If f(x)=sqrt{3} sin x-cos x-2 a x+b decreases | vedclass

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(A) Given the function $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$.To find the condition for $f(x)$ to be decreasing,we first find its derivative $f^{\prime

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$a \geq 1$

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(A) Given the function $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$.To find the condition for $f(x)$ to be decreasing,we first find its derivative $f^{\prime}(x)$.$f^{\prime}(x)=\sqrt{3} \cos x+\sin x-2 a$.We can rewrite this as $f^{\prime}(x)=2 \left( \frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x \right)-2 a$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $f^{\prime}(x)=2 \sin \left(x+\frac{\pi}{3}\right)-2 a$.For $f(x)$ to decrease for all $x$,we must have $f^{\prime}(x) \leq 0$ for all $x$.$2 \sin \left(x+\frac{\pi}{3}\right)-2 a \leq 0$.$2 a \geq 2 \sin \left(x+\frac{\pi}{3}\right)$.$a \geq \sin \left(x+\frac{\pi}{3}\right)$.Since the maximum value of $\sin \left(x+\frac{\pi}{3}\right)$ is $1$,for the inequality to hold for all $x$,we must have $a \geq 1$.

If $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$ decreases for all values of $x$,then

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