Question

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

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

Answer 1

The correct option is B

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

Explanation for the correct option:

Applying condition for increasing function:

Given function is $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$.

We have,
$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left({\sin }^{4}x+{\cos }^{4}x\right)\\ \therefore f\text{'}\left(x\right)& =& 4{\sin }^{3}x\cos x-4{\cos }^{3}x\sin x\left[\because \frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]=f\text{'}\left(g\left(x\right)\right)·g\text{'}\left(x\right)\right]\end{array}$

We know that a function increases if $f\text{'}\left(x\right)>0$
$\begin{array}{ccc}& 4{\sin }^3\cos x-4{\cos }^3\sin x>0& \\ \Rightarrow & 4\sin x\cos x\left({\sin }^{2}x-{\cos }^{2}x\right)>0& \\ \Rightarrow & -2·2\sin x\cos x\left({\cos }^{2}x-{\sin }^{2}x\right)>0& \\ \Rightarrow & -2\sin \left(2x\right)\cos \left(2x\right)>0& \\ \Rightarrow & -\sin \left(4x\right)>0& \\ \Rightarrow & \sin \left(4x\right)<0& \end{array}$

We know that $\sin \left(x\right)$ is negative in $III$ and $IV$ Quadrants $\Rightarrow x\in \left(\mathrm{\pi },2\mathrm{\pi }\right)$

$\begin{array}{ccc}\Rightarrow & \mathrm{\pi }<4\mathrm{x}<2\mathrm{\pi }& \\ \Rightarrow & \frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2}& \\ \Rightarrow & \frac{\mathrm{\pi }}{4}<x<\frac{4\mathrm{\pi }}{8}& \\ \Rightarrow & \frac{\mathrm{\pi }}{4}<x<\frac{3\mathrm{\pi }}{8}<\frac{4\mathrm{\pi }}{8}& \end{array}$

Therefore, the given function is increasing if $\frac{\mathrm{\pi }}{4}<x<\frac{3\mathrm{\pi }}{8}$ [according to the given options].

Hence, option(B) is correct.