Key Concepts and Formulas

• Monotonicity of Functions: A differentiable function $f(x)$ is increasing on an interval if $f'(x) > 0$ for all $x$ in that interval.
• Trigonometric Identity: $\sin^2 x + \cos^2 x = 1$
• Double Angle Identity: $\sin(2x) = 2\sin x \cos x$

Step-by-Step Solution

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

We are given $f(x) = \sin^4 x + \cos^4 x$. We want to simplify this expression using trigonometric identities to make differentiation easier. $f(x) = \sin^4 x + \cos^4 x = (\sin^2 x)^2 + (\cos^2 x)^2$ Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$ with $a = \sin^2 x$ and $b = \cos^2 x$: $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$ Since $\sin^2 x + \cos^2 x = 1$, we have: $f(x) = (1)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$ Using the double angle identity $\sin(2x) = 2\sin x \cos x$, we can rewrite $\sin x \cos x$ as $\frac{\sin(2x)}{2}$: $f(x) = 1 - 2\left(\frac{\sin(2x)}{2}\right)^2 = 1 - 2\left(\frac{\sin^2(2x)}{4}\right) = 1 - \frac{1}{2}\sin^2(2x)$ Simplifying, we get: $f(x) = 1 - \frac{1}{2}\sin^2(2x)$

Why this step? Simplifying the function before differentiation makes the differentiation process easier and less prone to errors.

Step 2: Calculate the first derivative, $f'(x)$

Now we differentiate the simplified function $f(x) = 1 - \frac{1}{2}\sin^2(2x)$ with respect to $x$. $f'(x) = \frac{d}{dx}\left(1 - \frac{1}{2}\sin^2(2x)\right) = 0 - \frac{1}{2} \frac{d}{dx}(\sin^2(2x))$ Using the chain rule, let $u = \sin(2x)$, so $\frac{d}{dx}(\sin^2(2x)) = \frac{d}{du}(u^2) \cdot \frac{d}{dx}(\sin(2x))$. $\frac{d}{du}(u^2) = 2u$ $\frac{d}{dx}(\sin(2x)) = \cos(2x) \cdot 2 = 2\cos(2x)$ Therefore, $\frac{d}{dx}(\sin^2(2x)) = 2\sin(2x) \cdot 2\cos(2x) = 4\sin(2x)\cos(2x)$ Substituting this back into the derivative of $f(x)$: $f'(x) = -\frac{1}{2} \left[4\sin(2x)\cos(2x)\right] = -2\sin(2x)\cos(2x)$ Using the double angle identity $\sin(2A) = 2\sin A \cos A$, with $A = 2x$, we have $2\sin(2x)\cos(2x) = \sin(4x)$. Therefore, $f'(x) = -\sin(4x)$

Why this step? Calculating the derivative is essential because the sign of the derivative tells us whether the function is increasing or decreasing.

Step 3: Determine the intervals where $f(x)$ is increasing

For $f(x)$ to be an increasing function, its derivative $f'(x)$ must be greater than zero. $f'(x) > 0$ Substituting our calculated derivative: $-\sin(4x) > 0$ Multiplying by -1 and reversing the inequality sign: $\sin(4x) < 0$

Why this step? We are applying the definition of an increasing function: $f'(x) > 0$.

Now we need to find the values of $x$ for which $\sin(4x) < 0$. The sine function is negative in the third and fourth quadrants. Therefore, for $4x$: $\pi + 2n\pi < 4x < 2\pi + 2n\pi$ where $n$ is an integer. Dividing by 4: $\frac{\pi}{4} + \frac{n\pi}{2} < x < \frac{\pi}{2} + \frac{n\pi}{2}$

Let's test some values of $n$: If $n = 0$: $\frac{\pi}{4} < x < \frac{\pi}{2}$ If $n = 1$: $\frac{3\pi}{4} < x < \frac{4\pi}{4} = \pi$ If $n = -1$: $-\frac{\pi}{4} < x < 0$

Now, let's check the options given to see which interval satisfies $\sin(4x) < 0$.

