If the function $f: R \rightarrow R$ be defined by $f(x)=\tan x-x$, then f(x)
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If the function $f: R \rightarrow R$ be defined by $f(x)=\tan x-x$, then $f(x)$

(A) Increases

(B) Decreases

(C) Remains Constant

(D) Becomes Zero

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SOLUTION —

$f(x)=\tan x-x$

$\Rightarrow f^{\prime}(x)=\sec ^2 x-1=\frac{1}{\cos ^2 x}-1=\frac{1-\cos ^2 x}{\cos ^2 x}$

Since, $0 \leq \cos ^2 x \leq 1$ for all values of $x$.

$\therefore f^{\prime}(x)>0$ for all values of $x$. Thus, $f(x)$ always increases

So, The correct option will be (A).

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