The value of k such that the function f(x) = | vedclass

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(D) For a function $f(x)$ to be strictly decreasing for all real $x$, its derivative must satisfy $f'(x) < 0$ for all $x \in \mathbb{R}$.First, find t

Vedclass · Accepted Answer

$k > \sqrt{2}$

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(D) For a function $f(x)$ to be strictly decreasing for all real $x$, its derivative must satisfy $f'(x) First, find the derivative: $f'(x) = \frac{d}{dx}(\sin x - \cos x - kx + b) = \cos x + \sin x - k$.We require $\cos x + \sin x - k  \cos x + \sin x$.To ensure this holds for all $x$, $k$ must be greater than the maximum value of the expression $(\cos x + \sin x)$.The expression $\cos x + \sin x$ can be written as $\sqrt{2}(\frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x) = \sqrt{2}\sin(x + \frac{\pi}{4})$.The maximum value of $\sqrt{2}\sin(x + \frac{\pi}{4})$ is $\sqrt{2}$.Therefore, we must have $k > \sqrt{2}$.

The value of $k$ such that the function $f(x) = \sin x - \cos x - kx + b$ is strictly decreasing for all real $x$ is

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