Question

The number of all the values of \( x \) for which the function \[ f(x) = \sin x + \frac{1 - \tan^2 x}{1 + \tan^2 x} \] attains its maximum value on \( [0, 2\pi] \).

• A. \( 4 \)
• B. \( 3 \)
• C. \( 2 \)
• D. {infinite}

Hint:

For trigonometric functions, express terms in a common form and use standard identities to simplify derivatives.

Answer 1

The Correct Option is C

Step 1: Expressing the given function 
We rewrite: \[ f(x) = \sin x + \frac{1 - \tan^2 x}{1 + \tan^2 x}. \] Using the identity: \[ \frac{1 - \tan^2 x}{1 + \tan^2 x} = \cos 2x, \] we obtain: \[ f(x) = \sin x + \cos 2x. \] Step 2: Finding the critical points 
Differentiate \( f(x) \): \[ f'(x) = \cos x - 2 \sin 2x. \] Setting \( f'(x) = 0 \): \[ \cos x - 2 \sin 2x = 0. \] \[ \cos x = 2 \sin 2x. \] Using \( \sin 2x = 2 \sin x \cos x \), we substitute: \[ \cos x = 2 (2 \sin x \cos x). \] \[ \cos x = 4 \sin x \cos x. \] Dividing by \( \cos x \) (except when \( \cos x = 0 \)): \[ 1 = 4 \sin x. \] \[ \sin x = \frac{1}{4}. \] Solving for \( x \) in \( [0, 2\pi] \): \[ x = \sin^{-1} \left(\frac{1}{4}\right) \quad {or} \quad x = \pi - \sin^{-1} \left(\frac{1}{4}\right). \] These give two values in \( [0,2\pi] \). 
Step 3: Verifying maximum values We check \( f''(x) \) or use the first derivative test. It turns out that \( f(x) \) attains a maximum at these two points. 
Step 4: Conclusion Thus, the number of values of \( x \) where \( f(x) \) attains its maximum is: \[ 2. \]

Answer 2

To solve for the number of \( x \) values where the function f(x) achieves its maximum within the interval \( [0, 2\pi] \), we first simplify the expression for f(x):

f(x) is given by:

\[ f(x) = \sin x + \frac{1 - \tan^2 x}{1 + \tan^2 x} \]

Note that the expression \(\frac{1 - \tan^2 x}{1 + \tan^2 x}\) simplifies to \(\cos 2x\) because of the double angle identity:

\[\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}\]

Thus, the function simplifies to:

\[ f(x) = \sin x + \cos 2x \]

Find maximum value:
To find the maximum value of \(f(x)\), we first find the derivative \(f'(x)\) and then set it to zero to find critical points:

\[ f'(x) = \cos x - 2\sin 2x \]

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

\[ f'(x) = \cos x - 4\sin x \cos x \]

\[ = \cos x(1 - 4\sin x) \]

Setting \(f'(x) = 0\):

\[ \cos x(1 - 4\sin x) = 0 \]

This gives us two cases:
1. \(\cos x = 0\) leading to \(x = \frac{\pi}{2}, \frac{3\pi}{2}\)
2. \(1 - 4\sin x = 0\) leading to \(\sin x = \frac{1}{4}\)
For \(\sin x = \frac{1}{4}\), \(x = \arcsin\left(\frac{1}{4}\right)\) which are within \( [0, 2\pi] \):

These two solutions are approximately:

\[ x_1 = \arcsin\left(\frac{1}{4}\right)\quad \text{and}\quad x_2 = \pi - \arcsin\left(\frac{1}{4}\right) \]

The function values at these points are evaluated to determine the maximum:

\[ f(x_1) = \sin(\arcsin\left(\frac{1}{4}\right)) + \cos(2\arcsin\left(\frac{1}{4}\right)) \]

\[ f(x_2) = \sin(\pi - \arcsin\left(\frac{1}{4}\right)) + \cos(2(\pi - \arcsin\left(\frac{1}{4}\right))) \]

Solving these and checking boundary points \(x=0\) and \(x=2\pi\) yield no larger values than at \(x_1\) and \(x_2\). Thus, there are only two points where f(x) achieves its maximum within \([0, 2\pi]\). Therefore, the number of such \(x\) values is \(2\).