If the function f(x) = cos |x| - 2ax + b is a | vedclass

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(B) Given $f(x) = \cos |x| - 2ax + b$. Since $\cos |x| = \cos x$,we have $f(x) = \cos x - 2ax + b$. For $f(x)$ to be an increasing function,$f'(x) \ge

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$a \le -1/2$

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(B) Given $f(x) = \cos |x| - 2ax + b$. Since $\cos |x| = \cos x$,we have $f(x) = \cos x - 2ax + b$. For $f(x)$ to be an increasing function,$f'(x) \ge 0$ for all $x \in \mathbb{R}$. $f'(x) = -\sin x - 2a$. For $f'(x) \ge 0$,we must have $-\sin x - 2a \ge 0$,which implies $2a \le -\sin x$,or $a \le -\frac{1}{2} \sin x$. Since this must hold for all $x \in \mathbb{R}$,$a$ must be less than or equal to the minimum value of $-\frac{1}{2} \sin x$. The minimum value of $\sin x$ is $-1$,so the maximum value of $-\frac{1}{2} \sin x$ is $\frac{1}{2}$. However,for the function to be non-decreasing for all $x$,we consider the condition $f'(x) \ge 0$. Since $-\sin x$ ranges from $-1$ to $1$,for $-\sin x - 2a \ge 0$ to hold for all $x$,we need $2a \le \min(-\sin x) = -1$. Thus,$a \le -1/2$.

If the function $f(x) = \cos |x| - 2ax + b$ is an increasing function on the entire real line,then what is the value of $a$?

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