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Question

If $$\displaystyle g\ (x) = \int_0^x \cos^4 t  dt$$, then $$g\ (x + \pi)$$ equals


A
g (x)+g (π)
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B
g (x)g (π)
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C
f (x)g (π)
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D
g (x)/g (π)
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Solution

The correct option is B $$g\ (x) + g\ (\pi)$$
We have
$$g(x) = \displaystyle \int_0^x \cos^4 t  dt$$

$$\therefore
g(x + \pi) = \displaystyle \int_0^{x + \pi} \cos^4 t   dt = \int_0^{\pi}
\cos^4  t.  dt + \int_{\pi}^{x + \pi} \cos^4 t  dt$$

$$\Rightarrow \displaystyle g(x+ \pi) = g(\pi) + \int_0^x \cos^4 tdt$$, 

$$\Rightarrow g(x + \pi) = g(\pi) + g(x)$$

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