Show that for a 1, f(x)=3 - -2ax+b is decreasing in R. - Vidyadip

Question

Show that for $\text{a}\geq1,\text{ f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}-2\text{ax}+\text{b}$ is decreasing in R.

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Answer

We have,$\text{ f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}-2\text{ax}+\text{b}$
$\Rightarrow\ \text{f(x)}=\sqrt{3}\cos\text{x}-(-\sin\text{x})-2\text{a}$
$=\sqrt{3}\cos\text{x}+\sin\text{x}-2\text{a}$
$=2\Big[\frac{\sqrt{3}}{2}\cdot\cos\text{x}+\frac{1}{2}\cdot\sin\text{x}\Big]-2\text{a}$
$=2\Big[\cos\frac{\pi}{6}\cdot\cos\text{x}+\sin\frac{\pi}{6}\cdot\sin\text{x}\Big]-2\text{a}$
$=2\cos\Big(\frac{\pi}{6}-\text{x}\Big)-2\text{a}$ $\big[\therefore\cos(\text{A}-\text{B})=\cos\text{A}\cdot\text{B}+\sin\text{A}\cdot\sin\text{B}\big]$
$=2\Big[\cos\Big(\frac{\pi}{6}-\text{x}\Big)-\text{a}\Big]$
Since, $\cos\text{x}\in[-1,1]\text{ and a}\geq1$
$\therefore\ 2\Big[\cos\Big(\frac{\pi}{6}-\text{a}\Big)-\text{a}\Big]\leq0$
We know tha if $\text{f(x)}\leq0,$ then f(x) is a decreasing function.
Thus, f(x) is a decreasing function in R.

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