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

Question

Accepted Answer

To determine why Option 1 is the correct answer, we need to analyze the behavior of the function $ f(x) = \sin^4 x + \cos^4 x $ over the given intervals.

### Explanation for Option 1: $\frac{\pi}{4}<x<\frac{\pi}{2}$

1. **Function Analysis**: The function $ f(x) = \sin^4 x + \cos^4 x $ can be rewritten using the identity $\sin^2 x + \cos^2 x = 1$. We have:
   $$
   f(x) = (\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x
   $$
   Using the double angle identity, $\sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x)$, we get:
   $$
   f(x) = 1 - \frac{1}{2} \sin^2(2x)
   $$
   The function $ f(x) $ increases when $\sin^2(2x)$ decreases.

2. **Interval Analysis**: In the interval $\frac{\pi}{4}<x<\frac{\pi}{2}$, the angle $2x$ ranges from $\frac{\pi}{2}$ to $\pi$. In this range, $\sin(2x)$ decreases from 1 to 0. Therefore, $\sin^2(2x)$ also decreases, which means $ f(x) $ increases.

### Explanation for Other Options:

- **Option 2: $0<x<\frac{\pi}{8}$**
  - In this interval, $2x$ ranges from $0$ to $\frac{\pi}{4}$. Here, $\sin(2x)$ increases from 0 to $\frac{\sqrt{2}}{2}$, so $\sin^2(2x)$ increases, causing $ f(x) $ to decrease.

- **Option 3: $\frac{5\pi}{8}<x<\frac{3\pi}{4}$**
  - In this interval, $2x$ ranges from $\frac{5\pi}{4}$ to $\frac{3\pi}{2}$. Here, $\sin(2x)$ decreases from $-\frac{\sqrt{2}}{2}$ to $-1$, so $\sin^2(2x)$ increases, causing $ f(x) $ to decrease.

- **Option 4: $\frac{3\pi}{8}<x<\frac{5\pi}{8}$**
  - In this interval, $2x$ ranges from $\frac{3\pi}{4}$ to $\frac{5\pi}{4}$. Here, $\sin(2x)$ decreases from $\frac{\sqrt{2}}{2}$ to $-\frac{\sqrt{2}}{2}$, so $\sin^2(2x)$ increases, causing $ f(x) $ to decrease.

Thus, the function $ f(x) $ increases only in the interval $\frac{\pi}{4}<x<\frac{\pi}{2}$, making Option 1 the correct answer.