Question

Let $f(x) = \sin^4 x + \cos^4 x$. Then $f$ is an increasing function in the interval

• A. $0, \frac{\pi}{4}$
• B. $\frac{\pi}{4}, \frac{\pi}{2}$
• C. $\frac{\pi}{2}, \frac{5\pi}{8}$
• D. $\frac{5\pi}{8}, \frac{3\pi}{4}$

Answer 1

Answer: (B)

B

Answer 2

To determine where the function $f(x) = \sin^4 x + \cos^4 x$ is increasing, we need to find its first derivative, $f'(x)$, and identify the intervals where $f'(x) > 0$.

Step 1: Simplify the function $f(x)$

We use the identity $a^2 + b^2 = (a+b)^2 - 2ab$. Let $a = \sin^2 x$ and $b = \cos^2 x$. $f(x) = (\sin^2 x)^2 + (\cos^2 x)^2$ $f(x) = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$ Since $\sin^2 x + \cos^2 x = 1$: $f(x) = 1^2 - 2 \sin^2 x \cos^2 x$ $f(x) = 1 - 2 \sin^2 x \cos^2 x$ We know that $\sin(2x) = 2 \sin x \cos x$, so $\sin^2(2x) = 4 \sin^2 x \cos^2 x$. Therefore, $2 \sin^2 x \cos^2 x = \frac{1}{2} (4 \sin^2 x \cos^2 x) = \frac{1}{2} \sin^2(2x)$. Substitute this back into $f(x)$: $f(x) = 1 - \frac{1}{2} \sin^2(2x)$

Step 2: Find the derivative $f'(x)$

Differentiate $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx} \left( 1 - \frac{1}{2} \sin^2(2x) \right)$ Using the chain rule, $\frac{d}{du}(u^n) = nu^{n-1} \frac{du}{dx}$ and $\frac{d}{dx}(\sin(ax)) = a\cos(ax)$: $f'(x) = 0 - \frac{1}{2} \cdot 2 \sin(2x) \cdot \cos(2x) \cdot \frac{d}{dx}(2x)$ $f'(x) = - \sin(2x) \cos(2x) \cdot 2$ $f'(x) = - 2 \sin(2x) \cos(2x)$ Using the double angle identity $\sin(2A) = 2 \sin A \cos A$: $f'(x) = - \sin(2 \cdot 2x)$ $f'(x) = - \sin(4x)$

Step 3: Determine the condition for $f(x)$ to be an increasing function

A function is increasing when its derivative is positive, i.e., $f'(x) > 0$. So, we need: $- \sin(4x) > 0$ $\sin(4x) < 0$

Step 4: Find the intervals where $\sin(\theta) < 0$

The sine function is negative in the third and fourth quadrants. In general, for an angle $\theta$, $\sin(\theta) < 0$ when: $2n\pi + \pi < \theta < 2n\pi + 2\pi$, where $n$ is an integer. Let $\theta = 4x$. So, $2n\pi + \pi < 4x < 2n\pi + 2\pi$. Divide the inequality by 4: $\frac{2n\pi + \pi}{4} < x < \frac{2n\pi + 2\pi}{4}$ $\frac{(2n+1)\pi}{4} < x < \frac{(n+1)\pi}{2}$

Step 5: Check the given options

Let's find the intervals for different integer values of $n$. For $n=0$: $\frac{(2(0)+1)\pi}{4} < x < \frac{(0+1)\pi}{2}$ $\frac{\pi}{4} < x < \frac{\pi}{2}$

This interval matches option B. Let's verify other options as well:

• A ]$0, \frac{\pi}{4}$[: For $x \in (0, \frac{\pi}{4})$, $4x \in (0, \pi)$. In this interval, $\sin(4x) > 0$. So $f'(x) = -\sin(4x) < 0$. Thus, $f(x)$ is decreasing.
• B ]$\frac{\pi}{4}, \frac{\pi}{2}$[: For $x \in (\frac{\pi}{4}, \frac{\pi}{2})$, $4x \in (\pi, 2\pi)$. In this interval, $\sin(4x) < 0$. So $f'(x) = -\sin(4x) > 0$. Thus, $f(x)$ is increasing.
• C ]$\frac{\pi}{2}, \frac{5\pi}{8}$[: For $x \in (\frac{\pi}{2}, \frac{5\pi}{8})$, $4x \in (2\pi, \frac{5\pi}{2})$. This is equivalent to $4x \in (0, \frac{\pi}{2})$ in terms of sine's sign (after one full rotation). In this interval, $\sin(4x) > 0$. So $f'(x) = -\sin(4x) < 0$. Thus, $f(x)$ is decreasing.
• D ]$\frac{5\pi}{8}, \frac{3\pi}{4}$[: For $x \in (\frac{5\pi}{8}, \frac{3\pi}{4})$, $4x \in (\frac{5\pi}{2}, 3\pi)$. This is equivalent to $4x \in (\frac{\pi}{2}, \pi)$ in terms of sine's sign (after one full rotation). In this interval, $\sin(4x) > 0$. So $f'(x) = -\sin(4x) < 0$. Thus, $f(x)$ is decreasing.

Only option B corresponds to an interval where $f(x)$ is increasing.