Question

If the function $f$ be given by
$f(x)=\tan ^{-1}(\sin x+\cos x), x>0$
The $f$ is always an strictly increasing function in the interval.

• A. $\left[0, \frac{\pi}{6}\right]$
• B. $\left(0, \frac{\pi}{4}\right)$
• C. $\left[0, \frac{\pi}{2}\right]$
• D. None of these

Answer 1

Answer: (B)

We have,
$f(x)=\tan ^{-1}(\sin x+\cos x), x>0$
Therefore, $ f^{\prime}(x)=\frac{1}{1+(\sin x+\cos x)^2}(\cos x-\sin x) $
$=\frac{\cos x-\sin x}{1+\sin ^2 x+\cos ^2 x+2 \sin x \cos x}$
$=\frac{\cos x-\sin x}{2+\sin 2 x} \left(\because \sin ^2 x+\cos ^2 x=1\right)$
Note that $2+\sin 2 x>0$ for all $x$ in $\left(0, \frac{\pi}{4}\right)$.
Therefore, $f^{\prime}(x)>0$ if $\cos x-\sin x>0$ or $f^{\prime}(x)>0$ if $\cos x>\sin x$ or $\cot x>1$
Now, $\cot x>1$ if $\tan x < 1$, i.e., if $0 < x < \frac{\pi}{4}$
Thus, $f^{\prime}(x)>0$ in $\left(0, \frac{\pi}{4}\right)$
Hence, $f$ is strictly increasing function in $\left(0, \frac{\pi}{4}\right)$.