Let h(x) = intlimits_0^x {g(t)dt},where g(x) | vedclass

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(B) Given $h(x) = \int\limits_0^x g(t)dt$. Since $g(x)$ is an odd function,$g(-t) = -g(t)$. Consider $h(-x) = \int\limits_0^{-x} g(t)dt$. Let $t = -u$

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Statement $2$ & Statement $3$

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(B) Given $h(x) = \int\limits_0^x g(t)dt$. Since $g(x)$ is an odd function,$g(-t) = -g(t)$. Consider $h(-x) = \int\limits_0^{-x} g(t)dt$. Let $t = -u$,then $dt = -du$. When $t=0, u=0$ and when $t=-x, u=x$. $h(-x) = \int\limits_0^x g(-u)(-du) = \int\limits_0^x -g(-u)du = \int\limits_0^x g(u)du = h(x)$. Thus,$h(-x) = h(x)$,which means $h(x) - h(-x) = 0$ and $h(x) + h(-x) = 2h(x) = 2\int\limits_0^x g(t)dt$. So,Statement $2$ is true. For Statement $3$,$h(3n) = \int\limits_0^{3n} g(t)dt$. Since $g(t)$ is periodic with period $3$ and odd,$\int_0^3 g(t)dt = 0$. Because the integral of an odd periodic function over its period is zero,$h(3n) = n \int_0^3 g(t)dt = n(0) = 0$. So,Statement $3$ is true. Therefore,Statement $2$ and Statement $3$ are true.

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