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The function f(x)+tan x-x

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

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 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 by thanvigandhi_1