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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Show that $f(x)=\tan^{-1}(\sin x+\cos x)$ is an increasing function in$\bigg(0,\Large \frac{1}{4}\normalsize \bigg).$

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  • 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 increasing on $(a,b)$
Step 1
$f(x)= \tan^{-1} ( \sin x+ \cos x)$
Differentiating w.r.t $x$ we get,
$ f'(x)=\large\frac{1}{1+ ( \sin x+ \cos x)^2} \times ( \cos x- \sin x)$
$ = \large\frac{\cos x- \sin x}{1+( \sin x+ \cos x)^2}= \large\frac{\cos x(1- \tan x)}{1+ ( \sin x+ \cos x)^2}$
Now let us consider the interval $ 0 < x < \large\frac{\pi}{4}$
$ \Rightarrow \cos x > 0, \large\frac{1}{1+(\sin x+ \cos x)^2}\: and \: (1- \tan x) > 0$
$ ( \because \tan x < 1 $ in the interval $ 0 < x < \large\frac{\pi}{4})$
$ \Rightarrow \large\frac{\cos x(1-\tan x)}{1+ ( \sin x+ \cos x)^2} > 0$
$ \Rightarrow f'(x) > 0$
Hence $ f'(x)>0$ for all $ x \in \bigg( 0, \large\frac{\pi}{4} \bigg)$
Hence $f(x)$ is increasing on $ \bigg( 0, \large\frac{\pi}{4} \bigg)$
answered Aug 5, 2013 by thanvigandhi_1
 

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