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

1 Answer

  • Let $f(x)$ be a function defined on $(a,b)$ if $f'(x) <0$ for all $ x \in (a,b)$ except for a finite number of points where $f'(x)=0,$ then $f(x)$ is decreasing on $(a,b)$
Step 1
$f(x) = \sqrt3\: \sin x-\cos x-2ax+b$
Differentiating w.r.t $x$ we get,
$f'(x)= \sqrt 3 \cos x-(- \sin x)-2a$
$ = \sqrt 3 \cos x+ \sin x-2a$
$ f'(x)=0$
$ \Rightarrow \sqrt 3 \cos x+ \sin x-2a=0$
$ \Rightarrow \sqrt 3 \cos x+ \sin x=2a$
divide throughout by 2
$ \large\frac{\sqrt 3}{2}\cos x+ \large\frac{1}{2}\sin x=a$
But $ \cos \large\frac{\pi}{6}=\large\frac{\sqrt 3}{2}\: and \: \sin \large\frac{\pi}{6}=\large\frac{1}{2}$
$ \Rightarrow \cos \large\frac{\pi}{6} \cos x+ \sin \large\frac{\pi}{6} \sin x=a$
This is of the form
$ \cos A \cos B+ \sin A \sin B = \cos (A-B)$
$ \Rightarrow \cos \bigg( x - \large\frac{\pi}{6} \bigg)=a$
If $ a < 1$
$ \cos \bigg( x- \large\frac{\pi}{6} \bigg) < 1$
indicating that it is a decreasing function.
answered Aug 5, 2013 by thanvigandhi_1

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