Question

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

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

Hint:

To determine where a function is increasing, find the derivative and solve for intervals where the derivative is positive.

Answer 1

The Correct Option is B

Step 1: We are given the function \( f(x) = \tan^{-1}(\sin x + \cos x) \). 
To determine the intervals where this function is increasing, we need to find the derivative of \( f(x) \). 
The derivative of \( f(x) \) can be found using the chain rule: \[ f'(x) = \frac{d}{dx} \left( \tan^{-1}(\sin x + \cos x) \right) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx} (\sin x + \cos x). \] 
Step 2: The derivative of \( \sin x + \cos x \) is: \[ \frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x. \] 
Therefore, the derivative of \( f(x) \) is: \[ f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}. \] 
Step 3: For \( f(x) \) to be increasing, \( f'(x)>0 \). 
This means that the numerator \( \cos x - \sin x \) must be positive. 
So, we need to solve: \[ \cos x - \sin x>0. \] 
Rewriting this inequality: 
\[ \cos x>\sin x. \] 
This inequality holds in the interval \( \left( -\frac{\pi}{2}, \frac{\pi}{4} \right) \), because in this interval, \( \cos x \) is greater than \( \sin x \). 
Step 4: Thus, the function \( f(x) = \tan^{-1}(\sin x + \cos x) \) is increasing in the interval \( \left( -\frac{\pi}{2}, \frac{\pi}{4} \right) \).