Understanding the Function and Problem

The question asks us to analyze the function \( f(x) = \sin x + \frac{1}{2} \cos 2x \) over the closed interval \( [0, \frac{\pi}{2}] \) and determine which given statements about its derivative, critical points, minimum value, and maximum value are correct. To do this, we need to calculate the derivative, find where the derivative is zero or undefined within the interval, evaluate the function at these points and the endpoints of the interval, and then compare the results with the given statements.

Analyzing Statement (A): The Derivative f'(x)

Statement (A) claims that \( f'(x) = \cos x - \sin 2x \). Let's find the derivative of the given function \( f(x) = \sin x + \frac{1}{2} \cos 2x \). We will use the rules of differentiation:

• The derivative of \( \sin x \) with respect to \( x \) is \( \cos x \).
• The derivative of \( \cos u \) with respect to \( x \) is \( -\sin u \cdot \frac{du}{dx} \) (using the chain rule). Here, \( u = 2x \), so \( \frac{du}{dx} = 2 \).

Applying these rules: \[ f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\frac{1}{2} \cos 2x) \] \[ f'(x) = \cos x + \frac{1}{2} \cdot (-\sin 2x) \cdot \frac{d}{dx}(2x) \] \[ f'(x) = \cos x + \frac{1}{2} \cdot (-\sin 2x) \cdot 2 \] \[ f'(x) = \cos x - \sin 2x \] Our calculated derivative matches the expression in statement (A). Therefore, statement (A) is correct.

Analyzing Statement (B): Critical Points of the Function

Statement (B) claims the critical points of the function are \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{2} \). Critical points of a function \( f(x) \) in an interval are the points \( x \) where \( f'(x) = 0 \) or \( f'(x) \) is undefined. We also need to consider the endpoints of the closed interval \( [0, \frac{\pi}{2}] \) when looking for absolute extrema, but critical points specifically refer to points within the open interval or where the derivative is zero/undefined. The derivative \( f'(x) = \cos x - \sin 2x \) is defined for all \( x \). So, we need to find the values of \( x \) in \( [0, \frac{\pi}{2}] \) where \( f'(x) = 0 \). Set \( f'(x) = 0 \): \[ \cos x - \sin 2x = 0 \] We know the double angle identity \( \sin 2x = 2 \sin x \cos x \). Substitute this into the equation: \[ \cos x - 2 \sin x \cos x = 0 \] Factor out \( \cos x \): \[ \cos x (1 - 2 \sin x) = 0 \] This equation is satisfied if either \( \cos x = 0 \) or \( 1 - 2 \sin x = 0 \). Case 1: \( \cos x = 0 \) In the interval \( [0, \frac{\pi}{2}] \), the value of \( x \) for which \( \cos x = 0 \) is \( x = \frac{\pi}{2} \). Case 2: \( 1 - 2 \sin x = 0 \) This implies \( 2 \sin x = 1 \), so \( \sin x = \frac{1}{2} \). In the interval \( [0, \frac{\pi}{2}] \), the value of \( x \) for which \( \sin x = \frac{1}{2} \) is \( x = \frac{\pi}{6} \). The critical points within the domain \( [0, \frac{\pi}{2}] \) where the derivative is zero are \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{2} \). Note that \( x = \frac{\pi}{2} \) is also an endpoint. Statement (B) lists these two points as critical points. Therefore, statement (B) is correct.

Analyzing Statements (C) and (D): Minimum and Maximum Values

Statement (C) claims the minimum value is 2. Statement (D) claims the maximum value is \( \frac{3}{4} \). To find the minimum and maximum values of a continuous function on a closed interval, we need to evaluate the function at the critical points within the open interval and at the endpoints of the interval. The points we need to evaluate \( f(x) \) at are the critical points found in the open interval \( (0, \frac{\pi}{2}) \) (which is \( x = \frac{\pi}{6} \)) and the endpoints \( x = 0 \) and \( x = \frac{\pi}{2} \). Note that \( x = \frac{\pi}{2} \) is an endpoint but also a critical point. The points to evaluate are \( x = 0, x = \frac{\pi}{6}, \) and \( x = \frac{\pi}{2} \). Let's evaluate \( f(x) = \sin x + \frac{1}{2} \cos 2x \) at these points: 
 

The values of the function at the points we checked are \( \frac{1}{2}, \frac{3}{4}, \frac{1}{2} \). Comparing these values:

• The minimum value is the smallest value among these, which is \( \frac{1}{2} \).
• The maximum value is the largest value among these, which is \( \frac{3}{4} \).

Statement (C) claims the minimum value is 2. This is incorrect, as the minimum value is \( \frac{1}{2} \). Statement (D) claims the maximum value is \( \frac{3}{4} \). This is correct.

Summary of Correct Statements

Based on our analysis:

• Statement (A) is correct.
• Statement (B) is correct.
• Statement (C) is incorrect.
• Statement (D) is correct.

The correct statements are (A), (B), and (D).

Revision Table: Function Analysis Summary

Additional Information: Extrema on a Closed Interval

To find the absolute maximum and minimum values of a continuous function \( f(x) \) on a closed interval \( [a, b] \), you should follow these steps:

1. Find the critical points of \( f(x) \) in the open interval \( (a, b) \). These are the points where \( f'(x) = 0 \) or \( f'(x) \) is undefined.
2. Evaluate the function \( f(x) \) at each critical point found in step 1 that is inside the interval \( (a,b) \).
3. Evaluate the function \( f(x) \) at the endpoints of the interval, \( a \) and \( b \).
4. Compare all the values obtained in steps 2 and 3. The largest value is the absolute maximum of \( f(x) \) on \( [a, b] \), and the smallest value is the absolute minimum of \( f(x) \) on \( [a, b] \).

In this problem, the critical points we found were \( \frac{\pi}{6} \) and \( \frac{\pi}{2} \). Both are in the closed interval \( [0, \frac{\pi}{2}] \). The endpoints are \( 0 \) and \( \frac{\pi}{2} \). The set of points to check is \( \{0, \frac{\pi}{6}, \frac{\pi}{2}\} \). Evaluating \( f(x) \) at these points gives us the candidates for absolute extrema.