Domain of derivative of function f(x) = left{ begin{array}{l}{tan ^{

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1.Relation and Function

hard

The domain of the derivative of the function $f(x) = \left\{ \begin{array}{l}{\tan ^{ - 1}}x\;\;\;\;\;,\;|x|\; \le 1\\\frac{1}{2}(|x|\; - 1)\;,\;|x|\; > 1\end{array} \right.$ is

$R - \{ 0\} $

$R - \{ 1\} $

$R - \{ - 1\} $

$R - \{ - 1,\;1\} $

(IIT-2002)

Solution

(c) $f(x) = \left\{ \begin{array}{l}\frac{1}{2}( – x – 1),\,\,\,x < – 1\\{\tan ^{ – 1}}x,\,\,\,\, – 1 \le x \le 1\\\frac{1}{2}(x + 1),\,\,\,x > 1\end{array} \right.$; $f'(x) = \left\{ \begin{array}{l} – \frac{1}{2},\,\,\,\,\,\,\,\,x < – 1\\\frac{1}{{1 + {x^2}}},\,\, – 1 < x < 1\\\frac{1}{2},\,\,\,\,\,\,\,\,\,\,\,\,x > 1\end{array} \right.$

$f'( – 1 – 0) = – \frac{1}{2};\,\,f'( – 1 + 0) = \frac{1}{{1 + {{( – 1 + 0)}^2}}} = \frac{1}{2}$

$f'(1 – 0) = \frac{1}{{1 + {{(1 – 0)}^2}}} = \frac{1}{2};\,\,f'(1 + 0) = \frac{1}{2}$

$\therefore$ $f'( – 1)$ does not exist;

$\therefore$ domain of $f'(x) = R – \{ – 1\} $.

Standard 12

Mathematics

Statement $-1$ : The equation $x\, log\, x = 2 – x$ is satisfied by at least one value of $x$ lying between $1$ and $2$

Statement $-2$ : The function $f(x) = x\, log\, x$ is an increasing function in $[1, 2]$ and $g (x) = 2 -x$ is a decreasing function in $[ 1 , 2]$ and the graphs represented by these functions intersect at a point in $[ 1 , 2]$

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The domain of the derivative of the function $f(x) = \left\{ \begin{array}{l}{\tan ^{ - 1}}x\;\;\;\;\;,\;|x|\; \le 1\\\frac{1}{2}(|x|\; - 1)\;,\;|x|\; > 1\end{array} \right.$ is

Solution

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