MCQ
If $g(x) = \int_0^x {{{\cos }^4}t\,dt,} $ then $g(x + \pi )$ equals
- ✓$g(x) + g(\pi )$
- B$g(x) - g(\pi )$
- C$g(x)g(\pi )$
- D$g(x)/g(\pi )$
$={ \int_0^\pi {{{\cos }^4}t\,dt + \int_\pi ^{x + \pi } {{{\cos }^4}t\,dt} } } $
$ = g(\pi ) + f(x)$
$f(x) = \int_0^x {{{\cos }^4}u\,du = g(x)} $, $(\because t = \pi + u)$
$\therefore \,\,g(x + \pi ) = g(x) + g(\pi )$.
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