(A) Given: $f(x) = \tan x - x$. Differentiating with respect to $x$, we get $f'(x) = \sec^2 x - 1$. We know that $\sec^2 x \geq 1$ for all $x$ in the

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(A) Given: $f(x) = \tan x - x$. Differentiating with respect to $x$, we get $f'(x) = \sec^2 x - 1$. We know that $\sec^2 x \geq 1$ for all $x$ in the domain $R^*$. Therefore, $f'(x) = \sec^2 x - 1 \geq 0$. Since the derivative $f'(x) \geq 0$ for all $x \in R^*$, the function $f(x)$ is an increasing function. Hence, option $A$ is correct.

Class 12 Mathematics Applications of Derivatives Increasing and Decreasing function

Easy

Let $R^* = R - \left\{ (2k - 1) \frac{\pi}{2} \mid k \in I \right\}$. The function $f: R^* \rightarrow R$ is defined as $f(x) = \tan x - x$, then $f(x)$ is

TS EAMCET 2021 All TS EAMCET

A
an increasing function
B
a decreasing function
C
minimum at $x = 0$
D
periodic function

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Applications of Derivatives

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Increasing and Decreasing function

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Let $R^* = R - \left\{ (2k - 1) \frac{\pi}{2} \mid k \in I \right\}$. The function $f: R^* \rightarrow R$ is defined as $f(x) = \tan x - x$, then $f(x)$ is

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