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

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(b) Odd extension from $[0, 1]$ to $[-1, 1]$ means we have to choose the function out of $(a), (b), (c), (d)$ which satisfies the condition $f( - x) =

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Odd extension from $[0, 1]$ to $[-1, 1]$ means we have to choose the function out of $(a), (b), (c), (d)$ which satisfies the condition $f( - x) = - f(x).$
Now $| - x|\,\, = \,\,|x|$
$f( - x) = {x^2} - x - \sin x - \cos x + \log \,(1 + |x|)$
$ = - \,[{m{function given in (b)]}}$
$	herefore \,\,\,({m{b}})$ is correct. No other given function satisfies this criteria of $f( - x) = - f(x).$

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

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