Question

The value of ${}^{\prime }{a}^{\prime }$ for which the function $f\left(x\right)=\sin x-\cos x-ax+b$ decreases for all real values of $x$, is

• A. $a\ge -\sqrt{2}$
• B. $a\le -\sqrt{2}$
• C. $a\le \sqrt{2}$
• D. $a\ge \sqrt{2}$

Answer 1

Answer: (D)

$a\ge \sqrt{2}$

Hence
$f^{\prime }\left(x\right)$
$=\cos x+\sin x-a$

$=\sqrt{2}\left[\frac{\cos x+\sin x}{\sqrt{2}}\right]-a$

$=\sqrt{2}\sin (\frac{\pi }{4}+x)-a$
$<0$
Hence
$\sqrt{2}\sin (\frac{\pi }{4}+x)<a$
Now
$|\sin \theta |<1$
Hence
$a>\sqrt{2}$ ...for strictly decreasing nature.
If only decreasing and not strictly decreasing, then
$a\ge \sqrt{2}$.