Question

If $f\left(x\right)=\cos (2x+\frac{\pi }{4})$ then show that f(x) is increasing in the interval $\frac{3\pi }{8}<x<\frac{7\pi }{8}$

Answer 1

We have $f\left(x\right)=\cos (2x+\frac{\pi }{4})$

$\frac{dy}{dx}=-2\sin (2x+\frac{\pi }{4})$

Since, $\frac{3\pi }{8}<x<\frac{7\pi }{8}$

$\Rightarrow \frac{3\pi }{4}<2x<\frac{7\pi }{4}$

$\Rightarrow \pi <2x+\frac{\pi }{4}<2\pi$

i.e $3^{rd}$ and $4^{th}$ quadrant

So, $\sin (2x+\frac{\pi }{4})$ is negative

$\Rightarrow \frac{dy}{dx}$ is positive

Hence, $f\left(x\right)$ is increasing in the given interval