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# The function f(x)+tan x-x

$\begin{array}{1 1}(A)\;\text{always increases} & (B)\;\text{always decreases}\\(C)\;\text{never increases} &(D)\;\text{sometimes increases and sometimes decreases}\end{array}$

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## 1 Answer

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Toolbox:
• 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
$f(x)= \tan x - x$
On differentiating w.r.t $x$ we get,
$f'(x)= \sec^2x-x$
Since $sec^2x$ is a square function, it is clear that if $x < 0; f'(x)>0$
if $x=0$ then $f'(x)>0$
if $x >0$ then $f'(x)>0$
In all cases $f'(x)>0$ implies that the given function is always increasing.
Hence the correct option is A.
answered Aug 11, 2013

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