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If g(x)=0xcos4tdt,, then g(x+π) equals

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a
g(x)±g(π)
b
g(π)−g(x)
c
-gx
d
g(x)g(π)

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detailed solution

Correct option is A

g(x+π)=∫0x+π cos⁡4tdt=∫0π cos⁡4tdt+∫πx+π cos⁡4tdt=g(π)+Iwhere I=∫πx+π cos⁡4tdt, Put t=π+θ so thatI=∫0x cos⁡(4π+4θ)dθ=∫0x cos⁡4θdθ=g(x)So g(x+π)=g(π)+g(x) butg(π)=∫0π cos⁡4tdt=14sin⁡4t0=0.∴ g(x+π)=g(x)−g(π) also.


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