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If $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$ decreases for all values of $x$, then
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Verified Answer
The correct answer is:
$a \geq 1$
$f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$
$\Rightarrow f^{\prime}(x)=\sqrt{3} \cos x+\sin x-2 a$
$\begin{aligned} & =2\left(\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x\right)-2 a \\ & =2 \sin \left(x+\frac{\pi}{3}\right)-2 a\end{aligned}$
When, $f(x)$ decreases.
Then, $f^{\prime}(x) \leq 0$
$\begin{aligned} & \Rightarrow \quad 2 \sin \left(x+\frac{\pi}{3}\right)-2 a \leq 0 \\ & \Rightarrow \quad 2 a \geq 2 \sin \left(x+\frac{\pi}{3}\right) \Rightarrow a \geq \sin \left(x+\frac{\pi}{3}\right)\end{aligned}$
$\begin{aligned} & \because \quad \sin \left(x+\frac{\pi}{3}\right) \leq 1 \\ & \therefore \quad a \geq 1\end{aligned}$
$\Rightarrow f^{\prime}(x)=\sqrt{3} \cos x+\sin x-2 a$
$\begin{aligned} & =2\left(\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x\right)-2 a \\ & =2 \sin \left(x+\frac{\pi}{3}\right)-2 a\end{aligned}$
When, $f(x)$ decreases.
Then, $f^{\prime}(x) \leq 0$
$\begin{aligned} & \Rightarrow \quad 2 \sin \left(x+\frac{\pi}{3}\right)-2 a \leq 0 \\ & \Rightarrow \quad 2 a \geq 2 \sin \left(x+\frac{\pi}{3}\right) \Rightarrow a \geq \sin \left(x+\frac{\pi}{3}\right)\end{aligned}$
$\begin{aligned} & \because \quad \sin \left(x+\frac{\pi}{3}\right) \leq 1 \\ & \therefore \quad a \geq 1\end{aligned}$
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