AIEEE 2007 (Offline)Question 1
The function $f(x)=\tan^{-1}(\sin x+\cos x)$ is an increasing function in
Differentiation and its Application
Solution
Concept. $f$ increases where $f'>0$, i.e. where $\cos x>\sin x$. Formulas used. (1) $\dfrac{d}{dx}\tan^{-1}u=\dfrac{u'}{1+u^2}$Step 1 - Derivative. By (1), $$f'(x)=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}.$$ Step 2 - Sign. $\cos x-\sin x=\sqrt2\cos\left(x+\frac\pi4\right)>0$ for $x\in\left(-\frac{3\pi}{4},\frac\pi4\right)$. Step 3 - Option. $\left(-\frac\pi2,\frac\pi4\right)$ lies inside this interval; the other options do not. Answer: (D)