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Show that for  a \geq 1, f (x) = \sqrt 3 \sin x - \cos x - 2ax + b is decreasing in R.


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Given: a \geq 1, f (x) = \sqrt 3 \sin x - \cos x - 2ax + b

To show: the above function is decreasing in R.

Explanation: Given f (x) = \sqrt 3 \sin x - cos x - 2ax + b

The first derivative is applied with respect to x,

\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\sqrt{3} \sin \mathrm{x}-\cos \mathrm{x}-2 \mathrm{ax}+\mathrm{b})}{\mathrm{dx}}
By using the sum rule of differentiation, we get

\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\sqrt{3} \sin \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}(2 \mathrm{ax})}{\mathrm{dx}}+0
Removing all the constant terms, we get
\mathrm{f}^{\prime}(\mathrm{x})=\sqrt{3} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}-2 \mathrm{a} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}

But the derivative of sin X = cos x and that of cos x = -sin x, so

\\ {f' (x) = \sqrt 3cos x-(-sin x)-2a}\\ {f' (x) = \sqrt 3\cos x+ \sin x-2a}\\

Multiplying and dividing RHS by 2,

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=2\left[\frac{\sqrt{3}}{2} \cos \mathrm{x}+\frac{1}{2} \sin \mathrm{x}\right]-2 \mathrm{a}\\ &\sin \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \text { and } \sin \frac{\pi}{6}=\frac{1}{2} \text { , putting these values in the above equation, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=2\left[\cos \frac{\pi}{6} \cos \mathrm{x}+\sin \frac{\pi}{6} \cdot \sin \mathrm{x}\right]-2 \mathrm{a}\\ &\text { But } \cos (A-B)=\cos A \cos B+\sin A \cdot \sin B \text { , placing the values in the above given equation, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=2\left[\cos \left(\frac{\pi}{6}-\mathrm{x}\right)\right]-2 \mathrm{a}\\ &\text { Now, we know cos } \mathrm{x} \text { always belong to }[-1,1] \text { for } \mathrm{a} \geq 1\\ &2\left[\cos \left(\frac{\pi}{6}-x\right)\right]-2 a \leq 0 \end{aligned}

And we know, if f'(x)\leq0, then f(x) is decreasing function.

Therefore, the given function is decreasing function in R.


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