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
    $g(x)/g(\pi )$

Answer

Correct option: A.
$g(x) + g(\pi )$
a
(a) $g(x + \pi ) = \int_0^{x + \pi } {{{\cos }^4}t\,dt }$

$={ \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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