Question

The set of values of 'a' for which the function f(x) = sin x - cos x - ax + b decreases for all the real values of x, is ___________.

Answer 1

The given function is f(x) = sin x − cos x − ax + b.

f(x) = sin x − cos x − ax + b

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\cos x-\left(-\sin x\right)-a$

$\Rightarrow f\text{'}\left(x\right)=\cos x+\sin x-a$

$\Rightarrow f\text{'}\left(x\right)=\sqrt{2}\left(\cos x\times \frac{1}{\sqrt{2}}+\sin x\times \frac{1}{\sqrt{2}}\right)-a$

$\Rightarrow f\text{'}\left(x\right)=\sqrt{2}\left(\cos x\cos \frac{\mathrm{\pi }}{4}+\sin x\sin \frac{\mathrm{\pi }}{4}\right)-a$

$\Rightarrow f\text{'}\left(x\right)=\sqrt{2}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)-a$

For f(x) to be decreasing for all x,

$f\text{'}\left(x\right)\le 0$

$\Rightarrow \sqrt{2}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)-a\le 0$

$\Rightarrow a\ge \sqrt{2}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)$

$\Rightarrow a\ge \sqrt{2}\left[-1\le \cos \left(x-\frac{\mathrm{\pi }}{4}\right)\le 1\right]$

$\therefore a\in [\sqrt{2},\infty )$

Thus, the set of values of 'a' for which the given function f(x) decreases for all the real values of x is $[\sqrt{2},\infty )$.


The set of values of 'a' for which the function f(x) = sin x − cos x − ax + b decreases for all the real values of x, is $\underline{[\sqrt{2},\infty )}$.