Question

Show that the function f(x) = sin (2x + π/4) is decreasing on (3π/8, 5π/8).

Answer 1

$$\begin{gathered} f\left(x\right)=\sin \left(2x+\frac{\mathrm{\pi }}{4}\right) \\ f\text{'}\left(x\right)=2\cos \left(2x+\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{Here}, \\ \frac{3\mathrm{\pi }}{8}<x<\frac{5\mathrm{\pi }}{8} \\ \Rightarrow \frac{3\mathrm{\pi }}{4}<2x<\frac{5\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{\pi }<2x+\frac{\mathrm{\pi }}{4}<\frac{3\mathrm{\pi }}{2} \\ \Rightarrow \cos \left(2x+\frac{\mathrm{\pi }}{4}\right)<0\left[\text{∵ Cos function is negative in}\mathrm{third}\text{quadrant}\right] \\ \Rightarrow 2\cos \left(2x+\frac{\mathrm{\pi }}{4}\right)<0 \\ \Rightarrow f\text{'}\left(x\right)<0,\forall x\in \left(\frac{3\mathrm{\pi }}{8},\frac{5\mathrm{\pi }}{8}\right) \\ \text{So,}f\left(x\right)\text{is decreasing on}\left(\frac{3\mathrm{\pi }}{8},\frac{5\mathrm{\pi }}{8}\right)\text{.} \end{gathered}$$