Question

The value of $K$ in order that $f(x)=\sin x-\cos x-K x+5$ decreases for all positive real values of $x$ is given by

• A. $K < 1$
• B. $K \ge 1$
• C. $K > \sqrt{2}$
• D. $K < \sqrt{2}$

Answer 1

Answer: (C)

We have,
$f(x) =\sin x-\cos x-K x+5$
$\Rightarrow f'(x) =\cos x+\sin x-K$
For decreasing, $f'(x)<0$
$\Rightarrow \cos x+\sin x-K<0 $
$\Rightarrow K>\cos x+\sin x$
$\Rightarrow K>\sqrt{2}\left[\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right]$
$\Rightarrow K>\sqrt{2} \sin \left(\frac{\pi}{4}+x\right) $
$\therefore K>\sqrt{2}$