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Toppr Admin · Accepted Answer

Since f-x-xA0-is an odd function-xA0-f-x-x2212-5-x2212-f-x2212-x-5-Also g-x-f-x-5-x2234-g-x2212-x-f-x2212-x-5-x2212-f-x-x2212-5-Hence- f-x-x2212-5-x2212-g-x2212-x-x2212-g-x- as g-x-xA0-is even-

If $f$ is odd function , $g$ be an even function and $g (x) = f (x + 5)$ then $f (x - 5)$ equals
$- g (x)$
$g (x)$
$f (5)$
None of these

Since $f (x)$ is an odd function, $f (x - 5) = - f (- x + 5)$
Also $g (x) = f (x + 5)$ ,
$∴ g (- x) = f (- x + 5) = - f (x - 5)$
Hence, $f (x - 5) = - g (- x) = - g (x)$ , as $g (x)$ is even.

If f is odd function , g be an even function and g(x)=f(x+5) then f(x−5) equals−g(x)g(x)f(5)None of these

Since f(x) is an odd function, f(x−5)=−f(−x+5)Also g(x)=f(x+5),∴g(−x)=f(−x+5)=−f(x−5)Hence, f(x−5)=−g(−x)=−g(x), as g(x) is even.

If $f$ is odd function , $g$ be an even function and $g (x) = f (x + 5)$ then $f (x - 5)$ equals
$- g (x)$
$g (x)$
$f (5)$
None of these

Since $f (x)$ is an odd function, $f (x - 5) = - f (- x + 5)$
Also $g (x) = f (x + 5)$ ,
$∴ g (- x) = f (- x + 5) = - f (x - 5)$
Hence, $f (x - 5) = - g (- x) = - g (x)$ , as $g (x)$ is even.