Question

The function $f\left(x\right)={\sin }^4\left(x\right)+{\cos }^4\left(x\right)$ increases in

• A. $0<x<\frac{\pi }{8}$
• B. $\frac{\pi }{4}<x<\frac{3\pi }{8}$
• C. $\frac{3\pi }{8}<x<\frac{5\pi }{8}$
• D. $\frac{5\pi }{8}<x<\frac{3\pi }{4}$

Answer 1

The correct option is B

$\frac{\pi }{4}<x<\frac{3\pi }{8}$

Find the interval in which the given function increases

We know that a function $f\left(x\right)$ is increasing then $f\text{'}\left(x\right)>0$

$$\begin{gathered} f\left(x\right)={\sin }^4\left(x\right)+{\cos }^4\left(x\right) \\ \Rightarrow f\text{'}\left(x\right)=4\cos \left(x\right){\sin }^3\left(x\right)-4{\cos }^3\left(x\right)\sin \left(x\right) \\ \Rightarrow f\text{'}\left(x\right)=4\cos \left(x\right)\sin \left(x\right)\left({\sin }^2\left(x\right)-{\cos }^2\left(x\right)\right) \\ \Rightarrow f\text{'}\left(x\right)=-2\sin \left(2x\right)\cos \left(2x\right)\left[\because {\sin }^2\left(\theta \right)-{\cos }^2\left(\theta \right)=\cos \left(2\theta \right)\mathrm{and}\sin \left(2\mathrm{\theta }\right)=2\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)\right] \\ \Rightarrow f\text{'}\left(x\right)=-\sin \left(4x\right)\left[\because \sin \left(2\mathrm{\theta }\right)=2\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)\right] \end{gathered}$$

For increasing function $f\text{'}\left(x\right)>0$

$$\begin{gathered} \Rightarrow -\sin \left(4x\right)>0 \\ \Rightarrow \sin \left(4x\right)<0 \\ \therefore \mathrm{\pi }<4\mathrm{x}<2\mathrm{\pi } \\ \Rightarrow \frac{\mathrm{\pi }}{4}<\mathrm{x}<\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence, option $\left(\mathrm{B}\right)$ is correct.