The value of a such that f(x) = sqrt{3} sin x | vedclass

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Question

(B) Given $f(x) = \sqrt{3} \sin x - \cos x - 2ax + b$. For $f(x)$ to be decreasing for all real values of $x$,we must have $f'(x) \le 0$ for all $x$.

Vedclass · Accepted Answer

$a \ge 1$

Vedclass · Answer

(B) Given $f(x) = \sqrt{3} \sin x - \cos x - 2ax + b$. For $f(x)$ to be decreasing for all real values of $x$,we must have $f'(x) \le 0$ for all $x$. $f'(x) = \sqrt{3} \cos x + \sin x - 2a$. Setting $f'(x) \le 0$,we get $\sqrt{3} \cos x + \sin x - 2a \le 0$. $\sqrt{3} \cos x + \sin x \le 2a$. Dividing by $2$,we get $\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x \le a$. Using the identity $\sin(x + \frac{\pi}{3}) = \sin x \cos \frac{\pi}{3} + \cos x \sin \frac{\pi}{3} = \frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x$,the inequality becomes $\sin(x + \frac{\pi}{3}) \le a$. Since the maximum value of $\sin(x + \frac{\pi}{3})$ is $1$,for the inequality to hold for all $x$,we must have $a \ge 1$.

The value of $a$ such that $f(x) = \sqrt{3} \sin x - \cos x - 2ax + b$ decreases for all real values of $x$ is given by:

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