Question

The function \(f(x) = tan^{-1}(sinx+cosx)\) is an increasing function in the interval.....

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

Hint:

The inverse tangent is increasing, so we need sin x + cos x increasing.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
\(\tan^{-1}u\) is an increasing function of \(u\). So \(f\) increases exactly where \(u = \sin x + \cos x\) increases.

Step 2: Derivative:
\[ \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]
We need \(\cos x - \sin x > 0\), i.e. \(\sqrt2\cos\left(x + \dfrac\pi4\right) > 0\).

Step 3: Solve:
\(x + \dfrac\pi4 \in \left(-\dfrac\pi2, \dfrac\pi2\right)\), so \(x \in \left(-\dfrac{3\pi}{4}, \dfrac\pi4\right)\).

Step 4: Match with the options:
The interval \(\left(-\dfrac\pi2, \dfrac\pi4\right)\) lies inside this range, so \(f\) increases there: option (D). Options (A), (B), (C) all contain points beyond \(\pi/4\) where \(\cos x < \sin x\).

Final Answer:
f increases where cos x exceeds sin x. \[ \boxed{\text{(D) }\left(-\dfrac{\pi}{2},\dfrac{\pi}{4}\right)} \]