The function f(x) = cos |x| - 2ax + b increas | vedclass

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(C) Given the function $f(x) = \cos |x| - 2ax + b$.Since $\cos |x| = \cos x$,we have $f(x) = \cos x - 2ax + b$.For the function to be increasing on th

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$a \leq -1/2$

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(C) Given the function $f(x) = \cos |x| - 2ax + b$.Since $\cos |x| = \cos x$,we have $f(x) = \cos x - 2ax + b$.For the function to be increasing on the entire real line,we must have $f'(x) \geq 0$ for all $x \in \mathbb{R}$.Calculating the derivative: $f'(x) = -\sin x - 2a$.Setting the condition $f'(x) \geq 0$ gives $-\sin x - 2a \geq 0$,which simplifies to $\sin x + 2a \leq 0$.For this inequality to hold for all $x \in \mathbb{R}$,the maximum value of $\sin x + 2a$ must be less than or equal to $0$.The maximum value of $\sin x$ is $1$,so we require $1 + 2a \leq 0$.Solving for $a$,we get $2a \leq -1$,which implies $a \leq -1/2$.

The function $f(x) = \cos |x| - 2ax + b$ increases on the entire real line. What is the range of $a$?

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