{\pi}{8})$$: $$\sin(4 \cdot \frac{\pi}{8}) = \sin(\frac{\pi}{2}) = 1 > 0$$ $$\cos(4 \cdot \frac{\pi}{8}) = \cos(\frac{\pi}{2}) = 0$$ So, $$f'(\frac{\pi}{8}) = -4 \times 1 \times (2 \times 0 + 3) = -4 \times 1 \times 3 = -12 < 0$$ Derivative is negative, so $$f$$ is decreasing on $$\left(0, \frac{\pi}{4}\right)$$. - For $$x = \frac{3\pi}{8}$$ in $$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$$: $$\sin(4 \cdot \frac{3\pi}{8}) = \sin(\frac{3\pi}{2}) = -1 < 0$$ $$\cos(4 \cdot \frac{3\pi}{8}) = \cos(\frac{3\pi}{2}) = 0$$ So, $$f'(\frac{3\pi}{8}) = -4 \times (-1) \times (2 \times 0 + 3) = 12 > 0$$ Derivative is positive, so $$f$$ is increasing on $$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$$. 5. **Conclusion:** - $$f(x)$$ is decreasing on $$\left]0, \frac{\pi}{4}\right[$$ - $$f(x)$$ is increasing on $$\left[\frac{\pi}{4}, \frac{\pi}{2}\right[$$ This matches option (d). **Final answer:** The function is decreasing on $$\left]0, \frac{\pi}{4}\right[$ and increasing on $$\left[\frac{\pi}{4}, \frac{\pi}{2}\right[$.