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, $f^{\prime }(x)=\frac{1}{1+(\sin x+\cos x{)}^2}(\cos x-\sin x)$
$=\frac{\cos x-\sin x}{1+{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}$
$=\frac{\cos x-\sin x}{2+\sin 2x}\left(\because {\sin }^{2}x+{\cos }^{2}x=1\right)$
Note that $2+\sin 2x>0$ 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 $\left(0,\frac{\pi }{4}\right)$.