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: Chain rule:
$f'(x) = \dfrac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}$. The denominator is always positive.

Step 2: Sign:
The sign of $f'$ is the sign of $\cos x - \sin x$, positive when $\tan x < 1$ on $(-\pi/2, \pi/2)$, i.e. for $x < \pi/4$.

Step 3: Pick the interval:
So $f$ is increasing on $(-\pi/2, \pi/4)$, option (D).

Final Answer:
The function increases on (-pi/2, pi/4). \[ \boxed{\text{(D) }\left(-\dfrac{\pi}{2},\dfrac{\pi}{4}\right)} \]