Question

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

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

Answer 1

Answer: (C)

$K\ge \sqrt{2}$

$f^{\prime }\left(x\right)$
$=\cos x+\sin x-k$
$=\sqrt{2}\sin (x+\frac{\pi }{4})-k$
$=\sqrt{2}\sin \theta -k$ ...(i)
Now it is a decreasing function for all $x$.
Hence
$f^{\prime }\left(x\right)<0$
Or
$\sqrt{2}\sin \theta -k\le 0$
$\sqrt{2}\sin \theta \le k$
Now
$\sin \theta ϵ[-1,1]$
Hence
$k\ge \sqrt{2}$ and $k\ge -\sqrt{2}$.
Thus from the above two, we get
$k\ge \sqrt{2}$.