A) \[g(x)+g(\pi )\]
B) \[g(x)-g(\pi )\]
C) \[g(x).g(\pi )\]
D) \[\frac{g(x)}{g(\pi )}\]
E) \[\frac{g(\pi )}{g(x)}\]
Correct Answer: A
Solution :
\[\because \]\[g(x)=\int_{0}^{x}{{{\cos }^{4}}t\,dt}\] \[\therefore \]\[g(x+\pi )=\int_{0}^{\pi +x}{{{\cos }^{4}}t}\,dt\] \[=\int_{0}^{\pi }{{{\cos }^{4}}t}\,dt+\int_{\pi }^{\pi +x}{{{\cos }^{4}}t}\,dt\] \[={{I}_{1}}+{{I}_{2}}\] \[\Rightarrow \] \[{{I}_{1}}=g(\pi )\]and\[{{I}_{2}}=\int_{\pi }^{\pi +x}{{{\cos }^{4}}t}\,dt\] \[\Rightarrow \] \[{{I}_{2}}=\int_{0}^{x}{{{\cos }^{4}}(y+\pi )}\,dy\] \[=\int_{0}^{x}{{{[\cos (\pi +y)]}^{4}}}dy\] \[=\int_{0}^{x}{{{\cos }^{4}}y\,}dy=g(x)\]
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