Question

The value of $K$ in order that $f\left(x\right)=\sin x-\cos x-Kx+5$ 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)

$K>\sqrt{2}$

Given : $f\left(x\right)=\sin x-\cos x-Kx+5$

$f\left(x\right)$ decrease for all $x$ if $f^{\prime }\left(x\right)<0$

$f^{\prime }\left(x\right)=\cos x+\sin x-K<0$

$\therefore k>\cos x+\sin x$

We know that, $\cos x+\sin x=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x\right)=\sqrt{2}(\sin \frac{\pi }{4}\cos x+\cos \frac{\pi }{4}\sin x)=\sqrt{2}\left(\sin \right(\frac{\pi }{4}+x\left)\right)$
and,
$-1\le \sin (\frac{\pi }{4}+x)\le 1\Rightarrow -\sqrt{2}\le \sqrt{2}\cdot \sin (\frac{\pi }{4}+x)\le \sqrt{2}$

$\therefore -\sqrt{2}\le \cos x+\sin x\le \sqrt{2}$

$max(\cos x+\sin x)=\sqrt{2}$

$\therefore K>\sqrt{2}$