Question

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

$a\ge \sqrt{2}$

Explanation for the correct option:

Find the value of $a$:

A function $f\left(x\right)=\sin x-\cos x-ax+b$ is given.

It is also given that the given function is a strictly decreasing function.

We know that, if a function is a decreasing function, then the derivative of the function is less than zero.

Therefore, $f\text{'}\left(x\right)<0$.

$$\begin{gathered} \cos x+\sin x-a<0 \\ \Rightarrow \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x\right)-a<0 \\ \Rightarrow a>\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x\right) \\ \Rightarrow a>\sqrt{2}\sin \left(\frac{\mathrm{\pi }}{4}+x\right) \end{gathered}$$

So, for strictly decreasing function $a\ge \sqrt{2}$.

Therefore, The value of $a$ in order that $f\left(x\right)=\sin x-\cos x-ax+b$ decreases for all real value of $x$ is given by $a\ge \sqrt{2}$.

Hence, option (A), $a\ge \sqrt{2}$ is the correct answer.