Let the function f(x) = x^2 + x + sin x - cos | vedclass

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(B) An odd extension $g(x)$ of a function $f(x)$ defined on $[0, 1]$ to $[-1, 1]$ must satisfy $g(-x) = -g(x)$ for all $x \in [-1, 1]$.For $x \in (0,

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$-x^2 + x + \sin x + \cos x - \log(1 + |x|)$

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(B) An odd extension $g(x)$ of a function $f(x)$ defined on $[0, 1]$ to $[-1, 1]$ must satisfy $g(-x) = -g(x)$ for all $x \in [-1, 1]$.For $x \in (0, 1]$,the odd extension is defined as $g(x) = -f(-x)$.Given $f(x) = x^2 + x + \sin x - \cos x + \log(1 + |x|)$.Then $f(-x) = (-x)^2 + (-x) + \sin(-x) - \cos(-x) + \log(1 + |-x|)$.Since $\sin(-x) = -\sin x$,$\cos(-x) = \cos x$,and $|-x| = |x|$,we have:$f(-x) = x^2 - x - \sin x - \cos x + \log(1 + |x|)$.Now,the odd extension $g(x)$ for $x \in (0, 1]$ is:$g(x) = -f(-x) = -(x^2 - x - \sin x - \cos x + \log(1 + |x|)) = -x^2 + x + \sin x + \cos x - \log(1 + |x|)$.Comparing this with the given options,option $B$ matches this result.

List-$I$	List-$II$
$A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$,where $[.]$ denotes the greatest integer function	$I$. Odd function
$B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$	$II$. $\left\{0, \frac{1}{2}\right\}$
$C$. $f(x+y)=f(x)+f(y) ; f(1)=5$	$III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$
$D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$	$IV$. $R$
	$V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$

Let the function $f(x) = x^2 + x + \sin x - \cos x + \log(1 + |x|)$ be defined over the interval $[0, 1]$. The odd extension of $f(x)$ to the interval $[-1, 1]$ is:

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