Question

The derivative of the function \(f(x) = cos^4x+sin^4x, 0\leq x\leq 2π\) is positive for

• A. \(0 < x < \frac{π}{8}\)
• B. \(\frac{π}{4} < x < \frac{π}{2}\)
• C. \(\frac{π}{2} < x < \frac{5π}{8}\)
• D. \(\frac{5π}{8} < x < \frac{3π}{4}\)

Hint:

Simplify \(f(x)=\frac34+\frac14\cos4x\) and find where \(f'>0\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Write \(\cos^4x+\sin^4x=1-2\sin^2x\cos^2x=1-\tfrac12\sin^22x\).

Step 2: Key Formula or Approach
\(\sin^22x=\dfrac{1-\cos4x}{2}\), so \(f(x)=\dfrac34+\dfrac14\cos4x\) and \(f'(x)=-\sin4x\).

Step 3: Detailed Explanation
\(f'(x)>0\) means \(\sin4x<0\), i.e. \(4x\in(\pi,2\pi)\) (within the first period), so \(x\in\left(\dfrac\pi4,\dfrac\pi2\right)\).
Check (A): \(4x\in(0,\pi/2)\), sine positive, so \(f'<0\). (C): \(4x\in(2\pi,5\pi/2)\), sine positive. (D): \(4x\in(5\pi/2,3\pi)\), sine positive. So only (B) works.

Final Answer:
The derivative is positive on \((\pi/4,\pi/2)\), option (B). \[ \boxed{\left(\dfrac\pi4,\dfrac\pi2\right)\ \text{(B)}} \]