Question

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

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

Answer 1

The correct option is A $a\ge \sqrt{2}$

Any function $f\left(x\right)$ is decreasing if $f^{\prime }\left(x\right)\le 0$
Given $f\left(x\right)=\sin x-\cos x-ax+b$
differentiating w.r.t $x$ we get
$f^{\prime }\left(x\right)=\cos x+\sin x-a$
Since $f^{\prime }\left(x\right)\le 0\Rightarrow \cos x+\sin x-a\le 0$
$\Rightarrow a\ge (\sin x+\cos x)$
$\Rightarrow a\ge \sqrt{2}(\because max(\sin x+\cos x)=\sqrt{2}atx=\frac{\pi }{4})$