Question

The maximum values of the function
\(\sin(x)+\cos(2x)\), are

• A. \((0, -2)\)
• B. \((0, \frac{9}{8})\)
• C. \((\frac{3}{8}, \frac{9}{8})\)
• D. \((\frac{9}{8}, \frac{9}{8})\)

Hint:

When dealing with trigonometric functions involving different angles (like \(x\) and \(2x\)), try to use identities to express the entire function in terms of a single trigonometric function and angle. This often simplifies the problem to finding extrema of a polynomial.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the maximum and minimum values (extrema) of a function, we can use calculus. This involves finding the first derivative, setting it to zero to find critical points, and then using the second derivative test to classify them as maxima or minima.

Step 2: Key Formula or Approach:
1. Let \(f(x) = \sin(x) + \cos(2x)\).
2. Use the identity \(\cos(2x) = 1 - 2\sin^2(x)\) to express \(f(x)\) as a function of \(\sin(x)\).
3. Let \(u = \sin(x)\) and find the maximum of the resulting quadratic function in \(u\) on the interval \([-1, 1]\).

Step 3: Detailed Explanation:
Let the function be \(f(x) = \sin(x) + \cos(2x)\).
Using the double angle identity for cosine, we have: \[ f(x) = \sin(x) + (1 - 2\sin^2(x)) \] Let \(u = \sin(x)\). Since the range of \(\sin(x)\) is \([-1, 1]\), we have \(-1 \leq u \leq 1\).
The function becomes a quadratic in \(u\): \[ g(u) = -2u^2 + u + 1 \] This is a downward-opening parabola. Its maximum value will occur either at its vertex or at the endpoints of the interval \([-1, 1]\).
The vertex of a parabola \(au^2+bu+c\) is at \(u = -\frac{b}{2a}\). \[ u_{vertex} = -\frac{1}{2(-2)} = \frac{1}{4} \] Since \(\frac{1}{4}\) is within the interval \([-1, 1]\), it is a candidate for an extremum.
Now we evaluate the function at the vertex and the endpoints:

At the vertex \(u = \frac{1}{4}\): \[ g(\frac{1}{4}) = -2(\frac{1}{4})^2 + \frac{1}{4} + 1 = -2(\frac{1}{16}) + \frac{1}{4} + 1 = -\frac{1}{8} + \frac{2}{8} + \frac{8}{8} = \frac{9}{8} \]
At the endpoint \(u = -1\): \[ g(-1) = -2(-1)^2 + (-1) + 1 = -2 - 1 + 1 = -2 \]
At the endpoint \(u = 1\): \[ g(1) = -2(1)^2 + 1 + 1 = 0 \] \end{itemize} The possible extremum values of the function are \(\frac{9}{8}\), \(-2\), and \(0\).
The absolute maximum value is \(\frac{9}{8}\) and the absolute minimum value is \(-2\). The value \(0\) is a local minimum.

Step 4: Final Answer:
The question asks for "maximum values" (plural). The options are pairs of numbers. Option (B) lists \(0\) (a local minimum) and \(\frac{9}{8}\) (the absolute maximum). This is the most plausible interpretation of the ambiguous question phrasing among the given choices.