Question

The function \( f(x) = \tan^{-1} (\sin x + \cos x) \) is an increasing function in:

• A. \( \left( 0, \frac{\pi}{2} \right) \)
• B. \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
• C. \( \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \)
• D. \( \left( -\frac{\pi}{2}, \frac{\pi}{4} \right) \)

Hint:

When differentiating inverse trigonometric functions, carefully check the sign of the derivative to determine where the function is increasing or decreasing.

Answer 1

The Correct Option is D

We are given the function: \[ f(x) = \tan^{-1} (\sin x + \cos x). \] Step 1: Differentiate the function.
The derivative of \( f(x) \) is: \[ f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x). \] Using a trigonometric identity, we can simplify: \[ \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right). \] Substituting this into the derivative: \[ f'(x) = \frac{\sqrt{2} \cos \left( x + \frac{\pi}{4} \right)}{1 + (\sin x + \cos x)^2}. \] Step 2: Find the interval where \( f'(x)>0 \).
To have \( f'(x)>0 \), the numerator \( \sqrt{2} \cos \left( x + \frac{\pi}{4} \right) \) must be positive. This implies: \[ \cos \left( x + \frac{\pi}{4} \right)>0. \] The cosine function is positive in the first and fourth quadrants. Solving: \[ -\frac{\pi}{2}<x + \frac{\pi}{4}<0. \] Rearranging the inequality: \[ -\frac{\pi}{2} - \frac{\pi}{4}<x<0 + \frac{\pi}{4}. \] Simplifying: \[ -\frac{3\pi}{4}<x<\frac{\pi}{4}. \] Step 3: Confirm the function's increasing behavior.
In the interval \( \left( -\frac{3\pi}{4}, \frac{\pi}{4} \right) \), \( f'(x)>0 \), which indicates that \( f(x) \) is an increasing function. Final Answer: \[ \boxed{\left( -\frac{3\pi}{4}, \frac{\pi}{4} \right)} \]