(A) $] 0, \frac{\pi}{4}[$: If $x = \frac{\pi}{8}$, $4x = \frac{\pi}{2}$, $\sin(\frac{\pi}{2}) = 1 > 0$. So this is not correct.

(B) $] \frac{\pi}{4}, \frac{\pi}{2}[$: If $x = \frac{3\pi}{8}$, $4x = \frac{3\pi}{2}$, $\sin(\frac{3\pi}{2}) = -1 < 0$. This interval satisfies the inequality.

Since the problem statement says the answer is (A), let's re-examine the simplification. $f(x) = 1 - \frac{1}{2}\sin^2(2x)$ $f'(x) = -\sin(4x)$ $-\sin(4x) > 0 \implies \sin(4x) < 0$ $\sin(4x) < 0$ when $\pi < 4x < 2\pi$, so $\frac{\pi}{4} < x < \frac{\pi}{2}$.

However, since the question indicates that the correct answer is $]0, \frac{\pi}{4}[$, there must be an error in the problem statement. Let's analyze when $f'(x) > 0$ for $]0, \frac{\pi}{4}[$. If $x \in (0, \frac{\pi}{4})$, then $4x \in (0, \pi)$. For $f'(x) = -\sin(4x) > 0$, we need $\sin(4x) < 0$. Since $\sin(4x) > 0$ for $4x \in (0, \pi)$, the function $f(x)$ is not increasing in the interval $(0, \frac{\pi}{4})$.

If we consider $f(x)$ to be decreasing when $f'(x) < 0$, then $-\sin(4x) < 0$, which means $\sin(4x) > 0$. For $0 < 4x < \pi$, $\sin(4x) > 0$. Then $0 < x < \frac{\pi}{4}$. So $f(x)$ is decreasing in the interval $(0, \frac{\pi}{4})$.

Let us consider differentiating $f(x) = \sin^4 x + \cos^4 x$ directly: $f'(x) = 4\sin^3 x \cos x - 4\cos^3 x \sin x = 4\sin x \cos x (\sin^2 x - \cos^2 x) = 2(2\sin x \cos x)(-\cos 2x) = -2\sin(2x)\cos(2x) = -\sin(4x)$ We want $f'(x) > 0$, which means $-\sin(4x) > 0$, or $\sin(4x) < 0$. Thus, $\pi < 4x < 2\pi$, so $\frac{\pi}{4} < x < \frac{\pi}{2}$.

Let's reconsider option A. If the function was $f(x) = \sin(4x)$, then $f'(x) = 4\cos(4x)$. Then $f'(x) > 0$ if $\cos(4x) > 0$. This occurs when $-\frac{\pi}{2} < 4x < \frac{\pi}{2}$, or $-\frac{\pi}{8} < x < \frac{\pi}{8}$.

Let's test the endpoint of option A. At $x=0$, $f'(0) = -\sin(0) = 0$. At $x = \frac{\pi}{4}$, $f'(\frac{\pi}{4}) = -\sin(\pi) = 0$.

The given answer is incorrect. The function $f(x)$ is increasing in the interval $] \frac{\pi}{4}, \frac{\pi}{2}[$.

Common Mistakes & Tips

• Remember to reverse the inequality sign when multiplying or dividing by a negative number.
• Be careful when applying the chain rule during differentiation.
• Double-check trigonometric identities and their applications.

Summary

We simplified the function $f(x) = \sin^4 x + \cos^4 x$ using trigonometric identities. Then, we calculated the first derivative $f'(x)$ and determined the intervals where $f'(x) > 0$, which indicates where the function is increasing. The correct interval where $f(x)$ is increasing is $] \frac{\pi}{4}, \frac{\pi}{2}[$. However, the provided answer is $] 0, \frac{\pi}{4}[$, which is incorrect based on our calculations. Option (B) is the correct interval.

Final Answer

The final answer is $] \frac{\pi}{4}, \frac{\pi}{2}[$, which corresponds to option (B).