Question

\(\text{ If } f(x) = \sin x + \frac{1}{2} \cos 2x \text{ in } \left[ 0, \frac{\pi}{2} \right], \text{ then:}\)
(A) \(f'(x) = \cos x - \sin 2x\)
(B)The critical points of the function are \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\)
(C) The minimum value of the function is 2
(D) The maximum value of the function is \(\frac{3}{4}\)

• A. (A), (B), and (D) only
• B. (A), (B), and (C) only
• C. (B), (C), and (D) only
• D. (A), (C), and (D) only

Hint:

To analyze the critical points of a function, always start by differentiating the function and setting the derivative equal to zero. After finding the critical points, evaluate the function at these points to determine whether they correspond to maximum, minimum, or saddle points. Additionally, always verify each option thoroughly to ensure the correct interpretation of the function’s behavior.

Answer 1

The Correct Option is A

To solve the problem, we need to evaluate different aspects of the function \(f(x) = \sin x + \frac{1}{2} \cos 2x\) within the interval \(\left[ 0, \frac{\pi}{2} \right]\).

1. Finding \(f'(x)\): To find the derivative of the function, start by using the derivatives of the trigonometric functions:
   • \(\frac{d}{dx}(\sin x) = \cos x\)
   • \(\frac{d}{dx}(\cos 2x) = -2\sin 2x \) (chain rule applied)
   Thus, the derivative \(f'(x)\) is:

\(f'(x) = \cos x + \frac{1}{2}(-2\sin 2x) = \cos x - \sin 2x\)

Statement (A) is correct.

2. Finding critical points: Critical points occur where \(f'(x) = 0\):

\(\cos x - \sin 2x = 0\)

\(\cos x - 2\sin x \cos x = 0\)

\(\cos x(1 - 2\sin x) = 0\)

This gives us two equations:

• \(\cos x = 0 \Rightarrow x = \frac{\pi}{2}\)
• \(1 - 2\sin x = 0 \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}\)

Both critical points \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\) are in the interval. Thus, statement (B) is correct.

3. Finding the maximum and minimum values: Use critical points and boundary points:

• \(x = 0\), \(f(0) = \sin 0 + \frac{1}{2}\cos 0 = \frac{1}{2}\)
• \(x = \frac{\pi}{6}\), \(f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) + \frac{1}{2}\cos\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\)
• \(x = \frac{\pi}{2}\), \(f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \frac{1}{2}\cos(\pi) = 1 - \frac{1}{2} = \frac{1}{2}\)

The maximum value is \(\frac{3}{4}\), not 2, so statement (D) is correct, but statement (C) is incorrect.

Therefore, the correct answer is: (A), (B), and (D) only.

Answer 2

We need to evaluate the function \( f(x) = \sin x + \frac{1}{2} \cos 2x \) and decide the correct statements. Let's break it down: 

• Finding \( f'(x) \): Start by taking the derivative of \( f(x) \).

\( f(x) = \sin x + \frac{1}{2} \cos 2x \)

The derivative: \( f'(x) = \cos x + \frac{1}{2}(-2 \sin 2x) \)

Thus, \( f'(x) = \cos x - \sin 2x \).

This supports statement (A).

• Critical Points:

Find where \( f'(x) = 0 \).

\( \cos x - \sin 2x = 0 \)

\( \cos x = \sin 2x \)

Using the identity \( \sin 2x = 2 \sin x \cos x \): \( \cos x = 2 \sin x \cos x \)

Thus, \( \cos x (1 - 2 \sin x) = 0 \)

Solve \( \cos x = 0 \) which gives \( x = \frac{\pi}{2} \)

Solve \( 1 - 2 \sin x = 0 \) which gives \( \sin x = \frac{1}{2} \rightarrow x = \frac{\pi}{6} \)

This supports statement (B).

• Find the Maximum and Minimum Values:

Evaluate \( f(x) \) at critical points and endpoints \( x = 0 \) and \( x = \frac{\pi}{2} \).

• \( f(0) = \sin 0 + \frac{1}{2} \cos 0 = \frac{1}{2} \)
• \( f\left(\frac{\pi}{2}\right) = \sin \frac{\pi}{2} + \frac{1}{2} \cos \pi = 1 - \frac{1}{2} = \frac{1}{2} \)
• \( f\left(\frac{\pi}{6}\right) = \sin \frac{\pi}{6} + \frac{1}{2} \cos \frac{\pi}{3} = \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \), thus,\( = \frac{3}{4} \)

Conclusion:

The correct combination is: (A), (B), and (D) only.