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: \[ 2. \]